RBSE Class 7 Maths Important Questions Chapter 11 Perimeter and Area

Rajasthan Board RBSE Class 7 Maths Important Questions Chapter 11 Perimeter and Area Important Questions and Answers.

RBSE Class 7 Maths Chapter 11 Important Questions Perimeter and Area

Multiple Choice Questions

Question 1.
Area of a square garden of side 2.1 metre:
(a) 8.4 m
(b) 8.4 m²
(c) 4.41 m
(d) 4.41 m²
Answer:
(d) 4.41 m²

Question 2.
The circumference of a circle of diameter 28 m is :
(a) 44 m
(b) 88 m
(c) 176 m
(d) 14 m
Answer:
(b) 88 m

Question 3.
The hypotenuse of a triangle measure 25 m and its base is 15 m. Its area will be:
(a) 150 m²
(b) 75 m²
(c) 187.5 m²
(d) 80 m²
Answer:
(a) 150 m²

Question 4.
The area of a circular rim of radius 7 m is:
(a) 154 m²
(b) 44 m²
(c) 145 m²
(d) 144 m²
Answer:
(d) 144 m²

Question 5.
If a piece of wire is 24 m long is bent in the shape of a square, then its side is:
(a) 4 m
(b) 8 m
(c) 12 m
(d) 6 m
Answer:
(d) 6 m

Question 6.
If the perimeter of a regular pentagon is 24 m, then length of each side will be
(a) 1.2 m
(b) 4.8 m
(c) 6 m
(d) 2 m
Answer:
(b) 4.8 m

Question 7.
If the circumference of a circle is 31.4 cm, then its radius will be :
(a) 10 cm
(b) 5 cm
(c) 15 cm
(d) 20 cm
Answer:
(b) 5 cm

Fill in the Blanks

Question 1.
The length of boundary of a figure is called as ____________ .
Answer:
Perimeter

Question 2.
The cost of fencing a square park of side 5 m @ ₹ 10 per m will be ____________ .
Answer:
₹ 200

Question 3.
The diameter of a circle of perimeter 616 cm is ____________ .
Answer:
196 cm

Question 4.
The area of a semi-circle of radius 4 m is ____________ .
Answer:
8 π

Question 5.
Circumference of a circle of diameter 35 cm is ____________ .
Answer:
110 cm

State whether True or False

Question 1.
The area of a square whose side is 2.6 cm is 12 cm².
Answer:
False

Question 2.
The circumference of a circle of diameter d is πd.
Answer:
True

Question 3.
1 hectare is equal to 10000 m².
Answer:
True

Question 4.
The perimeter of a square is 48 cm. Its area is 100 cm².
Answer:
False

Question 5.
The diameter of a circle is 7 cm. Its area is 22 cm².
Answer:
False

Very Short Answer Type Questions

Question 1.
Find the area of a parallelogram whose one side is 4 cm and corresponding height is 3 cm.
Answer:
Base = 4 cm and Height = 3 cm
∴ Area of Parallelogram = b × h
= 4 × 3 = 12 cm²

Question 2.
Find the circumference of a circle of diameter 10 cm. (Take % = 3.14)
Answer:
Diameter of circle = 10 cm
∴ Circumference of circle = πd
= 3.14 × 10 cm = 31.4 cm

Question 3.
Diameter of a circular garden is 9.8 m. Find its area.
Answer:
Diameter = 9.8 m
∴ Radius r = 9.8 ÷ 2 = 4.9 m
Area of circle (πr²) = \(\frac{22}{7} \)× 4.9 × 4.9 m²
= 75.46 m²

Question 4.
Convert 50 cm² into mm².
Answer:
1 cm² = 1 cm× 1 cm
= 10 mm × 10 mm = 100 mm²
∴ 50 cm² = 50 × 100 mm² = 5000 mm²

Question 5.
Convert 10 m² into cm².
Answer:
1 m² = 1 m × 1 m
= 100 cm × 100 cm = 10,000 cm²
∴ 10 m² = 10 × 10000 cm² = 100000 cm²

Short and Long Answer Type Questions

Question 1.
Find the area of a square park, whose perimeter is 420 m.
Answer:
Perimeter of square = 420 m
Side = \(\frac{\text { Perimeter }}{4}\) = \(\frac{420}{4}\)m = 105 m
∵ Area of square = (Side)² = (105)² = 11025
So, the area of square park is 11025 m².

Question 2.
The perimeter of a regular pentagon in 22.5 cm, find the measure of each side.
Answer:
Perimeter of a regular pentagon
= 5 × side
side = \(\frac{\text { Perimeter }}{5}\) = \(\frac{22.5}{5}\) = 4.5 cm
So, the measure of each side of the regular pentagon is 4.5 cm.

Question 3.
In parallelogram ABCD, if DE = 6 cm, FB = 8 cm and AD = 10 cm, then find the length of AB.

Answer:
Area of parallelogram = b × h
⇒ AD × BF = AB × DE
⇒ 10 × 8 = AB × 6
⇒ AB = \(\frac{80}{6}\) = 13.3 cm (approx.)
The length of AB is approximately 13.3 cm.

Question 4.
Find the area of the shaded part.

Answer:
Area of rectangle = l × b = 30 × 40
= 1200 cm²
Area of triangle I = \(\frac{1}{2}\) × 6 × h
= \(\frac{1}{2}\) × 10 × 40 = 200 cm²
Area of triange II = \(\frac{1}{2}\) × b × h
= \(\frac{1}{2}\) × 20 × 18 = 180 cm²
Area of triangle III
= \(\frac{1}{2}\) × 6 × h = \(\frac{1}{2}\) × 30 × 22 = 330 cm²
Area of shaded portion = Area of rectangle PQRS - (Area of I triangle + Area of II triangle + Area of III triangle)

= 1200 - (200 + 180 + 330)
= (1200 - 710) cm² = 490 cm²
So, the area of shaded portion is 490 cm².