In the figure 2.28 seg PS is the median of PQR and PT⊥QR. Prove that,
(1) QS = ½ QR ……(i) [S is the midpoint of QR]SR = ½ QR ……(ii)QS = SR [From (i) and (ii)]PT ⊥QR [Given]PSR is an obtuse angle. [From figure]PR2 = SR2+PS2+2SR×ST …..(iii) [Application of Pythagoras theorem]Substitute SR = ½ QR in (iii)PR2 =[(½)QR]2+PS2+2(1/2)QR×STPR2 =[(½)QR]2+PS2+QR×STPR2 = PS2 + QR×ST +(QR/ 2 )2Hence proved. (ii) PT⊥QS [Given] PSQ is an acute […]
In ABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn from point C to side AB
Let CD is the median drawn from C to AB.Given AB = 10AD = (1/2)×AB [D is the midpoint of side AB]AD = 10/2 = 5Since CD is the medianAC2+BC2 = 2CD2+2AD2 [Apollonius theorem]72+92 = 2 CD2+2×522 CD2 = 72+92-2×522 CD2 = 80CD2 = 40Taking square roots on both sidesCD = 2√10Hence the length of median drawn from point C to […]
In PQR, point S is the midpoint of side QR. If PQ = 11,PR = 17, PS =13, find QR.
Given , S is the midpoint of QR .PS is the median.PQ2+PR2 = 2 PS2+2SR2 [By Apollonius theorem]112+172 = 2×132+2×SR2121+289 = 2×169+2×SR22SR2 = 121+289-3382SR2 = 72SR2 = 72/2 = 36SR = 6Since S is the midpoint of QR , QR = 2SRQR = 2×6 = 12Hence QR = 12 units.
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, […]
In figure 1.75, A – D – C and B – E – C seg DE || side AB If AD = 5, DC = 3, BC = 6.4 then find
Given DE AB.AD/DC = BE/EC [Basic proportionality theorem]AD = 5, DC = 3, BC = 6.4 [Given]BE = x , EC = 6.4-x [Given]5/3 = x/(6.4-x)Cross multiplying we get5×(6.4-x) = 3× x32-5x = 3×32 = 8xx = 32/8 = 4Hence BE = 4 units.
MNT ~ QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio A( MNT) /A( QRS) .
Given MNT ~ QRSTMN SQR [corresponding angles of similar triangles]Construction:Draw altitude from T to MN meeting at L.Draw altitude from S to QR meeting at P.TLM = SPQ = 90˚In MLT and QPSTMN SQRTLM SPQMLT ~ QPS [AA test of similarity]MT/QS = TL/SPMT/QS = 5/9MNT ~ QRS [Given]A( MNT) /A( QRS) = MT2/QS2 [Theorem of areas […]