Question

In the adjoining figure,ABCD is a quadrilateral.A line through D,parallel to AC,meets BC produced in P.

Then, $ar\left(△ABP\right)$ = _____.

• A.
  ar(quad. ACPD)
• B.
  ar($△ACD)$
• C.
  ar($△ABC)$
• D.
  ar(quad. ABCD)

Answer 1

The correct option is D

ar(quad. ABCD)
We have:
ar(quad. ABCD) = ar(∆ACD) + ar(∆ABC)
ar(∆ABP ) = ar(∆ACP)​​ + ar(∆ABC)

∆ACD and ∆ACP are on the same base and between the same parallels AC and DP.
∴ ar(∆ACD) = ar(∆ACP)

By adding ar(∆ABC) on both sides, we get:
ar(∆ACD) + ar(∆ABC) = ar(∆ACP)​​ + ar(∆ABC)
⇒​ ar (quad. ABCD) = ar(∆ABP )