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Question
Rhombus PQRB is inscribed in ΔABC such that ∠B is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus.
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Solution
Let the side of the rhombus be “x”. Since PQRB is a
Rhombus PQ || BC
By basic proportionality theorem
`"AP"/"AB" = "PQ"/"BC" ⇒ (12 - x)/"BC" = x/6`
12x = 6(12 – x)
12x = 72 – 6x
12x + 6x = 72
18x = 72 ⇒ x = `72/18` = 4
Side of a rhombus = 4 cm
PQ = RB = 4 cm
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