Chapter 11, Problem 3MCQ

To determine

To Compute : The perimeter and area of triangle ABD.

Answer to Problem 3MCQ

The perimeter of triangle ABD is 70 cm and area of triangle ABD is 111 cm².

Explanation of Solution

Given information :

  

AD = 21 m in above triangle.

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 $△ACD$,

  $AC=\sqrt{A{D}^2-C{D}^2}$

  $AC=\sqrt{{21}^2-{17}^2}$

  $=\sqrt{152}$m

Using equation (2),

Perimeter of $(△ABD)$= Length of (AB + BD + AD )

= 31 + 18 + 21

= 70 m

Using equation (1),

Area of $(△ABD)$= $\frac{1}{2}\times AC\times BD$, where AC and BD are height and base of $(△ABD)$ respectively.

= $\frac{1}{2}\times 18\times \sqrt{152}$

  $=110.96m^2=111m^2$

Thus,

The perimeter of triangle ABD is 70 cm and area of triangle ABD is 111 cm².