CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A triangle PQR is drawn to circumscribe a circle of radius 8 cm such that the segments QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If area of ∆PQR is 336 cm2, find the sides PQ and PR.                           [CBSE 2014]


Solution



Here, T, S and U are the points of contact of the circle with the sides QR, PQ and PR, respectively.

OT = OS = OU = 8 cm      (Radii of the circle)

We know that the lengths of tangents drawn from an external point to a circle are equal.

∴ QS = QT = 14 cm

RU = RT = 16 cm

PS = PU = x cm (say)

So, QR = QT + TR = 14 cm + 16 cm = 30 cm

PQ = PS + SQ = x cm + 14 cm = (x + 14) cm

PR = PU + UR = x cm + 16 cm = (x + 16) cm

Also, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OT ⊥ QR, OS ⊥ PQ and OU ⊥ PR

Now,

ar(∆OQR) + ar(∆OPQ) + ar(∆OPR) = ar(∆PQR)

12×QR×OT+12×PQ×OS+12×PR×OU=336 cm212×30×8+12×x+14×8+12×x+16×8=336120+4x+56+4x+64=3368x+240=336 
8x=336-240=96x=12

∴ PQ = (x + 14) cm = (12 + 14) cm = 26 cm

PR = (x + 16) cm = (12 + 16) cm = 28 cm

Hence, the lengths of sides PQ and PR are 26 cm and 28 cm, respectively.

Mathematics
RD Sharma (2016)
Standard X

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More



footer-image