Question

What is the difference in perimeters of triangles ABC and ACD in the given figure ABCD? (Take $\sqrt{2}$ = 1.4 and $\sqrt{5}$ = 2.2)

• A. 0.8 m
• B. 0.3 m
• C. 0.6 m
• D. 0.4 m

Answer 1

The correct option is D

0.4 m

ABC is a right - angled isosceles triangle.

The sides are in the ratio 1 : 1 : $\sqrt{2}$

Let the sides be x, x , $\sqrt{2}$ x

Here, x = $\sqrt{2}$

$\therefore$ The Length of AC = $\sqrt{2}$x

= $\sqrt{2}\times \sqrt{2}$ = 2 m

Hence, using Pythagoras theorem we can say:

AD = $\sqrt{1^2+2^2}$ = $\sqrt{5}$ m

$\therefore$ Perimeter of the figure ABC = AB + BC + AC

= $\sqrt{2}+\sqrt{2}+2$

= 1.4 + 1.4 + 2

= 4.8 m

Perimeter of the figure ACD = AC+CD+AD

= $2+1+\sqrt{5}$

= 3+ 2.2

= 5.2 m

So, the required difference in perimeters =5.2 -4.8 = 0.4 m