Answer :

We have

** Given:** PQ = 24 cm

QR = 26 cm

PA = 6 cm

AR = 8 cm

∠PAR = 90°

We need to find ∠QPR.

We know that ∆PAR is a right-angled triangle, right angled at A.

So, by Pythagoras theorem, we can write as

PR^{2} = PA^{2} + AR^{2}

⇒ PR^{2} = 6^{2} + 8^{2}

⇒ PR^{2} = 36 + 64

⇒ PR^{2} = 100

⇒ PR = √100

⇒ PR = 10

Now, in ∆PQR apply Pythagoras theorem again,

PQ^{2} + PR^{2} = 24^{2} + 10^{2}

PQ^{2} + PR^{2} = 576 + 100 = 676

But QR^{2} = 26^{2} = 676 [∵ QR = 26 cm]

⇒ PQ^{2} + PR^{2} = QR^{2}

Therefore, by the converse of Pythagoras theorem

∠P = 90°

**⇒** **∠****QPR = 90°**

Rate this question :

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

In two congruent RD Sharma - Mathematics

In two triangles RD Sharma - Mathematics

*D, E, F* arRD Sharma - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics