Question

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

Answer 1

Join the mid points of AB, BC, CD, DA of a rhombus ABCD and name them M, N, O and P respectively to form a figure MNOP.

Joint the line PN, then $PN\parallel AB$ and $PN\parallel DC$

We know that if a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to one-half area of the parallelogram.

From the above result parallelogram, ABNP and triangle MNOP are on the same base PN and in between same parallel lines PN and AB.

$\therefore area\Delta MNP=\frac{1}{2}areaABPN$.................(1)

$area\Delta PON=\frac{1}{2}areaPNCD$........................................(2)

Then area of ABCD$=\frac{1}{2}\times d_2\times d_2$

From (i) and (ii) we get

$area\left(MNOP\right)=area\left(\Delta MNP\right)+area\left(\Delta PON\right)$

$=\frac{1}{2}areaABPN+\frac{1}{2}areaPNDC$

$=\frac{1}{2}\left(\frac{1}{2}areaABCD\right)$

$=\frac{1}{2}(\frac{1}{2}\times 12\times 16)=48c{m}^2$