In PQR- - Q - 90- PQ - 12 units- QR - 5 units and QS is a median- Find lQS-

In PQR, PR^2 = PQ^2 + QR^2 nbsp;(By Pythagoras theorem) ⇒ PR^2 = 12^2 + 5^2 ⇒ PR^2 = 144 + 25 ⇒ PR^2 = 13^2 ⇒ PR = 13 In a right angled triangle, the length ...

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Standard VIII

Mathematics

Altitude of a Triangle and Drawing Method

In PQR, ∠ Q =...

Question

In $△ P Q R, ∠ Q = 90^{\circ}$ , PQ = 12 units, QR = 5 units and QS is a median. Find l(QS).

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Solution

$In △ P Q R,$
$P R^{2} = P Q^{2} + Q R^{2}$ (By Pythagoras theorem)
$\Rightarrow P R^{2} = 12^{2} + 5^{2}$
$\Rightarrow P R^{2} = 144 + 25$
$\Rightarrow P R^{2} = 13^{2}$
$\Rightarrow P R = 13$
In a right angled triangle, the length of the median on its hypotenuse is half the length of the hypotenuse.
$∴ l (Q S) = \frac{1}{2} (P R) = \frac{1}{2} (13)$
= 6.5 units.
Hence, the length of QR is 6.5 units.

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In △PQR,∠Q=90∘, PQ = 12 units, QR = 5 units and QS is a median. Find l(QS).

In △PQR, PR2=PQ2+QR2 (By Pythagoras theorem) ⇒PR2=122+52 ⇒PR2=144+25 ⇒PR2=132 ⇒PR=13 In a right angled triangle, the length of the median on its hypotenuse is half the length of the hypotenuse. ∴l(QS)=12(PR)=12(13) = 6.5 units. Hence, the length of QR is 6.5 units.

In $△ P Q R, ∠ Q = 90^{\circ}$ , PQ = 12 units, QR = 5 units and QS is a median. Find l(QS).