Question

Find the sum of areas of triangle ABC and AEC as shown in the figure given below.

• A. $\frac{be+eh}{2}{\text{unit}}^2$
• B. $\frac{bd+ah}{2}{\text{unit}}^2$
• C. $\frac{b(d+c)+eh}{2}{\text{unit}}^2$
• D. $\frac{bd+eh}{2}{\text{unit}}^2$

Answer 1

The correct option is D $\frac{bd+eh}{2}{\text{unit}}^2$



$\to$ Area of a triangle ABC
The perpendicular line drawn from the vertex 'A' to the extended line BC is called height of the triangle ABC and it is equal to $b$.

Since, AD is perpendicular to extended line BC, BC is the base of triangle ABC.
Base of the triangle ABC $=BC=d$ unit
Height of the triangle $=AD=b$ unit


As we know,
$\text{Area of}△ABC=\frac{1}{2}\times \text{Base}\times \text{Height}$

$=\frac{1}{2}\times d\times b$

$=\frac{db}{2}{\text{unit}}^2$

$\to$ Area of a triangle AEC
The perpendicular line drawn from the vertex 'E' to the line AC is called height of the triangle ABC and it is equal to $h$.

Since, EF is perpendicular to line AC, AC is the base of triangle ABC.
Base of the triangle AEC $=AC=e$ unit
Height of the triangle $=EF=h$ unit

As we know,
$\text{Area of}△ABC=\frac{1}{2}\times \text{Base}\times \text{Height}$

$=\frac{1}{2}\times e\times h$

$=\frac{eh}{2}{\text{unit}}^2$

Hence, the sum of two triangles is
Area of a triangle ABC + Area of a triangle AEC $=\frac{bd}{2}+\frac{eh}{2}$

$=\frac{bd+eh}{2}{\text{unit}}^2$

Option (d) is correct.