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 = $\frac{1}{2}$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.MathematicsRD Sharma (2019)All

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