ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress

Question 1.
A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.
Solution:
Length of rectangular park (l) = 125 m
and breadth (b) = 65 m
Width of path around it = 3 m
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q1.1
Outer length (L) = 125 + 2 × 3 = 125 + 6 = 131 m
and breadth = 65 + 2 × 3 = 65 + 6 = 71 m
Area of path = L × B – l × b
= 131 × 71 – 125 × 65
= 9301 – 8125 = 1176 m2

Question 2.
In the given figure, all adjacent line segments are at right angles. Find:
(i) the area of the shaded region
(ii) the area of the unshaded region.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q2.1
Solution:
In the given figure,
Length (l) = 22 m
and breadth (b) = 14 m
Width of length wire region = 3 m
and breadth wire = 2 m
Area of shaded portion = 22 × 3 + 14 × 2 – 3 × 2 m2
= 66 + 28 – 6
= 88 m2
Total area = l × b = 22 × 14 = 308 m2
Area of unshaded region = 308 – 88 = 220 m2

Question 3.
Find the area of a triangle whose:
(i) base = 2 m, height = 1.5 m
(ii) base = 3.4 m and height = 90 cm
Solution:
(i) Base of a triangle (b) = 2 m
Height (h) = 1.5 m
Area = \(\frac { 1 }{ 2 }\) × b × h
= \(\frac { 1 }{ 2 }\) × 2 × 1.5 = 1.5 m2
(ii) Base of the triangle (b) = 3.4 m
and height (h) = 90 cm = \(\frac { 90 }{ 100 }\) = \(\frac { 9 }{ 10 }\) m
Area = \(\frac { 1 }{ 2 }\) × b × h
= \(\frac { 1 }{ 2 }\) × 3.4 × \(\frac { 9 }{ 10 }\)
= \(\frac { 30.6 }{ 20 }\) = 1.53 m2

Question 4.
In the given figure, PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm, PS = 8 cm and QM = 7.6 cm, find:
(i) the area of the parallelogram PQRS
(ii) the length of QN.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q4.1
Solution:
In the given figure,
PQRS is a parallelogram in which QM ⊥ SR and QN ⊥ PQ
SR = 12 cm, PS = 8 cm, QM = 7.6 cm
(i) Area of ||gm ABCD = b × h
= SR × QM
= 12 × 7.6 cm2
= 91.2 cm2
Base PS = 8 cm
Area of ||gm = 91.2 cm2
Heigh QN = \(\frac { Area }{ Base }\) = \(\frac { 91.2 }{ 8 }\) = 11.4 cm

Question 5.
From the given figure, find
(i) the area of ΔABC
(ii) length of BC
(iii) the length of altitude from A to BC.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q5.1
Solution:
In the given figure,
ABC is a right angled triangle in which
AB = 3 cm , AC = 4 cm
AD ⊥ BC
(i) Area ΔABC = \(\frac { 1 }{ 2 }\) × Base × Height
= \(\frac { 1 }{ 2 }\) × 3 × 4 = 6 cm2
(ii) BC2 = AB2 + AC2 (Pythagoras Theorem)
= 32 + 42 = 9 + 16 = 25 = (5)2
BC = 5 cm
(iii) Now length of altitude AD
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q5.2

Question 6.
In the given figure, the area of the right-angled triangle is 54 cm2. If one of its legs is 12 cm long, find its perimeter.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q6.1
Solution:
In the given figure,
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q6.2
Area of right-angled triangle = 54 cm2
Length of one leg AB = 12 cm
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q6.3
Now AC2 = AB2 + BC2 (Pythagoras Theorem)
= 122 + 92 = 144 + 81 = 225 = (15)2
AC = 15 cm
Now perimeter = AB + BC + AC = 12 + 9 + 15 = 36 cm

Question 7.
If the area of a circle is 78.5 cm2, find its circumference. (Take π = 3.14)
Solution:
Area of a circle = 78.5 cm2
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q7.1
Circumference = 2πr = 2 × 3.14 × 5 = 31.4 cm

Question 8.
Find the circumference of the circle whose area is 16 times the area of the circle with diameter 7 cm.
Solution:
Diameter of first circle = 7 cm
Radius (r) = \(\frac { 7 }{ 2 }\) cm
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q8.1
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q8.2

Question 9.
From square cardboard, a circle of the biggest area was cut out. If the area of the circle is 154 cm2, calculate the original area of the cardboard.
Solution:
From square cardboard, the biggest circle is cut out
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q9.1
Side of square = diameter of the circle = 2 × 7 = 14 cm
Area of square cardboard = (Side)2 = (14)2 = 196 cm2

Question 10.
A road 3.5 m wide surrounds a circular park whose circumference is 88 m. Find the cost of paving the road at the rate of ₹ 60 per square meter.
Solution:
Circumference of a circular park = 88 m
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q10.1
Width of road surounded it = 3.5 m
Outer radius (R) = 14 + 3.5 = 17.5 m
Area of road = πR2 – πr2 = π(R2 – r2)
= \(\frac { 1 }{ 2 }\) × (17.52 – 142)
= \(\frac { 1 }{ 2 }\) × 31.5 × 3.5
= 346.5 m2
Cost of paving the road = ₹ 60 per m2
Total cost = ₹ 60 × 346.5 = ₹ 20790

Question 11.
In the given figure, ABCD is a square of side 14 cm. Find the area of the shaded region.
Take π = \(\frac { 22 }{ 7 }\)
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q11.1
Solution:
In the given figure, ABCD is a square whose side is 14 cm.
Area of square = (Side)2 = (14)2 cm2 = 14 × 14 = 196 cm2
Radius of each circle in it = \(\frac { 7 }{ 2 }\) cm
Area of 4 circles = 4 × πr2
= 4 × \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\)
= 154 cm2
Area of shaded portion =196 – 154 = 42 cm2

Question 12.
The boundary of a shaded region in the given figure consists of three semicircles, the smaller being equal. If the diameter of the larger one is 28 cm, find
(i) the length of the boundary
(ii) the area of the shaded region.
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q12.1
Solution:
In the given figure,
There are a bigger semicircle and two small semicircles
Diameter of bigger semicircle = 28 cm
Radius (R) = \(\frac { 28 }{ 2 }\) = 14 cm
and radius of each of smaller semicircles = \(\frac { 14 }{ 2 }\) = 7 cm
Now the area of shaded portion
= Area of bigger semicircle + Area of one smaller on semicircle
– Area of another smaller semicircle
Both small semicircles have the same area
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Q12.2

ML Aggarwal Class 7 Solutions for ICSE Maths

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