Question

One diagonal of a rhombus exceeds the other by $4$ cm. If the area of the rhombus is $96$ $c{m}^2$, find the side of the rhombus.

• A. $5$
• B. $8$
• C. $10$
• D. $13$

Answer 1

Answer: (C)

$10$

Let the length of diagonals be $a$ and $b$
$a-b=4$
or $a=b+4$

Area of rhombus is $96c{m}^2$
$\therefore \frac{1}{2}ab=96$
$(b+4)b=192$
$b^2+4b-192=0$
$b^2-12b+16b-192=0$
$(b-12)(b+16)=0$
$b=12,-16$
Neglect the negative value.
Hence, length of diagonals are $12$ and $16$
Length of the side $=\frac{1}{2}\sqrt{{12}^2+{16}^2}$ $=10cm$