Practice Set 2.2 Geometry 10th Std Maths Part 2 Answers Chapter 2 Pythagoras Theorem
Question 1
Maharashtra Board Solution
In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.
Solution & Step-by-Step Answer:
In ∆PQR, point S is the midpoint of side QR. [Given] ∴ seg PS is the median. ∴ PQ2 + PR2 = 2 PS2 + 2 SR2 [Apollonius theorem] ∴ 112 + 172 = 2 (13)2 + 2 SR2 ∴ 121 + 289 = 2 (169)+ 2 SR2 ∴ 410 = 338+ 2 SR2 ∴ 2 SR2 = 410 – 338 ∴ 2 SR2 = 72 ∴ SR2 = = 36 ∴ SR = [Taking square root of both sides] = 6 units Now, QR = 2 SR [S is the midpoint of QR] = 2 × 6 ∴ QR = 12 units
Question 2
Maharashtra Board Solution
In ∆ABC, AB = 10, AC = 7, BC = 9, then find the length of the median drawn from point C to side AB.
Solution & Step-by-Step Answer:
Let CD be the median drawn from the vertex C to side AB. BD = AB [D is the midpoint of AB] = × 10 = 5 units In ∆ABC, seg CD is the median. [Given] ∴ AC2 + BC2 = 2 CD2 + 2 BD2 [Apollonius theorem] ∴ 72 + 92 = 2 CD2 + 2 (5)2 ∴ 49 + 81 = 2 CD2 + 2 (25) ∴ 130 = 2 CD2 + 50 ∴ 2 CD2 = 130 – 50 ∴ 2 CD2 = 80 ∴ CD2 = = 40 ∴ CD = [Taking square root of both sides] = 2 units ∴ The length of the median drawn from point C to side AB is 2 units.
Question 3
Maharashtra Board Solution
In the adjoining figure, seg PS is the median of APQR and PT ⊥ QR. Prove that, i. PR2 = PS2 + QR × ST + ()2 ii. PQ2 = PS2 – QR × ST + ()2
Solution & Step-by-Step Answer:
i. QS = SR = QR (i) [S is the midpoint of side QR] ∴ In ∆PSR, ∠PSR is an obtuse angle [Given] and PT ⊥ SR [Given, Q-S-R] ∴ PR2 = SR2 +PS2 + 2 SR × ST (ii) [Application of Pythagoras theorem] ∴ PR2 = ( QR)2 + PS2 + 2 ( QR) × ST [From (i) and (ii)] ∴ PR2 = ()2 + PS2 + QR × ST ∴ PR2 = PS2 + QR × ST + ()2
ii. In.∆PQS, ∠PSQ is an acute angle and [Given]
PT ⊥QS [Given, Q-S-R]
∴ PQ2= QS2+ PS2– 2 QS × ST (iii) [Application of Pythagoras theorem]
∴ PR2= ( QR)2+ PS2– 2 ( QR) × ST [From (i) and (iii)]
∴ PR2= ()2+ PS2– QR × ST
∴ PR2= PS2– QR × ST + ()2
Question 4
Maharashtra Board Solution
In ∆ABC, point M is the midpoint of side BC. If AB2 + AC2 = 290 cm, AM = 8 cm, find BC.
Solution & Step-by-Step Answer:
In ∆ABC, point M is the midpoint of side BC. [Given] ∴ seg AM is the median. ∴ AB2 + AC2 = 2 AM2 + 2 MC2 [Apollonius theorem] ∴ 290 = 2 (8)2 + 2 MC2 ∴ 145 = 64 + MC2 [Dividing both sides by 2] ∴ MC2 = 145 – 64 ∴ MC2 = 81 ∴ MC = [Taking square root of both sides] MC = 9 cm Now, BC = 2 MC [M is the midpoint of BC] = 2 × 9 ∴ BC = 18 cm
Question 5
Maharashtra Board Solution
In the adjoining figure, 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) Given: ꠸PQRS is a rectangle. Point T is in the interior of ꠸PQRS. To prove: TS2 + TQ2 = TP2 + TR2 Construction: Draw seg AB || side SR such that A – T – B.
Solution & Step-by-Step Answer:
Proof: ꠸PQRS is a rectangle. [Given] ∴ PS = QR (i) [Opposite sides of a rectangle] In ꠸ASRB, ∠S = ∠R = 90° (ii) [Angles of rectangle PQRS] side AB || side SR [Construction] Also ∠A = ∠S = 90° [Interior angle theorem, from (ii)] ∠B = ∠R = 90° ∴ ∠A = ∠B = ∠S = ∠R = 90° (iii) ∴ ꠸ASRB is a rectangle. ∴ AS = BR (iv) [Opposite sides of a rectanglel In ∆PTS, ∠PST is an acute angle and seg AT ⊥ side PS [From (iii)] ∴ TP2 = PS2 + TS2 – 2 PS.AS (v) [Application of Pythagoras theorem] In ∆TQR., ∠TRQ is an acute angle and seg BT ⊥ side QR [From (iii)] ∴ TQ2 = RQ2 + TR2 – 2 RQ.BR (vi) [Application of pythagoras theorem] TP2 – TQ2 = PS2 + TS2 – 2PS.AS -RQ2 – TR2 + 2RQ.BR [Subtracting (vi) from (v)] ∴ TP2 – TQ2 = TS2 – TR2 + PS2 – RQ2 -2 PS.AS +2 RQ.BR ∴ TP2 – TQ2 = TS2 – TR2 + PS2 – PS2 – 2 PS.BR + 2PS.BR [From (i) and (iv)] ∴ TP2 – TQ2 = TS2 – TR2 ∴ TS2 + TQ2 = TP2 + TR2
Question 1
Maharashtra Board Solution
In ∆ABC, ∠C is an acute angle, seg AD Iseg BC. Prove that: AB2 = BC2 + A2 – 2 BC × DC. (Textbook pg. no. 44) Given: ∠C is an acute angle, seg AD ⊥ seg BC. To prove: AB2 = BC2 + AC2 – 2BC × DC
Solution & Step-by-Step Answer:
Proof: ∴ LetAB = c, AC = b, AD = p, ∴ BC = a, DC = x BD + DC = BC [B – D – C] ∴ BD = BC – DC ∴ BD = a – x In ∆ABD, ∠D = 90° [Given] AB2 = BD2 + AD2 [Pythagoras theorem] ∴ c2 = (a – x)2 + [P2] (i) ∴ c2 = a2 – 2ax + x2 + [P2] In ∆ADC, ∠D = 90° [Given] AC2 = AD2 + CD2 [Pythagoras theorem] ∴ b2 = p2 + [X2] ∴ p2 = b2 – [X2] (ii) ∴ c2 = a2 – 2ax + x2 + b2 – x2 [Substituting (ii) in (i)] ∴ c2 = a2 + b2 – 2ax ∴ AB2 = BC2 + AC2 – 2 BC × DC
Question 2
Maharashtra Board Solution
In ∆ABC, ∠ACB is an obtuse angle, seg AD ⊥ seg BC. Prove that: AB2 = BC2 + AC2 + 2 BC × CD. (Textbook pg. no. 40 and 4.1) Given: ∠ACB is an obtuse angle, seg AD ⊥ seg BC. To prove: AB2 = BC2 + AC2 + 2BC × CD
Solution & Step-by-Step Answer:
Proof: Let AD = p, AC = b, AB = c, BC = a, DC = x BD = BC + DC [B – C – D] ∴ BD = a + x In ∆ADB, ∠D = 90° [Given] AB2 = BD2 + AD2 [Pythagoras theorem] ∴ c2 = (a + x)2 + p2 (i) ∴ c2 = a2 + 2ax + x2 + p2 Also, in ∆ADC, ∠D = 90° [Given] AC2 = CD2 + AD2 [Pythagoras theorem] ∴ b2 = x2 + p2 ∴ p2 = b2 – x2 (ii) ∴ c2 = a2 + 2ax + x2 + b2 – x2 [Substituting (ii) in (i)] ∴ c2 = a2 + b2 + 2ax ∴ AB2 = BC2 + AC2 + 2 BC × CD
Question 3
Maharashtra Board Solution
In ∆ABC, if M is the midpoint of side BC and seg AM ⊥seg BC, then prove that AB2 + AC2 = 2 AM2 + 2 BM2. (Textbook pg, no. 41) Given: In ∆ABC, M is the midpoint of side BC and seg AM ⊥ seg BC. To prove: AB2 + AC2 = 2 AM2 + 2 BM2
Solution & Step-by-Step Answer:
Proof: In ∆AMB, ∠M = 90° [segAM ⊥ segBC] ∴ AB2 = AM2 + BM2 (i) [Pythagoras theorem] Also, in ∆AMC, ∠M = 90° [seg AM ⊥ seg BC] ∴ AC2 = AM2 + MC2 (ii) [Pythagoras theorem] ∴ AB2 + AC2 = AM2 + BM2 + AM2 + MC2 [Adding (i) and (ii)] ∴ AB2 + AC2 = 2 AM2 + BM2 + BM2 [∵ BM = MC (M is the midpoint of BC)] ∴ AB2 + AC2 = 2 AM2 + 2 BM2