In the adjoining frigure , `trianglePQR` is a right angled triangle in which `anglePQR=90^(@), squreABCD` is a square of side 4 units and `squre GHIJ` is a square of side 7 units. EC and CG are the length and breadth of rectangle CEFG. Find the length EC.

Explanation

Let `angleQPR= theta`
`angleQRP= 90^(@)-theta ( angleQ= 90^(@))`
`angleEAB= theta` ( corresponding angle of ` angleQPR`)
`angleAEB=90^(@)-theta`
Also `angleFIH = 90^(@)-theta` ( corresponding angle of `angleQRP`)
Let EC= x units
AB= 4 units HG = 7 units.
EB = (x-4) units, FH = (x-7) units
now in `triangleEBA and triangleFHI`
`angleAEB=angleFIH (each 90-theta) `
`angleEBA= angleFHI (each 90^(@))`
`triangle EBA ~ triangleIHF` (AA corollary)
`(EB)/(IH)= (BA)/(HF)= (EA)/(IF)`
` Rightarrow (x-4)/7= 4/(x-7)`
` Rightarrow x^(2)-7x-4x+28= 28`
`x^(2)-11x=0`
`x^(2)-11x =0`
x(x-11)=0
x=0 or x=11
But x=0 is not possible as length of side cannot be zero.
so, x=11 units , Hence , EF = 11 units.