Question

urgent ...............pls anwer only if u know ...with full solution

Answer 1

since, /_QPR=90°

SO,∆QPR is right angle triangle

now by applying Pythagorean theoram in ∆QPR

we get
${qr}^{2} = {qp}^{2} + {pr}^{2} \\ {26}^{2} = {(24)}^{2} + {(pr)}^{2} \\ {(pr)}^{2} = 676 - 576 \\ {(pr)}^{2} = 100 \\ pr = 10$
Now on observing ∆ PKR

we concluded that:-

${(10)}^{2} = {8}^{2} + {6}^{2} \\ {(pr)}^{2} = {(pk)}^{2} + {(kr)}^{2}$
By applying Above theoram we can say that

/_PKR=90°.


HOPE IT WILL HELP YOU‼️

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Answer 2

The answer is provided in the attachment.
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