In triangle PQR right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find all the sides.
Answer
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Hint: We will apply the Pythgoras’ Theorem to find the length of the three sides. For triangle ABC right angled at B,
${AB}^2 + {BC}^2 = {AC}^2$
Complete step-by-step solution -
In this question, one side is known and two are unknown. So we need to form two equations to solve those unknowns. One of them is given, and the other can be formed using the Pythagoras’ theorem-
${PQ}^2 +{QR}^2 = {PR}^2$
Applying the value of PQ,
$5^2 + {QR}^2 = {PR}^2$
${PR}^2 - {QR}^2$ = 25
(PR - QR)(PR + QR) = 25
Using the value of PR + QR,
(PR - QR)25 = 25
PR - QR = 1 cm ………………………..(1)
Also, PR + QR = 25 cm …………....(2)
Adding equations (1) and (2),
2PR = 26
PR = 13 cm
QR = 12 cm
Hence the sides of the triangle are PQ = 5 cm, QR = 12 cm and PR = 13 cm.
Note: In this question, students may find it difficult to solve the equations. If we substitute the value of PR in the first step itself, a quadratic equation in QR will be formed, which is a very lengthy method. So, it is better to apply a2-b2=(a-b)(a+b), which makes it simpler.
${AB}^2 + {BC}^2 = {AC}^2$
Complete step-by-step solution -
In this question, one side is known and two are unknown. So we need to form two equations to solve those unknowns. One of them is given, and the other can be formed using the Pythagoras’ theorem-
${PQ}^2 +{QR}^2 = {PR}^2$
Applying the value of PQ,
$5^2 + {QR}^2 = {PR}^2$
${PR}^2 - {QR}^2$ = 25
(PR - QR)(PR + QR) = 25
Using the value of PR + QR,
(PR - QR)25 = 25
PR - QR = 1 cm ………………………..(1)
Also, PR + QR = 25 cm …………....(2)
Adding equations (1) and (2),
2PR = 26
PR = 13 cm
QR = 12 cm
Hence the sides of the triangle are PQ = 5 cm, QR = 12 cm and PR = 13 cm.
Note: In this question, students may find it difficult to solve the equations. If we substitute the value of PR in the first step itself, a quadratic equation in QR will be formed, which is a very lengthy method. So, it is better to apply a2-b2=(a-b)(a+b), which makes it simpler.
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