Chapter 11, Problem 2MCQ

To determine

To Compute : The perimeter and area of triangle ABD.

Answer to Problem 2MCQ

The perimeter of triangle ABD is 60 cm and area of triangle ABD is 150 cm².

Explanation of Solution

Given information :

  

Theory used :

  $\text{Area}\text{of}\text{a}\text{triangle}=\frac{1}{2}\text{bh}$.....(1)

  $\text{Perimeter}\text{of}\text{a}\text{triangle}\text{=}\text{Sum}\text{of}\text{all}\text{the}\text{sides}\text{of the}\text{triangle}\text{.}$........(2)

In $△ABD$ ,

  $AB=\sqrt{B{D}^2-A{D}^2}$ according to Pythagoras theorem.

Therefore,

  $AB=\sqrt{{25}^2-{15}^2}$

That is, $AB=\sqrt{400}=20$

As perimeter( $△ABD$ ) = AB + BD + AD, using equation (2)

Implies, P( $△ABD$ ) = 20 + 25 + 15

P( $△ABD$ ) = 60 cm.

According to Pythagoras theorem in $△ACD$,

  $AC=\sqrt{A{D}^2-C{D}^2}$

  $AC=\sqrt{{15}^2-9^2}$

  $AC=\sqrt{144}=12$cm

Therefore,

Area of $△ABD$= $\frac{1}{2}\times \text{BD}\times \text{AC}$

= $\frac{1}{2}\times 25\times 12$

  $=150c{m}^2$

Thus,

The perimeter and area of $△ABD$ are 60cm and 150cm²respectively.