## Maharashtra State Board Class 10 Maths Solutions Chapter 2 Pythagoras Theorem Practice Set 2.2

Question 1.

In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.

Solution:

In ∆PQR, point S is the midpoint of side QR. [Given]

∴ seg PS is the median.

∴ PQ^{2} + PR^{2} = 2 PS^{2} + 2 SR^{2} [Apollonius theorem]

∴ 11^{2} + 17^{2} = 2 (13)^{2} + 2 SR^{2}

∴ 121 + 289 = 2 (169)+ 2 SR^{2}

∴ 410 = 338+ 2 SR^{2}

∴ 2 SR^{2} = 410 – 338

∴ 2 SR^{2} = 72

∴ SR^{2} = \(\frac { 72 }{ 2 } \) = 36

∴ SR = \(\sqrt{36}\) [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.

In ∆ABC, AB = 10, AC = 7, BC = 9, then find the length of the median drawn from point C to side AB.

Solution:

Let CD be the median drawn from the vertex C to side AB.

BD = \(\frac { 1 }{ 2 } \) AB [D is the midpoint of AB]

= \(\frac { 1 }{ 2 } \) × 10 = 5 units

In ∆ABC, seg CD is the median. [Given]

∴ AC^{2} + BC^{2} = 2 CD^{2} + 2 BD^{2} [Apollonius theorem]

∴ 7^{2} + 9^{2} = 2 CD^{2} + 2 (5)^{2}

∴ 49 + 81 = 2 CD^{2} + 2 (25)

∴ 130 = 2 CD^{2} + 50

∴ 2 CD^{2} = 130 – 50

∴ 2 CD^{2} = 80

∴ CD^{2} = \(\frac { 80 }{ 2 } \) = 40

∴ CD = \(\sqrt { 40 }\) [Taking square root of both sides]

= 2 \(\sqrt { 10 }\) units

∴ The length of the median drawn from point C to side AB is 2 \(\sqrt { 10 }\) units.

Question 3.

In the adjoining figure, seg PS is the median of APQR and PT ⊥ QR. Prove that,

i. PR^{2} = PS^{2} + QR × ST + (\(\frac { QR }{ 2 } \))^{2}

ii. PQ^{2} = PS^{2} – QR × ST + (\(\frac { QR }{ 2 } \))^{2}

Solution:

i. QS = SR = \(\frac { 1 }{ 2 } \) QR (i) [S is the midpoint of side QR]

∴ In ∆PSR, ∠PSR is an obtuse angle [Given]

and PT ⊥ SR [Given, Q-S-R]

∴ PR^{2} = SR^{2} +PS^{2} + 2 SR × ST (ii) [Application of Pythagoras theorem]

∴ PR^{2} = (\(\frac { 1 }{ 2 } \) QR)^{2} + PS^{2} + 2 (\(\frac { 1 }{ 2 } \) QR) × ST [From (i) and (ii)]

∴ PR^{2} = (\(\frac { QR }{ 2 } \))^{2} + PS^{2} + QR × ST

∴ PR^{2} = PS^{2} + QR × ST + (\(\frac { QR }{ 2 } \))^{2}

ii. In.∆PQS, ∠PSQ is an acute angle and [Given]

PT ⊥QS [Given, Q-S-R]

∴ PQ^{2} = QS^{2} + PS^{2} – 2 QS × ST (iii) [Application of Pythagoras theorem]

∴ PR^{2} = (\(\frac { 1 }{ 2 } \) QR)^{2} + PS^{2} – 2 (\(\frac { 1 }{ 2 } \) QR) × ST [From (i) and (iii)]

∴ PR^{2} = (\(\frac { QR }{ 2 } \))^{2} + PS^{2} – QR × ST

∴ PR^{2} = PS^{2} – QR × ST + (\(\frac { QR }{ 2 } \))^{2}

Question 4.

In ∆ABC, point M is the midpoint of side BC. If AB^{2 }+ AC^{2} = 290 cm, AM = 8 cm, find BC.

Solution:

In ∆ABC, point M is the midpoint of side BC. [Given]

∴ seg AM is the median.

∴ AB^{2} + AC^{2} = 2 AM^{2} + 2 MC^{2} [Apollonius theorem]

∴ 290 = 2 (8)^{2} + 2 MC^{2}

∴ 145 = 64 + MC^{2} [Dividing both sides by 2]

∴ MC^{2} = 145 – 64

∴ MC^{2} = 81

∴ MC = \(\sqrt{81}\) [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.

In the adjoining figure, point T is in the interior of rectangle PQRS. Prove that, TS^{2} + TQ^{2} = TP^{2} + TR^{2}. (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: TS^{2} + TQ^{2} = TP^{2} + TR^{2}

Construction: Draw seg AB || side SR such that A – T – B.

Solution:

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)]

∴ TP^{2} = PS^{2} + TS^{2} – 2 PS.AS (v) [Application of Pythagoras theorem]

In ∆TQR., ∠TRQ is an acute angle

and seg BT ⊥ side QR [From (iii)]

∴ TQ^{2} = RQ^{2} + TR^{2} – 2 RQ.BR (vi) [Application of pythagoras theorem]

TP^{2} – TQ^{2} = PS^{2} + TS^{2} – 2PS.AS

-RQ^{2} – TR^{2} + 2RQ.BR [Subtracting (vi) from (v)]

∴ TP^{2} – TQ^{2} = TS^{2} – TR^{2} + PS^{2}

– RQ^{2} -2 PS.AS +2 RQ.BR

∴ TP^{2} – TQ^{2} = TS^{2} – TR^{2} + PS^{2}

– PS^{2} – 2 PS.BR + 2PS.BR [From (i) and (iv)]

∴ TP^{2} – TQ^{2} = TS^{2} – TR^{2}

∴ TS^{2} + TQ^{2} = TP^{2} + TR^{2}

Question 1.

In ∆ABC, ∠C is an acute angle, seg AD Iseg BC. Prove that: AB^{2} = BC^{2} + A^{2} – 2 BC × DC. (Textbook pg. no. 44)

Given: ∠C is an acute angle, seg AD ⊥ seg BC.

To prove: AB^{2} = BC^{2} + AC^{2} – 2BC × DC

Solution:

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]

AB^{2} = BD^{2} + AD^{2} [Pythagoras theorem]

∴ c^{2} = (a – x)^{2} + [P^{2}] (i)

∴ c^{2} = a^{2} – 2ax + x^{2} + [P^{2}]

In ∆ADC, ∠D = 90° [Given]

AC^{2} = AD^{2} + CD^{2} [Pythagoras theorem]

∴ b^{2} = p^{2} + [X^{2}]

∴ p^{2} = b^{2} – [X^{2}] (ii)

∴ c^{2} = a^{2} – 2ax + x^{2} + b^{2} – x^{2} [Substituting (ii) in (i)]

∴ c^{2} = a^{2} + b^{2} – 2ax

∴ AB^{2} = BC^{2} + AC^{2} – 2 BC × DC

Question 2.

In ∆ABC, ∠ACB is an obtuse angle, seg AD ⊥ seg BC. Prove that: AB^{2} = BC^{2} + AC^{2} + 2 BC × CD. (Textbook pg. no. 40 and 4.1)

Given: ∠ACB is an obtuse angle, seg AD ⊥ seg BC.

To prove: AB^{2} = BC^{2} + AC^{2} + 2BC × CD

Solution:

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]

AB^{2} = BD^{2} + AD^{2} [Pythagoras theorem]

∴ c^{2} = (a + x)^{2} + p^{2} (i)

∴ c^{2} = a^{2} + 2ax + x^{2} + p^{2}

Also, in ∆ADC, ∠D = 90° [Given]

AC^{2} = CD^{2} + AD^{2} [Pythagoras theorem]

∴ b^{2} = x^{2} + p^{2}

∴ p^{2} = b^{2} – x^{2 }(ii)

∴ c^{2} = a^{2} + 2ax + x^{2} + b^{2} – x^{2} [Substituting (ii) in (i)]

∴ c^{2} = a^{2} + b^{2} + 2ax

∴ AB^{2} = BC^{2} + AC^{2} + 2 BC × CD

Question 3.

In ∆ABC, if M is the midpoint of side BC and seg AM ⊥seg BC, then prove that

AB^{2} + AC^{2} = 2 AM^{2} + 2 BM^{2}. (Textbook pg, no. 41)

Given: In ∆ABC, M is the midpoint of side BC and seg AM ⊥ seg BC.

To prove: AB^{2} + AC^{2} = 2 AM^{2} + 2 BM^{2}

Solution:

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]

∴ AB^{2} + AC^{2 }= AM^{2} + BM^{2} + AM^{2} + MC^{2} [Adding (i) and (ii)]

∴ AB^{2} + AC^{2} = 2 AM^{2} + BM^{2} + BM^{2} [∵ BM = MC (M is the midpoint of BC)]

∴ AB^{2} + AC^{2} = 2 AM^{2} + 2 BM^{2}