Question

In the following figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, then:

• A. area of($△$BPC) = $\frac{1}{3}$ area of ($△$DPQ)
• B. area of($△$BPC) = $\frac{3}{8}$area of ($△$DPQ)
• C. area of($△$BPC) = $\frac{1}{4}$ area of ($△$DPQ)
• D. area of($△$BPC) = area of ($△$DPQ)

Answer 1

The correct option is D

area of($△$BPC) = area of ($△$DPQ)

(i)ABCD is a ||gm (given)

(ii)$\therefore$ AD =DC

(iii)But AD =CQ (given)

$\left(iv\right)\therefore AD=DC=CQ$

(v)AD||CQ

(vi)ADQC is a ||gm

(one pair opposite sides are equal and parallel)

(vii)area $\Delta ADC$ =area $\Delta \left(AQD\right)$

$({\Delta }^{\prime }s$ in the same base and between the same||'s)

$\therefore$ area $\Delta ADP$ +area $\Delta APC$ =are $\Delta ADP$ +area $\Delta DPQ$

$\therefore$ Area $\Delta APC$ =area $\Delta DPQ$

(viii)area $\Delta APC$ =area $\Delta BPC({\Delta }^{\prime }s$ in the same base and between the same || lines)

$\left(ix\right)\therefore$ area $\Delta APC$ =area $\Delta BPC$