Question

3. In adjoining figure 1.14
seg Ps 1 seg RQ seg QT I seg PR.
IfRQ = 6, PS = 6 and PR = 12,
then find QT.​

Answer 1

Given :

• seg Ps ⊥ seg RQ seg QT ⊥ seg PR. If RQ = 6, PS = 6 and PR = 12,

To find :

• QT

Solution :

PS ⊥ RQ

QT ⊥ PR

Area of triangle ΔPQR = ½ x b x h

= ½ x 6 x 6

= 3 x 6

= 18

Area of triangle ΔPQR is 18 sq.unit

Area of triangle ΔPQR = ½ x PR x QT

18 = ½ x 12 x QT

18 = 6 x QT

18/6 = QT

3 = QT

∴ Value of QT IS 3

Answer 2

Answer:

✡ Correct Question ✡

✏ In the adjoining figure seg PS $\perp$ seg RQ, seg QT $\perp$ seg PR. If RQ = 6, PS = 6 and PR = 12 . Find QT.

✡ Given ✡

• seg PS $\perp$ seg RQ, seg QT $\perp$ seg PR. If RQ= 6, PS= 6 and PR = 12.

✡ To Find ✡

• What is the value of QT

✡ Solution ✡

✏ In ∆PQR, PR is the base and QT is the corresponding height.

✏ Also, RQ is the base and PS is the corresponding height.

Given :-

• RQ = 6
• PS = 6
• PR = 12

⭐ ∆PRS,

=> RS² = PR² - PS²

=> RS² = 144 - 36 = 108

=> RS = √108

$\mapsto$ Area of ∆PRS = Area of∆PQR + QR(∆PSQ)

=> $\dfrac{1}{2}$ × 6$\sqrt{108}$ = $\dfrac{1}{2}$ × 12 × QT + $\dfrac{1}{2}$ × 6 × ($\sqrt{108}$ - 6)

=> 6 × 6$\sqrt{3}$ = 12 × QT + 6 × (6$\sqrt{3}$ - 6)

=> 6$\sqrt{3}$ = 2QT + 6($\sqrt{3}$ - 1)

=> 2QT = 6

=> QT = $\dfrac{6}{2}$

=> QT = 3

$\therefore$ Then the value of QT is 3

Step-by-step explanation:

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