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Median of Triangle

In Δ PQR , ∠ ...

Question

In $∆$ PQR, $∠$ Q = 90 $°$ and seg QS is the median. QS = 7.4 cm, find PR.

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Solution

Construction:
Draw a circumcircle of $△$ PQR.
We know:
$∠ PQR = 90 °$
Also, we know that the angle in a semicircle is 90°.
Hence, PR is the diameter of the circle .
QS, which is equal to 7.4 cm, is the radius of the circle.
PR = 2(Radius)
= 2(7.4)
= 14.8 cm

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