Question

The figure above shows six right triangles. What is the value of $x^2+y^2$?

• A. $21$
• B. $27$
• C. $33$
• D. $\sqrt{593}$ (approximately $24.35$)
• E. $\sqrt{611}$ (approximately $24.72$)

Answer 1

Answer: (A)

$21$

Using Pythagoras theorem on all the triangles we get,

In $△ABH$,

$B{H}^2=A{H}^2+A{B}^2=10+7=17$ ....(I)

In $△BCD$,

$B{D}^2=B{C}^2+C{D}^2=4+6=10$ ....(II)

In $△BHD$,

$H{D}^2=B{H}^2+B{D}^2=17+10=27$ ....(III) ...From (I) and (II)

In $△GHF$,

$H{F}^2=G{H}^2+G{F}^2=3+3=6$ ....(IV)

In $△HFD$,

$F{D}^2=H{D}^2-F{H}^2=27-6=21$ ....(V) ...From (III) and (IV)

In $△FED$,

$F{D}^2=F{E}^2+D{E}^2=x^2+y^2$

$x^2+y^2=21$ ....From (V)