Maharashtra Board Class 10 Math Part – 2 Solution Chapter 2 Practice Set 2.2 – Pythagoras Theorem
Balbharati Maharashtra Board Class 10 Math Part – 2 Solution Chapter 2: Pythagoras Theorem. Marathi or English Medium Students of Class 10 get here Pythagoras Theorem full Exercise Solution.
Std |
Maharashtra Class 10 |
Subject |
Math Part 2 Solution |
Chapter |
Pythagoras Theorem |
Practice Set |
2.2 |
Practice Set – 2.2
(1) In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, Find QR.
Solution:
Given, S mid point of QR
PQ = 11, PR = 17, PS = 13
∴ By Apollonius theorem,
PQ2 + PR2 = 2PS2 + 2QS2
Or, 112 + 172 = 2×132 + 2QS2
Or, 121 + 289 = 2×169 + 2QS2
Or, 2QS2 = 410 – 338
Or, QS2 = 72/2
Or, QS = √36 = 6
∴ QR = 2QS
∵ S is the midpoint of QR.
∴ QR = 2×6
= 12
(2) In ∆ABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn from point C to side AB
Solution:
In the adjacent fig.
M is the median of ∆ABC to side AB
∴ M is the midpoint of AB
∴ AM = ½ × AB = 10/2 5
∴ by Apollonius theorem,
AC2 + BC2 = 2CM2 + 2AM2
Or, 72 + 92 = 2CM2 + 2 × 52
Or, 49+81 = 2CM2 + 50
Or, CM2 = 130+50/2
Or, CM = √40
= 2√10
∴ Length of median CM = 2√10
(3) In the figure 2.28 seg PS is the median of ∆PQR and PT⊥QR. Prove that,
(i) PR2 = PS2 + QR × ST + (QR/2)2
(ii) PQ2 = PS2 – QR × ST + (QR/2)2
Solution:
In the adjacent fig,
PS is the median of ∆PQR PT⊥QR
∴ QS = SR [∵ S midpoint of QR]
Also SR = QR/2 [∵ S midpoint of QR]
Or, 2SR = QR —– (ii)
(i) Now, In∆PST
by Pythagoras theorem,
PS2 = PT2 + ST2
Or, PT2 = PS2 – ST2 —- (iii)
Now, In ∆PRT, by Pythagoras theorem,
PR2 = PT2 + RT2
Or, PR2 = PS2 – ST2 + (SR + ST)2 [From of on(iii) & T – S – R]
Or, PR2 = PS2 – ST2 + SR2 + 2SR × ST + ST2
Or, PR2 = PS2 + QR× ST × SR2 [From equation (ii)]
Or, PR2 = PS2 + QR × ST2 + (QR/2)2 [From equation (i)] Hence proved
(ii) Now, In∆PQR.
by Apollonius theorem,
PQ2 + PR2 = 2PS2 + 2QS2
Or, PQ2 = 2PS2 + 2 (QR/2)2 – PR2 [∵ S is midpoint of QR]
Or, PQ2 = 2PS2 + 2 (QR/2)2 – [PS2 + QR × ST + (QR/2)2]
Or, PQ2 = 2PS2 + 2 (QR/2)2 – PS2 – QR×ST – (QR/2)2 [Putting the value of PR from previous proof]
Or, PQ2 = PS2 – QR × ST + (QR/2)2(Hence proved)
(4) In ∆ABC, point M is the midpoint of side BC.
If, AB2 + AC2 = 290 cm2, AM = 8cm, find BC.
Solution:
In the adjacent fig it is given, M is the midpoint of BC.
AB2 + AC2 = 290cm2
AM = 8cm.
∴ By Apollonius theorem,
AB2 + AC2 = 2AM2 + 2BM2
Or, 290 = 2 × 82 + 2BM2
Or, 2BM2 = 290 – 2×64
Or, BM = √162/2
= √81 = 9cm
∴ BM = 9cm
∴ BC = 2BM
= 2 × 9
= 18 cm
∵ M is midpoint of BC.
(5) In figure 2.30, point T is in the interior of rectangle PQRS,
Prove that, TS2 + TQ2 = TP2 + TR2 (As shown in the figure, draw seg AB||side SR and A-T-B)
Solution:
Given, PQRS is a rectangle.
∴ Interior angle are 90°
AB||SR, A – T – B.
∵ AB is || to SR.
∴ As = BR & AP = BQ [Parallel line drawn in the interior bisect the two opposite sides]
In ∆TPA
TP2 = AP2 + AT2 — (i)
By Pythagoras theorem.
In ∆TAS
By Pythagoras theorem,
TS2 = AT2 + AS2 —- (ii)
In ∆TBQ
by Pythagoras theorem,
TQ2 = TB2 + BQ2 —- (iii)
In ∆TBR
TR2 = TB2 + BR2 —– (iv)
By Pythagoras theorem
Now, Adding equation (ii) & (iii) we get,
TS2 + TQ2 = AT2 + AS2 + TB2 + BQ2
Or, TS2 + TQ2 = AT2 + BR2 + TB2 + AP2 [∵ AS = BR, AP = BQ]
Or, TS2 + TQ2 = AT2 + AP2 + TB2 + BR2
Or, TS2 + TQ2 = TP2 + TR2[From equation (i) & equation (iv)] Hence proved
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 2 Pythagoras Theorem Practice Set 2.2
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