Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm ?
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 4 | Page 44
Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm ?
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solution
[text{Area} = text{Length} times text{Breadth}]
[ Rightarrow 192 = 16 times text{BC}]
[ Rightarrow text{BC} = frac{192}{16}]
[ Rightarrow text{BC} = 12 text{cm} . . . left( 1 right)]
According to Pythagoras theorem,
In ∆ABC
[{text{AB}}^2 + {text{BC}}^2 = {text{AC}}^2 ]
[ Rightarrow left( 16 right)^2 + left( 12 right)^2 = {text{AC}}^2 ]
[ Rightarrow 256 + 144 = {text{AC}}^2 ]
[ Rightarrow {text{AC}}^2 = 400]
[ Rightarrow text{AC} = 20 text{cm}]
Hence, the length of a diagonal of the rectangle is 20 cm.
Explanation:-
We know that the area of a rectangle is given by:
area = length x width
We are given that the length of the rectangle is 16 cm and its area is 192 sq.cm. So we can find the width of the rectangle as follows:
width = area / length width = 192 sq.cm / 16 cm width = 12 cm
Now we can use the Pythagorean theorem to find the length of the diagonal of the rectangle. The diagonal is the hypotenuse of a right-angled triangle whose legs are the length and width of the rectangle. Therefore, we have:
diagonal^2 = length^2 + width^2 diagonal^2 = 16^2 + 12^2 diagonal^2 = 256 + 144 diagonal^2 = 400
Taking the square root of both sides, we get:
diagonal = sqrt(400) diagonal = 20 cm
Therefore, the length of the diagonal of the rectangle is 20 cm.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 4 | Page 44