Question

in triangle PQR,P=90°PQ=4.5cm and PR=5.2cm.
calculate QR.
if QR is extended to S so that PS=3.5cm. findSR
pls, show workings. thanks​

Answer 1

Step-by-step explanation:

1) Draw the base QR = 6 cm.

At point Q draw a ray QX making an ∠ QXR = 60

o

.

Here, PR-PQ = 2cm

PR > PQ

The side containing the base angle Q is less than third side.

2) Cut the line segment QS equal to PR-PQ = 2 cm, from the ray QX extended on opposite side of base QR.

3) Join SR and draw its perpendicular bisector ray AB which intersect SR at M.

4) Let P be the intersection point of SX and perpendicular bisector AB. Then join PR.

Thus, △ PQR is the required triangle.

solution

Answer 2

Step-by-step explanation:

Given,

<P=90°

PQ=4.5cm

PR=5.2cm

QR=?

BY PYTHAGORAS THEOREM:

${QR}^{2}={PQ}^{2}+{PR}^{2}$

${QR}^{2}={4.5}^{2}+{5.2}^{2}$

${QR}^{2}=20.25+27.04$

${QR}^{2}=47.29$

$QR=6.8$