#### Question

Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm ?

#### Solution

It is given that, area of rectangle is 192 sq.cm.

\[\text{Area} = \text{Length} \times \text{Breadth}\]

\[ \Rightarrow 192 = 16 \times \text{BC}\]

\[ \Rightarrow \text{BC} = \frac{192}{16}\]

\[ \Rightarrow \text{BC} = 12 \text{cm} . . . \left( 1 \right)\]

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.

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Solution Find the Diagonal of a Rectangle Whose Length is 16 Cm and Area is 192 Sq.Cm. Concept: Application of Pythagoras Theorem in Acute Angle and Obtuse Angle.