IN TRIANGLE PQR , Q = 90 A CIRCLE WITH CENTRE O AND RADIUS 2 CM - Maths - Circles

Question

IN TRIANGLE PQR , < Q = 90 A CIRCLE WITH CENTRE O AND RADIUS
2 CM HAS BEEN INSCRIBED IN THE TRIANGLE IF QR = 6 CM , FIND THE
PERIMETER OF TRIANGLE PQR

Ankush Jain · Accepted Answer

Given: Triangle PQR is a right angled triangle at Q Also, QR = 6 cm â‡’ x + y = 6 (1) Again, OA = OB = OC = 2 cm [radii of the same circle] Again, QA = QB = x, RB = RC = y and PA = PC= z [As tangents drawn from an exterior point to a circle are equal] Now, âˆ PQR = 90Â° Also, âˆ OAQ = âˆ OBQ = 90Â° [The radius of a circle is perpendicular to the tangent at the point of contact] Again, QA = QB and OA = OB [radii of same circle] Therefore, OABQ is a square â‡’ OA = OB = QA = QB = x = 2 cm Now, QR = 6 cm â‡’ x + y = 6 â‡’ 2 cm + y = 6 cm â‡’ y = 4 cm â‡’ BR = CR = 4 cm Also, Area of âˆ†PQR = Area âˆ†POQ + Area âˆ†OQR + Area âˆ†POR $\Rightarrow \frac{1}{2} \times Q R \times P Q = \frac{1}{2} \times O A \times P Q + \frac{1}{2} \times O B \times Q R + \frac{1}{2} \times O C \times P R \Rightarrow \frac{1}{2} \times 6 \times (x + z) = \frac{1}{2} \times 2 \times (x + z) + \frac{1}{2} \times 2 \times 6 + \frac{1}{2} \times 2 \times (y + z)] \Rightarrow \frac{1}{2} \times 6 \times (x + z) = \frac{1}{2} \times 2 \times [(x + z) + 6 + (y + z)] \Rightarrow 3 (x + z) = [x + y + 6 + 2 z] \Rightarrow 3 (2 + z) = 12 + 2 z [u \sin g : x + y = 6] \Rightarrow 6 + 3 z = 12 + 2 z \Rightarrow z = 6 c m$ Now, perimeter of triangle PQR = PQ + QR + PR = (x + z) + 6 + (y + z) = 6 + 6 + 2z [using: x + y = 6 cm] = 12 +2(6) = 24 cm

IN TRIANGLE PQR , < Q = 90 . A CIRCLE WITH CENTRE O AND RADIUS 2 CM HAS BEEN INSCRIBED IN THE TRIANGLE . IF QR = 6 CM , FIND THE PERIMETER OF TRIANGLE PQR .