Mathematics
PQR is a right-angled triangle with PQ = 3 cm and QR = 4 cm. A circle which touches all the sides of the triangle is inscribed in the triangle. Calculate the radius of the circle.
Circles
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Answer
Let the inscribed circle touch the sides PQ, QR and PR at A, B and C respectively.
PQR is a right-angled triangle with PQ = 3 cm and QR = 4 cm.
By pythagoras theorem,
⇒ PR2 = PQ2 + QR2
⇒ PR2 = 32 + 42
⇒ PR2 = 9 + 16
⇒ PR2 = 25
⇒ PR = 5 cm.
By tangent property we have,
∠OAQ = ∠OBQ = 90°
∠AQB = 90°
Since, all the angles of OAQB equals to 90°.
From figure,
OAQB is a square.
OA = OB = AQ = BQ = x (let)
PA = PQ - AQ = (3 - x) cm
PA = PC = (3 - x) cm.[∵ Tangents from exterior point are equal in length.]
RB = RQ - QB = (4 - x) cm
RC = RB = (4 - x) cm.[∵ Tangents from exterior point are equal in length.]
PR = PC + RC = 3 - x + 4 - x = 7 - 2x
5 = 7 - 2x
2x = 7 - 5
2x = 2
x = 1 cm.
Hence, radius of the circle inscribed in the triangle equals to 1 cm.
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