Question

Pls help me with this question​

Answer 1

QS=√(14^2-10^2)

=4√6

***PQ=√(5^2+4√6^2)

=√25+96

=√121

=11

I think this will help you.

Answer 2

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

The length of PQ is 11 cm.

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given : In ΔPQR

S is the point on the PR Such that RSQ = 90 °

RQ = 14 cm

RS= 10 cm

PS = 5 cm

To find : Length of the PQ

Solution :

First find the length of SQ

Consider ΔRSQ

Since one of the angle of the triangle is right-angled. So, ΔRSQ is a right-angled triangle.

$\boxed{\sf{So,\:{Hypotenuse}^{2}={Base}^{2}+{Height}^{2} }}$

Here base = SR, Height = SQ, Hypotenuse = QR

${qr}^{2} = {(sr)}^{2} + {(sq)}^{2}$

${14}^{2} = {10}^{2} + {(sq)}^{2}$

$196 = 100 + {(sq)}^{2}$

$196 - 100 = {(sq)}^{2} \\ \\ 96 = {(sq)}^{2} \\ \\ sq = \sqrt{96} \\ \\ sq = \sqrt{16} \times \sqrt{6} \\ \\ sq = 4 \times \sqrt{6} \\ \\ sq = 4 \sqrt{6}$

Now consider, ΔPSQ

Since one of the angle of the triangle is right-angled. So, ΔRSQ is a right-angled triangle.

$\boxed{\sf{So,\:{Hypotenuse}^{2}={Base}^{2}+{Height}^{2} }}$

Here base = PS, Height = SQ, Hypotenuse = QP

${(qp)}^{2} = {(ps)}^{2} + {(sq)}^{2}$

${(qp)}^{2} = {5}^{2} + {(4 \sqrt{6}) }^{2}$

${(qp)}^{2} = 25 - 16(6) \\ \\ {(qp)}^{2} = 25 + 96 \\ \\ {(qp)}^{2} = 121 \\ \\ qp = \sqrt{121} \\ \\ qp = 11 \: cm$

Therefore the length of PQ is 11 cm.