Question

Match the images on the left with the corresponding unknown sides.

• A. $2\sqrt{pq}$
• B. $2p^2{q}^2$
• C. $2pq$

Answer 1

From Pythagoras' theorem, we have $A{B}^2+B{C}^2=A{C}^2$
Now in the first case, we have $(p-q{)}^2+B{C}^2=(p+q{)}^2$
$⟹B{C}^2=(p+q{)}^2-(p-q{)}^2=p^2+2pq+q^2-(p^2-2pq+q^2)=4pq$
$⟹BC=\sqrt{4pq}=2\sqrt{pq}$
In the second case, $A{B}^2+(p^4-q^4{)}^2=(p^4+q^4{)}^2$
$⟹A{B}^2=(p^4+q^4{)}^2-(p^4-q^4{)}^2=p^8+2p^4{q}^4+q^8-(p^8-2p^4{q}^4+q^8)=4p^4{q}^4$
$⟹AB=\sqrt{4p^4{q}^4}=2p^2{q}^2$
Similarly, in the third case, $(p^2-q^2{)}^2+B{C}^2=(p^2+q^2{)}^2$
$⟹B{C}^2=(p^2+q^2{)}^2-(p^2-q^2{)}^2=p^4+2p^2{q}^2+q^4-(p^4-2p^2{q}^2+q^4)=4p^2{q}^2$
$⟹AB=\sqrt{4p^2{q}^2}=2pq$