# Mark (✓) against the correct answer:

Question:

Mark (✓) against the correct answer:

The area of a rhombus is 120 cm2 and one of its diagonals is 24 cm. Each side of the rhombus is

(a) 10 cm

(b) 13 cm

(c) 12 cm

(d) 15 cm

Solution:

(b) 13 cm

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 $24 \mathrm{~cm}$.

$A$ rea of the rhombus $=\left(\frac{1}{2} \times A C \times B D\right) \mathrm{cm}^{2}$

But the area of the rohmbus is $120 \mathrm{~cm}^{2}$ (given)

$\therefore \frac{1}{2} \times A C \times B D=120$

or $\frac{1}{2} \times 24 \times B D=120$

or $12 \times B D=120$

or $B D=\frac{120}{12}=10 \mathrm{~cm}$

$\mathrm{OB}=\frac{\mathrm{BD}}{2}=\frac{10}{2}=5 \mathrm{~cm}$

And $O A=\frac{A C}{2}=\frac{24}{2}=12 \mathrm{~cm}$

Now, in right triangle $A O B$ :

$(A B)^{2}=(O A)^{2}+(O B)^{2}$

or $(A B)^{2}=12^{2}+5^{2}$

$=144+25$

$=169$

or $A B=\sqrt{169}=13 \mathrm{~cm}$

Therefore, each side of the rhombus is $13 \mathrm{~cm}$.