# Angles Q and R of a ∆PQR are 25º and 65º. Write which of the following is true:

(i)PQ^{2} + QR^{2} = RP^{2}

(ii)PQ^{2 }+ RP^{2 }= QR^{2}

(iii)RP^{2} + QR^{2} = PQ^{2}

**Solution:**

From the figure we see that the two angles of a triangle are given and we must find out the third angle by using the angle sum property that is the sum of three interior angles of a triangle is 180°.

We know that, sum of interior angles of a triangle is 180°.

∠P + ∠Q + ∠R = 180°

∠P + 25° + 65° = 180°

∠P + 90° = 180°

∠P = 180° - 90°

∠P = 90°

Thus, triangle PQR is a right angled at P

As one of the angles is 90° that means it is a right-angled triangle and the square of the hypotenuse is equal to the sum of the square of the other two sides.

Therefore, by Pythagoras theorem,

(Perpendicular)^{2} + (Base)^{2 }= (Hypotenuse )^{2}

Here, Perpendicular = QP, Base = PR and Hypotenuse = QR

(QP)^{2 }+ (PR)^{2} = (QR)^{2}

Hence, option (ii) is correct.

**☛ Check: **NCERT Solutions for Class 7 Maths Chapter 6

**Video Solution:**

## Angles Q and R of a ∆PQR are 25º and 65º. Write which of the following is true: (i)PQ² + QR² = RP²^{ }(ii)PQ²^{ }+ RP²^{ }= QR²^{ }(iii)RP² + QR² = PQ²

NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.5 Question 6

**Summary:**

Angles Q and R of a ∆PQR are 25º and 65º. Option (ii) PQ^{2 }+ RP^{2 }= QR^{2 }is the correct result.

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