Question : A circle is inscribed in $\triangle $PQR touching the sides QR, PR and PQ at the points S, U and T, respectively. PQ = (QR + 5) cm, PQ = (PR + 2) cm. If the perimeter of $\triangle $PQR is 32 cm, then PR is equal to:
Option 1: 10 cm
Option 2: 13 cm
Option 3: 8 cm
Option 4: 11 cm
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Correct Answer: 11 cm
Solution : ⇒ PQ = (QR + 5) cm ⇒ PQ = (PR + 2) cm From the above two equations, ⇒ PR = (QR +3) cm The perimeter of $\triangle $PQR is 32 cm. ⇒ PQ + QR + PR = 32 cm ⇒ PQ + QR + PR = 32 cm ⇒ (QR + 5) + QR + (QR +3) = 32 cm ⇒ 3QR + 8 = 32 ⇒ QR = 8 cm The length of PR = (QR +3) = 8 + 3 = 11 cm Hence, the correct answer is 11 cm.
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Question : Let A, B, and C be the mid-points of sides PQ, QR, and PR, respectively, of PQR. If the area of $\triangle$ PQR is 32 cm2, then find the area of $\triangle$ ABC.
Question : In the given figure, a circle is inscribed in $\triangle$PQR, such that it touches the sides PQ, QR and RP, at points D, E, and F, respectively. If the lengths of the sides PQ = 18 cm, QR = 13 cm, and RP = 15 cm, find the length of PD.
Question : In a triangle PQR, S, and T are the points on PQ and PR, respectively, such that ST || QR and $\frac{\text{PS}}{\text{SQ}} = \frac{3}{5}$ and PR = 6 cm. Then PT is:
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Question : In a $\triangle ABC$, if $\angle A=90^{\circ}, AC=5 \mathrm{~cm}, BC=9 \mathrm{~cm}$ and in $\triangle PQR, \angle P=90^{\circ}, PR=3 \mathrm{~cm}, QR=8$ $\mathrm{cm}$, then:
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