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.
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.
Related Questions
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