RBSE Solutions for Class 7 Maths Chapter 16 Perimeter and Area Additional Questions

RBSE Solutions for Class 7 Maths Chapter 16 Perimeter and Area Additional Questions is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 16 Perimeter and Area Additional Questions.

Rajasthan Board RBSE Class 7 Maths Chapter 16 Perimeter and Area Additional Questions

Multiple Choice Questions

Question 1
Formula to find out (RBSESolutions.com) area of circle is :
(A) πr²
(B) 2πr²
(C) 2πr
(D) πr

Question 2
The radius of circular shape field is 14 cm then circumference of field will be :
(A) 48 cm
(B) 88 cm
(C) 68 cm
(D) 98 cm

Question 3
Formula of area (RBSESolutions.com) of rectangle is :
(A) l x b
(B) l²
(C) l x b²
(D) b²

Question 4
The area of 10 cm side square will be :
(A) 101 sq. m
(B) 100 sq. cm
(C) 100 sq. m
(D) 100 km

Question 5
The base of a triangle is 12 cm and height is 5 cm (RBSESolutions.com) then area of triangle will be :
(A) 30 sq. cm
(B) 60 sq. cm
(C) 120 sq. cm
(D) 28 sq. cm

Question 6
The formula of the area of right angle tri¬angle will be :
(A) \(\frac { 1 }{ 2 }\) x Base x Height
(B) Base x Height
(C) 2 x Base x Height
(D) 2 x Base + Height
Answer
1. (A), 2. (B), 3. (A), 4. (B), 5. (A), 6. (A).

True/False
(i) Half circumference (RBSESolutions.com) of circle = πr.
(ii) Area of parallelogram = \(\frac { 1 }{ 2 }\) x Base x Height
(iii) 1 km = 1000 m
(iv) π = \(\frac { 22 }{ 7 }\)
Ans
(i) True,
(ii) False,
(iii) True,
(iv) True.

Fill in the blanks :
(i) The perimeter of rectangle = ………….
(ii) Area of square = ……………
(iii) 1 m = ……….. cm
Answer
(i) 2(length + breadth) unit,
(ii) side²
(iii) 100

Very Short Answers Type Questions

Question 1
Find out the circumference of circle (RBSESolutions.com) whose radius is 21 cm. (π = \(\frac { 22 }{ 7 }\) )
Solution:
Radius of circle (r) = 21 cm
Circumference of circle = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x 21= 2 x 22 x 3 = 2 x 66 = 132 cm

Question 2
Find the area of a circle with radius 10 cm.
Solution:
Radius (RBSESolutions.com) of circle (r) = 10 cm
Area of circle = πr²
= 3.14 x (10)²
= 3.14 x 100
= 314 sq. cm.

Question 3
A rectangle’s length is 10 m and (RBSESolutions.com) breadth 5 m then find its perimeter.
Solution:
Perimeter of rectangle = 2 (length + breadth)
= 2(10 + 5) = 2(15) = 30 m

Question 4
A square whose sides is 12 m. Find its perimeter.
Solution:
Side of square = 12 m
Perimeter of square = 4 x side
= 4 x 12 = 48 m.

Long Answer Questions

Question 1
The area of rectangular sheet is 500 cm². If length (RBSESolutions.com) of sheet is 25 cm then what will be its breadth. Also find out perimeter of the sheet.
Solution:
Area of rectangular sheet = 500 cm
Length (l) = 25 cm
Area of rectangle sheet = l x b
(where b = breadth of sheet)
∴ breadth (b) = \(\frac { Area }{ Length }\) = \(\frac { 500 }{ 25 }\)
= 20 cm
Perimeter of (RBSESolutions.com) rectangular sheet
= 2 x (l + b)
= 2 x (25 + 20) cm
= 90 cm
So breadth of sheet is 20 cm and its perimeter is 90 cm.

Question 2
Find the perimeter of given shape.

Solution:
In this shape we have to find out perimeter, All (RBSESolutions.com) around of a sqaure, semi circles are joined.
The diameter of every semi circle is 14 cm
Circumference of a cirlce = πd (∴2r = d)
Half circumference = \(\frac { 1 }{ 2 }\)πd
= ( \(\frac { 1 }{ 2 }\) x \(\frac { 22 }{ 7 }\) x 14) = 22cm
So, the perimeter of given shape = 4 x 22 = 88 cm.

Question 3
Sudhanshu divides a 7 cm radius plate in two equal parts (RBSESolutions.com) find out the perimeter of every half plate take (π = \(\frac { 22 }{ 7 }\))

Solution:
Radius (r) = 7 cm
Circumference = 2πr
∴ Circumference of (RBSESolutions.com) half plate = \(\frac { 1 }{ 2 }\) x 2πr = πr
= ( \(\frac { 22 }{ 7 }\) x 7) = 22 cm
Diameter = 2r
= (2 x 7) cm = 14 cm
Perimeter of every half plate
= 22 cm + 14 cm = 36 cm.

Question 4
In a 100 side square park has 5 m wide path inside it bound. Find (RBSESolutions.com) out the area of path. Calculate the expenditure @ ₹250/- per 10m² of cementing it.
Solution:
Let ABCD is 100 m side square park and shaded region is 5 m wide park.

PQ = 100 -(5+5) = 90m
Area of ABCD = (side)²
= (100)² sq. m = 10000 sq. m
Area of PQRS = (side)²
= (90)² sq. m
= 8100 sq. m
∴ Area of path = (10000 – 8100) sq. m
= 1900 sq. m
10 m² cementing = ₹250
∴ Expense on 1900 m² cementing 250
= \(\frac { 250 }{ 10 }\) x 1900
= ₹47,500

We hope the RBSE Solutions for Class 7 Maths Chapter 16 Perimeter and Area Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 16 Perimeter and Area Additional Questions, drop a comment below and we will get back to you at the earliest.