In rhombus PQRS,PR = 40 , cm and QS = 42 , cm | vedclass

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(B) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.Let the diagonals $PR$ and $QS$ intersect at point $O$.Therefore,$PO =

Vedclass · Accepted Answer

$29$

Vedclass · Answer

(B) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.Let the diagonals $PR$ and $QS$ intersect at point $O$.Therefore,$PO = \frac{PR}{2} = \frac{40}{2} = 20 \, cm$.And $QO = \frac{QS}{2} = \frac{42}{2} = 21 \, cm$.In the right-angled triangle $	riangle POQ$,by the Pythagorean theorem:$PQ^2 = PO^2 + QO^2$$PQ^2 = 20^2 + 21^2$$PQ^2 = 400 + 441$$PQ^2 = 841$$PQ = \sqrt{841} = 29 \, cm$.

In rhombus $PQRS$,$PR = 40 \, cm$ and $QS = 42 \, cm$,then $PQ = \dots \dots \dots \, cm$.

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