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Mensuration

This document provides examples of calculating perimeter and area for different shapes. It includes 4 examples of finding the perimeter of various shapes like a right triangle, rectangular path, triangular garden, and rectangular park. It also defines area as the amount of space an object has and notes there is one additional example of calculating area.

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Dhiman Dey
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111 views2 pages

Mensuration

This document provides examples of calculating perimeter and area for different shapes. It includes 4 examples of finding the perimeter of various shapes like a right triangle, rectangular path, triangular garden, and rectangular park. It also defines area as the amount of space an object has and notes there is one additional example of calculating area.

Uploaded by

Dhiman Dey
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Mensuration

Perimeter of a rectangle, square, triangle and other shapes.

We have given some problems on perimeter below.

Ex 1. Find the perimeter of a right triangle of altitude 4 cm and base 3 cm.


Sol. Let the triangle be ABC.
A

4 cm

B 3 cm C
By Pythagoras Theorem, we have
AB2 + BC2 = AC2
 42 + 32 = AC2  16 cm + 9 cm = AC2
 AC2 = 25 cm  AC = 5 cm
Perimeter = (4 + 3 + 5) cm = 12 cm.

Ex 2. A man runs on a rectangular path of length 25 cm and breadth 18 cm at a speed of 43 m/s.


Find the time taken by the man to complete 4 rounds on the path.
Sol. Total distance covered by the man on first round = Perimeter of the rectangular path
= 2 × (25 + 18) cm
= 2 × 43 cm = 86 cm.
Speed of the man = 43 m/s.
86
 Time taken by the man to cover 1 round = 43 sec.
86×4
 Time taken by the man to cover 4 rounds = sec = 8 seconds.
43

Ex 3. Found the cost of fencing a triangular garden of length 8 m, 9 m and 11 m if the rate is Rs.
1680 per m.
Sol. Perimeter of the garden = (8 + 9 + 11) m = 28 m.
Cost required for fencing the garden = Rs. (1680 × 28) = Rs. 47040.

Ex 4. The length of a rectangular part is 3 times subtracted by 50 m from the breadth. Find the
area of the park, if perimeter is 492 m.
Sol. Let the breadth of the park be x metres. Then we have,
∴ Length of the park = (3x - 50) metres.
According to the question,
 2 × (3x - 50 + x) = 492 m
 6x - 100 + 2x = 492 m
 6x + 2x = 492 m + 100 m
 8x = 592 m  x = 592 × 1/8 m  x = 74 m
Length of the park = (74 × 3) + 50 m = 222 m + 50 m = 272 m.
Area = Length × breadth = (272 × 74) m2 = 20128 m2
Area

Area is the amount of space and object has.


Solved Examples.

Ex 5.

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