In triangle PQR, right angled at Q, PR+QR=25cm and PQ=5cm. {fill_interger},[[1]],[[2]]are all the sides of the triangle.

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Detailed Solution

First, let us draw the given right angled triangle PQR.

https://www.vedantu.com/question-sets/f82e1523-410f-4df8-8af7-38fdaa99c6eb7878106478273177028.png

PR+QR = 25 cm.
⇒PR = 25−QR
In triangle PQR,

Hypotenus e^{2} = Perpendicula r^{2} + Bas e^{2} .

\Rightarrow P R^{2} = P Q^{2} + Q R^{2}

Substituting PQ = 5 cm and PR = 25−QR

\Rightarrow {(25 - QR)}^{2} = 5^{2} + Q R^{2}

\Rightarrow {(25)}^{2} + Q R^{2} - 2 (25) (QR) = 5^{2} + Q R^{2}

\Rightarrow 625 + Q R^{2} - 2 (25) (QR) = 25 + Q R^{2}

\Rightarrow 625 + Q R^{2} - 50 QR = 25 + Q R^{2}

\Rightarrow 625 + Q R^{2} - 50 QR - Q R^{2} = 25 + Q R^{2} - Q R^{2}

\Rightarrow 625 - 50 QR = 25

⇒ 625−50QR−25 = 25−25
⇒ 600−50QR = 0

\Rightarrow 50 QR = 600

Dividing both sides of the equation by 50, we get

\Rightarrow \frac{50 QR}{50} = \frac{600}{50}

∴ QR=12
Thus, we get the length of base QR is 12 cm.
Now, substituting QR=12cm in the equation PR=25−QR, we get
⇒PR=25−12
∴ PR=13
Thus, we get the length of hypotenuse PR as 13 cm.
Therefore, the sides of the triangle are of lengths 5 cm, 12 cm, and 13 cm.

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