Question

In $△PQR,$ $PD\perp QR$ such that $D$ lies on $QR.$ If $PQ=a,PR=b,QD=c$ and $DR=d,$ then

• A. $(a-d)(a+d)=(b-c)(b+c)$
• B. $(a-c)(b-d)=(a+c)(b+d)$
• C. $(a-b)(a+b)=(c+d)(c-d)$
• D. $(a-b)(c-d)=(a+b)(c+d)$

Answer 1

Answer: (C)

$(a-b)(a+b)=(c+d)(c-d)$

Use Pythagoras' Theorem to find $PD$ from $△PQD$ and $△PDR.$

First take $△PDQ$

Now, $P{Q}^2=Q{D}^2+P{D}^2$

$\Rightarrow P{D}^2=P{Q}^2-Q{D}^2$ ....(1)

Now take $△PDR$

We have $P{R}^2=P{D}^2+D{R}^2$

$\Rightarrow P{D}^2=P{R}^2-D{R}^2$ ....(2)

Now equate (1) and (2) equations, we get

$P{Q}^2-Q{D}^2=P{R}^2-D{R}^2$

$\Rightarrow a^2-c^2=b^2-d^2$

$\Rightarrow a^2-b^2=c^2-d^2$

$\Rightarrow (a-b)(a+b)=(c-d)(c+d)$
Hence, option C is correct.