The number of students who absented from the class during a week are given below:
Day
Monday
Tuesday
Wednesday
Thursday
Friday
No. of absentees
6
2
4
2
8
Take the scale Figure = 2 absentees. Draw the pictograph.
ANSWER:
Day
Number of absentees
Monday
Tuesday
Wednesday
Thursday
Friday
Page No 238:
Question 2:
The number of stools in five rooms of a school are given below:
Room number
I
II
III
IV
V
Number of stools
30
40
60
50
20
Taking the scale Figure = 10 stools, draw the pictograph.
ANSWER:
Room Number
Number of stools
I
II
III
IV
V
Page No 238:
Question 3:
In a class test, the number of students passed in various subjects are given below.
Subject
English
Mathematics
Hindi
Drawing
Number of students passed
15
25
10
20
Taking the scale Figure = 5 successful students, draw the pictograph.
ANSWER:
Subject
Number of students passed
English
Mathematics
Hindi
Drawing
Page No 238:
Question 4:
The number of fans sold by a shopkeeper during 6 months are given below:
Month
March
April
May
June
July
August
Number of fans sold
30
40
60
50
20
30
Taking the scale Figure = 10 fans sold, draw the pictograph.
ANSWER:
Month
Number of fans sold
March
April
May
June
July
August
Page No 239:
Question 5:
The following pictograph shows different kinds of trees planted in a park. Each symbol represents 8 trees. Look at the pictograph adn answer the questions given below.
Banyan tree
Figure
Neem tree
Figure
Mango tree
Figure
(i) How many mango trees are there? (ii) How many banyan trees are there? (iii) How many neem trees are there? (iv) How many trees are there in all?
ANSWER:
(i) Number of mango trees = 3××8 = 24
(ii) Number of banyan trees = 4××8 = 32
(iii) Number of neem trees = 5××8 = 40
(iv) Total number of trees = Number of mango trees + Number of banyan trees + Number of neem trees = 24+32+40 = 96
Page No 239:
Question 6:
The following pictograph shows the number of scooters sold by a company during a week.
Scale used is Figure = 6 scooters sold.
Day
Figure
Monday
Figure
Tuesday
Figure
Wednesday
Figure
Thursday
Figure
Friday
Figure
Saturday
Figure
Study the pictograph carefully and answer the questions given below. (i) How many scooters were sold on Monday? (ii) How many scooters were sold on Tuesday? (iii) On what day of the week was the sale of the scooters maximum? How many scooters were sold on that day? (iv) On what day of the week was the sale of the scooters minimum? How many scooters were sold on that day?
ANSWER:
(i) Number of scooters sold on Monday = 5××6 = 30
(ii) Number of scooters sold on Tuesday = 4××6 = 24
Look at the bar graph given below. Figure Read it carefully and answer the questions given below: (i) What information does the bar graph give? (ii) In which subject is the student poorest? (iii) In which subject is the student best? (iv) In which subjects did he get more than 40 marks?
ANSWER:
(i) The given bar graph provides information about the marks obtained by a student in an examinations in four different subjects.
(ii) The student is poorest in science because the height of the bar representing the marks obtained in science is the lowest.
(iii) The student is best in mathematics because the height of the bar representing the marks obtained in mathematics is the highest.
(iv) In Hindi and mathematics he got more than 40 marks. He scored 55 marks in Hindi and 70 marks in mathematics.
Page No 243:
Question 2:
In a survey of 60 families of a colony, the number of members in each family was recorded and the data has been represented by the bar graph given below: Figure Read the bar graph carefully and answer the following questions: (i) What information does the bar graph give? (ii) How many families have 3 members? (iii) How many couples have no child? (iv) Which type of family is the most common?
ANSWER:
(i) The given bar graph provides information about the number of members in 60 families of a colony.
(ii) There are 10 families that have 3 members.
(iii) There are 5 couples that do not have children.
(iv) The families having two children is most common because the bar of the families having four members has the highest height.
Page No 244:
Question 3:
Look at the bar graph given below: Figure Study the bar graph carefully and answer the questions given below: (i) In which week was the production maximum? (ii) In which week was the production minimum? (iii) What is the average production during these five weeks? (iv) How many cycles were produced in the first 3 weeks?
ANSWER:
(i) Production was maximum in the 2nd week.
(ii) Production was minimum in the 4th week.
(iii) Average production of these five weeks =Total production of all weeksNumber of weeksTotal production of all weeksNumber of weeks
(iv) Number of cycles produced in the first 3 week = Production in the 1st week + Production in the 2nd week + Production in the 3rd week
= 600 + 1000 + 800 = 2400 cycles
Page No 244:
Question 4:
51 students from a locality use different modes of transport to school, as shown by the bar graph given below: Figure Look at the bar graph given above and answer the questions given below: (i) What does the above bar graph show? (ii) Which mode of transport is used by maximum number of students? (iii) How many students use bus for going to school? (iv) How many students of the locality do not use bus for going to school?
ANSWER:
(i) The given bar graph gives information about the different modes of transport used by the students for going to school.
(ii) Since the height of the bar representing the bicycles is the highest, bicycles are used by maximum number of students. 16 students use bicycles for going to school.
(iii) 14 students use bus for going to school.
(iv) Number of students who do not use bus = Total number of students − Number of students who use bus = 51 − 14 = 37 students
Mark (✓) against the correct answer in each of Q.1 to Q.6. A cuboid has (a) length (b) length and breadth only (c) length, breadth and height (d) thickness only
ANSWER:
A cuboid is a 3-dimensional figure. So, a cuboid has length, breadth and height.
Hence, the correct answer is option (c).
Page No 213:
Question 2:
Mark (✓) against the correct answer in each of Q.1 to Q.6. A dice is an example of a (a) cuboid (b) cube (c) cone (d) cylinder
ANSWER:
(b) cube
Page No 213:
Question 3:
Mark (✓) against the correct answer in each of Q.1 to Q.6. A gas pipe is an example of a (a) cuboid (b) cube (c) cone (d) cylinder
ANSWER:
(d) cylinder
Page No 213:
Question 4:
Mark (✓) against the correct answer in each of Q.1 to Q.6. A football is an example of a (a) cylinder (b) cone (c) sphere (d) none of these
ANSWER:
(c) sphere
Page No 213:
Question 5:
Mark (✓) against the correct answer in each of Q.1 to Q.6. A brick is an example of a (a) cube (b) cuboid (c) prism (d) cylinder
ANSWER:
(b) cuboid
Page No 213:
Question 6:
Mark (✓) against the correct answer in each of Q.1 to Q.6. An ice-cream cone is an example of a (a) cuboid (b) cube (c) pyramid (d) none of these
ANSWER:
(d) None of these Its a cone.
Page No 213:
Question 7:
Fill in the blanks: (i) An object that occupies space is called a …… . (ii) A cuboid has …… faces, …… edges and …… vertices. (iii) The …… faces of a cuboid are identical. (iv) A …… has no vertex and no edge. (v) All the faces of a …… are identical. (vi) A square pyramid has …… lateral triangular faces and …… edges. (vii) A triangular pyramid has …… triangular lateral faces and …… edges. (viii) A triangular prism has …… vertices, …… rectangular lateral faces, …… triangular bases and …… edges.
ANSWER:
(i) An object that occupies space is called a solid. (ii) A cuboid has 6 rectangular faces, 12 edges and 8 vertices. (iii) The opposite faces of a cuboid are identical. (iv) A sphere has no vertex and no edge. (v) All the faces of a cube are identical. (vi) A square pyramid has 4 lateral triangular faces and 8 edges. (vii) A triangular pyramid has 3 triangular lateral faces and 6 edges. (viii) A triangular prism has 6 vertices, 3 rectangular lateral faces, 2 triangular bases and 9 edges.
Page No 213:
Question 8:
Give examples of four objects which are in the shape of: (a) a cone (b) a cuboid (c) a cylinder
ANSWER:
(a) An ice cream cone, a conical tent, a clown’s cap and a conical vessel are in the shape of a cone.
(b) A wooden box, a match box, a brick and an almirah are in the shape of a cuboid.
(c) A measuring jar, a gas cylinder, a test tube and a circular pillar are in the shape of a cylinder.
Mark (✓) against the correct answer in each of Q.1 to Q.8. A square has (a) one line of symmetry (b) two lines of symmetry (c) three lines of symmetry (d) four lines of symmetry
ANSWER:
A square has four lines of symmetry. Hence, the correct answer is option (d).
Page No 218:
Question 2:
Mark (✓) against the correct answer in each of Q.1 to Q.8. A rectangle is symmetrical about (a) each one of its sides (b) each one of its diagonals (c) a line joining the midpoints of its opposite sides (d) none of these
ANSWER:
(c) a line joining the midpoints of its opposite sides
Page No 218:
Question 3:
Mark (✓) against the correct answer in each of Q.1 to Q.8. A rhombus is symmetrical about (a) the line joining the midpoints of its opposite sides (b) each of its diagonals (c) perpendicular bisector of each of its sides (d) none of these
ANSWER:
(b) each of its diagonals
Page No 218:
Question 4:
Mark (✓) against the correct answer in each of Q.1 to Q.8. A circle has (a) no line of symmetry (b) one line of symmetry (c) two lines of symmetry (d) an unlimited number of lines of symmetry
ANSWER:
(d) an unlimited number of lines of symmetry
This is because a circle has infinite number of diameters. Also, a circle is symmetrical about each of its diameter.
Page No 218:
Question 5:
Mark (✓) against the correct answer in each of Q.1 to Q.8. A scalene triangle has (a) no line of symmetry (b) one line of symmetry (c) two lines of symmetry (d) three lines of symmetry
ANSWER:
(a) no line of symmetry
Page No 219:
Question 6:
Mark (✓) against the correct answer in each of Q.1 to Q.8. ABCD is a kite in which AB = AD and BC = DC. The kite is symmetrical about (a) the diagonal AC (b) the diagonal BD (c) none of these Figure
ANSWER:
(a) the diagonal AC
Page No 219:
Question 7:
Mark (✓) against the correct answer in each of Q.1 to Q.8. The letter O of the English alphabet has (a) no line of symmetry (b) one line of symmetry (c) two lines of symmetry (d) none of these
ANSWER:
(c) two lines of symmetry
Page No 219:
Question 8:
Mark (✓) against the correct answer in each of Q.1 to Q.8. The letter Z of the English alphabet has (a) no line of symmetry (b) one line of symmetry (c) two lines of symmetry (d) none of these
ANSWER:
(a) no line of symmetry
Page No 219:
Question 9:
Mark (✓) against the correct answer in each of Q.1 to Q.8. Draw the line (or lines) of symmetry of each of the following figures. Figure
ANSWER:
(i)
(ii)
(iii) (iv)
Page No 219:
Question 10:
Mark (✓) against the correct answer in each of Q.1 to Q.8. Which of the following statements are true and which are false? (i) A parallelogram has no line of symmetry. (ii) An angle with equal arms has its bisector as the line of symmetry. (iii) An equilateral triangle has three lines of symmetry. (iv) A rhombus has four lines of symmetry. (v) A square has four lines of symmetry. (vi) A rectanle has two lines of symmetry. (vii) Each one of the letters H, I, O, X of the English alphabet has two lines of symmetry.
ANSWER:
(i) True
(ii) True
(iii) True An equilateral triangle is symmetrical about each one of the bisectors of its interior angle. Also, it has three bisectors.
(iv) False A rhombus has two lines of symmetry. It is symmetrical about each one of its diagonals.
(v) True A square is symmetrical about each one of its diagonals and the lines joining the midpoints of the opposite sides.
(vi) True A rectangle is symmetrical about the lines joining the midpoints of the opposite sides.
Find the perimeter of a rectangle in which: (i) length = 16.8 cm and breadth = 6.2 cm (ii) length = 2 m 25 cm and breadth = 1 m 50 cm (iii) length = 8 m 5 dm and breadth = 6 m 8 dm
ANSWER:
We know: Perimeter of a rectangle = 2×(Length+Breadth)2×(Length+Breadth)
(i) Length = 16.8 cm Breadth = 6.2 cm Perimeter = 2×(Length+Breadth)2×(Length+Breadth) = 2×(16.8+6.2) =46 cm2×(16.8+6.2) =46 cm (ii) Length = 2 m 25 cm =(200+25) cm (1 m = 100 cm ) = 225 cm Breadth =1 m 50 cm = (100+50) cm (1 m = 100 cm ) = 150 cm Perimeter = 2×(Length+Breadth)2×(Length+Breadth) = 2×(225+150) =750 cm2×(225+150) =750 cm (iii) Length = 8 m 5 dm = (80+5) dm (1 m = 10 dm ) = 85 dm Breadth = 6 m 8 dm = (60+8) dm (1 m = 10 dm ) = 68 dm Perimeter = 2×(Length+Breadth)2×(Length+Breadth) = 2×(85+68) =306 dm2×(85+68) =306 dm
Page No 222:
Question 2:
Find the cost of fencing a rectangular field 62 m long and 33 m wide at Rs 16 per metre.
ANSWER:
Length of the field = 62 m Breadth of the field = 33 m Perimeter of the field = 2(l + b) units = 2(62 + 33) m =190 m Cost of fencing per metre = Rs 16 Total cost of fencing = Rs (16××190) = Rs 3040
Page No 222:
Question 3:
The length and the breadth of a rectangular field are in the ratio 5 : 3. If its perimeter is 128 m, find the dimensions of the field.
ANSWER:
Let the length of the rectangle be 5x m. Breadth of the rectangle = 3x m Perimeter of the rectangle = 2(l + b) = 2(5x + 3x) m = (16x) m It is given that the perimeter of the field is 128 m.
The cost of fencing a rectangular field at Rs 18 per metre is Rs 1980. If the width of the field is 23 m, find its length.
ANSWER:
Total cost of fencing = Rs 1980 Rate of fencing = Rs 18 per metre
Perimeter of the field = Total costRate=Rs 1980Rs 18/m=(198018) m=110 mTotal costRate=Rs 1980Rs 18/m=(198018) m=110 m
Let the length of the field be x metre. Perimeter of the field = 2(x + 23) m
∴2(x+23)=110⇒(x+23)=55x=(55−23)=32∴2(x+23)=110⇒(x+23)=55x=(55-23)=32 Hence, the length of the field is 32 m.
Page No 222:
Question 5:
The length and the breadth of a rectangular field are in the ratio 7 : 4. The cost of fencing the field at Rs 25 per metre is Rs 3300. Find the dimensions of the field.
ANSWER:
Total cost of fencing = Rs 3300 Rate of fencing = Rs 25/m Perimeter of the field = Total cost Rate of fencing=(Rs 3300Rs 25/m)=330025 m=132 mTotal cost Rate of fencing=(Rs 3300Rs 25/m)=330025 m=132 m
Let the length and the breadth of the rectangular field be 7x and 4x, respectively. Perimeter of the field = 2(7x + 4x) = 22x
It is given that the perimeter of the field is 132 m.
∴ 22x=132⇒x=13222=6∴Length of the field =(7 × 6) m=42 mBreadth of the field =(4 × 6) m=24 m∴ 22x=132⇒x=13222=6∴Length of the field =(7 × 6) m=42 mBreadth of the field =(4 × 6) m=24 m
Page No 222:
Question 6:
Find the perimeter of a square, each of whose sides measures: (i) 3.8 cm (ii) 4.6 cm (iii) 2 m 5 dm
ANSWER:
(i) Side of the square = 3.8 cm Perimeter of the square = (4××side) = (4××3.8) = 15.2 cm
(ii) Side of the square = 4.6 cm Perimeter of the square = (4××side) = (4××4.6) = 18.4 cm
(iii) Side of the square = 2 m 5 dm = (20+5) dm (1 m = 10 dm) = 25 dm Perimeter of the square = (4××side) = (4××25) = 100 dm
Page No 222:
Question 7:
The cost of putting a fence around a square field at Rs 35 per metre is Rs 4480. Find the length of each side of the field.
ANSWER:
Total cost of fencing = Rs 4480 Rate of fencing = Rs 35/m Perimeter of the field = Total costRate=Rs 4480Rs 35/m=448035 m=128 mTotal costRate=Rs 4480Rs 35/m=448035 m=128 m
Let the length of each side of the field be x metres. Perimeter = (4x) metres ∴4x=128⇒x=1284=32 ∴4x=128⇒x=1284=32
Hence, the length of each side of the field is 32 m.
Page No 222:
Question 8:
Each side of a square field measures 21 m. Adjacent to this field, there is a rectangular field having its sides in the ratio 4 : 3. If the perimeters of both the fields are equal, find the dimensions of the rectangular field.
ANSWER:
Side of the square field = 21m Perimeter of the square field = (4××21) m = 84 m
Let the length and the breadth of the rectangular field be 4x and 3x, respectively. Perimeter of the rectangular field = 2(4x+ 3x) = 14x
Perimeter of the rectangular field = Perimeter of the square field
∴14x=84 ⇒x=8414=6∴14x=84 ⇒x=8414=6 ∴ Length of the rectangular field =(4 × 6) m=24 mBreadth of the rectangular field = (3 × 6) m=18 m∴ Length of the rectangular field =(4 × 6) m=24 mBreadth of the rectangular field = (3 × 6) m=18 m
Page No 222:
Question 9:
Find the perimeter of (i) a triangle of sides 7.8 cm, 6.5 cm and 5.9 cm, (ii) an equilateral triangle of side 9.4 cm, (iii) an isosceles triangle with equal sides 8.5 cm each and third side 7 cm.
ANSWER:
(i) Sides of the triangle are 7.8 cm, 6.5 cm and 5.9 cm. Perimeter of the triangle = (First side + Second side + Third Side) cm = (7.8 + 6.5 + 5.9) cm = 20.2 cm
(ii) In an equilateral triangle, all sides are equal. Length of each side of the triangle = 9.4 cm ∴∴ Perimeter of the triangle = (3 ×× Side) cm = (3 ×× 9.4) cm = 28.2 cm
(iii) Length of two equal sides = 8.5 cm Length of the third side = 7 cm ∴∴ Perimeter of the triangle = {(2 ×× Equal sides) + Third side} cm = {(2 ×× 8.5) + 7} cm = 24 cm
Page No 222:
Question 10:
Find the perimeter of (i) a regular pentagon of side 8 cm, (ii) a regular octagon of side 4.5 cm, (iii) a regular decagon of side 3.6 cm,
ANSWER:
(i) Length of each side of the given pentagon = 8 cm ∴∴ Perimeter of the pentagon = (5××8) cm = 40 cm
(ii) Length of each side of the given octagon = 4.5 cm ∴∴ Perimeter of the octagon = (8××4.5) cm = 36 cm
(iii) Length of each side of the given decagon = 3.6 cm ∴∴ Perimeter of the decagon = (10××3.6) cm = 36 cm
Page No 222:
Question 11:
Find the perimeter of each of the following figures: Figure
ANSWER:
(i) Perimeter of the figure = Sum of all the sides =(27 + 35 + 35 + 45) cm = 142 cm (ii) Perimeter of the figure = Sum of all the sides =(18 + 18 + 18 + 18) cm = 72 cm (iii) Perimeter of the figure = Sum of all the sides =(8 + 16 + 4 + 12 + 12 + 16 + 4) cm = 72 cm
Page No 224:
Exercise 21B
Question 1:
Find the circumference of a circle whose radius is (i) 28 cm (ii) 10.5 cm (iii) 3.5 m
ANSWER:
(i) Radius, r = 28 cm
∴ Circumference of the circle, C=2πr =(2×227×28) =176 cmHence, the circumference of the given circle is 176 cm.∴ Circumference of the circle, C=2πr =(2×227×28) =176 cmHence, the circumference of the given circle is 176 cm.
(ii) Radius, r = 10.5 cm ∴ Circumference of the circle, C=2πr=(2×227×10.5)=66 cmHence, the circumference of the given circle is 66 cm.∴ Circumference of the circle, C=2πr=(2×227×10.5)=66 cmHence, the circumference of the given circle is 66 cm.
(iii) Radius, r = 3.5 m ∴ Circumference of the circle, C=2πr=(2×227×3.5)=22 mHence, the circumference of the given circle is 22 m.∴ Circumference of the circle, C=2πr=(2×227×3.5)=22 mHence, the circumference of the given circle is 22 m.
Page No 224:
Question 2:
Find the circumference of a circle whose diameter is (i) 14 cm (ii) 35 cm (iii) 10.5 m
ANSWER:
(i)
Circumference=2πr =π(2r) =π× Diameter of the circle (d) (Diameter=2×radius)⇒Circumference=Diameter×π Diameter of the given circle is 14 cm.Circumference of the given circle=14×π⇒(14×22 7)=44 cmCircumference of the given circle is 44 cm.Circumference=2πr =π(2r) =π× Diameter of the circle (d) (Diameter=2×radius)⇒Circumference=Diameter×π Diameter of the given circle is 14 cm.Circumference of the given circle=14×π⇒(14×22 7)=44 cmCircumference of the given circle is 44 cm.
(ii) Circumference=2πr =π(2r) =π×Diameter of the circle(d) (Diameter=2×Radius)⇒Circumference=Diameter×πDiameter of the given circle is 35 cm.⇒Circumference of the given circle=35×π ⇒(35×22 7)=110 cmCircumference of the given circle is 110 cm.Circumference=2πr =π(2r) =π×Diameter of the circle(d) (Diameter=2×Radius)⇒Circumference=Diameter×πDiameter of the given circle is 35 cm.⇒Circumference of the given circle=35×π ⇒(35×22 7)=110 cmCircumference of the given circle is 110 cm.
(iii) Circumference=2πr =π(2r) =π×Diameter of the circle(d) (Diameter=2×Radius)⇒Circumference=Diameter×πDiameter of the given circle is 10.5 m.Circumference of the given circle=10.5×π⇒(10.5×22 7)=33 mCircumference of the given circle is 33 m.Circumference=2πr =π(2r) =π×Diameter of the circle(d) (Diameter=2×Radius)⇒Circumference=Diameter×πDiameter of the given circle is 10.5 m.Circumference of the given circle=10.5×π⇒(10.5×22 7)=33 mCircumference of the given circle is 33 m.
Page No 224:
Question 3:
Find the radius of a circle whose circumference is 176 cm.
ANSWER:
Let the radius of the given circle be r cm. Circumference of the circle = 176 cm Circumference = 2πr2πr ∴ 2πr=176 ⇒r=1762π⇒r=(1762×722) ⇒r=28 The radius of the given circle is 28 cm.∴ 2πr=176 ⇒r=1762π⇒r=(1762×722) ⇒r=28 The radius of the given circle is 28 cm.
Page No 224:
Question 4:
Find the diameter of a wheel whose circumference is 264 cm.
ANSWER:
Let the radius of the circle be r cm.Diameter=2×Radius=2r cmCircumference of the wheel=264 cmCircumference of the wheel=2πr∴2πr=264⇒2r=264π⇒2r=(264×722)⇒2r=84 Diameter of the given wheel is 84 cm.Let the radius of the circle be r cm.Diameter=2×Radius=2r cmCircumference of the wheel=264 cmCircumference of the wheel=2πr∴2πr=264⇒2r=264π⇒2r=(264×722)⇒2r=84 Diameter of the given wheel is 84 cm.
Page No 224:
Question 5:
Find the distance covered by the wheel of a car in 500 revolutions if the diameter of the wheel is 77 cm.
ANSWER:
Radius of the wheel =Diameter of the wheel2Diameter of the wheel2 ⇒r=772cm⇒r=772cm Circumference of the wheel =2π r2π r =(2×227×772)=242 cm=(2×227×772)=242 cm
In 1 revolution the wheel covers a distance equal to its circumference.
∴ Distance covered by the wheel in 1 revolution=242 cm∴ Distance covered by the wheel in 500 revolutions=(500 × 242) cm =121000 cm =1210 m (100 cm= 1m ) =1.21 km (1000 m=1 km )∴ Distance covered by the wheel in 1 revolution=242 cm∴ Distance covered by the wheel in 500 revolutions=(500 × 242) cm =121000 cm =1210 m (100 cm= 1m ) =1.21 km (1000 m=1 km )
Page No 224:
Question 6:
The diameter of the wheel of a car is 70 cm. How many revolutions will it make to travel 1.65 km?
ANSWER:
Radius of the wheel(r)=Diameter of the wheel2r=702cm=35 cmCircumference of the wheel = 2πr =(2×227×35) =220 cmIn one revolution, the wheel covers the distance equal to its circumference.∴ 220 cm distance =1 revolution∴1 cm distance =1220 revolution∴1km (or 100000 cm) distance =1×100000220 revolution (∴ 1 km=100000 cm)∴1.65 km distance = 1.65×100000 220 revolutions = 750 revolutions Thus,the wheel will make 750 revolutions to travel 1.65 km.Radius of the wheel(r)=Diameter of the wheel2r=702cm=35 cmCircumference of the wheel = 2πr =(2×227×35) =220 cmIn one revolution, the wheel covers the distance equal to its circumference.∴ 220 cm distance =1 revolution∴1 cm distance =1220 revolution∴1km (or 100000 cm) distance =1×100000220 revolution (∴ 1 km=100000 cm)∴1.65 km distance = 1.65×100000 220 revolutions = 750 revolutions Thus,the wheel will make 750 revolutions to travel 1.65 km.
Page No 226:
Exercise 21C
Question 1:
Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm2. Figure
ANSWER:
The figure contains 12 complete squares. Area of 1 small square = 1 sq cm ∴ Area of the figure = Number of complete squares ×× Area of the square =(12×1) sq cm(12×1) sq cm =12 sq cm
Page No 226:
Question 2:
Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm2. Figure
ANSWER:
The figure contains 18 complete squares. Area of 1 small square = 1 sq cm ∴ Area of the figure = Number of complete squares ×× Area of the square =(18×1) sq cm(18×1) sq cm =18 sq cm
Page No 226:
Question 3:
Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm2. Figure
ANSWER:
The figure contains 14 complete squares and 1 half square. Area of 1 small square = 1 sq cm ∴ Area of the figure = Number of squares ×× Area of the square =[(14 × 1) + (1 × 12)]sq cm(14 × 1) + (1 × 12)sq cm =14121412 sq cm
Page No 226:
Question 4:
Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm2. Figure
ANSWER:
The figure contains 6 complete squares and 4 half squares. Area of 1 small square = 1 sq cm ∴ Area of the figure = Number of squares ×× Area of the square =[(6×1)+(4×12)] sq cm(6×1)+(4×12) sq cm =8 sq cm
Page No 226:
Question 5:
Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm2. Figure
ANSWER:
The figure contains 9 complete squares and 6 half squares. Area of 1 small square = 1 sq cm ∴ Area of the figure = Number of squares ×× Area of the square =[(9 × 1) + (6 × 12)] sq cm(9 × 1) + (6 × 12) sq cm =12 sq cm
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Question 6:
Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm2. Figure
ANSWER:
The figure contains 16 complete squares. Area of 1 small square = 1 sq cm ∴ Area of the figure = Number of squares ×× Area of a square =(16×1) sq cm(16×1) sq cm =16 sq cm
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Question 7:
Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm2. Figure
ANSWER:
In the given figure, there are 4 complete squares, 8 more than half parts of squares and 4 less than half parts of squares. We neglect the less than half parts and consider each more than half part of the square as a complete square.
∴ Area = (4 + 8) sq cm = 12 sq cm
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Question 8:
Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm2. Figure
ANSWER:
In the given figure, there are 9 complete squares, 5 more than half parts of squares and 7 less than half parts of squares. We neglect the less than half parts of squares and consider the more than half squares as complete squares. ∴ Area of the figure = (9 + 5) sq cm = 14 sq cm
Page No 226:
Question 9:
Count the number of squares enclosed by figure and find its area, taking the area of each square as 1 cm2. Figure
ANSWER:
The figure contains 14 complete squares and 4 half squares. Area of 1 small square = 1 sq cm Area of the figure = Number of squares ×× Area of one square =[(14×1)+(4×12) ]sq cm(14×1)+(4×12) sq cm =16 sq cm
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Exercise 21D
Question 1:
Find the area of a rectangle whose (i) length = 46 cm and breadth = 25 cm (ii) length = 9 m and breadth = 6 m (iii) length = 14.5 m and breadth = 6.8 m (iv) length = 12 m 5 cm and breadth = 60 cm (v) length = 3.5 km and breadth = 2 km
ANSWER:
(i) Length = 46 cm Breadth = 25 cm Area of the rectangle = (Length ××Breadth) sq units = (46××25) cm2 = 1150 cm2
(ii) Length = 9 m Breadth = 6 m Area of the rectangle = (Length ××Breadth) sq units = (9××6) m2 = 54 m2
(iii) Length = 14.5 m Breadth = 6.8 m Area of the rectangle = (Length ××Breadth) sq units = (14510×681014510×6810) m2 = 98601009860100 m2 =98.60 m2
(iv) Length = 2 m 5 cm = (200+5) cm (1 m = 100 cm ) =205cm Breadth = 60 cm Area of the rectangle = (Length ××Breadth) sq units = (205××60) cm2 = 12300 cm2
(v) Length = 3.5 km Breadth = 2 km Area of the rectangle = (Length ××Breadth) sq units = (3.5××2) km2 = (3510×2)(3510×2) km2 =7 km2
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Question 2:
Find the area of a square plot of side 14 m.
ANSWER:
Side of the square plot = 14 m Area of the square plot = (Side)2 sq units = (14)2 m2 = 196 m2
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Question 3:
The top of a table measures 2 m 25 cm by 1 m 20 cm. Find its area in square metres.
ANSWER:
Length of the table = 2 m 25 cm = (2 + 0.25) m (100 cm = 1 m) = 2.25 m Breadth of the table = 1 m 20 cm = (1 + 0.20) m (100 cm = 1 m) =1.20 m Area of the table = (Length × Breadth) sq units = (2.25 × 1.20) m2
= (225100×120100)(225100×120100) m2 = 2.7 m2
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Question 4:
A carpet is 30 m 75 cm long and 80 cm wide. Find its cost at Rs 150 per square metre.
ANSWER:
Length of the carpet = 30 m 75 cm =(30 + 0.75) cm (100 cm = 1 m) = 30.75 m Breadth of the carpet = 80 cm = 0.80 m (100 cm = 1 m)
Cost of 1 m2 carpet= Rs 150 Cost of 24.6 m2 carpet = Rs (24.6××150) = Rs 3690
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Question 5:
How many envelopes can be made out of a sheet of paper 3 m 24 cm by 1 m 72 cm. If each envelope requires a piece of paper of size 18 cm by 12 cm?
ANSWER:
Length of the sheet of paper = 3 m 24 cm = 324 cm Breadth of the sheet of paper = 1 m 72 cm = 172 cm Area of the sheet = (Length ×× Breadth) =(324×172) cm2 =55728 cm2=(324×172) cm2 =55728 cm2
Length of the piece of paper required to make 1 envelope = 18 cm Breadth of the piece of paper required to make 1 envelope = 12 cm Area of the piece of paper required to make 1 envelope = (18××12) cm2 = 216 cm2
No. of envelopes that can be made =Area of the sheetArea of the piece of paper required to make 1 envelope⇒No. of envelopes that can be made =55728216=258 envelopesNo. of envelopes that can be made =Area of the sheetArea of the piece of paper required to make 1 envelope⇒No. of envelopes that can be made =55728216=258 envelopes
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Question 6:
A room is 12.5 m long and 8 m wide. A square carpet of side 8 m is laid on its floor. Find the area of the floor which is not carpeted.
ANSWER:
Length of the room = 12.5 m Breadth of the room = 8 m Area of the room = (Length××Breadth) =(12.5×8) m2 = 100 m2=(12.5×8) m2 = 100 m2 Side of the square carpet = 8 m Area of the carpet = (Side)2 = 82 m2 = 64 m2
Area of the floor which is not carpeted = Area of the room − Area of the carpet = (100 − 64) m2 = 36 m2
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Question 7:
A lane, 150 m long and 9 m wide, is to be paved with bricks, each measuring 22.5 cm by 7.5 cm. Find the number of bricks required.
ANSWER:
Length of the road = 150 m = 15000 cm Breadth of the road = 9 m = 900 cm Area of the road = (Length××Breadth) =15000×900 cm2=13500000 cm2=15000×900 cm2=13500000 cm2 Length of the brick = 22.5 cm Breadth of the brick = 7.5 cm Area of one brick = (Length××Breadth) =(22.5×7.5) cm2=168.75 cm2=(22.5×7.5) cm2=168.75 cm2
Number of bricks = Area of the roadArea of one brick=13500000168.75=80000 bricksNumber of bricks = Area of the roadArea of one brick=13500000168.75=80000 bricks
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Question 8:
A room is 13 m long and 9 m broad. Find the cost of carpeting the room with a carpet 75 cm broad at the rate of Rs 65 per metre.
ANSWER:
Length of the room = 13 m Breadth of the room = 9 m Area of the room = (13××9) m2 = 117 m2
Let length of required carpet be x m. Breadth of the carpet = 75 cm = 0.75 m (100 cm = 1 m) Area of the carpet = (0.75××x) m2 = 0.75x m2 For carpeting the room: Area covered by the carpet = Area of the room ⇒0.75x=117⇒x=1170.75⇒x=117×43⇒x=156 m⇒0.75x=117⇒x=1170.75⇒x=117×43⇒x=156 m
So, the length of the carpet is 156 m. Cost of 1 m carpet = Rs 65 Cost 156 m carpet = Rs (156××65) = Rs 10140
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Question 9:
The length and the breadth of a rectangular park are in the ratio 5 : 3 and its perimeter is 128 m. Find the area of the park.
ANSWER:
Let the length of the rectangular park be 5x. ∴ Breadth of the rectangular park = 3x Perimeter of the rectangular field = 2(Length + Breadth) =2(5x + 3x) = 16x
It is given that the perimeter of rectangular park is 128 m. ⇒16x=128⇒x=12816⇒x=8 Length of the park=(5×8) m =40 mBreadth of the park=(3×8) m =24 m⇒16x=128⇒x=12816⇒x=8 Length of the park=(5×8) m =40 mBreadth of the park=(3×8) m =24 m
Area of the park = (Length ×× Breadth) sq units
=(40×24) m2=960 m2=(40×24) m2=960 m2
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Question 10:
Two plots of land have the same perimeter. One is a square of side 64 m and the other is a rectangle of length 70 m. Find the breadth of the rectangular plot. Which plot has the greater area and by how much?
ANSWER:
Side of the square plot = 64 m Perimeter of the square plot = (4×Side) m =(4×64) m=256 m(4×Side) m =(4×64) m=256 m Area of the square plot = (Side)2 = 642 m2 = 4096 m2
Let the breadth of the rectangular plot be x m. Perimeter of the rectangular plot = 2(l+b) m = 2(70+x) m
Perimeter of the rectangular plot = Perimeter of the square plot (Given) ⇒2(70+x)=256⇒140+2x=256⇒2x=256−140⇒2x=116⇒x=1162=58⇒2(70+x)=256⇒140+2x=256⇒2x=256-140⇒2x=116⇒x=1162=58 So, the breadth of the rectangular plot is 58 m. Area of the rectangular plot = (Length × Breadth)=(70 × 58) m2=4060 m2(Length × Breadth)=(70 × 58) m2=4060 m2 Area of the square plot − Area of the rectangular plot = (4096 − 4060) = 36 m2 Area of the square plot is 36 m2 greater than the rectangular plot.
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Question 11:
The cost of cultivating a rectangular field at Rs 35 per square metre is Rs 71400. If the width of the field is 40 m, find its length. Also, find the cost of fencing the field at Rs 50 per metre.
ANSWER:
Total cost of cultivating the field = Rs 71400 Rate of cultivating the field = Rs 35/m2
Area of the field=Total cost of cultivating the fieldRate of cultivating=Rs 71400Rs 35/m2=2040 m2Area of the field=Total cost of cultivating the fieldRate of cultivating=Rs 71400Rs 35/m2=2040 m2
Let the length of the field be x m.
Area of the field = (Length × Width) m2=(x × 40) m2 =40x m2It is given that the area of the field is 2040 m2.⇒40x=2040⇒x=204040=51∴Length of the field = 51 m Area of the field = (Length × Width) m2=(x × 40) m2 =40x m2It is given that the area of the field is 2040 m2.⇒40x=2040⇒x=204040=51∴Length of the field = 51 m
Perimeter of the field = 2(l+b) = 2(51+40) m = 182 m Cost of fencing 1 m of the field = Rs 50 Cost of fencing 182 m of the field = Rs (182××50) = Rs 9100
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Question 12:
The area of a rectangle is 540 cm2 and its length is 36 cm. Find its width and perimeter.
ANSWER:
Let the width of the rectangle be x cm. Length of the rectangle = 36 cm Area of the rectangle = (Length × WidthLength × Width) = (36 × x36 × x) cm2 It is given that the area of the rectangle is 540 cm2.
⇒36 × x= 540⇒x=54036⇒x=15∴ Width of the rectangle =15 cm⇒36 × x= 540⇒x=54036⇒x=15∴ Width of the rectangle =15 cm
Perimeter of the rectangle = 2(Length + Width) cm = 2(36 + 15) cm = 102 cm
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Question 13:
A marble tile measures 12 cm × 10 cm. How many tiles will be required to cover a wall of size 4 m by 3 m? Also, find the total cost of the tiles at Rs 22.50 per tile.
ANSWER:
Length of the wall = 4 m = 400 cm Breadth of the wall = 3 m = 300 cm Area of the wall = (400×300) cm2 = 120000 cm2
Length of the tile = 12 cm Breadth of the tile = 10 cm Area of one tile = (12×10) cm2 = (120) cm2
Number of tiles required to cover the wall=Area of the wallArea of one tile=120000120=1000 tilesNumber of tiles required to cover the wall=Area of the wallArea of one tile=120000120=1000 tiles Cost of 1 tile = Rs 22.50 Cost of 1000 tiles = (1000 × 22.50) = Rs 22500
Thus, the total cost of the tiles is Rs 22500.
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Question 14:
Find the perimeter of a rectangle whose area is 600 cm2 and breadth is 25 cm.
ANSWER:
Let the length of the rectangle be x cm. Breadth of the rectangle is 25 cm. Area of the rectangle = (Length × Breadth) cm2 = (x×25) cm2 =25x cm2
It is given that the area of the rectangle is 600 cm2. ⇒25x=600⇒x=60025=24⇒25x=600⇒x=60025=24 So, the length of the rectangle is 24 cm. Perimeter of the rectangle = 2(Length + Breadth) units = 2(25 + 24) cm = 98 cm
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Question 15:
Find the area of a square whose diagonal is 52–√52 cm.
ANSWER:
Area of the square ={12× (Diagonal)2} sq units 12× (Diagonal)2 sq units = {12×(52–√)2} cm2={12×(5)2×(2–√)2} cm2={12×25×2} cm2=(12×50) cm2= 25 cm2 = 12×(52)2 cm2=12×(5)2×(2)2 cm2=12×25×2 cm2=(12×50) cm2= 25 cm2
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Question 16:
Calculate the area of each one of the shaded regions given below: Figure
ANSWER:
(i) Area of rectangle ABDC = Length ×× Breadth = AB××AC (AC = AE − CE) = (1×8)m21×8m2 = 8 m2 Area of rectangle CEFG = Length ×× Breadth = CG××GF (CG = GD + CD) = (9×2)m29×2m2 = 18 m2 Area of the complete figure = Area of rectangle ABDC + Area of rectangle CEFG = (8 + 18) m2 = 26 m2
(ii) Area of rectangle AEDC = Length ×× Breadth = ED ×× CD = (12×2)m212×2m2 = 24 cm2 Area of rectangle FJIH = Length ×× Breadth = HI ×× IJ = (1×9)m21×9m2 = 9 m2 Area of rectangle ABGF = Length ×× Breadth = AB ×× AF {(AB = FJ − GJ) and AF = EH − (EA + FH)} = (7×1.5)m27×1.5m2 = 10.5 m2
Area of the complete figure = Area of rectangle AEDC + Area of rectangle FJIH + Area of rectangle ABGF = (24 + 9 + 10.5) m2 = 43.5 m2
(iii) Area of the shaded portion = Area of the complete figure − Area of the unshaded figure = Area of rectangle ABCD − Area of rectangle GBFE =(CD××AD) − (GB××BF) ={(12×9)−(7.5×10)}m2(12×9)-(7.5×10)m2 (BF = BC − FC) =(108 − 75) m2
=33 m2
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Question 17:
Calculate the area of each one of the shaded regions given below (all measures are given in cm): Figure
ANSWER:
(i) Area of square BCDE= (Side)2 = (CD)2 = (3)2 cm2 = 9 cm2 Area of rectangle ABFK = Length × BreadthLength × Breadth = AK××AB [(AB = AC − BC) and (AK = AL + LK) = (5××1) cm2 = 5 cm2
Area of rectangle MLKG = Length × BreadthLength × Breadth = ML ×× MG = (2 ×× 3) cm2 = 6 cm2 Area of rectangle JHGF= Length × BreadthLength × Breadth = JH××HG = (2××4) cm2 = 8 cm2 Area of the figure = Area of rectangle ABFK + Area of rectangle MLKG + Area of rectangle JHGF + Area of square BCDE = (9 + 5 + 6 + 8) cm2 = 28 cm2 (ii) Area of rectangle CEFG= Length × BreadthLength × Breadth = EF××CE = (1××5) cm2 (CE = EA − AC) = 5 cm2 Area of rectangle ABDC = Length × BreadthLength × Breadth = AB××BD = (1××2) cm2 = 2 cm2
Area of rectangle HIJG = Length × BreadthLength × Breadth = HI ×× IJ = (1××2) cm2 = 2 cm2 Area of the figure = Area of rectangle CEFG + Area of rectangle HIJG + Area of rectangle ABDC = (5+2+2) cm2 = 9 cm2 (iii) In the figure, there are 5 squares, each of whose sides are 6 cm in length. Area of the figure = 5 ×× Area of square = 5××(side)2 = 5××(6)2 cm2 = 180 cm2
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Exercise 21E
Question 1:
The sides of a rectangle are in the ratio 7 : 5 and its perimeter is 96 cm. The length of the rectangle is (a) 21 cm (b) 28 cm (c) 35 cm (d) 14 cm
ANSWER:
(b) 28 cm
Let the length and the breadth of the rectangle be 7x cm and 5x cm, respectively. It is given that the perimeter of the rectangle is 96 cm. Perimeter of the rectangle = 2(7x+5x) cm
⇒2(7x+5x)=96=2(12x)=96=24x=96⇒x=9624=4∴ Length =(7×4)cm=28 cm⇒2(7x+5x)=96=2(12x)=96=24x=96⇒x=9624=4∴ Length =(7×4)cm=28 cm
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Question 2:
The area of a rectangle is 650 cm2 and its breadth is 13 cm. The perimeter of the rectangle is (a) 63 cm (b) 130 cm (c) 100 cm (d) 126 cm
ANSWER:
(d) 126 cm Let length of the rectangle be L cm. Area of the rectangle = 650 cm2 Area of the rectangle = (L×13L×13) cm2 ⇒(L×13)=650⇒L=65013=50 Length of the rectangle is 50 cm⇒(L×13)=650⇒L=65013=50 Length of the rectangle is 50 cm
Perimeter of the rectangle = 2(Length + Breadth) cm = 2(50+13) cm = 126 cm
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Question 3:
The cost of fencing a rectangular field 34 m long and 18 m wide at Rs 22.50 per metre is (a) Rs 2430 (b) Rs 2340 (c) Rs 2400 (d) Rs 3340
ANSWER:
(b) Rs 2340 Perimeter of the rectangular field = 2(Length + Breadth) = 2(34 + 18) m = 104 m Cost of fencing 1 metre = Rs 22.50 Cost of fencing 104 m = Rs (22.50××104) = Rs 2340
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Question 4:
The cost of fencing a rectangular field at Rs 30 per metre is Rs 2400. If the length of the field is 24 m, then its breadth is (a) 8 m (b) 16 m (c) 18 m (d) 24 m
ANSWER:
(b) 16 m Total cost of fencing = Rs 2400 Rate of fencing = Rs 30/m Perimeter of the rectangular field = Total costRate=Rs 2400Rs 30/m=80 mTotal costRate=Rs 2400Rs 30/m=80 m Let the breadth of the rectangular field be x m. Perimeter of the rectangular field = 2(24 + x) m ⇒2(24+x)=80⇒48+2x=80⇒2x=(80−48)⇒2x=32⇒x=322=16So, the breadth of the rectangular field is 16 m.⇒2(24+x)=80⇒48+2x=80⇒2x=(80-48)⇒2x=32⇒x=322=16So, the breadth of the rectangular field is 16 m.
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Question 5:
The area of a rectangular carpet is 120 m2 and its perimeter is 46 m. The length of its diagonal is (a) 15 m (b) 16 m (c) 17 m (d) 20 m
ANSWER:
(c) 17 m Let the length and the breadth of the rectangle be L m and B m, respectively.
Area of the rectangular carpet = (L×BL×B) m2 ⇒LB=120 … (i)⇒LB=120 … (i) Perimeter of the rectangular carpet = 2(L+B)2(L+B) ⇒2(L+B)=46⇒(L+B)=462⇒(L+B)=23 …(ii)⇒2(L+B)=46⇒(L+B)=462⇒(L+B)=23 …(ii)
Diagonal of the rectangle = L2+B2−−−−−−−√ m =(L+B)2−2LB−−−−−−−−−−−−−√ m =(23)2−240−−−−−−−−−√ m (from equations (i) and (ii)) =529−240−−−−−−−−√ m =289−−−√ m=17 mDiagonal of the rectangle = L2+B2 m =(L+B)2-2LB m =(23)2-240 m (from equations (i) and (ii)) =529-240 m =289 m=17 m
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Question 6:
The length of a rectangle is three times its width and the length of its diagonal is 610−−√610 cm. The perimeter of the rectangle is (a) 48 cm (b) 36 cm (c) 24 cm (d) 2410 −−−√2410 cm
ANSWER:
(a) 48 cm Let the width and the length of the rectangle be x cm and 3x cm, respectively.
Applying Pythagoras theorem:
(Diagonal)2=(Length)2+(Width)2⇒(610−−√)2=(3x)2+(x)2⇒360=9×2+x2⇒360=10×2⇒x2=36010⇒x2=36⇒x=±6Since the width cannot be negative, we will neglect −6.(Diagonal)2=(Length)2+(Width)2⇒(610)2=(3x)2+(x)2⇒360=9×2+x2⇒360=10×2⇒x2=36010⇒x2=36⇒x=±6Since the width cannot be negative, we will neglect -6.
So, width of the rectangle is 6 cm. Length of the rectangle = (3×6)=18 cm3×6=18 cm Perimeter of the rectangle = 2(Length + Breadth) = 2(18 + 6) = 48 cm
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Question 7:
If the ratio between the length and perimeter of a rectangular plot is 1 : 3, then the ratio between the length and breadth of the plot is (a) 1 : 2 (b) 2 : 1 (c) 3 : 2 (d) 2 : 3
ANSWER:
(b) 2 : 1 Let the breadth of the plot be b cm.
Let the length of the plot be x cm. Perimeter of the plot = 3xcm
Perimeter of the plot =2(Length + Breadth)= 2(x + b) cm ⇒2(x+b)=3x2x+2b=3x⇒2b=3x−2x⇒2b=x⇒b=x2∴Ratio of the length and the breadth of the plot =x(x2)=xx × 2 = 21∴Ratio of the length and the breadth of the plot =2:1⇒2(x+b)=3x2x+2b=3x⇒2b=3x-2x⇒2b=x⇒b=x2∴Ratio of the length and the breadth of the plot =x(x2)=xx × 2 = 21∴Ratio of the length and the breadth of the plot =2:1
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Question 8:
The length of the diagonal of a square is 20 cm. It area is (a) 400 cm2 (b) 200 cm2 (c) 300 cm2 (d) 1002–√1002 cm2
ANSWER:
(b) 200 cm2 Area of the square = {12×(Diagonal)2 } sq units12×(Diagonal)2 sq units ={12×(20)2 } cm2={12×(20)×(20)} cm2=(20×10) cm2=200 cm2=12×(20)2 cm2=12×(20)×(20) cm2=(20×10) cm2=200 cm2
Page No 232:
Question 9:
The cost of putting a fence around a square field at Rs 25 per metre is Rs 2000. The length of each side of the field is (a) 80 m (b) 40 m (c) 20 m (d) none of these
ANSWER:
(c) 20 m Let one side of the square field be x m. Total cost of fencing a square field = Rs 2000 Rate of fencing the field = Rs 25/m
Perimeter of the square field=Total cost of fencing the fieldRate of fencing the field=Rs 2000Rs 25/m=200025 m=80 mPerimeter of the square field=Total cost of fencing the fieldRate of fencing the field=Rs 2000Rs 25/m=200025 m=80 m
Perimeter of the square field = (4×side4×side) = 4x m ⇒4x=80⇒x=804⇒x=20Each side of the field is 20 m.⇒4x=80⇒x=804⇒x=20Each side of the field is 20 m.
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Question 10:
The diameter of a circle is 7 cm. Its circumference is (a) 44 cm (b) 22 cm (c) 28 cm (d) 14 cm
ANSWER:
(b) 22 cm Radius=Diameter 2=72 cmCircumference of the circle=2πr=(2×227×72) cm=22 cmRadius=Diameter 2=72 cmCircumference of the circle=2πr=(2×227×72) cm=22 cm
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Question 11:
The circumference of a circle is 88 cm. Its diameter is (a) 28 cm (b) 42 cm (c) 56 cm (d) none of these
ANSWER:
(a) 28 cm Circumference of the circle is 88 cm.Let the radius be r cm.It is given that the circumference of the circle is (2πr) cm.⇒2πr=88⇒2×227×r=88⇒r=12×722×88⇒r=14∴ Radius=14 cmDiameter=(2× Radius)=(2 × 14) cm =28 cmCircumference of the circle is 88 cm.Let the radius be r cm.It is given that the circumference of the circle is (2πr) cm.⇒2πr=88⇒2×227×r=88⇒r=12×722×88⇒r=14∴ Radius=14 cmDiameter=(2× Radius)=(2 × 14) cm =28 cm
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Question 12:
The diameter of a wheel of a car is 70 cm. How much distance will it cover in making 50 revolutions? (a) 350 m (b) 110 m (c) 165 m (d) 220 m
ANSWER:
(b) 110 m Radius of the wheel=Diameter2=702=35 cmCircumference of the wheel =2πr=(2×227×35) cm =220 cmThe distance covered by the wheel in one revolution is equal to its circumference.Distance covered by the wheel in 1 revolution = 220 cm∴Distance covered by the wheel in 50 revolution = (50 × 220)cm=11000 cm=110 mRadius of the wheel=Diameter2=702=35 cmCircumference of the wheel =2πr=(2×227×35) cm =220 cmThe distance covered by the wheel in one revolution is equal to its circumference.Distance covered by the wheel in 1 revolution = 220 cm∴Distance covered by the wheel in 50 revolution = (50 × 220)cm=11000 cm=110 m
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Question 13:
A lane 150 m long and 9 m wide is to be paved with bricks, each measuring 22.5 cm by 7.5 cm. How many bricks are required? (a) 65000 (b) 70000 (c) 75000 (d) 80000
ANSWER:
(d) 80000 Length of the road = 150 m = 15000 cm Breadth of the road = 9 m = 900 cm Area of the road = (Length × Breadth) = (15000 × 900) cm2 = 13500000 cm2
Length of the brick = 22.5 cm Breadth of the brick = 7.5 cm Area of one brick = (Length × Breadth) = ( 22.5 × 7.5 ) cm2 = 168.75 cm2
Number of bricks = Area of the roadArea of one brick =13500000 cm2168.75 cm2=80000 bricksNumber of bricks = Area of the roadArea of one brick =13500000 cm2168.75 cm2=80000 bricks
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Question 14:
A room is 5 m 40 cm long and 4 m 50 cm broad. Its area is (a) 23.4 m2 (b) 24.3 m2 (c) 25 m2 (d) 98.01 m2
ANSWER:
(b) 24.3 m2
Length of the room = 5 m 40 cm = 5.40 m Breadth of the room = 4 m 50 cm = 4.50 m
Area of the room = (Length × Breadth)=(5.40 × 4.50) m2=(540100×450100)m2=(275×92)m2=24310m2=24.3 m2Area of the room = (Length × Breadth)=(5.40 × 4.50) m2=540100×450100m2=275×92m2=24310m2=24.3 m2
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Question 15:
How many envelopes can be made out of a sheet of paper 72 cm by 48 cm, if each envelope requires a paper of size 18 cm by 12 cm? (a) 4 (b) 8 (c) 12 (d) 16
ANSWER:
(d) 16
Length of the sheet of paper = 72 cm Breadth of the sheet of paper = 48 cm Area of the sheet = (Length × Breadth) ⇒ ( 72 × 48 ) cm2 = 3456 cm2
Length of the piece of paper required to make 1 envelope = 18 cm Breadth of the piece of paper required to make 1 envelope = 12 cm Area of the piece of paper required to make 1 envelope = (18 × 12) cm2 = 216 cm2
No. of envelopes that can be made = Area of the sheetArea of the piece of paper required to make 1 envelope⇒No. of envelopes that can be made =3456216=16 envelopesNo. of envelopes that can be made = Area of the sheetArea of the piece of paper required to make 1 envelope⇒No. of envelopes that can be made =3456216=16 envelopes
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Exercise 21F
Question 1:
Find the perimeter of the following shapes: (i) a triangle whose sides are 5.4 cm, 4.6 cm and 6.8 cm (ii) a regular hexagon of side 8 cm (iii) an isosceles triangle with equal sides 6 cm each and third side 4.5 cm.
ANSWER:
(i) Sides of the triangle are 5.4 cm, 4.6 cm and 6.8 cm. Perimeter of the triangle = (First side + Second side + Third Side) = (5.4 + 4.6 + 6.8) cm = 16.8 cm
(ii) Length of each side of the given hexagon = 8 cm ∴ Perimeter of the hexagon = (6 × 8) cm = 48 cm
(iii) Length of the two equal sides = 6 cm Length of the third side = 4.5 cm ∴ Perimeter of the triangle = {(2 × equal sides) + third side} cm = (2 × 6) + 4.5 = 16.5 cm
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Question 2:
The perimeter of a rectangular field is 360 m and its breadth is 75 m. Find its length.
ANSWER:
Let the length of the rectangle be x m. Breadth of the rectangle = 75 m Perimeter of the rectangle = 2(Length + Breadth) = 2(x + 75) m = (2x+ 150) m It is given that the perimeter of the field is 360 m. ⇒2x+150=360⇒2x=360−150⇒2x=210⇒x=2102=105 ⇒2x+150=360⇒2x=360-150⇒2x=210⇒x=2102=105 So, the length of the rectangle is 105 m.
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Question 3:
The length and breadth of a rectangular field are in the ratio 5 : 4. If its perimeter is 108 m, find the dimensions of the field.
ANSWER:
Let the length of the rectangle be 5x m. Breadth of the rectangle = 4x m Perimeter of the rectangle = 2(Length + Breadth) = 2(5x+ 4x) m = 18x m It is given that the perimeter of the field is 108 m. ∴ 18 x = 108 ⇒ x = 10818=610818=6 ∴ Length of the field = ( 5 × 6 )m = 30 m Breadth of the field = ( 4 × 6 )m = 24 m
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Question 4:
Find the area of a square whose perimeter is 84 cm.
ANSWER:
Let one side of the square be x cm. Perimeter of the square = (4×side)=(4×x) cm =4x cm(4×side)=(4×x) cm =4x cm It is given that the perimeter of the square is 84 cm. ⇒4x=84⇒x=844=21Thus, one side of the square is 21 cm.Area of the square = (Side)2=(21)2 cm2= 441cm2⇒4x=84⇒x=844=21Thus, one side of the square is 21 cm.Area of the square = (Side)2=(21)2 cm2= 441cm2
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Question 5:
The area of a room is 216 m2 and its breadth is 12 m. Find the length of the room.
ANSWER:
Let the length of the room be x m. Breadth of the room = 12 m Area of the room = (Length × Breadth) = (x × 12) m2 It is given that the area of the room is 216 m2. ⇒ x × 12 = 216 ⇒ x = 21612=1821612=18 ∴ Length of the rectangle = 18 m
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Question 6:
Find the circumference of a circle of radius 7 cm. [Take π = 227π = 227]
ANSWER:
Radius(r) of the given circle = 7 cm Circumference of the circle, C = 2 πr = (2×227×7) cm= 44 cm= 2×227×7 cm= 44 cm Hence, the circumference of the given circle is 44 cm.
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Question 7:
The diameter of a wheel of a car is 77 cm. Find the distance covered by the wheel in 500 revolutions.
ANSWER:
Radius of the wheel =Diameter of the wheel2Diameter of the wheel2 ⇒ r = 772772cm Circumference of the wheel = 2 πr = (2×227×772)2×227×772 cm = 242 cm
In 1 revolution, the wheel covers a distance equal to its circumference. ∴ Distance covered by the wheel in 1 revolution = 242 cm ∴ Distance covered by the wheel in 500 revolutions = ( 500 × 242 ) cm = 121000 cm (100 cm =1 m) = 1210 m
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Question 8:
Find the diameter of a wheel whose circumference is 176 cm.
ANSWER:
Let the radius be r cm. Diameter = 2 × Radius(r) = 2r cm Circumference of the wheel = 2πr ∴ 2πr = 176 ⇒ 2r=176π⇒ 2r=176×722=56⇒ 2r=176π⇒ 2r=176×722=56 ⇒ 2r = 56 Thus, the diameter of the given wheel is 56 cm.
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Question 9:
Find the area of a rectangle whose length is 36 cm and breadth 15 cm.
ANSWER:
Length of the rectangle = 36 cm Breadth of the rectangle = 15 cm Area of the rectangle = (Length × Breadth) sq units = (36 × 15) cm2 = 540 cm2
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Question 10:
Perimeter of a square of side 16 cm is (a) 256 cm (b) 64 cm (c) 32 cm (d) 48 cm
ANSWER:
(b) 64 cm Side of the square = 16 cm Perimeter of the square = (4 × side) = (4 × 16) cm = 64 cm
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Question 11:
The area of a rectangle is 240 m2 and its length is 16 m. Then, its breadth is (a) 15 m (b) 16 m (c) 30 m (d) 40 m
ANSWER:
(a) 15 m
Let the breadth of the rectangle be x m. Length of the rectangle = 16 m Area of rectangle = (Length × Breadth) = (16 × x) m2 It is given that the area of the rectangle is 240 m2. ⇒ 16 × x = 240 ⇒ x = 24016=1524016=15 So, the breadth of the rectangle is 15 m.
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Question 12:
The area of a square lawn of side 15 m is (a) 60 m2 (b) 225 m2 (c) 45 m2 (d) 120 m2
ANSWER:
(b) 225 m2
Side of the square lawn = 15 mArea of the square lawn = (Side)2 sq units=(15)2 m2=225 m2Side of the square lawn = 15 mArea of the square lawn = (Side)2 sq units=(15)2 m2=225 m2
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Question 13:
The area of a square is 256 cm2. The perimeter of the square is (a) 16 cm (b) 32 cm (c) 48 cm (d) 64 cm
ANSWER:
(a) 16 cm Let one side of the square be x cm. Area of the square = (Side )2 cm2 = x2 cm2 It is given that the area of the square is 256 cm2. ⇒ x2 = 256 ⇒ x = 256−−−√=±16256=±16 We know that the side of a square cannot be negative. So, we will neglect −16. Therefore, the side of the square is 16 cm.
Perimeter of the square = (4×side)=(4×16)cm=64 cm4×side=4×16cm=64 cm
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Question 14:
The area of a rectangle is 126 m2 and its length is 12 m. The breadth of the rectangle is (a) 10 m (b) 10.5 m (c) 11 m (d) 11.5 m
ANSWER:
(b) 10.5 m Let the breadth of the rectangle be x m. Length of the rectangle = 12 m Area of the rectangle = 126 m2 Area of the rectangle = (length×breadth)sq units=(12×x)m2=12x m2length×breadthsq units=(12×x)m2=12x m2 It is given that the area of the rectangle is 126 m2. ⇒12x=126⇒x=12612=10.5So, the breadth of the rectangle is 10.5 m.⇒12x=126⇒x=12612=10.5So, the breadth of the rectangle is 10.5 m.
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Question 15:
Fill in the blanks. (i) A polygon having all sides equal and all angles equal is called a …… polygon. (ii) Perimeter of a square = …… × side. (iii) Area of a rectangle = (……) × (……). (iv) Area of a square = …… . (v) If the length of a rectangle is 5 m and its breadth is 4 m, then its area is …… .
ANSWER:
(i) A polygon having all sides equal and all angles equal is called a regular polygon (ii) Perimeter of a square = 4 × side (iii) Area of a rectangle = (length) × (breadth) (iv) Area of a square = (side)2 (v) If the length of a rectangle is 5 m and its breadth is 4 m, then its area is 20 m2 Area of a rectangle = (length) × (breadth) = (5×4) m2 = 20 m2
Define the terms: (i) Data (ii) Raw data (iii) Array (iv) Tabulation of data (v) Observations (vi) Frequency of an observation (vii) Statistics
ANSWER:
(i) Data: It refers to the information in the form of numerical figures.
The marks obtained by 5 students of a class in a unit test are 34, 45, 65, 67, 87.
We call it the data related to the marks obtained by 5 students of a class in a unit test.
(ii) Raw Data: Data obtained in the original form is called raw data.
(iii) Array: Arranging the numerical figures in an ascending or a descending order is called an array.
(iv) Tabulation of data: Arranging the data in a systematic form in the form of a table is called tabulation or presentation of the data.
(v) Observations: Each numerical figure in a data is called an observation.
(vi) Frequency of an observation: The number of times a particular observation occurs is called its frequency.
(viii) Statistics: It is the science that deals with the collection, presentation, analysis and interpretation of numerical data.
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Question 2:
The number of children in 25 families of a colony are give below: 2, 0, 2, 4, 2, 1, 3, 3, 1, 0, 2, 3, 4, 3, 1, 1, 1, 2, 2, 3, 2, 4, 1, 2, 2. Represent the above data in the form of a frequency distribution table.
ANSWER:
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Question 3:
The sale of shoes of various sizes at a shop on a particular day is given below: 6, 9, 8, 5, 5, 4, 9, 8, 5, 6, 9, 9, 7, 8, 9, 7, 6, 9, 8, 6, 7, 5, 8, 9, 4, 5, 8, 7. Represent the above data in the form of a frequency distribution table.
ANSWER:
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Question 4:
Construct a frequency table for the following: 3, 2, 5, 4, 1, 3, 2, 2, 5, 3, 1, 2, 1, 1, 2, 2, 3, 4, 5, 3, 1, 2, 3.
ANSWER:
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Question 5:
Construct a frequency table for the following: 7, 8, 6, 5, 6, 7, 7, 9, 8, 10, 7, 6, 7, 8, 8, 9, 10, 5, 7, 8, 7, 6.
ANSWER:
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Question 6:
Fill in the blanks: (i) Data means information in the form of …… . (ii) Data obtained in the …… form is called raw data. (iii) Arranging the numerical figures in an ascending or a descending order is called an ……. . (iv) The number of times a particular observation occurs is called its …… . (v) Arranging the data in the form of a table is called …… .
ANSWER:
(i) Data means information in the form of numerical figures. (ii) Data obtained in the original form is called raw data. (iii) Arranging the numerical figures in an ascending or a descending order is called an array. (iv) The number of times a particular observation occurs is called its frequency. (v) Arranging the data in the form of a table is called tabulation of data.
Which of the following are simple closed figures? Figure
ANSWER:
(a) It is a simple closed figure because it does not intersect itself (b) It is a simple closed figure because it does not intersect itself. (c) It is not a simple closed figure because it intersects itself. (d) It is a simple closed figure because it does not intersect itself. (e) It is not a simple closed figure because it intersects itself. (f) It is a simple closed figure because it does not intersect itself.
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Question 2:
Which of the following are polygons? Figure
ANSWER:
(a) It is formed by four line segments and we know that a simple closed figure formed by three or more line segments is called a polygon. So, it is a polygon named quadrilateral.
(b) It is formed by three line segments and we know that a simple closed figure formed by three or more line segments is called a polygon. So, it is polygon named triangle.
(c) It is formed by twelve line segments and we know that a simple closed figure formed by three or more line segments is called a polygon. So, it is polygon.
(d) It is not a polygon because it contains curves.
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Question 3:
Fill in the blanks: (i) A polygon is a simple closed figure formed by more than …… line segments. (ii) A polygon formed by three line segments is called a ……. . (iii) A polygon formed by four line segments is called a …… . (iv) A triangle has …… sides and …… angles. (v) A quadrilateral has …… sides and …… angles. (vi) A figure which ends at the starting point is called a ………. .
ANSWER:
(i) A polygon is a simple closed figure formed by more than two line segments. (ii) A polygon formed by three line segments is called a triangle. (iii) A polygon formed by four line segments is called a quadrilateral . (iv) A triangle has three sides and three angles. (v) A quadrilateral has four sides and four angles. (vi) A figure which ends at the starting point is called a closed figure.
Take three noncollinear points A, B and C on a page of your notebook. Join AB, BC and CA. What figure do you get? Name: (i) the side opposite to ∠C (ii) the angle opposite to the side BC (iii) the vertex opposite to the side CA (iv) the side opposite to the vertex B Figure
ANSWER:
We get a triangle by joining the three non-collinear points A, B, and C. (i) The side opposite to ∠C is AB. (ii) The angle opposite to the side BC is ∠A. (iii) The vertex opposite to the side CA is B. (iv) The side opposite to the vertex B is AC.
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Question 2:
The measures of two angles of a triangle are 72° and 58°. Find the measure of the third angle.
ANSWER:
The measures of two angles of a triangle are 72° and 58°. Let the third angle be x. Now, the sum of the measures of all the angles of a triangle is 180o. ∴∴ x + 72o + 58o = 180o ⇒ x + 130o = 180o ⇒ x = 180o −- 130o ⇒ x = 50o The measure of the third angle of the triangle is 50o.
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Question 3:
The angles of a triangle are in the ratio 1 : 3 : 5. Find the measure of each of the angles.
ANSWER:
The angles of a triangle are in the ratio 1:3:5. Let the measures of the angles of the triangle be (1x), (3x) and (5x) Sum of the measures of the angles of the triangle = 180o ∴ 1x + 3x + 5x = 180o ⇒ 9x = 180o ⇒ x = 20o 1x = 20o 3x = 60o 5x = 100o The measures of the angles are 20o, 60o and 100o.
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Question 4:
One of the acute angles of a right triangle is 50°. Find the other acute angle.
ANSWER:
In a right angle triangle, one of the angles is 90o. It is given that one of the acute angled of the right angled triangle is 50o. We know that the sum of the measures of all the angles of a triangle is 180o. Now, let the third angle be x. Therefore, we have: 90o + 50o + x = 180o ⇒ 140o + x = 180o ⇒ x = 180o −- 140o ⇒ x = 40o The third acute angle is 40o.
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Question 5:
One of the angles of a triangle is 110° and the other two angles are equal. What is the measure of each of these equal angles?
ANSWER:
Given: ∠A = 110o and ∠B = ∠C Now, the sum of the measures of all the angles of a traingle is 180o . ∠A + ∠B + ∠C = 180o ⇒ 110o + ∠B + ∠B = 180o ⇒ 110o + 2∠B = 180o ⇒ 2∠B = 180o −- 110o ⇒ 2∠B = 70o ⇒ ∠B = 70o / 2 ⇒ ∠B = 35o
∴ ∠C = 35o The measures of the three angles: ∠A = 110o, ∠B = 35o, ∠C = 35o
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Question 6:
If one angle of a triangle is equal to the sum of other two, show that the triangle is a right triangle.
Look at the figures given below. State for each triangle whether it is acute, right or obtuse. Figure
ANSWER:
(i) It is an obtuse angle triangle as one angle is 130o, which is greater than 90o.
(ii) It is an acute angle triangle as all the angles in it are less than 90o.
(iii) It is a right angle triangle as one angle is 90o.
(iv) It is an obtuse angle triangle as one angle is 92o, which is greater than 90o.
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Question 9:
In the given figure some triangles have been given. State for each triangle whether it is scalene, isosceles or equilateral. Figure
ANSWER:
Equilateral Triangle: A triangle whose all three sides are equal in length and each of the three angles measures 60o. Isosceles Triangle: A triangle whose two sides are equal in length and the angles opposite them are equal to each other. Scalene Triangle: A triangle whose all three sides and angles are unequal in measure.
(i) Isosceles AC = CB = 2 cm (ii) Isosceles DE = EF = 2.4 cm (iii) Scalene All the sides are unequal. (iv) Equilateral XY = YZ = ZX = 3 cm (v) Equilateral All three angles are 60o. (vi) Isosceles Two angles are equal in measure. (vii) Scalene All the angles are unequal.
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Question 10:
Draw a ∆ABC. Take a point D on BC. Join AD. How many triangles do you get? Name them. Figure
ANSWER:
In ∆ABC, if we take a point D on BC, then we get three triangles, namely ∆ADB, ∆ADC and ∆ABC.
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Question 11:
Can a triangle have (i) two right angles? (ii) two obtuse angles? (iii) two acute angles? (iv) each angle more than 60°? (v) each angle less than 60°? (vi) each angle equal to 60°?
ANSWER:
(i) No If the two angles are 90o each, then the sum of two angles of a triangle will be 180o, which is not possible. (ii) No For example, let the two angles be 120o and 150o. Then, their sum will be 270o, which cannot form a triangle. (iii) Yes For example, let the two angles be 50oand 60o, which on adding, gives 110o. They can easily form a triangle whose third angle is 180o −- 110o = 70o. (iv) No For example, let the two angles be 70o and 80o, which on adding, gives 150o. They cannot form a triangle whose third angle is 180o −- 150o = 30o, which is less than 60o. (v) No For example, let the two angles be 50oand 40o, which on adding, gives 90o. Thus, they cannot form a triangle whose third angle is 180o−- 90o = 90o, whichis greater than 60o. (vi) Yes Sum of all angles = 60o + 60o + 60o = 180o
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Question 12:
Fill in the blanks. (i) A triangle has …… sides, …… angles and …… vertices. (ii) The sum of the angles of a triangle is …… . (iii) The sides of a scalene triangle are of ……. lengths. (iv) Each angle of an equilateral triangle measures …… . (v) The angles opposite to equal sides of an isosceles triangle are ……. . (vi) The sum of the lengths of the sides of a triangle is called its ………. .
ANSWER:
(i) A triangle has 3 sides 3 angles and 3 vertices. (ii) The sum of the angles of a triangle is 180o. (iii) The sides of a scalene triangle are of different lengths. (iv) Each angle of an equilateral triangle measures 60o. (v) The angles opposite to equal sides of an isosceles triangle are equal. (vi) The sum of the lengths of the sides of a triangle is called its perimeter.
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Exercise 16B
Question 1:
How many parts does a triangle have? (a) 2 (b) 3 (c) 6 (d) 9
ANSWER:
Correct option: (c) A triangle has 6 parts: three sides and three angles.
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Question 2:
With the angles given below, in which case the construction of triangle is possible? (a) 30°, 60°, 70° (b) 50°, 70°, 60° (c) 40°, 80°, 65° (d) 72°, 28°, 90°
ANSWER:
Correct option: (b) (a) Sum = 30° + 60° + 70° = 160o This is not equal to the sum of all the angles of a triangle. (b) Sum = 50° + 70° + 60° = 180o Hence, it is possible to construct a triangle with these angles. (c) Sum = 40° + 80° + 65° = 185o This is not equal to the sum of all the angles of a triangle. (d) Sum = 72° + 28° + 90° = 190o This is not equal to the sum of all the angles of a triangle.
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Question 3:
The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle is (a) 60° (b) 80° (c) 76° (d) 84°
ANSWER:
(b) 80o Let the measures of the given angles be (2x)o, (3x)o and (4x)o. ∴∴ (2x)o + (3x)o + (4x)o = 180o ⇒ (9x)o = 180o ⇒ x = 180 / 9 ⇒ x = 20o ∴∴ 2x = 40o, 3x = 60o, 4x = 80o
Hence, the measures of the angles of the triangle are 40o, 60o, 80o. Thus, the largest angle is 80o.
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Question 4:
The two angles of a triangle are complementary. The third angle is (a) 60° (b) 45° (c) 36° (d) 90°
ANSWER:
Correct option: (d) The measure of two angles are complimentary if their sum is 90o degrees. Let the two angles be x and y, such that x + y = 90o. Let the third angle be z. Now, we know that the sum of all the angles of a triangle is 180o. x + y + z = 180o ⇒ 90o + z = 180o ⇒ z = 180o −- 90o = 90o The third angle is 90o.
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Question 5:
One of the base angles of an isosceles triangle is 70°. The vertical angle is (a) 60° (b) 80° (c) 40° (d) 35°
ANSWER:
Correct option: (c) Let ∠A = 70o The triangle is an isosceles triangle. We know that the angles opposite to the equal sides of an isosceles triangle are equal. ∴∴ ∠B = 70o We need to find the vertical angle ∠C. Now, sum of all the angles of a triangle is 180o. ∠A + ∠B + ∠C = 180o ⇒ 70o + 70o + ∠C = 180o ⇒ 140o + ∠C = 180o ⇒ ∠C = 180o−- 140o ⇒ ∠C = 40o
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Question 6:
A triangle having sides of different lengths is called (a) an isosceles triangle (b) an equilateral triangle (c) a scalene triangle (d) a right triangle
ANSWER:
Correct option: (c) A triangle having sides of different lengths is called a scalene triangle.
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Question 7:
In an isosceles ∆ABC, the bisectors of ∠B and ∠C meet at a point O. If ∠A = 40°, then ∠BOC = ? (a) 110° (b) 70° (c) 130° (d) 150°
ANSWER:
Correct option: (a) In the isosceles ABC, the bisectors of ∠B and ∠C meet at point O. Since the triangle is isosceles, the angles opposite to the equal sides are equal. ∠B = ∠C ∴∴ ∠A + ∠B + ∠C = 180o ⇒ 40o+ 2∠B = 180o ⇒ 2∠B = 140o ⇒ ∠B = 70o Bisectors of an angle divide the angle into two equal angles. So, in ∆BOC: ∠OBC = 35o and ∠OCB = 35o ∠BOC + ∠OBC + ∠OCB = 180o ⇒ ∠BOC + 35o + 35o= 180o ⇒ ∠BOC = 180o – 70o ⇒ ∠BOC = 110o
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Question 8:
The sides of a triangle are in the ratio 3 : 2 : 5 and its perimeter is 30 cm. The length of the longest side is (a) 20 cm (b) 15 cm (c) 10 cm (d) 12 cm
ANSWER:
Correct option: (b) The sides of a triangle are in the ratio 3:2:5. Let the lengths of the sides of the triangle be (3x), (2x), (5x). We know: Sum of the lengths of the sides of a triangle = Perimeter (3x) + (2x) + (5x) = 30 ⇒ 10x = 30 ⇒ x = 30 10 ⇒ x = 3 First side = 3x = 9 cm Second side = 2x = 6 cm Third side = 5x = 15 cm The length of the longest side is 15 cm.
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Question 9:
Two angles of a triangle measure 30° and 25° respectively. The measure of the third angle is (a) 35° (b) 45° (c) 65° (d) 125°
ANSWER:
Correct option: (d) Two angles of a triangle measure 30° and 25°, respectively. Let the third angle be x. x + 30o+ 25o= 180o x = 180o −- 55o x = 125o
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Question 10:
Each angle of an equilateral triangle measures (a) 30° (b) 45° (c) 60° (d) 80°
ANSWER:
Correct option: (c) Each angle of an equilateral triangle measures 60o.
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Question 11:
In the adjoining figure, the point P lies (a) in the interior of ∆ABC (b) in the exterior of ∆ABC (c) on ∆ABC (d) outside ∆ABC Figure
In the adjacent figure, a quadrilateral has been shown. Name: (i) its diagonals, (ii) two pairs of opposite sides, (iii) two pairs of opposite angles, (iv) two pairs of adjacent sides, (v) two pairs of adjacent angles. Figure
ANSWER:
(i) The diagonals are AC, and BD. (ii) AB and CD, and AD and BC are the two pairs of opposite sides. (iii) ∠A and ∠C, and ∠B and ∠D are the two pairs of opposite angles. (iv) AB and BC, and AD and DC are the two pairs of adjacent sides. (v) ∠A and ∠B, and ∠C and ∠D are the two pairs of adjacent angles.
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Question 2:
Draw a parallelogram ABCD in which AB = 6.5 cm, AD = 4.8 cm and ∠BAD = 70°. Measure its diagonals.
ANSWER:
Since ABCD is a parallelogram, AB = DC = 6.5 cm and AD = BC = 4.8 cm. Given: ∠A = 70°∠A = 70° Steps of construction : 1) Draw AD equal to 4.8 cm. 2) Make an angle of 70°at A and cut an arc of 6.5 cm. Name it B. 3) Cut an arc of 4.8 cm from B and 6.5 cm from D. Name it C. 4) Join AB, BC and CD. 5) Measuring the diagonals AC and BD, we get AC equal to 9.2 cm and BD equal to 6.6 cm.
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Question 3:
Two sides of a parallelogram are in the ratio 4 : 3. If its perimeter is 56 cm, find the lengths of its sides.
ANSWER:
Two sides of a parallelogram are in the ratio 4:3. Let the two sides be 4x and 3x. In a parallelogram, opposite sides are equal and parallel. So, they are also in the ratio of 4:3, i.e. 4x and 3x. Perimeter = 4x + 3x + 4x +3x 56 = 14x x = 56145614
x = 4
∴ 4x = 16 3x = 12 Length of its sides are 16cm, 12 cm, 16cm and 12cm.
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Question 4:
Name each of the following parallelograms: (i) The diagonals are equal and the adjacent sides are unequal. (ii) The diagonals are equal and the adjacent sides are equal. (iii) The diagonals are unequal and the adjacent sides are equal.
ANSWER:
(i) Rectangle (ii) Square (iii) Rhombus
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Question 5:
What is a trapezium? When do you call a trapezium an isosceles trapezium? Draw an isosceles trapezium. Measure its sides and angles.
ANSWER:
A trapezium has only one pair of parallel sides. A trapezium is said to be an isosceles trapezium if its non-parallel sides are equal. Following are the measures of the isosceles trapezium: AB = 5.4 cm BC = 3 cm DC = 7.4 cm AD = 3 cm
Which of the following statements are true and which are false? (a) The diagonals of a parallelogram are equal. (b) The diagonals of a rectangle are perpendicular to each other. (c) The diagonals of a rhombus are equal.
ANSWER:
(a) False (b) False (c) False
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Question 7:
Give reasons for the following: (a) A square can be thought of as a special rectangle. (b) A square can be thought of as a special rhombus. (c) A rectangle can be thought of as a special parallelogram. (d) a square is also a parallelogram.
ANSWER:
(a) This is because a rectangle with equal sides becomes a square. (b) This is because a rhombus with each angle a right angle becomes a square. (c) This is because a parallelogram with each angle a right angle becomes a rectangle. (d) This is because in a square opposite sides are parallel.
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Question 8:
A figure is said to be regular if its sides are equal in length and angles are equal in measure. What do you mean by a regular quadrilateral?
ANSWER:
A square is a regular quadrilateral all of whose sides are equal in length and all of whose angles are equal in measure.
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Exercise 17 B
Question 1:
The sum of all the angles of a quadrilateral is (a) 180° (b) 270° (c) 360° (d) 400°
ANSWER:
(c) 360° The sum of all the angles of a quadrilateral is 360°.
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Question 2:
The three angles of a quadrilateral are 80°, 70° and 120°. The fourth angle is (a) 110° (b) 100° (c) 90° (d) 80°
ANSWER:
(c) 90°
The three angles of a quadrilateral are 80°, 70° and 120°. Let the fourth angle be x. We know that the sum of all the angles of a quadrilateral is 360°. 80° + 70° + 120° + x = 360° ⇒ 270° + x = 360° ⇒ x = 360° − 270° ⇒ x = 90° Thus, the fourth angle is 90°.
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Question 3:
The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. The largest of these angles is (a) 90° (b) 120° (c) 150° (d) 102°
ANSWER:
Let the angles of a quadrilateral be (3x)°, (4x)°, (5x)° and (6x)°. Sum of all the angles of a quadrilateral is 360°.
∴ 3x + 4x + 5x + 6x = 360° ⇒ 18x = 360° ⇒ x = 3601836018 ⇒ x = 20° So, 3x = 60° 4x = 80° 5x = 100° 6x = 120° The largest of these angles is 120°. So, the correct answer is given in option (b).
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Question 4:
A quadrilateral having one and only one pair of parallel sides is called (a) a parallelogram (b) a kite (c) a rhombus (d) a trapezium
ANSWER:
(d) a trapezium A trapezium is a quadrilateral that has only one pair of parallel sides.
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Question 5:
A quadrilateral whose opposite sides are parallel is called (a) rhombus (b) a kite (c) a trapezium (d) a parallelogram
ANSWER:
(d) a parallelogram A parallelogram is a quadrilateral whose opposite sides are parallel.
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Question 6:
An isosceles trapezium has (a) equal parallel sides (b) equal nonparallel sides (c) equal opposite sides (d) none of these
ANSWER:
(b) equal nonparallel sides The non-parallel sides of an isosceles trapezium are equal.
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Question 7:
If the diagonals of a quadrilateral bisect each other at right angles, then this quadrilateral is (a) a rectangle (b) a rhombus (c) a kite (d) none of these
ANSWER:
(b) a rhombus The diagonals of a rhombus bisect each other at right angle.
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Question 8:
A square has (a) all sides equal and diagonals unequal (b) all sides equal and diagonals equal (c) all sides unequal and diagonals equal (d) none of these
ANSWER:
(b) all sides equal and diagonals equal In a square, all the sides are equal. All of its diagonals are also equal.
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Question 9:
A quadrilateral having two pairs of equal adjacent sides but unequal opposite sides, is called a (a) trapezium (b) parallelogram (c) kite (d) rectangle
ANSWER:
(c) kite
A kite has two pairs of equal adjacent sides, but unequal opposite sides.
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Question 10:
What do you mean by a regular quadrilateral? (a) A rectangle (b) A rhombus (c) A square (d) A trapezium
ANSWER:
(c) A square The only regular quadrilateral is a square. This is because all of its sides and angles are equal.
Take a point O on your notebook and draw circles of radii 4 cm, 5.3 cm and 6.2 cm, each having the same centre O.
ANSWER:
This is the required diagram as asked in the question.
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Question 2:
Draw a circle with centre C and radius 4.5 cm. Mark points P, Q, R such taht P lies in the interior of the circle, Q lies on the circle, and R lies in the exterior of the circle.
ANSWER:
This is the required diagram as asked in the question.
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Question 3:
Draw a circle with centre O and radius 4 cm. Draw a chord AB of the circle. Indicate by marking points X and Y, the minor arc AXB and the major arc AYB of the circle.
ANSWER:
This is the required diagram as asked in the question.
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Question 4:
Which of the following statements are true and which are false? (i) Each radius of a circle is also a chord of the circle. (ii) Each diameter of a circle is also a chord of the circle. (iii) The centre of a circle bisects each chord of the circle. (iv) A secant of a circle is a segment having its end points on the circle. (v) A chord of a circle is a segment having its end points on teh circle.
ANSWER:
(i) False Diameter of a circle is a chord of the circle, not radius. (ii) True It is the longest chord of the circle. (iii) False A perpendicular drawn from the centre of the circle to the chord, bisects the chord. (iv) False It is a line passing through the circle that intersects the circle at two points. (v) True.
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Question 5:
Draw a circle with centre O and radius 3.7 cm. Draw a sector having the angle 72°.
ANSWER:
Therefore, the required arc is arc OACB.
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Question 6:
Fill in the blanks by using <, >, = or ≤. (i) OP …… OQ, where O is the centre of the circle, P lies on the circle and Q is in the interior of the circle. (ii) OP …… OR, where O is the centre of the circle, P lies on the circle and R lies in the exterior of the circle. (iii) Major arc …… minor arc of the circle. (iv) Major arc …… semicircumference of the circle.
ANSWER:
(i) > (ii) < (iii) > (iv) > This is because the major arc covers more than half of the circumference of the circle.
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Question 7:
Fill in the blanks: (i) A diameter of a circle is a chord that ………. the centre. (ii) A radius of a circle is a line segment with one end point ……….. and the other end point …… . (iii) If we join any two points of a circle by a line segment, we obtain a ……….. of the circle. (iv) Any part of a circle is called an ……….. of the circle. (v) The figure bounded by an arc and the two radii joining the end points of the arc with the centre is called a ……….. of the circle.
ANSWER:
(i) passes through (ii) on the circle, at the centre of the circle (iii) chord (iv) arc (v) sector
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Exercise 18B
Question 1:
Define each of the following: (a) Closed figures (b) Open figures (c) Polygons
ANSWER:
(i) A closed figure is a figure that can be traced with the same starting and ending points, and that too without crossing or retracing any part of the figure. For example: Polygon, circle, etc.
(ii) A figure having no boundary and no starting or ending points is called as an open figure.
(iii) A polygon is a plane shape with 3 or more straight sides. It is a 2 dimensional closed figure.
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Question 2:
Define each of the following: (a) A scalene triangle (b) An isosceles triangle (c) An obtuse triangle
ANSWER:
(a) A triangle having no sides or angles equal is called a scalene triangle.
(b) A triangle having two sides and the corresponding opposite angles equal is called an isosceles triangle.
(c) A triangle having one of the angles more than 90°° is called an obtuse triangle.
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Question 3:
(i) What do you mean by a convex quadrilateral? (ii) Define a regular polygon.
ANSWER:
(i) A quadrilateral with no interior angles greater than 180° is known as a convex quadrilateral.
(ii) A regular polygon is a polygon all of whose sides are of the same lengths and all of whose interior angles are of the same measures.
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Question 4:
The angles of a triangle are in the ratio 3 : 5 : 7. Find the measures of these angles.
ANSWER:
The angles are in ratio 3:5:7. Suppose the angles are 3x, 5x and 7x.
∴ 3x + 5x + 7x = 180 (angle sum property of a triangle) 15x =180 x = 12o
Therefore, the angles are of the measures 36°, 60° and 84°.
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Question 5:
The angles of a quadrilateral are in the ratio 2 : 3 : 4 : 6. Find the measures of these angles.
ANSWER:
Suppose the angles are 2x, 3x, 4x, and 6x. We know that the sum of the angles of a quadrilateral is 360°. ∴ 2x + 3x + 4x + 6x = 360 15x = 360 x = 24
Therefore, the measures of the angles are 48°, 72°, 96° and 144°.
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Question 6:
State the properties of a rhombus.
ANSWER:
A rhombus is a parallelogram whose opposite sides are parallel.
All four of its sides are equal in length. Also, the opposite angles are equal.
The diagonals bisect each other at right angles.
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Question 7:
Define (i) a trapezium (ii) a kite.
ANSWER:
(i) A trapezium is a quadrilateral with only one pair of parallel sides.
(ii) A kite is a quadrilateral that has two pairs of equal adjacent sides, but unequal opposite sides.
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Question 8:
Draw a circle with centre O and radius 3 cm. Draw a sector having an angle of 54°.
ANSWER:
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Question 9:
A quadrilateral having two pairs of equal adjacent sides but unequal opposite sides is called a (a) parallelogram (b) rectangle (c) trapezium (d) kite
ANSWER:
(d) kite
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Question 10:
If the diagonals of a quadrilateral bisect each other at right angles, then this quadrilateral is a (a) rectangle (b) parallelogram (c) rhombus (d) kite
ANSWER:
(c) rhombus
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Question 11:
A quadrilateral having one and only one pair of parallel sides is called a (a) parallelogram (b) a kite (c) a trapezium (d) a rhombus
ANSWER:
(c) a trapezium
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Question 12:
One of the base angles of an isosceles triangle is 70°. The vertical angle is (a) 35° (b) 40° (c) 70° (d) 80°
ANSWER:
(b) 40°
Since one base angle is 70°, the other base angle will also be 70° because the triangle is isosceles. Vertical angle: 180 − 70 − 70 = 40°° (angle sum property of a triangle)
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Question 13:
Write ‘T’ for true and ‘F’ for false for each of the statements given below: (i) The diagonals of a rhombus are equal. (ii) The diagonals of a parallelogram bisect each other. (iii) The centre of a circle bisects each chord of a circle. (iv) Each diameter of a circle is a chord of the circle. (v) The diagonals of a rhombus bisect each other at right angles.
ANSWER:
(i) False Diagonals are perpendicular and bisect each other.
(ii) True The diagonals of a parallelogram bisect each other.
(iii) False A perpendicular drawn from the centre of a circle to the chord, bisects the chord.
(iv) True It divides the circle in two equal parts.
(v) True The diagonals of a rhombus bisect each other at right angles.
