## For finding AB and BC with the help of information given in figure 2.20, complete following activity.

AB = AC [Given ]BAC = BCA [Angles opposite to equal sides of an isosceles triangle are equal]AB = BC = 1/√2× AC [By 45˚ – 45˚-90˚ theorem]=(1/√2) ×√8=(1/√2) × 2√2= 2

## See figure 2.19. Find RP and PS using the information given in PSR.

In PSR , P = 30˚ , S = 90˚R = 180 – (90+30) = 60˚ [Angle Sum property of triangle]PSR is a 30˚ – 60˚ – 90˚ triangle.SR = (½)×PR [side opposite to 30˚]6 = (½)× PRPR = 12PS = √(PR2 -SR2 [Pythagoras theorem]PS = √(122-62)PS = √(144-36)PS = √108PS = 6√3Hence RP = 12 units […]

## In figure 2.18, QPR = 90°, seg PM ⊥ seg QR and Q-M-R, PM = 10, QM = 8, find QR.

In PQR , QPR = 90˚seg PM ⊥ seg QR [Given]PM2 = QM×MR [Theorem of geometric mean]102 = 8×MRMR = 100/8 = 12.5QR = QM + MRQR = 8+12.5 = 20.5Hence measure of QR is 20.5 units.

## In figure 2.17, MNP = 90°, seg NQ ⊥ seg MP, MQ = 9, QP = 4, find NQ.

In MNP , MNP = 90˚, seg NQ ⊥ seg MPMQ = 9 , QP = 4 [Given]NQ = √(MQ×QP) [Theorem of geometric mean]NQ = √(9×4) = √36 = 6Hence NQ = 6 units.

## Identify, with reason, which of the following are Pythagorean triplets.

(i)(3, 5, 4)32 = 942 = 1652 = 25Here 9+16 = 2552 = 32+42The square of largest number is equal to sum of squares of the other two numbers.(3, 5, 4) is a Pythagorean triplet. (ii)(4, 9, 12)42 = 1692 = 81122 = 144Here 42+92 ≠ 122The square of largest number is not equal to sum of squares of the other two numbers.(4, […]