Question

The area of an isosceles triangle whose base = 10 cm and equal sides = 13 cm is (in $c{m}^2$).

• A. 60
• B. 30
• C. 12

Answer 1

Answer: (A), (B)

The correct options are
A 60
B 30

Given: base = 10 cm
Equal sides = 13 cm
Construction: Draw a perpendicular to the base from the opposite vertex.

Using pythagoras theorem we know that,
$A{C}^2=A{D}^2+D{C}^2{13}^2=A{D}^2+5^{2}169-25=A{D}^{2}AD=\sqrt{144}AD=12$

Area of $△ABC$ = Area of $△ADC$ + Area of $△ADB$
Since $△ABC$ is isosceles, $△ADC$ is congruent to $△ADB$ .
$\Rightarrow$Area of $△ABC$ = $2\times$ Area of $△ADC$
= $2\times \{\frac{1}{2}\times \text{base}\times \text{height}\}$
= $2\times \{\frac{1}{2}\times \text{5}\times \text{12}\}$
$\therefore$ Area of $△ABC=60c{m}^2$