Question

# What is the difference in perimeters of triangles ABC and ACD in the given figure ABCD? (Take √2 = 1.4 and √5 = 2.2)

A

0.8 m

B

0.3 m

C

0.6 m

D

0.4 m

Solution

## The correct option is D 0.4 m ABC is a right - angled isosceles triangle. The sides are in the ratio 1 : 1 : √2 Let the sides be x, x , √2 x Here,  x = √2 ∴ The Length of AC =  √2x                            =  √2×√2 = 2 m Hence, using Pythagoras theorem we can say: AD = √12+22 = √5 m ∴ Perimeter of the figure ABC = AB + BC + AC  = √2+√2+2 = 1.4 + 1.4 + 2 = 4.8 m Perimeter of the figure ACD = AC+CD+AD = 2+1+√5 = 3+ 2.2 = 5.2 m So, the required difference in perimeters =5.2 -4.8 = 0.4 m

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