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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\[\text{Area} = \text{Length} \times \text{Breadth}\]
\[ \Rightarrow 192 = 16 \times \text{BC}\]
\[ \Rightarrow \text{BC} = \frac{192}{16}\]
\[ \Rightarrow \text{BC} = 12 \text{cm} . . . \left( 1 \right)\]
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