Question

In triangle PQR angle Q=90 degree and QR = 4cm more than PQ. If area of triangle PQR =96cm^2 , find the sides of the triangle​

Answer 1

Answer

PQ = 12 cm

QR = 16 cm

PR = 20 cm

Explanation

Given that in ΔPQR, ∠Q = 90°

⇒ ΔPQR is right angled triangle, right angled at Q

⇒ PR is hypotenuse and other two sides are legs of the triangle (base and height)

Also, given that QR = 4 + PQ and area of triangle PQR = 96 cm²

area of triangle PQR = 1/2 × base × height

⇒ 96 = 1/2 × PQ × QR

⇒ 96 = 1/2 × PQ × (4 + PQ)

⇒ 96 × 2 = 4PQ + PQ²

⇒ 192 = 4PQ + PQ²

⇒ PQ² + 4PQ - 192 = 0

⇒ PQ² + 16PQ - 12PQ - 192 = 0 (Splitting the middle terms)

⇒ PQ(PQ + 16) - 12(PQ + 16) = 0

⇒ (PQ - 12)(PQ + 16) = 0

⇒ PQ = 12 or PQ = -16

Since length can't be negative, PQ = 12 cm

⇒ QR = 4 + PQ = 4 + 12 = 16 cm

Now, since the triangle is right angled triangle, apply Pythagoras theorem to find PR

PR² = PQ² + QR²

⇒ PR² = 12² + 16²

⇒ PR² = 144 + 256

⇒ PR² = 400

⇒ PR = 20 cm

Answer 2

Step-by-step explanation:

$</p><p>Area \: of \: Δ PQR = \frac{1}{2} \times QR \times PQ \\ \\ \frac{1}{2} \times (4 +PQ)\times PQ = 96 \\ \\ 4PQ + PQ{}^{2}= {96 \times 2} \\ \\4PQ + PQ{}^{2} = 192$

PQ² + 4PQ - 192 = 0

PQ² + 16PQ - 12PQ - 192

( By Middle Term Split )

PQ(PQ+16) - 12(PQ+16)

(PQ-12)(PQ+16)

PQ - 12 = 0 and PQ + 16 = 0

PQ = 12 cm and PQ = - 16

• Length Can't Be Negative

QR = 4 + PQ

QR = 4 + 12

QR = 16 cm

(PR)² = (PQ)² + (QR)²

( By Pythagoras Theorem )

(PR)² = (12)² + (16)²

(PR)² = 144 + 256

(PR)² = 400

PR = √400

PR = 20 cm