Equilateral triangles are drawn on the sides of a right triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.

Explanation

Let a right triangle BAC in which `angleA` is right angle and AC=y, AB=x
Three equilateral triangles `DeltaAEC,DeltaAFB and DeltaCBD` are drawn on the three sides of `DeltaABC`
Again let area of triangles made on AC,AB and BC are `A_(1),A_(2) and A_(3)`, respectively.
To prove `A_(3)=A_(1)+A_(2)`
Proof In `DeltaCAB`, by pythagoras theorem,
`BC^(2)=AC^(2)+AB^(2)`
`rArrBC^(2)=y^(2)+x^(2)`
`rArrBC=sqrt(y^(2)+x^(2))`
We know that, area of an equilateral triangle=`(sqrt3)/4("side")^(@)`
`therefore` Area of equilateral `DeltaAEC,A_(1)=(sqrt3)/4(AC)^(2)`
`rArr A_(1)=(sqrt3)/4y^(2)`
and equilateral `DeltaAFB,A_(2)=(sqrt3)/4(AB)^(2)=(sqrt3)/4sqrt((y^(2)+x^(2)))`
`=(sqrt3)/4(y^(2)+x^(2))=(sqrt3)/4y^(2)+(sqrt(3))/4x^(2)`
`=A_(1)+A_(2)` [From Eqs (i) and (ii)]
Hence proved