Question

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

Answer 1

Given:
A rhombus ABCD with diagonals 12 cm and 16 cm
i.e., AC = 16 cm and BD = 12 cm
And a quadrilateral PQRS formed by joining the mid-points of the adjacent sides of ABCD.



Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆ABC,
PQ || AC
PQ = $\frac{1}{2}$AC
⇒ PQ = $\frac{1}{2}$(16)
⇒ PQ = 8 cm

In ∆ADC,
RS || AC
RS = $\frac{1}{2}$AC
⇒ RS = $\frac{1}{2}$(16)
⇒ RS = 8 cm

In ∆BCD,
RQ || BD
RQ = $\frac{1}{2}$BD
⇒ RQ = $\frac{1}{2}$(12)
⇒ RQ = 6 cm

In ∆BAD,
SP || BD
SP = $\frac{1}{2}$BD
⇒ SP = $\frac{1}{2}$(12)
⇒ SP = 6 cm

Since, PQ = 8 cm = RS and RQ = 6 cm = SP
and Diagonals of a rhombus intersect at right angle.
⇒ angle between AC and BD is 90°
⇒ angle between PQ and QR is 90°
Therefore, PQRS is a rectangle
Thus, Area of rectangle = PQ × QR
= 8 × 6
= 48 cm²

Hence, the area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm is 48 cm².