Question

Prove that the area of rhombus on the hypotenuse of a right angled triangle with one of the angle as 60° is equal to the sum of areas of rhombuses with one of their angles as 60°,drawn on the other two sides.

Answer 1

ΔABC is a right angle with ∠B as right ∠.

let AB = a and BC = b

By Pythagoras theorem AC = √ (a2 + b2 )

ACDE is rhombus with 60° inclination

Draw a perpendicular EF from E on base AC.

EF = AE sin 60°

AE = AC = √ (a^2 + b^2 )

EF = √ (a^2 + b^2 ) sin 60°

= √ (a^2 + b^2 ) √3 /2

Area of rhombus ( ACDE ) = base * height

ACDE = AC * EF

= √ (a^2 + b^2 ) * √ (a^2 + b^2 ) √3/2

= √3/2 (a^2 + b^2 ) ....................( Equation 1)

similarly in rhombus BCJG

Side BC = a

angle BCJ = 60° (GIVEN)

height = a sin 60° = a √3/2

area of rhombus BCGJ = a^2 √3/2 .................( equation 2)

similarly area of rhombus ABHI = b^2 √3/2 ..................( equation 3)

from equation (1) (2) and (3) we have

area of rhombus ACDE = area of rhombus BCJG + area of rhombus ABHI

√3/2 (a^2 + b^2 ) = a2 √3/2 + b^2 √3/2

√3/2 (a^2 + b^2 ) = √3/2 (a^2 + b^2 )

1 = 1

Hence proved...