Question

Find the area of a figure formed by joining the midpoints of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm.

Answer 1

Let ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join the diagonals, AC and BD.
In ∆ ABC, we have:
PQ ∣∣ AC and PQ = $\frac{1}{2}$AC [By midpoint theorem]
$\mathrm{PQ}=\frac{1}{2}\times 16=8\mathrm{cm}$
Again, in ∆DAC, the points S and R are the midpoints of AD and DC, respectively.
∴ SR ∣∣ AC and SR = $\frac{1}{2}$AC [By midpoint theorem]
$\mathrm{SR}=\frac{1}{2}\times 12=6\mathrm{cm}$
$\mathrm{Area}\mathrm{of}\mathrm{PQRS}=\mathrm{length}\times \mathrm{breadth}=6\times 8=48{\mathrm{cm}}^2$