A right angled $Δ P Q R$ , where $∠ P Q R = 90^{\circ}$ . A circle with centre O is inscribed in this triangle. Find the radius of circle. Length of side PQ = 8 cm and PR is 15 cm.

3 cm

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4 cm

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5 cm

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4.2 cm

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Solution

The correct option is A
3 cm

By Pythagoras theorem, QR = 17 cm

Area of $Δ P Q R = \frac{1}{2} \times Q P \times P R$

OT = OS = OU = r (radii of the same circle)

Area of $Δ Q O R = \frac{1}{2} \times Q R \times r$

Area of $Δ P O Q = \frac{1}{2} \times Q P \times r$

Area of $Δ P O R = \frac{1}{2} \times P R \times r$

Area of $Δ P Q R = A r e a o f Δ Q O R + A r e a o f Δ P O Q + A r e a o f Δ P O R$

Or, $\frac{1}{2} (8 \times 15) = \frac{1}{2} r (8 + 15 + 17)$

Or, 3 cm = r

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A right angled ΔPQR, where ∠PQR=90∘. A circle with centre O is inscribed in this triangle. Find the radius of circle. Length of side PQ = 8 cm and PR is 15 cm.

The correct option is A 3 cm By Pythagoras theorem, QR = 17 cm Area of ΔPQR=12×QP×PR OT = OS = OU = r (radii of the same circle) Area of ΔQOR=12×QR×r Area of ΔPOQ=12×QP×r Area of ΔPOR=12×PR×r Area of ΔPQR=Area of ΔQOR+Area of ΔPOQ+Area of ΔPOR Or, 12(8×15)=12r(8+15+17) Or, 3 cm = r

A right angled $Δ P Q R$ , where $∠ P Q R = 90^{\circ}$ . A circle with centre O is inscribed in this triangle. Find the radius of circle. Length of side PQ = 8 cm and PR is 15 cm.