Question

One diagonal of a rhombus is greater than the other by 4 cm . If the area of the rhombus is 96 cm² , find the side of the rhombus .

Answer 1

Let the length of the second diagonal of the rhombus be x cm. Then, the length of the first diagonal of the rhombus will be (x + 4) cm.
We know that the area of a rhombus is A = $\frac{1}{2}$ × d₁ × d₂, where d₁ and d₂ are the diagonals of the rhombus.
Thus, by the given condition, we have:
$\frac{1}{2}$ × (x + 4) × x = 96
$\Rightarrow$ x² + 4x = 192
$\Rightarrow$ x² + 4x – 192 = 0
On splitting the middle term 4x as 16x – 12x, we get:
x² + 16x – 12x – 192 = 0
$\Rightarrow$ x(x + 16) – 12(x + 16) = 0
$\Rightarrow$ (x + 16)(x – 12) = 0
$\Rightarrow$x + 16 = 0 or x – 12 = 0
$\Rightarrow$x = –16 or x = 12
Since the diagonal cannot be negative, x = 12.
Thus, the first diagonal of the rhombus is 12 cm and the second diagonal is 16 cm.
Now, draw the figure of the rhombus ABCD whose diagonals are AC = 16 cm and BD = 12 cm.

We know that the diagonals of a rhombus bisect each other and the angle made between them is 90°. Thus, $\mathrm{OA}=\frac{\mathrm{AC}}{2}=\frac{16}{2}=8\mathrm{cm}\mathrm{and}\mathrm{OB}=\frac{\mathrm{BD}}{2}=\frac{12}{2}=6\mathrm{cm}$

Thus, ΔAOB is a right-angled triangle.
On using Pythagoras' Theorem, we get:
AB² = OA² + OB²
$\Rightarrow$ AB² = (8)² + (6)²
$\Rightarrow$ AB² = 64 +36
$\Rightarrow$ AB² = 100 = 10²
$\Rightarrow$ AB = 10 cm
Thus, the side of the rhombus is 10 cm.