Points M and N are on the sides PQ and QR respectively of a triangle PQR, right angled at Q. If PN = 9 cm, MR = 7 cm, and MN = 3 cm, then find the length of PR (in cm).
Answer & explanation
Correct answer: option 1
Let the measure of the side PM, MQ, QN, NR be d, a, b and c respectively.
From \(\Delta \)PQR, \( { (a\; + \;d)}^{2 } \) + \( { (b\; +\; c)}^{2 } \) = \( { PR}^{2 } \)
So, \(\Delta \)PQN, \(\Delta \)MQN & \(\Delta \)MQR are all right-angled triangles.
Now,
From \(\Delta \)PQN,
= \( { (a\; +\; d)}^{2 } \) + \( { b}^{2 } \) = \( { 9}^{2 } \)
= \( { (a\; +\; d)}^{2 } \) + \( { b}^{2 } \) = 81 ..(1)
From \(\Delta \)MQN,
\( { a}^{2 } \) + \( { b}^{2 } \) = \( { 3}^{2 } \)
= \( { a}^{2 } \) + \( { b}^{2 } \) = 9 ..(2)
From \(\Delta \)MQR,
= \( { a}^{2 } \) + \( {b\;+\;c}^{2 } \) = \( { 7}^{2 } \)
= \( { a}^{2 } \) + \( {b\;+\;c}^{2 } \) = 49 ..(3.)
(1) + (3)
= \( { (a\; +\; d)}^{2 } \) + \( { b}^{2 } \) + \( { a}^{2 } \) + \( {b\;+\;c}^{2 } \) = 81 + 49
= \( { (a\; +\; d)}^{2 } \) + \( { (b\; +\; c)}^{2 } \) + \( { b}^{2 } \) + \( { a}^{2 } \) = 130
= \( { (a\; +\; d)}^{2 } \) + \( { (b\; +\; c)}^{2 } \) + 9 = 130
= \( { (a\; +\; d)}^{2 } \) + \( { (b\; +\; c)}^{2 } \) = 121
= \( { PR}^{2 } \) = 121
= PR = 11 cm
Therefore, PR is 11 cm.