Mark (✓) against the correct answer:
Each side of a rhombus is 15 cm and the length of one of its diagonals is 24 cm. The area of the rhombus is
(a) 432 cm2
(b) 216 cm2
(c) 180 cm2
(d) 144 cm2
(b) 216 cm2
Let $A B C D$ be a rhombus whose diagonals $A C$ and $B D$ intersect at a point $O$.
Let the length of the diagonal $A C$ be $24 \mathrm{~cm}$ and the side of the rhombus be $15 \mathrm{~cm}$.
We know that the diagonals of the rhombus bisect each other at right angles.
$\therefore A O=\frac{1}{2} A C$
$\Rightarrow A O=\left(\frac{1}{2} \times 24\right) \mathrm{cm}$
$\Rightarrow A O=12 \mathrm{~cm}$
From right $\Delta A O B$, we have :
$B O^{2}=A B^{2}-A O^{2}$
$\Rightarrow B O^{2}=\left\{(15)^{2}-(12)^{2}\right\}$
$\Rightarrow B O^{2}=(225-144)$
$\Rightarrow B O^{2}=81$
$\Rightarrow B O=\sqrt{81}$
$\Rightarrow B O=9 \mathrm{~cm}$
$\therefore B D=2 \times B O$
$B D=(2 \times 9) \mathrm{cm}$
$B D=18 \mathrm{~cm}$
Hence, the length of the other diagonal is $18 \mathrm{~cm} .$
Area of the rhombus $=\left(\frac{1}{2} \times 24 \times 18\right) \mathrm{cm}^{2}$
$=216 \mathrm{~cm}^{2}$