# Ex.6.5 Q6 The Triangle and Its Properties - NCERT Maths Class 7

## Question

Angles \(Q\) and \(R\) of a \(\Delta\,PQR\) are \(25^\circ\) and \(65^\circ\)Write which of the following is true:

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

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

(iii) **\(RP^2 + QR^2 = PQ^2\)**

## Text Solution

**What is known?**

Measure of two angles of a triangle i.e. \(25°\)and \(65°.\)

**What is the unknown?**

Which option is correct.

**Reasoning:**

This question is straight forward, in this question it is clear from the figure two angles of a triangle are given and we must find out the third angle by using angle sum property i.e. the sum of three interior angles of a triangle is \(180^\circ\) By using this property, we got the value of third angle i.e. \(P = 90^\circ\) that means side opposite to \(P\) is hypotenuse i.e. \(QR\). As one of the angles is \( 90^\circ\) that means it is a right-angled triangle and the square of hypotenuse is equal to the sum of square of other two sides.

**Steps:**

We know that, sum of interior angles of a triangle is \(180^\circ\).

\begin{align}\angle P + \angle Q + \angle R &= 180^\circ \\\angle P + 25^\circ + 65^\circ &= 180^\circ \\\angle P + 90^\circ &= 180^\circ \\\angle P &= 180^\circ - 90^\circ \\\angle P &= 90^\circ \end{align}

Thus, triangle \(PQR\) is a right angled at \(P \)

Therefore, by Pythagoras theorem

\[\begin{align}{(P)^2} + {(B)^2} &= {(H)^2}\\{(QP)^2} + {(PR)^2} &= {(QR)^2}\end{align}\]

Hence, option (ii) is correct.

**Useful Tip:**

Whenever you encounter problem of this kind, it is best to think of the Pythagoras property if one of the three angles is \(90^\circ\) then the square of hypotenuse or greater side is equal to the sum of square of other two sides.