RS Agarwal Solution | Class 9th | Chapter-14 |   Areas of Triangles and Quadrilaterals | Edugrown

Exercise MCQ

Question 1

In a ∆ABC it is given that base = 12 cm and height = 5 cm. Its area is

(a) 60 cm²

(b) 30 cm²

(d) 45 cm²Solution 1

Question 2

The lengths of three sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the triangle is

(a) 96 cm²

(b) 120 cm²

(c) 144 cm²

(d) 160 cm²Solution 2

Question 3

Each side of an equilateral triangle measures 8 cm. The area of the triangle is

Solution 3

Question 4

The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle is

Solution 4

Question 5

The base of an isosceles triangle is 6 cm and each of its equal sides is 5 cm. The height of triangle is

Solution 5

Question 6

Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is

Solution 6

Question 7

Each side of an equilateral triangle is 10 cm long. The height of the triangle is

Solution 7

Question 8

The height of an equilateral triangle is 6 cm. Its area is

Solution 8

Question 9

The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is

(a) 480 m²

(b) 320m²

(c) 384 m²

(d) 360m²Solution 9

Question 10

The sides of a triangle are in the ratio 5:12:13 and its perimeter is 150 cm. The area of the triangle is

(a) 375 cm²

(b) 750 cm²

(c) 250 cm²

(d) 500 cm²Solution 10

Question 11

The lengths of the three sides of a triangle are 30 cm, 24 cm and 18 cm respectively. The length of the altitude of the triangle corresponding to the smallest side is

(a) 24 cm

(b) 18 cm

(c) 30 cm

(d) 12 cmSolution 11

Question 12

The base of an isosceles triangle is 16 cm and its area is 48 cm². The perimeter of the triangle is

(a) 41 cm

(b) 36 cm

(c) 48 cm

(d) 324 cmSolution 12

Question 13

Solution 13

Question 14

Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is

(a)156 cm²

(b)78 cm²

(c) 60 cm²

(d) 120 cm²Solution 14

Question 15

The base of a right triangle is 48 cm and its hypotenuse is 50 cm long. The area of the triangle is

(a) 168 cm²

(b) 252 cm²

(c) 336 cm²

(d) 504 cm²Solution 15

Question 16

Solution 16

Exercise Ex. 14

Question 1

Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.Solution 1

Question 2

The base of the triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height.Solution 2

Question 3

Find the area of triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side.Solution 3

Question 4

Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.Solution 4

Question 5

Find the area of a triangular field whose sides are 91m, 98 m, 105m in length. Find the height corresponding to the longest side.Solution 5

Question 6

The sides of a triangle are in the ratio 5: 12:13 and its perimeter is 150 m. Find the area of the triangle.Solution 6

Question 7

The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the field. Also, find the cost of ploughing the field at Rs. 5 per m².Solution 7

It is given that the sides a, b, c of the triangle are in the ratio 25 : 17 : 12,

i.e. a : b : c = 25 : 17 : 12

⇒ a = 25x, b = 17x and c = 12x

Given, perimeter = 540 m

⇒ 25x + 17x + 12x = 540

⇒ 54x = 540

⇒ x = 10

So, the sides of the triangle are

a = 25x = 25(10) = 250 m

b = 17x = 17(10) = 170 m

c = 12x = 12(10) = 120 m

Cost of ploughing the field = Rs. 5/m²

⇒ Cost of ploughing 9000 m² = Rs. (5 × 9000) = Rs. 45,000Question 8

Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find (i) the area of the field and (ii) the length of the perpendicular from the opposite vertex on the side measuring 154 m.Solution 8

Question 9

Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measure is 20 cm.Solution 9

Question 10

The base of the isosceles triangle measures 80 cm and its area is 360cm².Find the perimeter of the triangle.Solution 10

Question 11

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle.

HINT Ratio of sides = 3 : 3 : 2.

* Back answer incorrectSolution 11

It is given that the ratio of equal side to its base is 3 : 2.

⇒ Ratio of sides of isosceles triangle = 3 : 3 : 2

i.e. a : b : c = 3 : 3 : 2

⇒ a = 3x, b = 3x and c = 2x

Given, perimeter = 32 cm

⇒ 3x + 3x + 2x = 32

⇒ 8x = 32

⇒ x = 4

So, the sides of the triangle are

a = 3x = 3(4) = 12 cm

b = 3x = 3(4) = 12 cm

c = 2x = 2(4) = 8 cm

Question 12

The perimeter of a triangle is 50 cm. One side of the triangle is 4 cm longer than the smallest side and the third side is 6 cm less than twice the smallest side. Find the area of the triangle.Solution 12

Let the three sides of a triangle be a, b and c respectively such that c is the smallest side.

Then, we have

a = c + 4

And, b = 2c – 6

Given, perimeter = 50 cm

⇒ a + b + c = 50

⇒ (c + 4) + (2c – 6) + c = 50

⇒ 4c – 2 = 50

⇒ 4c = 52

⇒ c = 13

So, the sides of the triangle are

a = c + 4 = 13 + 4 = 17 cm

b = 2c – 6 = 2(13) – 6 = 20 cm

c = 13 cm

Question 13

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m, 15 m. The advertisements yield an earning of Rs.2000 per m² a year. A company hired one of its walls for 6 months. How much rent did it pay?Solution 13

Three sides of a wall are 13 m, 14 m and 15 m respectively.

i.e.

a = 13 m, b = 14 m and c = 15 m

Rent for a year = Rs. 2000/m²

⇒ Rent for 6 months = Rs. 1000/m²

Thus, total rent paid for 6 months = Rs. (1000 × 84) = Rs. 84,000 Question 14

The perimeter of the isosceles triangle is 42 cm and its base is times with each of the equal sides. Find (i) the length of the equal side of the triangle (ii) the area of the triangle (iii)the height of the triangle. (Given, )Solution 14

Question 15

If the area of an equilateral triangle is cm², find its perimeter.Solution 15

Question 16

If the area of an equilateral triangle is cm², find its height.Solution 16

Question 17

Each side of the equilateral triangle measures 8 cm. Find (i) the area of the triangle, correct to 2 places of decimal and (ii) the height of the triangle, correct to 2 places of decimal. Take =1.732.Solution 17

(i) Area of an equilateral triangle=

Where a is the side of the equilateral triangle

Question 18

The height of an equilateral triangle measures 9 cm. Find its area, correct to 2 places of decimal. Take =1.732.Solution 18

Question 19

The base of the right -angled triangle measures 48 cm and its hypotenuse measures 50 cm; find the area of the triangle.Solution 19

Question 20

Find the area of the shaded region in the figure given below.

Solution 20

In right triangle ADB, by Pythagoras theorem,

AB² = AD² + BD² = 12² + 16² = 144 + 256 = 400

⇒ AB = 20 cm

For ΔABC, 

Thus, area of shaded region

= Area of ΔABC – Area of ΔABD

= (480 – 96) cm²

= 384 cm²Question 21

The sides of a quadrilateral ABCD taken in order are 6 cm, 8 cm, 12 cm and 14 cm respectively and the angle between the first two sides is a right angle. Find its area. (Given, )Solution 21

Let ABCD be the given quadrilateral such that ∠ABC = 90° and AB = 6 cm, BC = 8 cm, CD = 12 cm and AD = 14 cm.

In ΔABC, by Pythagoras theorem,

AC² = AB² + BC² = 6² + 8² = 36 + 64 = 100

⇒ AC = 10 cm

In ΔACD, AC = 10 cm, CD = 12 cm and AD = 14 cm

Let a = 10 cm, b = 12 cm and c = 14 cm

Thus, area of quadrilateral ABCD

= A(ΔABC) + A(ΔACD)

= (24 + 58.8) cm²

= 82.8 cm²Question 22

Find the perimeter and area of a quadrilateral ABCD in which BC = 12 cm, CD = 9 cm, BD = 15 cm, DA = 17 CM and ∠ABD = 90°.

Solution 22

In ΔABD, by Pythagoras theorem,

AB² = AD² – BD² = 17² – 15² = 289 – 225 = 64

⇒ AB = 8 cm

∴ Perimeter of quadrilateral ABCD = AB + BC + CD + AD

= 8 + 12 + 9 + 17

= 46 cm

In ΔBCD, BC = 12 cm, CD = 9 cm and BD = 15 cm

Let a = 12 cm, b = 9 cm and c = 15 cm

Thus, area of quadrilateral ABCD

= A(ΔABD) + A(ΔBCD)

= (60 + 54) cm²

= 114 cm² Question 23

Find the perimeter and area of the quadrilateral ABCD in which AB = 21 cm, ∠BAC = 90°, AC = 20 cm, CD = 42 cm and AD = 34 cm.

Solution 23

In ΔBAC, by Pythagoras theorem,

BC² = AC² + AB² = 20² + 21² = 400 + 441 = 841

⇒ BC = 29 cm

∴ Perimeter of quadrilateral ABCD = AB + BC + CD + AD

= 21 + 29 + 42 + 34

= 126 cm

In ΔACD, AC = 20 cm, CD = 42 cm and AD = 34 cm

Let a = 20 cm, b = 42 cm and c = 34 cm

Thus, area of quadrilateral ABCD

= A(ΔABC) + A(ΔACD)

= (210 + 336) cm²

= 546 cm² Question 24

Find the area of the quadrilateral ABCD in which BCD is an equilateral triangle, each of whose side is 26 cm, AD=24 cm and.Also, find the perimeter of the quadrilateral [Given =1.73]

Solution 24

Perimeter of quad. ABCD = AB + BC + CD + DA = 10 + 26 + 26 + 24 = 86 cmQuestion 25

Find the area of a parallelogram ABCD in which AB= 28 cm , BC=26 cm and diagonal AC=30 cm.

Solution 25

Question 26

Find the area parallelogram ABCD in which AB=14 cm, BC=10 cm and AC= 16 cm. [Given =1.73]

Solution 26

Question 27

In the given figure ABCD is a quadrilateral in which diagonal BD=64 cm, AL  BD and CM BD such that AL= 16.8 cm and CM=13.2 cm. Calculate the area of quadrilateral ABCD.

Solution 27

Question 28

The area of a trapezium is 475 cm² and its height is 19 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other.Solution 28

Let the smaller parallel side of trapezium = x cm

Then, larger parallel side = (x + 4) cm

Thus, the lengths of two parallel sides are 23 cm and 27 cm respectively.Question 29

In the given figure, a ΔABC has been given in which AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC, a parallelogram DBCE of the same area as that of ΔABC is constructed. Find the height DL of the parallelogram.

Solution 29

In ΔABC, AB = 7.5 cm, BC = 7 cm and AC = 6.5 cm

Let a = 7.5 cm, b = 7 cm and c = 6.5 cm

Question 30

A field is in the shape of a trapezium having parallel sides 90 m and 30 m. These sides meet the third side at right angles. The length of the fourth side is 100 m. If it costs Rs.5 to plough 1 m² of the field, find the total cost of ploughing the field.Solution 30

Construction: Draw BT ⊥ CD

In ΔBTC, by Pythagoras theorem,

BT² = BC² – CT² = 100² – 60² = 10000 – 3600 = 6400

⇒ BT = 80 m

⇒ AD = BT = 80 m

Cost of ploughing 1 m² field = Rs. 5

⇒ Cost of ploughing 4800 m² field = Rs. (5 × 4800) = Rs. 24,000Question 31

A rectangular plot is given for constructing a house, having a measurement of 40 m long and 15 m in the front. According to the laws, a minimum of 3-m-wide space should be left in the front and back each and 2 m wide space on each of the sides. Find the largest area where house can be constructed.Solution 31

Length of rectangular plot = 40 m

Width of rectangular plot = 15 m

Keeping 3 m wide space in the front and back,

length of rectangular plot = 40 – 3 – 3 = 34 m

Keeping 2 m wide space on both the sides,

width of rectangular plot = 15 – 2 – 2 = 11 m

Thus, largest area where house can be constructed

= 34 m × 11 m

= 374 m²Question 32

A rhombus -shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs.5 per cm². Find the cost of painting.Solution 32

Let ABCD be the rhombus-shaped sheet.

Perimeter = 40 cm

⇒ 4 × Side = 40 cm

⇒ Side = 10 cm

⇒ AB = BC = CD = AD = 10 cm

Let diagonal AC = 12 cm

Since diagonals of a rhombus bisect each other at right angles,

AO = OC = 6 cm

In right ΔAOD, by Pythagoras theorem,

OD² = AD² – AO² = 10² – 6² = 100 – 36 = 64

⇒ OD = 8 cm

⇒ BD = 2 × OD = 2 × 8 = 16 cm

Now,

Cost of painting = Rs. 5/cm²

∴ Cost of painting rhombus on both sides = Rs. 5 × (96 + 96)

= Rs. 5 × 192

= Rs. 960Question 33

The difference between the semiperimeter and sides of a ΔABC are 8 cm, 7 cm, and 5 cm respectively. Find the area of the triangle.Solution 33

Let the sides of a triangle be a, b, c respectively and ‘s’ be its semi-perimeter.

Then, we have

s – a = 8 cm

s – b = 7 cm

s – c = 5 cm

Now, (s – a) + (s – b) + (s – c) = 8 + 7 + 5

⇒ 3s – (a + b + c) = 20

⇒ 3s – 2s = 20

⇒ s = 20

Thus, we have

a = s – 8 = 20 – 8 = 12 cm

b = s – 7 = 20 – 7 = 13 cm

c = s – 5 = 20 – 5 = 15 cm

Question 34

A floral design on a floor is made up of 16 tiles , each triangular in shape having sides 16 cm, 12 cm and 20 cm. Find the cost of polishing the tiles at Re 1 per sq cm.

Solution 34

Question 35

An umbrella is made by stitching by 12 triangular pieces of cloth, each measuring ((50 cmx20 cm x50 cm).Find the area of the cloth used in it.

Solution 35

Question 36

In the given figure, ABCD is a square with diagonal 44 cm. How much paper of each shade is needed to make a kite given in the figure?

Solution 36

 In ΔAEF, AE = 20 cm, EF = 14 cm and AF = 20 cm

Let a = 20 cm, b = 14 cm and c = 20 cm

Question 37

A rectangular lawn, 75 m by 60 m, has two roads, each road 4 m wide, running through the middle of the lawn, one parallel to length and the other parallel to breadth, as shown in the figure. Find the cost of gravelling the roads at Rs.50 per m².

Solution 37

For road ABCD, i.e. for rectangle ABCD,

Length = 75 m

Breadth = 4 m

Area of road ABCD = Length × Breadth = 75 m × 4m = 300 m²

For road PQRS, i.e. for rectangle PQRS,

Length = 60 m

Breadth = 4 m

Area of road PQRS = Length × Breadth = 60 m × 4 m = 240 m²

For road EFGH, i.e. for square EFGH,

Side = 4 m

Area of road EFGH = (Side)² = (4)² = 16 m²

Total area of road for gravelling

= Area of road ABCD + Area of road PQRS – Area of road EFGH

= 300 + 240 – 16

= 524 m²

Cost of gravelling the road = Rs. 50 per m²

∴ Cost of gravelling 524 m² road = Rs. (50 × 524) = Rs. 26,200Question 38

The shape of the cross section of canal is a trapezium. If the canal is 10 m wide at the top, 6 m wide at the bottom and the area of its cross section is 640 m², find the depth of the canal.Solution 38

Area of cross section = Area of trapezium = 640 m²

Length of top + Length of bottom

= sum of parallel sides

= 10 m + 6 m

= 16 m

Thus, the depth of the canal is 80 m.Question 39

Find the area of a trapezium whose parallel sides are 11 m and 25 m long, and the nonparallel sides are 15 m and 13 m long.Solution 39

From C, draw CE ∥ DA.

Clearly, ADCE is a parallelogram having AD ∥ EC and AE ∥ DC such that AD = 13 m and D = 11 m.

AE = DC = 11 m and EC = AD = 13 m

⇒ BE = AB – AE = 25 – 11 = 14 m

Thus, in ΔBCE, we have

BC = 15 m, CE = 13 m and BE = 14 m

Let a = 15 m, b = 13 m and c = 14 m

Question 40

The difference between the lengths of the parallel sides of a trapezium is 8 cm, the perpendicular distance between these sides is 24 cm and the area of the trapezium is 312 cm². Find the length of each of the parallel sides.Solution 40

Let the smaller parallel side = x cm

Then, longer parallel side = (x + 8) cm

Height = 24 cm

Area of trapezium = 312 cm²

Thus, the lengths of parallel sides are 9 cm and 17 cm respectively.Question 41

A parallelogram and a rhombus are equal in area. The diagonals of the rhombus measure 120 m and 44 m. If one of the sides of the parallelogram measures 66 m, find its corresponding altitude.Solution 41

Area of parallelogram = Area of rhombus

Question 42

A parallelogram and a square have the same area. If the sides of the square measures 40 m and altitude of the parallelogram measures 25 m, find the length of the corresponding base of the parallelogram.Solution 42

Area of parallelogram = Area of square

Question 43

Find the area of a rhombus one side of which measures 20 cm and one of whose diagonals is 24 cm.Solution 43

Let ABCD be a rhombus and let diagonals AC and BD intersect each other at point O.

We know that diagonals of a rhombus bisect each other at right angles.

Thus, in right-angled ΔAOD, by Pythagoras theorem,

OD² = AD² – OA² = 20² – 12² = 400 – 144 = 256

⇒ OD = 16 cm

⇒ BD = 2(OD) = 2(16) = 32 cm

Question 44

The area of a rhombus is 480 cm², and one of its diagonals measures 48 cm. Find (i) the length of the other diagonal, (ii) the length of each of its sides, and (iii) its perimeter.Solution 44

(i) Area of a rhombus = 480 cm²

(ii) Let diagonal AC = 48 cm and diagonal BD = 20 cm

We know that diagonals of a rhombus bisect each other at right angles.

Thus, in right-angled ΔAOD, by Pythagoras theorem,

AD² = OA² + OD² = 24² + 10² = 576 + 100 = 676

⇒ AD = 26 cm

⇒ AD = BC = CD = AD = 26 cm

Thus, the length of each side of rhombus is 26 cm.

(iii) Perimeter of a rhombus = 4 × side = 4 × 26 = 104 cm