Question

The base of a triangle is $15$ cm and its height is $12$ cm. The height of another triangle of double the area and the base of which is $20$ cm, is given by

• A. $9cm$
• B. $18cm$
• C. $8cm$
• D. $22.5cm$

Answer 1

The correct option is B $18cm$

Given that,

Sides of 1^st triangle $base=15cm,height=12cm$

So,

Area of 1^st triangle

$A_1=\frac{1}{2}\times b\times h$

$=\frac{1}{2}\times 15\times 12$

$=90c{m}^2$

According to question ,

Area of 2^nd triangle is double .

So,

$A_2=90\times 2=180c{m}^2$

Given side base of 2^nd triangle is$b=20c{m}^2$

So,

Area of 2^nd triangle

$A_2=\frac{1}{2}\times b\times h$

$180=\frac{1}{2}\times 20\times h$

$h=\frac{180}{10}$

$h=18cm.$

Hence,

This is the required height of 2^nd triangle.