Question bank class 7

Fill in the blanks using < or >.
(a) -3 …… -4
(b) 6 ……. -20
(c) -8 …… -2
(d) 5 …… -7
Solution:
(a) -3 > -4
(b) 6 > -20
(c) -8 < -2
(d) 5 > -7

Question 2.
Solve the following:
(i) (-8) × (-5) + (-6)
(ii) [(-6) × (-3)] + (-4)
(iii) (-10) × [(-13) + (-10)]
(iv) (-5) × [(-6) + 5]
Solution:
(i) (-8) × (-5) + (-6)
= (-) × (-) × [8 × 5] + (-6)
= 40 – 6
= 34

Question 1.
In the given figure, name
(a) the side opposite to vertex A
(b) the vertex opposite A to side AB
(c) the angle opposite to side AC
(d) the angle made by the sides CB and CA.
Congruence of Triangles Class 7 Extra Questions Maths Chapter 7

Solution:
(a) The side opposite to vertex A is BC.
(b) The vertex opposite to side AB is C.
(c) The angle opposite to side AB is ∠ACB.
(d) The angle made by the sides CB and CA is ∠ACB.

Question 2.
Examine whether the given triangles are congruent or not.
Solution:
Here,
AB = DE = 3 cm
BC = DF = 3.5 cm
AC = EF = 4.5 cm
ΔABC = ΔEDF (By SSS rule)
Congruence of Triangles Class 7 Extra Questions Maths Chapter 7

question 3:
In the given congruent triangles under ASA, find the value of x and y, ΔPQR = ΔSTU.
Congruence of Triangles Class 7 Extra Questions Maths Chapter 7

Solution:
Given: ΔPQR = ΔSTU (By ASA rule)
∠Q = ∠T = 60° (given)
$\bar { QR }$ = $\bar { TU }$ = 4 cm (given)
∠x = 30° (for ASA rule)
Now in ΔSTU,
∠S + ∠T + ∠U = 180° (Angle sum property)
∠y + 60° + ∠x = 180°
∠y + 60° + 30° = 180°
∠y + 90° = 180°
∠y = 180° – 90° = 90°
Hence, x = 30° and y = 90°.

Question 4.
In the following figure, show that ΔPSQ = ΔPSR.
Congruence of Triangles Class 7 Extra Questions Maths Chapter 7

Solution:
In ΔPSQ and ΔPSR
$\bar { PQ }$ = $\bar { PR }$ = 6.5 cm (Given)
$\bar { PS }$ = $\bar { PS }$ (Common)
∠PSQ = ∠PSR = 90° (Given)
ΔPSQ = ΔPSR (By RHS rule)

Question 5.
Can two equilateral triangles always be congruent? Give reasons.
Solution:
No, any two equilateral triangles are not always congruent.
Reason: Each angle of an equilateral triangle is 60° but their corresponding sides cannot always be the same.

Question 6.
In the given figure, AP = BQ, PR = QS. Show that ΔAPS = ΔBQR
Congruence of Triangles Class 7 Extra Questions Maths Chapter 7
Solution:
In ΔAPS and ΔBQR
AP = BQ (Given)
PR = QS (Given)
PR + RS = QS + RS (Adding RS to both sides)
PS = QR
∠APS = ∠BQR = 90° (Given)
ΔAPS = ΔBQR (by SAS rule)

Perimeter and Area Class 7

Question 1.
The side of a square is 2.5 cm. Find its perimeter and area.
Solution:
Side of the square = 2.5 cm
Perimeter = 4 × Side = 4 × 2.5 = 10 cm
Area = (side)² = (4)² = 16 cm²

Question 2.
If the perimeter of a square is 24 cm. Find its area.
Solution:
Perimeter of the square = 24 cm
Side of the square = $\frac { 24 }{ 4 }$ cm = 6 cm
Area of the square = (Side)² = (6)² cm² = 36 cm²

Question 3.
If the length and breadth of a rectangle are 36 cm and 24 cm respectively. Find
(i) Perimeter
(ii) Area of the rectangle.
Solution:
Length = 36 cm, Breadth = 24 cm
(i) Perimeter = 2(l + b) = 2(36 + 24) = 2 × 60 = 120 cm
(ii) Area of the rectangle = l × b = 36 cm × 24 cm = 864 cm²

Question 4.
The perimeter of a rectangular field is 240 m. If its length is 90 m, find:
(i) it’s breadth
(ii) it’s area.
Solution:
(i) Perimeter of the rectangular field = 240 m
2(l + b) = 240 m
l + b = 120 m
90 m + b = 120 m
b = 120 m – 90 m = 30 m
So, the breadth = 30 m.
(ii) Area of the rectangular field = l × b = 90 m × 30 m = 2700 m²Question 5.

The length and breadth of a rectangular field are equal to 600 m and 400 m respectively. Find the cost of the grass to be planted in it at the rate of ₹ 2.50 per m².
Solution:
Length = 600 m, Breadth = 400 m
Area of the field = l × b = 600 m × 400 m = 240000 m²
Cost of planting the grass = ₹ 2.50 × 240000 = ₹ 6,00,000
Hence, the required cost = ₹ 6,00,000.

Question 6.
The perimeter of a circle is 176 cm, find its radius.
Solution:
The perimeter of the circle = 176 cm
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q6

Question 7.
The radius of a circle is 3.5 cm, find its circumference and area.
Solution:
Radius = 3.5 cm
Circumference = 2πr

Question 8.
Area of a circle is 154 cm², find thecircumference.
Solution:
Area of the circle = 154 cm²
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q8

Question 9.
Find the perimeter of the figure given below.
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q9
Solution:
Perimeter of the given figure = Circumference of the semicircle + diameter
= πr + 2r
= $\frac { 22 }{ 7 }$ × 7 + 2 × 7
= 22 + 14
= 36 cm
Hence, the required perimeter = 36 cm.

Question 10.

The length of the diagonal of a square is 50 cm, find the perimeter of the square.
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q10
Solution:
Let each side of the square be x cm.
x² + x² = (50)² [Using Pythagoras Theorem]
2x² = 2500
x² = 1250

x = √1250 = $\sqrt { 2\times 5\times 5\times 5\times 5 }$
x = 5 × 5 × √2 = 25√2
The side of the square = 25√2 cm
Perimeter of the square = 4 × side = 4 × 25√2 = 100√2 cm

Perimeter and Area Class 7 Extra Questions Short Answer Type

Question 11.
A wire of length 176 cm is first bent into a square and then into a circle. Which one will have more area?
Solution:
Length of the wire = 176 cm
Side of the square = 176 ÷ 4 cm = 44 cm
Area of the square = (Side)² = (44)² cm² = 1936 cm²
Circumference of the circle = 176 cm
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q11
Since 2464 cm² > 1936 cm²
Hence, the circle will have more area.

Question 12.
In the given figure, find the area of the shaded portion.
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q12
Solution:
Area of the square = (Side)² = 10 cm × 10 cm = 100 cm²
Area of the circle = πr²
= $\frac { 22 }{ 7 }$ × 3.5 × 3.5
= $\frac { 77 }{ 2 }$ cm²
= 38.5 cm²
Area of the shaded portion = 100 cm² – 38.5 cm² = 61.5 cstion 13.

Find the area of the shaded portion in the figure given below.
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q13
Solution:
Area of the rectangle = l × b = 14 cm × 14 cm = 196 cm²
Radius of the semicircle = $\frac { 14 }{ 2 }$ = 7 cm
Area of two equal semicircle = 2 × $\frac { 1 }{ 2 }$ πr²
= πr²
= $\frac { 22 }{ 7 }$ × 7 × 7
= 154 cm²
Area of the shaded portion = 196 cm² – 154 cm² = 42 cm²

Question 14.
A rectangle park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q14
Solution:
Length of the rectangular park = 45 m
Breadth of the park = 30 m
Area of the park = l × 6 = 45m × 30m = 1350 m²
Length of the park including the path = 45 m + 2 × 2.5 m = 50 m
Breadth of the park including the path = 30 m + 2 × 2.5 m = 30m + 5m = 35m
Area of the park including the path = 50 m × 35 m = 1750 m²
Area of the path = 1750 m² – 1350 m² = 400 m²
Hence, the required area = 400 m².uestion 15.

In the given figure, calculate:
(а) the area of the whole figure.
(b) the total length of the boundary of the field.
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q15
Solution:
Area of the rectangular portions = l × b = 80 cm × 42 cm = 3360 cm²
Area of two semicircles = 2 × $\frac { 1 }{ 2 }$ πr² = πr²
= $\frac { 22 }{ 7 }$ × 21 × 21
= 22 × 3 × 21
= 1386 cm²
Total area = 3360 cm² + 1386 cm² = 4746 cm²
Total length of the boundary of field = (2 × 80 + πr + πr) cm
= (160 + $\frac { 22 }{ 7 }$ × 21 + $\frac { 22 }{ 7 }$ × 21)
= (160 + 132) cm
= 292 cm
Hence, the required (i) area = 4746 cm² and (ii) length of boundary = 292 cm.

Question 16.
How many times a wheel of radius 28 cm must rotate to cover a distance of 352 m?
(Take π = $\frac { 22 }{ 7 }$ )
Solution:
Radius of the wheel = 28 cm
Circumference = 2πr = 2 × $\frac { 22 }{ 7 }$ × 28 = 176 cm
Distance to be covered = 352 m or 352 × 100 = 35200 m
Number of rotation made by the wheel to cover the given distance = $\frac { 35200 }{ 176 }$ = 200
Hence, the required number of rotations = 200.

Perimeter and Area Class 7 Extra Questions Long Answer Type

Question 1
A nursery school playground is 160 m long and 80 m wide. In it 80 m × 80 m is kept for swings and in the remaining portion, there are 1.5 m wide path parallel to its width and parallel to its remaining length as shown in Figure. The remaining area is covered by grass. Find the area covered by grass. (NCERT Exemplar)
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q17
Solution:
Area of the playground = l × b = 160 m × 80 m = 12800 m²
Area left for swings = l × b = 80m × 80m = 6400 m²
Area of the remaining portion = 12800 m² – 6400 m² = 6400 m²
Area of the vertical road = 80 m × 1.5 m = 120 m²
Area of the horizontal road = 80 m × 1.5 m = 120 m²
Area of the common portion = 1.5 × 1.5 = 2.25 m²
Area of the two roads = 120 m² + 120 m² – 2.25 m² = (240 – 2.25) m² = 237.75 m²
Area of the portion to be planted by grass = 6400 m² – 237.75 m² = 6162.25 m²
Hence, the required area = 6162.25 m².

Question 2
Rectangle ABCD is formed in a circle as shown in Figure. If AE = 8 cm and AD = 5 cm, find the perimeter of the rectangle. (NCERT Exemplar)
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q18
Solution:
DE (Radius) = AE + AD = 8 cm + 5 cm = 13 cm
DB = AC = 13 cm (Diagonal of a rectangle are equal)
In right ∆ADC,
AD² + DC² = AC² (By Pythagoras Theorem)
⇒ (5)² + DC² = (13)²
⇒ 25 + DC² = 169
⇒ DC² = 169 – 25 = 144
⇒ DC = √144 = 12 cm
Perimeter of rectangle ABCD = 2(AD + DC)
= 2(5 cm + 12 cm)
= 2 × 17 cm
= 34 cm

Question 3
Find the area of a parallelogram-shaped shaded region. Also, find the area of each triangle. What is the ratio of the area of shaded portion to the remaining area of the rectangle?
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q19
Solution:
Here, AB = 10 cm
AF = 4 cm
FB = 10 cm – 4 cm = 6 cm
Area of the parallelogram = Base × Height = FB × AD = 6 cm × 6 cm = 36 cm²
Hence, the required area of shaded region = 36 cm².
Area ∆DEF = $\frac { 1 }{ 2 }$ × b × h
= $\frac { 1 }{ 2 }$ × AF × AD
= $\frac { 1 }{ 2 }$ × 4 × 6
= 12 cm²
Area ∆BEC = $\frac { 1 }{ 2 }$ × b × h
= $\frac { 1 }{ 2 }$ × GC × BC
= $\frac { 1 }{ 2 }$ × 4 × 6
= 12 cm²
Area of Rectangle ABCD = l × b = 10 cm × 6 cm = 60 cm²
Remaining area of Rectangle = 60 cm² – 36 cm² = 24 cm²
Required Ratio = 36 : 24 = 3 : 2

Question 4
A rectangular piece of dimension 3 cm × 2 cm was cut from a rectangular sheet of paper of dimensions 6 cm × 5 cm. Find the ratio of the areas of the two rectangles. (NCER T Exemplar)
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q20
Solution:
Length of the rectangular piece = 6 cm
Breadth = 5 cm
Area of the sheet = l × b = 6 cm × 5 cm = 30 cm²
Area of the smaller rectangular piece = 3 cm × 2 cm = 6 cm²
Ratio of areas of two rectangles = 30 cm² : 6 cm² = 5 : 1

Perimeter and Area Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 5
In the given figure, ABCD is a square of side 14 cm. Find the area of the shaded region.
(Take π = $\frac { 22 }{ 7 }$ )
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q21
Solution:
PQ = $\frac { 1 }{ 2 }$ AB = $\frac { 1 }{ 2 }$ × 14 = 7 cm
PQRS is a square with each side 7 cm
Radius of each circle = $\frac { 7 }{ 2 }$ cm
Area of the quadrants of each circle = $\frac { 1 }{ 4 }$ × πr²
Area of the four quadrants of all circles
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q21.1
Area of the square PQRS = Side × Side = 7 cm × 7 cm = 49 cm²
Area of the shaded portion = 49 cm² – 38.5 cm² = 10.5 cm²
Hence, the required area = 10.5 cm².

Question 6
Find the area of the following polygon if AB = 12 cm, AC = 2.4 cm, CE = 6 cm, AD = 4.8 cm, CF = GE = 3.6 cm, DH = 2.4 cm.
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q22
Solution:
BE = AB – AE
= 12 cm – (AC + CE)
= 12 cm – (2.4 cm + 6 cm)
= 12 cm – 8.4 cm
= 3.6 cm

Area of the polygon AFGBH = Area of ∆ACF + Area of rectangle FCEG + Area of ∆GEB + Area of ∆ABH
= 3.6 cm² + 4.32 cm² + 21.6 cm² + 6.48 cm² + 14.4 cm²
= 50.40 cm²
Hence, the required area = 50.40 cm².

Find the perimeter of the figure given below.
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q9
Solution:
Perimeter of the given figure = Circumference of the semicircle + diameter
= πr + 2r
= $\frac { 22 }{ 7 }$ × 7 + 2 × 7
= 22 + 14
= 36 cm
Hence, the required perimeter = 36 cm.

Question 10.

Perimeter and Area Class 7 Extra Questions Short Answer Type

Question 13.

Question 15.

Perimeter and Area Class 7 Extra Questions Long Answer Type

Question 17.
A nursery school playground is 160 m long and 80 m wide. In it 80 m × 80 m is kept for swings and in the remaining portion, there are 1.5 m wide path parallel to its width and parallel to its remaining length as shown in Figure. The remaining area is covered by grass. Find the area covered by grass. (NCERT Exemplar)
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q17
Solution:
Area of the playground = l × b = 160 m × 80 m = 12800 m²
Area left for swings = l × b = 80m × 80m = 6400 m²
Area of the remaining portion = 12800 m² – 6400 m² = 6400 m²
Area of the vertical road = 80 m × 1.5 m = 120 m²
Area of the horizontal road = 80 m × 1.5 m = 120 m²
Area of the common portion = 1.5 × 1.5 = 2.25 m²
Area of the two roads = 120 m² + 120 m² – 2.25 m² = (240 – 2.25) m² = 237.75 m²
Area of the portion to be planted by grass = 6400 m² – 237.75 m² = 6162.25 m²
Hence, the required area = 6162.25 m².

Question 18.
Rectangle ABCD is formed in a circle as shown in Figure. If AE = 8 cm and AD = 5 cm, find the perimeter of the rectangle. (NCERT Exemplar)
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q18
Solution:
DE (Radius) = AE + AD = 8 cm + 5 cm = 13 cm
DB = AC = 13 cm (Diagonal of a rectangle are equal)
In right ∆ADC,
AD² + DC² = AC² (By Pythagoras Theorem)
⇒ (5)² + DC² = (13)²
⇒ 25 + DC² = 169
⇒ DC² = 169 – 25 = 144
⇒ DC = √144 = 12 cm
Perimeter of rectangle ABCD = 2(AD + DC)
= 2(5 cm + 12 cm)
= 2 × 17 cm
= 34 cm

Question 19.
Find the area of a parallelogram-shaped shaded region. Also, find the area of each triangle. What is the ratio of the area of shaded portion to the remaining area of the rectangle?
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q19
Solution:
Here, AB = 10 cm
AF = 4 cm
FB = 10 cm – 4 cm = 6 cm
Area of the parallelogram = Base × Height = FB × AD = 6 cm × 6 cm = 36 cm²
Hence, the required area of shaded region = 36 cm².
Area ∆DEF = $\frac { 1 }{ 2 }$ × b × h
= $\frac { 1 }{ 2 }$ × AF × AD
= $\frac { 1 }{ 2 }$ × 4 × 6
= 12 cm²
Area ∆BEC = $\frac { 1 }{ 2 }$ × b × h
= $\frac { 1 }{ 2 }$ × GC × BC
= $\frac { 1 }{ 2 }$ × 4 × 6
= 12 cm²
Area of Rectangle ABCD = l × b = 10 cm × 6 cm = 60 cm²
Remaining area of Rectangle = 60 cm² – 36 cm² = 24 cm²
Required Ratio = 36 : 24 = 3 : 2

Question 20.
A rectangular piece of dimension 3 cm × 2 cm was cut from a rectangular sheet of paper of dimensions 6 cm × 5 cm. Find the ratio of the areas of the two rectangles. (NCER T Exemplar)
Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q20
Solution:
Length of the rectangular piece = 6 cm
Breadth = 5 cm
Area of the sheet = l × b = 6 cm × 5 cm = 30 cm²
Area of the smaller rectangular piece = 3 cm × 2 cm = 6 cm²
Ratio of areas of two rectangles = 30 cm² : 6 cm² = 5 : 1

Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q22.2
Area of the polygon AFGBH = Area of ∆ACF + Area of rectangle FCEG + Area of ∆GEB + Area of ∆ABH
= 3.6 cm² + 4.32 cm² + 21.6 cm² + 6.48 cm² + 14.4 cm²
= 50.40 cm²

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