Question

ABCD is a parallelogram in which BC is produced to P such that CP = BC, as shown in the adjoining figure. AP intersects CD at M. If ar(DMB) = 7 cm², find the area of parallelogram ABCD.

Answer 1

In $△$MDA and $△$MCP,
$\angle$DMA = $\angle$CMP (Vertically opposite angles)
$\angle$MDA = $\angle$MCP (Alternate interior angles)
AD = CP (Since AD = CB and CB = CP)
So, $△$MDA $\cong$ $△$MCP (ASA congruency)
$\Rightarrow$DM = MC (CPCT)
$\Rightarrow$M is the midpoint of DC
$\Rightarrow$BM is the median of $△$BDC
$\Rightarrow$ar($△$BMC) = ar($△$DMB) = 7 cm²
ar($△$BMC) + ar($△$DMB) = ar($△$DBC) = 7 + 7 = 14 ${\mathrm{cm}}^2$
Area of parallelogram ABCD = 2 $\times$ ar($△$DBC) = 2 $\times$ 14 = 28 ${\mathrm{cm}}^2$