Question

6.
In a right triangle PQR right angled at Q, if PR=10cm and QR=8cm then PQ=
b)10cm
c)6cm
d) 118
a)8cm
2
bas​

Answer 1

Given :

• In a right angle triangle PQR at Q = 90°
• PR = 10 cm and QR = 8 cm .

To Find :

• Measure of PQ .

Solution :

$\longmapsto\tt{PR=10\:cm}$

$\longmapsto\tt{QR=8\:cm}$

Using Pythagoras Theorem :

$\longmapsto\tt\bf{{(PR)}^{2}={(QR)}^{2}+{(PQ)}^{2}}$

$\longmapsto\tt{{(10)}^{2}={(8)}^{2}+{(PQ)}^{2}}$

$\longmapsto\tt{100=64+{(PQ)}^{2}}$

$\longmapsto\tt{100-64={(PQ)}^{2}}$

$\longmapsto\tt{36={(PQ)}^{2}}$

$\longmapsto\tt{\sqrt{36}=PQ}$

$\longmapsto\tt\bf{6\:cm=PQ}$

So , The Measure of PQ is 6 cm ....

Answer 2

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C)6cm

Given :

• In a right angle triangle PQR at Q = 90°
• PR = 10 cm and QR = 8 cm .

To Find :

• Side PQ .

Solution :

★Here As side PR is opposite of the right then its th hypotenuse. let side QR be the base And PQ be the Height.

★And ,we know that in a right triangle

${Hypotenuse}^{2} = {Height}^{2} + {Base}^{2}$

i.e Pythagoras theorem

★Therefore

${10}^{2} = {8}^{2} + {pq}^{2}$

${pq}^{2} = 100 - 64$

$pq = \sqrt{3}$

$= > pq = 6cm$

★So , The Measure of PQ is 6 cm

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