Question

in triangle PQR, right angled at q ,PR-QR=2CM and PQ=10CM. Determine the values of secR

Answer 1

Answer:

Step-by-step explanation:

We know that in a right angled triangle PQR, by pythagoras theorem,

$\\ pr {}^{2} = pq {}^{2} + qr {}^{2} \\ pr {}^{2} = 10 {}^{2} + qr {}^{2} \\ pr {}^{2} - qr {}^{2} = 100 \\( pr - qr)(pr + qr) = 100$

$= 2(pr + qr) = 100 \\ pr + qr = 50$

Now we know the values of

(pr - qr) and (pr + qr)

By elimination method,

PR = 26cm

QR = 50 - 26 = 24cm

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secR = hypotnuse / adjacent side

= PR/QR

= 26/24 = 13/12.

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* Elimination method;

pr + qr = 50

pr - qr = 2

==> 2pr = 52 ==> pr = 26

*{Due to some error I could not attach a diagram (> . < *sorry*). So, please draw a diagram before practicing the sum.}

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Hope it helps....!

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Answer 2

Answer:

7/6

Step-by-step explanation:

PR-QR=2.

PR=QR+2

Given,∆PQR is a right angle ∆ at Q

according to pithagoras therom,