Question

Side of given rhombus is $5\text{units}.$ Find area of the rhombus if height of the rhombus is is $4\text{units}.$

• A. $16{\text{unit}}^2$
• B. $18{\text{unit}}^2$
• C. $20{\text{unit}}^2$
• D. None of the above.

Answer 1

The correct option is C $20{\text{unit}}^2$

Properties of rhombus:

• All sides are equal in length.
• Opposite angles are equal.
• Diagonals bisect each other at ${90}^o.$

$BE=4\text{units}=$ height of the rhombus.
As height is perpendicular to the base, $△BEA$ is a right triangle.

$B{A}^2=A{E}^2+E{B}^2$
$\Rightarrow 5^2=A{E}^2+4^2$
$\Rightarrow 5^2-4^2=A{E}^2$
$\Rightarrow AE=3\text{units}$

Area of $△BEA$
$=\frac{1}{2}\times \text{base}\times \text{height}$
$=\frac{1}{2}\times AE\times BE$
$=\frac{1}{2}\times 3\times 4=\frac{12}{2}{\text{unit}}^2=6{\text{unit}}^2$

$△BEA$ and $△DFC$ are congruent triangles (Hypotenuse-Leg-congruency)
Hence, area of $△DFC=\frac{12}{2}{\text{unit}}^2=6{\text{unit}}^2$

$AD=AE+ED$
$\Rightarrow 5=3+ED$
$\Rightarrow ED=2\text{units}$

Area of the rectangle BFDE $=\text{BF}\times \text{FD}$
$=4\times 2{\text{unit}}^2$
$=8{\text{unit}}^2$

Area of the rhombus

$=\text{area of}(△BEA+△DFC+\text{rectangle}BFDE)$
$=6+6+8{\text{unit}}^2$
$=12+8{\text{unit}}^2$
$=20{\text{unit}}^2$