Question

ABCD is a Square of side 2√2 cm and M, N are mid points of sides AB and AD respectively. Then the perimeter (in cm) of trapezium BDNM so formed, is
1. 2(3 + √2)
2. 2(2 + √2)
3. 2(1 + √2)
4. 2(1 + 3√2)

Answer 1

Correct Answer - Option 1 : 2(3 + √2)

Given,

ABCD is a square of side = 2√ 2 cm 

And M,N are mid points of sides AB and AD 

Then the  perimeter (in cm ) of trapezium BDNM

Calculation

A square = ABCD 

AB  = 2√2 = AD

M and N are mid points of sides AB and AD respectively 

so AM = MB =AN = ND = √2 

in triangle ANM

NM² = AN² + AM²        by Pythagoras theorem 

NM = √(√b)² + (√2)²  = 2cm 

In  triangle ABD

BD² = AD² + AB²

⇒ √ (2√2)² + (2√2)²

⇒ 4 cm 

Perimeter of Trapezium BDMN

⇒ Sum of parallel side + non parallel side

⇒ 4 + 2 + √2 + √2

⇒ 6 + 2√2

⇒ 2(3 + √2)   

∴ Perimeter of trapezium BDMN is 2(3 + √2).