R
4. In figure, ZPQR = 90° , QSI PR
if QS = 12, RS = 16,
find (1) PS (2) QR
S
P
3/
Answers
Given:
- In triangle PQR,
- ∠Q = 90 degree
- seg SQ perpendicular hypotenuse PR
- PS = 16 cm
To find:
- The length of QS
Solution:
- SR = 9cm
- 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
⇒ QS² = QR² - 9² ___ (ii)
substituting SR = 9 cm
⇒ QS² + 9² = QR²
From equation (i) & (ii), we get
PQ² - 16² = QR² - 9²
⇒ PQ² - QR² = 256 - 81
In rt-angled Δ PQR, using Pythagoras theorem,
substituting the value of PS = 16 cm & SR = 9 cm
⇒ PQ² + QR² = (PS + SR)²
⇒ PQ² - QR² = 175 ___ (iii)
PQ² + QR² = PR²
⇒ PQ² + QR² = (16 + 9)²
PQ² + QR² = 625
Now, adding equations (iii) & (iv), we get
⇒ PQ² + QR² = 625 ___ (iv)
⇒ PQ² + QR² = 25²
PQ² - QR² = 175
_____________
2PQ² = 800
_____________
∴ PQ² =
= 400
800/2 Subst.it.uting the value of PQ² = 400 in eq. (i), we get
⇒ QS² = 400 - 16²
QS² = PQ² - 16²
⇒ QS² = 400 - 256
⇒ QS = √144
⇒ QS = 12 cm
{QS = 12cm
Thus,
⇒ QS² = 144