Author Archives: Rohit

Question 1.
What cross-sections do you get when you give a
(i) vertical cut
(ii) horizontal cut to the following solids
(a) A brick
(b) A round apple
(c) A die
(d) A circular pipe
(e) An ice cream cone
Solution:

Question 1.
Simplify combining like terms:
(i) 21b -32 + 7b- 206
(ii) -z² + 13z2 -5z + 7z³ – 15²
(iii) p – (p – q) – q – (q – p)
(iv) 3a – 2b — ab – (a – b + ab) + 3ab + 6 – a
(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² -3y²
(vi) (3y² + 5y – 4) – (8y – y² – 4)
Solution:
(i) 21b – 32 + 7b – 206
Re-arranging the like terms, we get
216 + 7b – 206 – 32
= (21 + 7 – 20)b – 32
= 8b – 32 which is required.

(ii) -z² + 13z² – 5z – 15z
Re-arranging the like terms, we get
7z³ – z² + 13z² – 5z + 5z – 15z
= 7z³ + (-1 + 13)z² + (-5 – 15)z
= 7z³ + 12z² – 20z which is required.

(iii) p – (p – q) – q – (q – p)
=p – p + q – q – q + p
Re-arranging the like terms, we get

= p – q which is required.

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b – a
Re-arranging the like terms, we get
= 3a – a – a – 2b + b + b – ab – ab + 3ab

= a + ab which is required.

(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
Re-arranging the like terms, we get
5x²y + 3x²y + 8xy² – 5x² + x² – 3y² – y² – 3y²
= 8x²y + 8xy² – 4x² – 7y² which is required.

(vi) (3y² + 5y – 4) – (8y – y² – 4)
= 3y² + 5y – 4 – 8y + y² + 4 (Solving the brackets)
Re-arranging the like terms, we get
= 3y² + y² + 5y – 8y – 4 + 4
= 4y² – 3y which is required.

Ex 12.2 Class 7 Maths Question 2.
Add:
(i) 3mn, -5mn, 8mn, -4mn
(ii) t – 8tz, 3tz, -z, z – t
(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
(iv) a + b – 3, b – a + 3, a – b + 3
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(vii) 4x²y, -3xy², -5xy², 5x²y
(viii) 3p²q² – 4pq + 5, -10p²q², 15 + 9pq + 7p²q²
(ix) ab – 4a, 4b – ab, 4a – 4b
(x) x² – y² – 1, y² – 1 – x², 1 – x² – y²
Solution:
(i) 3mn, -5mn, 8mn, -4mn
= (3 mn) + (-5 mn) + (8 mn) + (- 4 mn)
= (3 – 5 + 8 – 4)mn
= 2mn which is required.

(ii) t – 8tz, 3tz – z, z – t
t – 8tz + 3tz – z + z – t
Re-arranging the like terms, we get
t – t – 8tz + 3tz – z + z
⇒ 0 – 5 tz + 0
⇒ -5tz which is required.

(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
= -7mn + 5 + 12mn + 2 + 9mn – 8 + (-2mn) – 3
Re-arranging the like terms, we get
-7mn + 12 mn + 9mn – 2 mn + 5 + 2 – 8 – 3

= 12 mn – 4 which is required.

(iv) a + b – 3, b – a + 3, a – b + 3
⇒ a + b – 3 + b – a + 3 + a – b + 3
Re-arranging the like terms, we get
a – a + a + b + b – b – 3 + 3 + 3
⇒ a + b + 3 which is required.

(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
∴ 14x + 14y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
Re-arranging the like terms, we get
-12xy + 8xy + 4xy + 14x – 7x + 10y – 10y – 13 + 18

= 0 + 7x + 0 + 5
= 7x + 5 which is required

(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5 5m -In + 3n — 4m + 2 + 2m – 3mn – 5
Re-arranging the like terms, we get
5m – 4m + 2m – 7n + 3n- 3mn + 2 – 5
= 3m – 4n – 3mn – 3 which is required.

(vii) 4x²y, -3xy², -5xy², 5x²y
Re-arranging the like terms and adding, we get
4x²y – 5xy² – 3xy² + 5x²y
=9x²y — 8xy² which is required.

(viii) 3p²q² – 4pq + 5, -10p²q², 15 + 9pq + 7p²q²
= (3p²q² – 4pq + 5) + (-10p²q²) + (15 + 9pq + 7p²q²)
= 3p²q² – 4pq + 5 – 10p²q² + (15 + 9pq + 7p²q² )
= 3p²q² + 7p²q² – 10p²q² – 4pq + 9pq + 5 + 15
= 10p²q² – 10p²q² + 5pq + 20
= 0 + 5pq + 20
= 5pq + 20 which is required.

(ix) ab – 4a, 4b – ab, 4a – 4b
= ab – 4a + 4b – ab + 4a – 4b

= 0 + 0 + 0 = 0 which is required.

(x) x² – y² – 1, y² – 1 – x², 1 – x² – y²
= x² – y² – 1 + y² – 1 – x² + 1 – x² – y²

= -x² – y² – 1
= -(x² + y² + 1) which is required.

Ex 12.2 Class 7 Maths Question 3.
Subtract:
(i) -5y² from y²
(ii) 6xy from -12xy
(iii) (a – b) from (a + b)
(iv) a(b – 5) from b(5 – a)
(v) -m² + 5mn from 4m² – 3mn + 8
(vi) -x² + 10x – 5 from 5x – 10
(vii) 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
(viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
Solution:
(i) -5y² from y² = y² – (-5y²)
= y² + 5y² = 6y²

(ii) 6ry from -12ry = -12xy – 6xy = -18xy which is required.
(iii) (a – b) from (a + b)
= (a + b) – (a – b)
= a + b – a + b = 2b which is required

(iv) a(b – 5) from b(5 – a)
= b(5 – a) – a(b – 5)
= 5b – ab – ab + 5a
= 5a – 2ab + 5b
= 5a + 5b – 2ab which is required.

(v) -m² + 5mn from 4m² – 3mn + 8
= (4m² – 3mn + 8) – (-m² + 5mn)
= 4m² – 3mn + 8 + m² – 5mn
= 4m² + m² – 3mn – 5mn + 8
= 5m² – 8mn + 8 which is required.

(vi) -x² + 10x – 5 from 5x – 10
= (5x – 10) – (-x² + 10x – 5)
= 5x – 10 + x² – 10x + 5
= x² + 5x – 10x – 10 + 5
= x² – 5x – 5 which is required.

(vii) 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
= (3ab – 2a² – 2b²) – (5a² – 7ab + 5b²)
= 3ab – 2a² – 2b² – 5a² + 7ab – 5b²
= 3ab + 7ab – 2a² – 5a² – 2b² – 5b²
= 10ab – 7a² – 7b²
which is required.

(viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
= (5p² + 3q² – pq) – (4pq – 5q² – 3p²)
= 5p² + 3q² – pq – 4pq + 5q² + 3p²
= 5p² + 3p² + 3q² + 5q² – pq – 4pq
= 8p² + 8q² – 5pq
which is required.

Ex 12.2 Class 7 Maths Question 4.
(a) What should be added to x² + xy + y²to obtain 2x² + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get -3a + 7b + 16?
Solution:
(a) (2x² + 3xy) – (x² + xy + y²)
= 2x² + 3xy – x² – xy – y²
= 2x² – x² + 3xy – xy – y²
= x² + 2xy – y² is required expression.

(b) (2a + 8b + 10) – (-3a + 7b + 16)
= 2a + 8b + 10 + 3a – 7b – 16
= 2a + 3a + 8b – 7b + 10 – 16
= 5a + b – 6 is required expression.

Ex 12.2 Class 7 Maths Question 5.
What should be taken away from 3x² – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20?
Solution:
Let A be taken away.
∴ (3x² – 4y² + 5 xy + 20)-A
= -x² – y² + 6xy + 20
⇒ A = (3x² – 4y² + 5xy + 20) – (-x²2 – y² + 6xy + 20)
= 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20
= 3x² + x² – 4y² + y² + 5xy – 6xy + 20 – 20
= 4x² – 3y² – xy is required expression.

Ex 12.2 Class 7 Maths Question 6.
(a) From the sum of 3x – y + 11 and -y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and -x² + 2x + 5.
Solution:
(a) Sum of 3x – y + 11 and -y – 11
= (3x – y + 11) + (-y – 11)
= 3x – y + 11 – y – 11
∴ 3x – 2y – (3x – 2y) – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= 3x – 3x – 2y + y + 11
= -y + 11 is required solution.

(b) Sum of (4 + 3x) and (5 – 4x + 2x²)
= 4 + 3x + 5 – 4x + 2x²
= 2x² – 4x + 3x + 9 = 2x² – x + 9
Sum of (3x² – 5x) and (-x² + 2x + 5)
= (3x² – 5x) + (-x² + 2x + 5)
= 3x² – 5x – x² + 2x + 5 = 2x² – 3x + 5
Now (2x² – x + 9) – (2x² – 3x + 5)
= 2x² – x + 9 – 2x² + 3x – 5
= 2x² – 2x² + 3x – x + 4
= 2x + 4 is required expression.

NCERT Solutions

Question 1.
If m = 2, find the value of:
(i) m – 2
(ii) 3m – 5
(iii) 9 – 5m
(iv) 3m² – 2m – 7
(v) 5m2−4
Solution:
(i) m – 2
Putting m = 2, we get
2 – 2 = 0

(ii) 3m – 5
Putting m = 2, we get
3 × 2 – 5 = 6 – 5 = 1

(iii) 9 – 5m
Putting m = 2, we get
9 – 5 × 2 = 9 – 10 = -1

(iv) 3m² – 2m – 7 Putting m = 2, we get
3(2)² – 2(2) – 7 = 3 × 4 – 4 – 7
=12 – 4 – 7 = 12 – 11 = 1

(v) 5m2−4
Putting m = 2, we get
5×22−4=5−4=1

Ex 12.3 Class 7 Maths Question 2.
If p = -2, find the value of:
(i) 4p + 7
(ii) -3p² + 4p + 7
(iii) -2p³ – 3p² + 4p + 7
Solution:
(i) 4p + 7
Putting p = -2, we get 4(-2) + 7 = -8 + 7 = -1

(ii) -3p² + 4p + l
Putting p = -2, we get
-3(-2)² + 4(-2) + 7
= -3 × 4 – 8 + 7 = -12 – 8+ 7 = -13

(iii) -2p³ – 3p² + 4p + 7
Putting p = -2, we get
– 2(-2)³ – 3(-2)² + 4(-2) + 7
= -2 × (-8) – 3 × 4 – 8 + 7
= 16 – 12 – 8 + 7 = 3

Ex 12.3 Class 7 Maths Question 3.
If a = 2, b = -2, find the value of:
(i) a² + b²
(ii) a² + ab + b²
(iii) a² – b²
Solution:
(i) a² + b²
Putting a = 2 and b = -2, we get
(2)² + (-2)² = 4 + 4 = 8

(ii) a² + ab + b²
Putting a = 2 and b = -2, we get
(2)² + 2(-2) + (-2)² = 4 – 4 + 4 = 4

(iii) a² – b²
Putting a = 2 and b = -2, we get
(2)² – (-2)² = 4 – 4 = 0

Ex 12.3 Class 7 Maths Question 4.
When a = 0, b = -1, find the value of the given expressions:
(i) 2a + 2b
(ii) 2a² + b² + 1
(iii) 2a²b + 2ab² + ab
(iv) a² + ab + 2
Solution:
(i) 2a + 2b = 2(0) + 2(-1)
= 0 – 2 = -2 which is required.

(ii) 2a² + b² + 1
= 2(0)² + (-1)² + 1 =0 + 1 + 1 = 2 which is required.

(iii) 2a²b + 2ab² + ab
= 2(0)² (-1) + 2(0)(-1)² + (0)(-1)
=0 + 0 + 0 = 0 which is required.

(iv) a² + ab + 2
= (0)² + (0)(-1) + 2
= 0 + 0 + 2 = 0 which is required.

Ex 12.3 Class 7 Maths Question 5.
Simplify the expressions and find the value if x is equal to 2.
(i) x + 7 +4(x – 5)
(ii) 3(x + 2) + 5x – 7
(iii) 6x + 5(x – 2)
(iv) 4(2x – 1) + 3x + 11
Solution:
(i) x + 7 + 4(x – 5) = x + 7 + 4x – 20 = 5x – 13
Putting x = 2, we get
= 5 × 2 – 13 = 10 – 13 = -3
which is required.

(ii) 3(x + 2) + 5x – 7 = 3x + 6 + 5x -7 = 8x – 1
Putting x = 2, we get
= 8 × 2 – 1 = 16 – 1 = 15
which is required.

(iii) 6x + 5(x – 2) = 6x + 5x – 10
= 11 × – 10
Putting x = 2, we get
= 11 × 2 – 10 = 22 – 10 = 12
which is required.

(iv) 4(2x – 1) + 3x + 11 = 8x – 4 + 3x + 11
= 11x + 7
Putting x = 2, we get
= 11 × 2 + 7 = 22+ 7 = 29

Ex 12.3 Class 7 Maths Question 6.
Simplify these expressions and find their values if x = 3, a = -1, b = -2.
(i) 3x – 5 – x + 9
(ii) 2 – 8x + 4x + 4
(iii) 3a + 5 – 8a + 1
(iv) 10 – 3b – 4 – 55
(v) 2a – 2b – 4 – 5 + a
Solution:
(i) 3x – 5 – x + 9 = 2x + 4
Putting x = 3, we get
2 × 3 + 4 = 6 + 4 = 10
which is required.

(ii) 2 – 8x + 4x + 4 = -8x + 4x + 2 + 4 = -4x + 6
Putting x = 2, we have
= -4 × 2 + 6 = -8 + 6 =-2
which is required.

(iii) 3a + 5 – 8a +1 = 3a – 8a + 5 + 1 = -5a + 6
Putting a = -1, we get
= -5(-1) + 6 = 5 + 6 = 11
which is required.

(iv) 10 – 3b – 4 – 5b = -3b – 5b + 10 – 4
= -8b + 6
Putting b = -2, we get
= -8(-2) + 6 = 16 + 6 = 22
which is required.

(v) 2a – 2b – 4 – 5 + a = 2a + a – 2b – 4 – 5
= 3a – 26 – 9
Putting a = -1 and b = -2, we get
= 3(-1) – 2(-2) – 9
= -3 + 4 – 9 = 1 – 9 = -8
which is required.

Ex 12.3 Class 7 Maths Question 7.
(i) If z = 10, find the value of z² – 3(z – 10).
(ii) If p = -10, find the value of p² -2p – 100.
Solution:
(i) z² – 3(z – 10)
= z² – 3z + 30
Putting z = 10, we get
= (10)² – 3(10) + 30
= 1000 – 30 + 30 = 1000 which is required.

(ii) p² – 2p – 100
Putting p = -10, we get
(-10)² – 2(-10) – 100
= 100 + 20 – 100 = 20 which is required.

Ex 12.3 Class 7 Maths Question 8.
What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0?
Solution:
2x² + x – a = 5
Putting x = 0, we get
2(0)² + (0) – a = 5
0 + 0 – a = 5
-a = 5
⇒ a = -5 which is required value.

Ex 12.3 Class 7 Maths Question 9.
Simplify the expression and find its value when a = 5 and b = -3.
2(a² + ab) + 3 – ab
Solution:
2(a² + ab) + 3 – ab = 2a2 + 2ab + 3 – ab
= 2a² + 2ab – ab + 3
= 2ab + ab + 3
Putting, a = 5 and b = -3, we get
= 2(5)² + (5)(-3) + 3
= 2 × 25 – 15 + 3
= 50 – 15 + 3
= 53 – 15 = 38
Hence, the required value = 38.

Question 1.
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.

If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern. How many segments are required to form 5, 10, 100 digits of the kind 6, 4, 3.
Solution:
(i) The number of line segments required to form
n digits is given by the expressions.

For 5 figures, the number of line segments = 5 × 5 + 1 = 25 + 1 = 26
For 10 figures, the number of line segments = 5 × 10 + 1
= 50 + 1 = 51
For 100 figures, the number of line segments = 5 × 100 + 1
= 500 + 1 = 501

(ii) 
For 5 figures, the number of line segments =3 ×5 + 1
= 15 + 1 = 16
For 10 figures, the number of line segments = 3 × 10 + 1
= 30 + 1 = 31
For 100 figures, the number of line segments = 3 × 100 + 1
= 300 + 1 = 301

(iii) 
For 5 figures, the number of line numbers = 5 × 5 + 2
= 25 + 2 = 27
For 10 figures, the number of line segments = 5 × 10 + 2
= 50 + 2 = 52
For 100 figures, the number of line segments = 5 × 100 + 2
= 500 + 2 = 502

Ex 12.4 Class 7 Maths Question 2.
Use the given algebraic expression to complete the table of number patterns:

Solution:
(i) Given expression is 2n – 1
For n = 100, 2 × 100 – 1
= 200 – 1 = 199

(ii) Given expression is 3n + 2
For n = 5, 3 × 5 + 2 = 15 + 2 = 17
For n = 10, 3 × 10 + 2 = 30 + 2 = 32
For n = 100, 3 × 100 + 2 = 300 + 2 = 302

(iii) Given expression is 4n + 1
For n = 5, 4 × 5 + 1 = 20 + 1 = 21
For n = 10, 4 × 10 + 1 = 40 + 1 = 41
For n = 100, 4 × 100 + 1 = 400 + 1 = 401

(iv) Given expression is 7n + 20
For n = 5, 7 × 5 + 20 = 35 + 20 = 55
For n = 10, 7 × 10 + 20 = 70 + 20 = 90
For n = 100, 7 × 100 + 20 = 700 + 20 = 720

(v) Given expression is n² + 1
For n = 5, 5² + 1 = 25 + 1 = 26
For n = 10, 10² + 1 = 100 + 1 = 101

Question 1.
The length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. Find
(i) its area
(ii) the cost of the land if 1 m² of the land costs ₹ 10,000.
Solution:
(i) For a rectangular piece of land,
Length = 500 m
Breadth = 300 m
∴ Area of the rectangular piece of land = Length × Breadth , = 500 × 300 m² = 150,000 m²

(ii) 1 m² of the land costs = ₹ 10,000
∵ 150,000 m² of the land costs = ₹ 10,000 × 150,000 = ₹ 1,500,000,000.

Question 2.
Find the area of a square park whose perimeter is 320 m.
Solution:
We have, perimeter of a square park = 320 m
∴ Side of the square park = ([late×]\frac { 320 }{ 4 } [/late×]) m = 80 m
∴ Area of square park = (Side)² = (80)² m² = 6400 m²

Question 3.
Find the breadth of a rectangular plot of land, if its area is 440 m² and the length is 22 m. Also find its perimeter.
Solution:
Let the breadth of the rectangular plot of land be b.
It is given that length of the plot = 22 m
and the area of the plot = 440 m²

Question 4.
The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also, find the area.
Solution:
Let the breadth of the rectangular sheet be b cm.
It is given that its length is 35 cm and perimeter is 100 cm.

Question 5.
The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Solution:

Question 6.
A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side. Also find which shape encloses more area?
Solution:


Hence, the measure of the side of the square is 31 cm and the square shape encloses more area.

Question 7.
The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also find the area of the rectangle.
Solution:

Question 8.
A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m. Find the cost of white washing the wall, if the rate of white washing the wall is ₹ 20 per m².

Solution:

Question 1.
Find the area of each of the following parallelograms :

Solution:
(a) Area of the parallelogram = base × height = (7 × 4) cm² = 28 cm²

(b) Area of the parallelogram = base × height = (5 × 3) cm² = 15 cm²

(c) Area of the karallelogram = base × height = (2.5 × 3.5) cm² = 8.75 cm²

(d) Area of the parallelogram = base × height = (5 × 4.8) cm² = 24 cm²

(e) Area of the parallelogram = base × height = (2 × 4.4) cm2 = 8.8 cm²

Question 2.
Find the area of each of the following triangles :

Solution:

Question 3.
Find the missing values :

Solution:
We know that
Area of a parallelogram = Base × Height
Therefore, the missing values are calculated as shown :

Question 4.
Find the missing values :

Solution:

Question 5.
PQRS is a parallelogram (in the given figure). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find
(a) the area of the parallelogram PQRS.
(b) QN, if PS = 8 cm.

Solution:

Question 6.
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (in the given figure). If the area of the parallelogram is 1470 cm², AB = 35 cm, AD = 49 cm, find the length of BM and DL.

Solution:

Question 7.
∆ ABC is right angled at A (in the given figure). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of ∆ABC. Also find the length of AD.

Solution:

Question 8.
∆ ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (in the given figure). The height AD from A to BC is 6 cm. Find the area of A ABC. What will be the height from C to AB i.e., CE ?
Solution:

Question 1.
Find the circumference of the circles with the following radius : ( Take π 227)
(a) 14 cm
(b) 28 mm
(c) 21 cm
Solution:

Question 2.
Find the area of the following circles, given that :
(a) radius = 14 mm (Take π = 227)
(b) diameter = 49 m
(c) radius = 5cm
Solution:

Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet (Take π = 227)
Solution:

Question 4.
A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs ₹ 4 per metre (Take π = 227)

Solution:

Question 5.
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)
Solution:
Here, Outer radius, r = 4 cm
Inner radius, r = 3 cm
Area of the remaining sheet = Outer area – Inner area
= π (R² – r²) = 3.14 (4² – 3²) cm²
= 3.14 (16 – 9) cm²
= 3.14 × 7 cm² = 21.98 cm²

Question 6.
Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also its cost if one metre of the lace costs ₹ 15. (Take π = 3.14)
Solution:

Question 7.
Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
Solution:

Question 8.
Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15/m² (Take π = 3.14)
Solution:

Question 9.
Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle of the square ? (Take π = 227)
Solution:

Question 10.
From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed (as shown in the adjoining figure). Find the area of the remaining sheet (Take π = 227)

Solution:

Question 11.
A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)
Solution:
Area of the square aluminium sheet = (6)² cm² = 36 cm²
Area of the circle cut out from the sheet = (3.14 × 2 × 2) cm² = 12.56 cm²
Area of the sheet left over = (36 – 12.56) cm² = 23.44 cm²

Question 12.
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take π = 3.14)
Solution:

Question 13.
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)

Solution:

Question 14.
A circular flower garden has an area of 314 m². A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Take π = 3.14)
Solution:
Circular area of the sprinkler = πr²
= 3.14 × 12 × 12
=3.14 × 144 = 452.16 m²
Area of the circular flower garden = 314 m²
Since, area of the circular flower garden is smaller than by sprinkler. Therefore, the sprinkler will water the entire garden.

Question 15.
Find the circumference of the inner and the outer circles, shown in the adjoining figure. (Take π = 3.14)

Solution:

Question 16.
How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 227)
Solution:

Question 17.
The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 3.14)
Solution:
We know that the minute hand describes one complete revolution in one hour.
∴ Distance covered by its tip
= Circumference of the circle of radius 15 cm
= (2 × 3.14 × 15) cm
= 94.2 cm

Ex 10.1 Class 7 Maths Question 1.
Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.
Solution:
Steps of Construction :

1. Take any point P on the line AB.
2. Take any point C outside AB and join CP.
3. With P as centre, draw an arc cutting AB and PC at X and Y respectively.
4. With centre C and the same radius as in step 3, draw an arc on the opposite side of the PC to cut the PC at Q.
5. With centre Q and radius equal to XY, draw an arc cutting the arc drawn in step 4 at R.
6. Join CP and produce it in both directions to obtain the required line.

Ex 10.1 Class 7 Maths Question 2.
Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l.
Solution:
Steps of Construction:

1. Draw a line l and take any point P on it.
2. With P as centre and any radius, draw an arc to intersect line l at A and B.
3. With A as centre and radius greater than PA, draw in an arc.
4. With centre B and the same radius, as in step 2, draw another arc to intersect the arc drawn in step 2 at C.
5. Join PC and produce it to Q. Then PQ ⊥ l
6. With P as centre and radius equal to 4 cm, draw an arc to intersect PQ at X such than PX = 4 cm.
7. At X, make ∠RXP = ∠BPX.
8. Join XR to obtain the required line m.

Validity: Since ∠BXP = ∠BPX and these are alternate angles, therefore, .XR || l, i.e., m || l and contain X such that PX = 4 cm and ∠XPB = 90°.

Ex 10.1 Class 7 Maths Question 3.
Let l be a line and P be a point on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R draw a line parallel to PQ. Let this meet l at S. What shape do the two sets of parallel lines enclose?
Solution:
Steps of Construction :

1. Draw a line l and take any point P outside it.
2. Take any point Q on line l.
3. Join PQ.
4. With a centre, draw an arc cutting l and PQ at C and D respectively.
5. With centre P and the same radius as in step 4, draw an arc on the opposite side of PQ to cut PQ at E.
6. With centre E and radius equal to CD,’ draw an arc cutting the arc of step 5 at F.
7. Join PF and produce it in both directions to obtain the required line m.
8. Take any point R on m.
9. Through P, draw a line PS || PQ by following the steps already explained.
10. The shape of the figure endorsed by these lines is a parallelogram RPQS.

Ex 10.2 Class 7 Maths Question 1.
Construct ∆XYZ in which XY = 45 cm, YZ = 5 cm and ZX = 6 cm.
Solution:
Steps of Construction :

1. Draw a line segment YZ = 5 cm.
2. With Y centre and draw an arc radius cm.
3. With Z as centre and draw another arc intersecting the arc radius = 6 cm, at X.
4. Join XY and XZ to obtain the required triangle.

Ex 10.2 Class 7 Maths Question 2.
Construct an equilateral triangle of side 5.5 cm.
Solution:
Steps of Construction :

1. Draw a line segment BC = 5.5 cm.
2. With centre B and radius = 5.5 cm, draw an arc.
3. With centre C and radius = 5.5 cm, draw another arc intersecting the arc drawn in step 2 at A.
4. Join AB and AC to obtain the required triangle.

Ex 10.2 Class 7 Maths Question 3.
Draw ∆PQR with PQ = 4, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?
Solution:
Steps of Construction :

1. Draw a line segment QR = 3.5 cm.
2. With centre Q and radius = 4 cm, draw an arc.
3. With R as centre and radius = 4 cm, draw another arc intersecting the arc drawn in step 2 at P.
4. Join PQ and PR to obtain the required triangle. ∆PQR is an isosceles triangle.

Ex 10.2 Class 7 Maths Question 4.
Construct AABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
Solution:
Steps of Construction :

1. Draw a line segment BC = 6 c.m
   
2. With centre B and radius = 2.5 cm, draw an arc.
3. With centre C and radius =6.5 cm, draw another arc intersecting the arc drawn in step 2 at A.
4. Join AB and AC to obtain the required triangle.
   On measuring, we find that ∠B = 90°.

Ex 10.3 Class 7 Maths Question 1.
Construct ∆DEF such that DE = 5 cm, DF = 3 cm and m ∠EDF = 90°
Solution:
Steps of Constipation :

1. Draw a line segmerit DE = 5cm.
2. Draw ∠EDX = 90°.
3. With centre D and radius = 3 cm, draw an are to intersect DX at F.
4. Join EF to obtain the required triangle DBF.

Ex 10.3 Class 7 Maths Question 2.
Construct an isosceles triangle in which the lengths of each of its equal Sides is 6.5 cm find the angle between them is 110°
Solution:
Steps of Construction :

1. Draw a line segment BC = 6.5 cm.
2. Draw ∠CBX = 110°.
3. With B as centre and radius = 6.5 cm, draw an arc intersecting BX at A.
4. Join AC to obtain the required ∆ABC.

Ex 10.3 Class 7 Maths Question 3.
Construct ∆ABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.
Solution:
Steps of Construction :

1. Draw a line segment BC =7.5 cm.
2. Draw ∠BCX = 60°.
3. With C as centre and radius = 5 cm, draw an arc intersecting CX at A.
4. Join AB to obtain the required ∆ABC.

Ex 10.4 Class 7 Maths Question 1.
Construct ∆ABC, given m∠A = 60°, m∠B = 30° and AB = 5.8 cm.
Solution:
Steps of Construction :

1. Draw a line segment AB = 5.8 cm.
2. Draw ∠BAX = 60°.
3. Draw ∠ABY, with Y on the same side of AB
such that ∠ABY = 30°.
Let AX and BY interest at C.
Then, ∆ABC is the required triangle.

Ex 10.4 Class 7 Maths Question 2.
Construct ∆PQR if PQ = 5cm, m ∠PQR = 105° and m ∠QRP = 40°.
Solution:
Here, we are given the side PQ, ∠Q and ∠R. But to draw the triangle, we require ∠P

Ex 10.4 Class 7 Maths Question 3.
Examine whether you can construct
∆DEF such that EF = 7.2 cm, m∠E = 110° and m∠F = 80°. Justify your answer.
Solution:
Since m∠E +m∠F = 110° + 80° = 190°, so the ∆DEF cannot be drawn as the’sum of all the angles of a triangle is 180°.

Ex 10.5 Class 7 Maths Question 1.
Construct the. right-angled ∆PQR, where m∠Q = 90°, QR = 8 cm and PR = 10 cm.
Solution:
Steps of Construction :

1. Draw a line segment QR = 8 cm.
2. Draw ∠XQR = 90°.
3. With R as centre and radius =10 cm, draw an arc to intersect ray QX at P.
4. Join RP to obtain the required ∆PQR.

Ex 10.5 Class 7 Maths Question 2.
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
Solution:
Steps of Construction :

1. Draw a line segment QR = 4 cm.
2. Draw ∠XQR = 90°.
3. With R as centre and radius equal to hypotenuse 6 cm, draw an arc to intersect ray QX at P.
4. Join RP to obtain the required ∆PQR.

Ex 10.5 Class 7 Maths Question 3.
Construct an isosceles right-angled triangle ABC, where m ∠ACB = 90° and AC = 6 cm.
Solution:
Steps of Construction :

1. Draw a line segment CB = 6 cm (∵ CB = AC = 6 cm)
2. Draw ∠BCX = 90°.
3. With C as centre and radius = 6 cm, draw an arc to intersect ray CX at A.
4. Join BA to obtain the required triangle ABC.

Ex 9.1 Class 7 Maths Question 1.
List five rational numbers between
(i) -1 and 0
(ii) -2 and -1
(iii) −45 and −23
(iv) –12 and 23
Solution:

Ex 9.1 Class 7 Maths Question 2.
Write four more rational numbers in each of the following patterns :

Solution:

Ex 9.1 Class 7 Maths Question 3.
Give four rational numbers equivalent to
(i) −27
(ii) 5−3
(iii) 49
Solution:

Ex 9.1 Class 7 Maths Question 4.
Draw the number line and represent the following rational numbers on it :
(i) 34
(ii) −58
(iii) −74
(iv) 78
Solution:
(i) In order to represent 34 on the number line, we first draw a number line and mark a point O on it to represent zero. Now, mark the point P representing 3 on the number line as shown. Now, divide the segment OP into four equal parts. Let A, B, C be the points of division so that OA = AB =BC = CP. By construction, OA is three- fourth of OP. So, A represents the rational number 34

(ii) In order to represent −58 on the number line, we first draw a number line and mark a point O on it to represent zero. Now, mark the point P representing -5 on it as shown. Now, divide the segment OP into eight equal parts. Let A, B, C, D, E, F, G be the points of division such that OA = AB = BC = CD = DE = EF = FG = GP. By construction, OA is one-eighth of OP. So, A represents the rational number −58.

(iii) In order to represent −74 on the number line, we first draw a number line and mark a point O on it to represent zero. Now, mark a point P to represent -7 on the number line. Now,divide the segment OP into 4 equal parts. Let A, B, C be the points of division so that OA = AB = BC = CP. By construction, OA is one-fourth of OP. Therefore, A represents the rational number −74.

(iv) In order to represent 78 on the number line, we first draw a number line and mark a point O on it to represent zero. Now, mark the point P to represent 7 on the number line. Now, divide the segment OP into 8 equal parts. Let A, B, C, D, E, F, G be the points of division such that OA = AB = BC = CD = DE = EF = FG = GP. By construction, OA is one-eighth of OP. Therefore, A represents the rational number 78.

Ex 9.1 Class 7 Maths Question 5.
The points P, Q, R, S, T, U, A, and B on the number line are such that, TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R, and S.

Solution:

Ex 9.1 Class 7 Maths Question 6.
Which of the following pairs represents the same rational number?

Solution:

Ex 9.1 Class 7 Maths Question 7.
Rewrite the following rational numbers in the simplest form :
(i) −86
(ii) 2545
(iii) −4472
(iv) −810
Solution:

Ex 9.1 Class 7 Maths Question 8.
Fill in the boxes with the correct symbol out of >, <, and =.

Solution:
(i) Clearly, −57 is a negative rational number and 23 is a positive rational number. We know that every negative rational number is less than every positive rational number. Therefore,

(ii) Clearly, denominators of the given rational numbers are positive. The denominators are 5 and 7. Their L.C.M. is 35. So, we first express each rational number with 35 as a common denominator.

(iii) First- we write each one of the given rational numbers with a positive denominator.
Clearly, denominator of −78 is positive.
The denominator of 14−16 is negative.
So, we express it with a positive denominator as follows:

(iv) Clearly, the denominators of the given rational numbers are positive. The denominators are 5 and 4.
Their L.C.M. is 20. So, we first express each rational number with 20 as a common denominator.

(v) First we write each one of the given rational numbers with a positive denominator.
Clearly, denominator of 1−3 is negative.
So, expressing it with a positive denominator as follows:

(vi) First we write each one of the given rational numbers with a positive denominator.

(vii) Since every negative rational number is less than 0,

Ex 9.1 Class 7 Maths Question 9.
Which is greater in each of the following :
(i) 23,52
(ii) −56,−43
(iii) −34,2−3
(iv) −14,14
(v) -327,-345
Solution:

Ex 9.1 Class 7 Maths Question 10.
Write the following rational numbers in ascending order :

Solution:

Ex 9.2 Class 7 Maths Question 1.
Find the sum :

Solution:

Ex 9.2 Class 7 Maths Question 2.
Find :

Solution:

Ex 9.2 Class 7 Maths Question 3.
Find the product :

Solution:

Ex 9.2 Class 7 Maths Question 4.
Find the value of :

Solution:

Ex 8.1 Class 7 Maths Question 1.
Find the ratio of:
(a) ₹5 to 50 paise
(b) 15 kg to 210 kg
(c) 9 m to 27 cm
(d) 30 days to 36 hours
Solution:

Ex 8.1 Class 7 Maths Question 2.
In a computer lab, there are 3 computers for every 6 students. How many computers are needed for 24 students?
Solution:
∵ For every 6 students, there are 3 computers
∴ For 1 student there are 36 computer
∴ For 24 students there are 36 × 24 computers = 12 computers
Hence, 12 computers are needed for 24 students.

Ex 8.1 Class 7 Maths Question 3.
Population of Rajasthan = 570 lakhs and population of U.P. = 1660lakhs. Area of Rajasthan = 3 lakh km2 and area of U.P. = 2 lakh km².
(i) How many people are there per km² in both these states?
(ii) Which state is less populated?
Solution:

Ex 8.2 Class 7 Maths Question 1.
Convert the given fractional number to per cents :
(a) 18
(b) 54
(c) 340
(d) 27
Solution:

Ex 8.2 Class 7 Maths Question 2.
Convert the given decimal fractions to per cents.
(a) 0.65
(b) 2.1
(c) 0.02
(d) 12.35
Solution:

Ex 8.2 Class 7 Maths Question 3.
Estimate what part of the figures is coloured and hence find the per cent which is coloured.

Solution:

Ex 8.2 Class 7 Maths Question 4.
Find :
(a) 15% of 250
(b) 1% of 1 hour
(c) 20% of ₹ 2500
(d) 75% of 1 kg
Solution:

Ex 8.2 Class 7 Maths Question 5.
Find the whole quantity if
(a) 5% of it is 600
(b) 12% of it is ? 1080
(c) 40% of it is 500 km
(d) 70% of it is 14 minutes
(e) 8% of it is 40 litres
Solution:

Ex 8.2 Class 7 Maths Question 6.
Convert given per cents to decimal fractions and also to fractions in simplest forms :
(a) 25%
(b) 150%
(c) 20%
(d) 5%
Solution:

Ex 8.2 Class 7 Maths Question 7.
In a city, 30% are females, 40% are males and remaining are children. What per cent are children?
Solution:
In a city,
Percentage of females = 30%
Percentage of males = 40%
Percentage of children = (100 – 30 – 40)% = 30%

Ex 8.2 Class 7 Maths Question 8.
Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
Solution:
Percentage of voters who voted = 60%
Percentage of voters who did not vote = (100 – 60)% = 40%
Total number of voters = 15000
Number of voters who did not vote
= 40% of 15000
= ( 40100 × 15000 ) = 6000

Ex 8.2 Class 7 Maths Question 9.
Meeta saves ₹ 400 from her salary. If this is 10% of her salary, what is her salary?
Solution:

Ex 8.2 Class 7 Maths Question 10.
A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
Solution:
Out of 100, 25% matches are won. Then, out of 20, the number of matches that the team won
= 25100 × 20 = 14 × 20 = 5

Ex 8.3 Class 7 Maths Question 1.
Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.
(a) Gardening shears bought for ₹ 250 and sold for ₹ 325.
(b) A refrigerator bought for ₹ 12,000 and sold at ₹ 13,500.
(c) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000.
(d) A skirt bought for ₹ 250 and sold at ₹ 150.
Solution:

Ex 8.3 Class 7 Maths Question 2.
Convert each part of the ratio to percentage :
(a) 3 : 1
(b) 2 : 3 : 5
(c) 1 : 4
(d) 1 : 2 : 5
Solution:

Ex 8.3 Class 7 Maths Question 3.
The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solution:
Original population of the city = 25000
Decreased population of the city = 24500
∴ Decrease in population = 25000 – 24500 = 500
∴ Percentage of decrease = ( 50025000 × 100 ) % = 2%

Ex 8.3 Class 7 Maths Question 4.
Arun bought a car for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the percentage of price increase?
Solution:
Original cost of the car = ₹ 3,50,000
Increased cost of the car = ₹ 3,70,000
∴ Increase in price = ₹ (370000 – 350000)
= ₹ 20,000
∴ Percentage increase = ( 20000350000 × 100 )% = 407 % = 5 57%

Ex 8.3 Class 7 Maths Question 5.
I buy a T.V. for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it?
Solution:

Ex 8.3 Class 7 Maths Question 6.
Juhi sells a washing machine for ? 13,500. She loses 20% in the bargain. What was the price at which she bought it?
Solution:
We have, S.P. = ₹13,500 and loss = 20%.

Ex 8.3 Class 7 Maths Question 7.
(i) Chalk contains calcium, carbon and oxygen in the ratio 10:3:12. Find the percentage of carbon in chalk.
(ii) If in a stick of chalk, carbon in 3g, what is the weight of the chalk stick?
Solution:

Ex 8.3 Class 7 Maths Question 8.
Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?
Solution:

Ex 8.3 Class 7 Maths Question 9.
Find the amount to be paid at the end of 3 years in each case :
(a) Principal = ₹ 1,200 at 12% p.a.
(b) Principal = ₹ 7,500 at 5% p.a.
Solution:

Ex 8.3 Class 7 Maths Question 10.
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years?
Solution:

Ex 8.3 Class 7 Maths Question 11.
If Meena gives an interest of ₹ 45 for one year at 9% rate p.a. What is the sum she has borrowed?
Solution:

Ex 7.1 Class 7 Maths Question 1.
Complete the following statements:

(a) Two line segments are congruent if
(b) Among two congruent angles, one has a measure of 70°; the measure of the other angle is
(c) When we write ∠A = ∠B, we actually mean

Solution:

(a) Two line segments are congruent if they have the same length.
(b) Among two congruent angles, one has a measure of 70° the measure of the other angle is 70°.
(c) When we write ∠A = ∠B, we actually mean m ∠A = m ∠B.

Ex 7.1 Class 7 Maths Question 2.
Give any two real-life examples for congruent shapes.
Solution:
Two one-rupee coins, two ten-rupee notes.

Ex 7.1 Class 7 Maths Question 3.
If ∆ABC = FED under the correspondence ABC ↔ FED, write all the corresponding congruent parts of the triangles.
Solution:
∆ABC = ∆FED means that ∆ABC superposes on ∆FED exactly such that the vertices of ∆ABC fall on the vertices of ∆FED in the following order A ↔ F, B ↔ E and C ↔ D.

Ex 7.1 Class 7 Maths Question 4.
If ∆DEF = BCA, write the part(s) of ∆BCA that correspond to

Solution:

Ex 7.2 Class 7 Maths Question 1.
Which congruence criterion do you use in the following ?

Solution:

Ex 7.2 Class 7 Maths Question 2.
You want to show that ∆ART = ∆PEN.

(a) If you have to use SSS criterion, then you need to

1. AR =
2. RT =
3. AT =

(b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have

1. RT = and,
2. PN =

(c) If it is given that AT = PN and you are to use ASA criterion, you need to have

1. ?
2. ?

Solution:
(a) In order to show that ∆ART = ∆PEN using SSS criterion, we need to show that

1. AR = PE
2. RT = EN
3. AT = PN

(b) If m∠T = m∠N and to use SAS criterion, we need to show that

1. RT = EN and
2. PN = AT

(c) If AT = PN and to use ASA criterion, we need to show that

1. ∠RAT = ∠EPN and
2. ∠ATR = ∠PNE

Ex 7.2 Class 7 Maths Question 3.

Solution:

Ex 7.2 Class 7 Maths Question 4.
In ∆ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ∆PQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°.
A student says that ∆ABC = ∆PQR by AAA congruence criterion. Is he justified? Why or why not?
Solution:
In two triangles, if the three angles of one triangle are respectively equal to the three angles of the other, then the triangles are not necessarily congruent.

Ex 7.2 Class 7 Maths Question 5.
In the figure, the two triangles are congruent The corresponding parts are ∆ RAT = ?

Solution:

Ex 7.2 Class 7 Maths Question 6.
Complete the congruence statement :

Solution:

Ex 7.2 Class 7 Maths Question 7.
In a squared sheet, draw two triangles of equal areas such that
(i) the triangles are congruent.
(ii) the triangles are not congruent.
What can you say about their perimeters?
Solution:

Ex 7.2 Class 7 Maths Question 8.
Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
Solution:
In some special cases (which depend on the lengths of the sides and the size of the angle involved),

SSA is enough to show congruence. However, it is not always enough. Consider the following triangles :

Here side AB is congruent to side DE (S) side AC is congruent to side DF(S) angle C is congruent to angle F(A)

But the triangles are not congruent, as we can see.

What happens is this : If we draw a vertical line through point A in the first triangle, we can sort of “flip” side AB around this line to get the second triangle. If we were to lay one triangle on top of the other and draw the vertical line, this how it would look.

Clearly, side DE is just side AB flipped around the line. So, we have not changed the length of the side, and the other side AC (or DF) is unchanged, as is angle C (or F). So, these two triangles that have the same SSA information, but they are not congruent.

Ex 7.2 Class 7 Maths Question 9.
If ∆ABC and ∆PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Solution:
In order prove that ∆ABC ≅ ∆PQR, the additional pair of corresponding parts are named as BC = QR.
Criterion used is ASA rule of congruence

Ex 7.2 Class 7 Maths Question 10.
Explain, why ∆ABC ≅ ∆FED

Solution:

Ex 6.1 Class 7 Maths Question 1.

Solution:

Ex 6.1 Class 7 Maths Question 2.
Draw rough sketches for the following :
(a) In ∆ABC, BE is a median.
(b) in ∆PQR, PQ and PR are altitudes of the triangle.
(c) In ∆XYZ, YL is an altitude In the exterior of the triangle.
Solution:
(a) Rough sketch of median BE of ∆ABC is as shown.

(b) Rough sketch of altitudes PQ and PR of ∆PQR is as shown.

(c) Rough sketch of an exterior altitude YL of ∆XYZ is as shown.

Ex 6.1 Class 7 Maths Question 3.
Verify by drawing a diagram if ‘the median and altitude of an isosceles triangle can be same.
Solution:

Draw a line segment BC. By paper folding, locate the perpendicular bisector of BC. The folded crease meets BC at D, its mid-point.
Take any point A on this perpendicular bisector. Join AB and AC. The triangle thus obtained is an isosceles ∆ABC in which AB = AC.
Since D is the mid-point of BC, so AD is its median. Also, AD is the perpendicular bisector of BC. So, AD is the altitude of ∆ABC.
Thus, it is verified that the median and altitude of an isosceles triangle are the same.

Ex 6.2 Class 7 Maths Question 1.
Find the value of the unknown exterior angle x in the following diagrams :

Solution:
Since, in a triangle an exterior angle is equal to the sum of the two interior opposite angles, therefore,

1. x = 50°+ 70° = 120°
2. x = 65°+ 45° = 110°
3. x = 30°+ 40°= 70°
4. x = 60° + 60° = 120c
5. x = 50° + 50° = 100c
6. x = 30°+ 60° = 90°

Ex 6.2 Class 7 Maths Question 2.
Find the value of the unknown interior angle x in the following figures :

Solution:
We know that in a triangle, an exterior angle is equal to the sum of the two interior opposite angles. Therefore,

Ex 6.3 Class 7 Maths Question 1.
Find the value of the unknown x in the following diagrams :


Solution:

Ex 6.3 Class 7 Maths Question 2.
Find the values of the unknowns x and y in the following diagrams :


Solution:

Ex 6.4 Class 7 Maths Question 1.
Is it possible to have a triangle with the following sides?
(i) 2 cm, 3 cm, 5 cm
(ii) 3 cm, 6 cm, 7 cm
(iii) 6 cm, 3 cm, 2 cm
Solution:
(i) Since, 2 + 3 > 5
So the given side lengths cannot form a triangle.
(ii) We have, 3 + 6 > 7, 3 + 7 > 6 and 6 + 7 > 3
i. e., the sum of any two sides is greater than the third side.
So, these side lengths form a triangle.
(iii) We have, 6 + 3 > 2, 3 + 2 Undefined control sequence \ngtr 6
So, the given side lengths cannot form a triangle.

Ex 6.4 Class 7 Maths Question 2.
Take any point O in the interior of a triangle PQR. Is

(i) OP + OQ > PQ?
(ii) OQ + OR > QR ?
(iii) OR + OP > RP ?
Solution:
(i) Yes, OP + OQ > PQ because on joining OP and OQ, we get a ∆OPQ and in a triangle, sum of the lengths of any two sides is always greater than the third side.

(ii) Yes, OQ + OR > QR, because on joining OQ and
OR, we get a ∆OQR and in a triangle, sum of the length of any two sides is always greater than the third side.

(iii) Yes, OR + OP > RP, because on joining OR and OP, we get a ∆OPR and in a triangle, sum of the lengths of any two sides is always greater than the third side.

Ex 6.4 Class 7 Maths Question 3.
AM is median of a triangle ABC. Is AB + BC + CA > 2AM?
(Consider the sides of triangles ∆ABM and ∆AMC.)

Solution:
Using triangle inequality property in triangles ABM and AMC, we have
AB + BM > AM …(1) and, AC + MC > AM …(2)
Adding (1) and (2) on both sides, we get
AB + (BM + MC) + AC > AM + AM
⇒ AB + BC + AC > 2AM

Ex 6.4 Class 7 Maths Question 4.
ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?

Solution:

Ex 6.4 Class 7 Maths Question 5.
ABCD is a quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)?
Solution:
Let ABCD be a quadrilateral and its diagonals AC and BD intersect at O. Using triangle inequality property, we have
In ∆OAB
OA + OB > AB ……(1)

Ex 6.4 Class 7 Maths Question 6.
The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?
Solution:
Let x cm be the length of the third side.
Thus, 12 + 15 > x, x + 12 > 15 and x + 15 > 12
⇒ 27 > x, x > 3 and x > -3
The numbers between 3 and 27 satisfy these.
∴ The length of the third side could be any length between 3 cm and 27 cm.

Ex 6.5 Class 7 Maths Question 1.
PQR is a triangle, right-angled at P. If PQ = 10 cm and PR? = 24 cm, find QR.
Solution:

Ex 6.5 Class 7 Maths Question 2.
ABC is a triangle right-angled atC. If AB = 25 cm and AC = 7cm, find BC.
Solution:

Ex 6.5 Class 7 Maths Question 3.
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Solution:

Hence, the distance of the foot of the ladder from the wall is 9 m.

Ex 6.5 Class 7 Maths Question 4.
Which of the following can be the sides of a right triangle?
(i) 2.5 em, 6.5 cm, 6 cm.
(ii) 2 cm, 2 cm, 5 cm.
(iii) 1.5 cm, 2 cm, 2.5 cm.
In the case of right-angled triangles, identify the right angles.
Solution:


Thus, the given sides form a right-triangle and the right-angle is opposite to the side of length 2.5 cm.

Ex 6.5 Class 7 Maths Question 5.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
Solution:

Let ACB be the tree before it broke at the point C. and let its top A touch the ground at A^‘ after it broke. Then, ∆A^‘BC is a right triangle, right-angled at B such that A’ B = 12 m, BC = 5 m. By Pythagoras theorem, we have

Ex 6.5 Class 7 Maths Question 6.
Angles Q and ii of a APQR are 25° and 65°. Write which of the following is true :
(i) PQ² + QR² = RP²
(ii) PQ² + RP² = QR²
(iii) RP² + QR² = PQ²

Solution:

Ex 6.5 Class 7 Maths Question 7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Solution:
Let ABCD be a rectangle such that AB = 40 m and AC = 41 m.
In right-angled ∆ABC, right-angled at B, by Pythagoras theorem, we have BC² = AC² – AB²

Ex 6.5 Class 7 Maths Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Solution:
Let ABCD be the rhombus such that AC = 30 cm and BD = 16 cm.
We know that the diagonals of a rhombus bisect each other at right angles.