Question

In an isosceles triangle, $ABC(AB=AC)$ the altitude to the base and to a lateral side are equal to $10$ cm and $12$ cm respectively. The length of the base is

• A. $12.5$
• B. $15$
• C. $16$
• D. $18$

Answer 1

Answer: (B)

$15$

The figure is as shown in the attached image.

$sinB=\frac{10}{AB}=\frac{12}{BC}$

$\Rightarrow BC=1.2AB=1.2AC$

$sin\frac{A}{2}=\frac{BD}{AB}=\frac{\frac{BC}{2}}{AB}=\frac{0.6AB}{AB}=0.6$

$\Rightarrow cos\frac{A}{2}=0.8andtan\frac{A}{2}=\frac{0.6}{0.8}=\frac{3}{4}$

$\Rightarrow \frac{BD}{AD}=\frac{3}{4}\Rightarrow BD=10\times \frac{3}{4}=7.5$

$BC=2\times BD=15$