Question

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

• A. $(c-d)$
• B. $(c+d)(c-d)$
• C. $(c+d)$
• D. $0$

Answer 1

Answer: (B)

$(c+d)(c-d)$

In $\Delta PQR$

$PD\perp QR$

$PQ=a,PR=b,QD=c,DR=d$

In right $\Delta PQD$

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

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

$P{D}^2=a^2-c^2$ ___(1)

In right $\Delta PDR$

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

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

$P{D}^2=b^2-d^2$ ___(2)

from (1) & (2)

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

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

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