Question

In triangle PQR angle Q = 90 degree & seg SQ SQ perpendicular hypotenuse PR , If PS = 16 cm & SR = 9cm then find QS ?​

Answer 1

Given:

In triangle PQR,

∠Q = 90 degree

seg SQ perpendicular hypotenuse PR

PS = 16 cm

SR = 9cm

To find:

The length of QS

Solution:

Since it is given that seg SQ ⊥ PR

∴ ∠PSQ = ∠QSR = 90°

⇒ ΔPSQ & ΔQSR are right-angled triangles

In rt-angled Δ PSQ, using Pythagoras theorem,

PS² + QS² = PQ²

substituting PS = 16 cm

⇒ 16² + QS² = PQ²

⇒ QS² = PQ² - 16² ..... (i)

In rt-angled Δ QSR, using Pythagoras theorem,

QS² + SR² = QR²

substituting SR = 9 cm

⇒ QS² + 9² = QR²

⇒ QS² = QR² - 9² ..... (ii)

From equation (i) & (ii), we get

PQ² - 16² = QR² - 9²

⇒ PQ² - QR² = 256 - 81

⇒ PQ² - QR² = 175 ...... (iii)

In rt-angled Δ PQR, using Pythagoras theorem,

PQ² + QR² = PR²

⇒ PQ² + QR² = (PS + SR)²

substituting the value of PS = 16 cm & SR = 9 cm

⇒ PQ² + QR² = (16 + 9)²

⇒ PQ² + QR² = 25²

⇒ PQ² + QR² = 625 ..... (iv)

Now, adding equations (iii) & (iv), we get

PQ² + QR² = 625

PQ² - QR² = 175

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2PQ² = 800

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∴ PQ² = $\frac{800}{2}$ = 400

Substituting the value of PQ² = 400 in eq. (i), we get

QS² = PQ² - 16²

⇒ QS² = 400 - 16²

⇒ QS² = 400 - 256

⇒ QS² = 144

⇒ QS = √144

⇒ QS = 12 cm

Thus, $\boxed{\bold{\underline{QS = 12 \:cm}}}$.

$\star$ Note:- the figure is given in the attachment below $\star$

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