**Question 1.In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.**

Solution:

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

= 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.

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

∴ The length of the median drawn from point C to side AB is 2√10 units.

**Question 3.In the adjoining figure, seg PS is the median of APQR and PT ⊥ QR. Prove that,**

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

PT ⊥QS [Given, Q-S-R]

**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 = 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 × DCSolution:**

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 × CDSolution:**

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 thatAB**

^{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}