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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 _________.


Solution

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 = 12AC
⇒ PQ = 
12(16)
⇒ PQ = 8 cm

In ∆ADC,
RS || AC
RS = 12AC
⇒ RS = 
12(16)
⇒ RS = 8 cm

In ∆BCD,
RQ || BD
RQ 12BD
⇒ RQ = 
12(12)
⇒ RQ = 6 cm

In ∆BAD,
SP || BD
SP 12BD
⇒ SP = 
12(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.

Mathematics
RD Sharma (2019)
All

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