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Given: ∆ABC is an isosceles triangle. 

G is the centroid.

AB = AC = 13 cm, BC = 10 cm. 

To find: AG 

Construction: Extend AG to intersect side BC at D, B – D – C.

Centroid G of ∆ABC lies on AD 

∴ seg AD is the median. (i) 

∴ D is the midpoint of side BC. 

∴ DC = 1/2 BC 

= 1/2 × 10 = 5

In ∆ABC, seg AD is the median. [From (i)] 

∴ AB² + AC² = 2AD² + 2DC² [Apollonius theorem] 

∴ 13² + 13² = 2AD² + 2(5)² 

∴ 2 × 13² = 2AD² + 2 × 25 

∴ 169 = AD² + 25 [Dividing both sides by 2]

∴ AD² = 169 – 25 

∴ AD² = 144 

∴ AD = √144  [Taking square root of both sides] 

= 12 cm We know that, the centroid divides the median in the ratio 2 : 1.

∴ AG/GD = 2/1 

∴ GD/AG = 1/2  [By invertendo] 

∴ (GD + AG)AG = (1 + 2)/2 [By componendo] 

∴ AD/AG = 3/2  = [A – G – D] 

∴ 12/AG = 3/2 

∴ AG = (12 x 2)/3 

= 8cm

∴ The distance between the vertex oppesite to the base and the centroid id 8 cm.

Proof: In ∆PQR, ∠PRQ = 90° [Given]

PQ² = PR² + QR² (i) [Pythagoras theorem] 

RM = QR [M is the midpoint of QR] 

∴ 2RM = QR (ii) 

∴ PQ² = PR² + (2RM)² [From (i) and (ii)] 

∴ PQ² = PR² + 4RM² (iii) 

Now, in ∆PRM, ∠PRM = 90° [Given] 

∴ PM² = PR² + RM² [Pythagoras theorem] 

∴ RM² = PM² – PR² (iv) 

∴ PQ² = PR² + 4 (PM² – PR² ) [From (iii) and (iv)] 

∴ PQ² = PR² + 4 PM² – 4 PR² 

∴ PQ² = 4 PM² – 3 PR²

The correct answer is : (B) (3,4,5)

DB = 3CD (i) [Given] 

In ∆ADB, ∠ADB = 90° [Given] 

∴ AB² = AD² + DB² [Pythagoras theorem] 

∴ AB² = AD² + (3CD)² [From (i)] 

∴ AB² = AD² + 9CD² (ii) 

In ∆ADC, ∠ADC = 90° [Given] 

∴ AC² = AD² + CD² [Pythagoras theorem] 

∴ AD² = AC² – CD² (iii)

AB² = AC² – CD² + 9CD² [From (ii) and(iii)] 

∴ AB² = AC² + 8CD² (iv) 

CD + DB = BC [C – D – B] 

∴ CD + 3CD = BC [From (i)] 

∴ 4CD = BC 

∴ CD = (BC/4) (v) 

AB² = AC² + 8( BC/4)² [From (iv) and (v)] 

∴ AB² = AC² + 8 × BC²/16

∴ AB² = AC² + BC²/2

∴ 2AB² = 2AC² + BC² [Multiplying both sides by 2]

i. In ∆DEF, ∠DFE = 90° and FG ⊥ ED [Given]

∴ FG² = GD × EG [Theorem of geometric mean]

∴ 122 = 8 × EG.

∴ EG = \(\frac{144}{8}\)

∴ EG = 18 units

ii. In ∆FGD, ∠FGD = 90° [Given]

∴ FD² = FG² + GD² [Pythagoras theorem]

= 122 + 82 = 144 + 64 

= 208

∴ FD \(=\sqrt{208}\) [Taking square root of both sides]

∴ FD \(=4\sqrt{13}\) units

iii. In ∆EGF, ∠EGF = 90° [Given] 

∴ EF² = EG² + FG² [Pythagoras theorem] 

= 182 + 122 = 324 + 144

= 468

∴ EF \(=\sqrt{468}\) [Taking square root of both sides]

∴ EF \(=6\sqrt{13}\) units.

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 MC² 

∴ 145 = 64 + MC² [Dividing both sides by 2] 

∴ MC² = 145 – 64 

∴ MC² = 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

i. Here, 5² = 25 

3² + 4² = 9 + 16 = 25 

∴ 5² = 3² + 4² 

The square of the largest number is equal to the sum of the squares of the other two numbers. 

∴ (3, 5, 4) is a Pythagorean triplet.

ii. Here, 12² = 144 

4² + 9² = 16 + 81 =97 

∴ 12² ≠ 4² + 9²

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (4, 9, 12) is not a Pythagorean triplet.

iii. Here, 13² = 169 

5² + 12² = 25 + 144 = 169 

∴ 13² = 5² + 12² 

The square of the largest number is equal to the sum of the squares of the other two numbers. 

∴ (5, 12, 13) is a Pythagorean triplet.

iv. Here, 74² = 5476 

24² + 70² = 576 + 4900 = 5476 

∴ 74² = 24² + 70² 

The square of the largest number is equal to the sum of the squares of the other two numbers. 

∴ (24, 70, 74) is a Pythagorean triplet.

v. Here, 27² = 729 

10² + 24² = 100 + 576 = 676 

∴ 27² ≠ 10² + 24² 

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (10, 24, 27) is not a Pythagorean triplet.

vi. Here, 61² = 3721 

11² + 60² = 121 + 3600 = 3721 

∴ 61² = 11² + 60² 

The square of the largest number is equal to the sum of the squares of the other two numbers. 

∴ (11, 60, 61) is a Pythagorean triplet.

In ∆MNP, ∠MNP = 90° and [Given] 

seg NQ ⊥ seg MP 

NQ² = MQ × QP [Theorem of geometric mean] 

∴ NQ = \({ \ \sqrt{MQ\times QP} }\) [Taking square root of both sides] 

= \({ \ \sqrt{9\times 4} }\)= 3 × 2 

∴NQ = 6 units

In ∆MNP, ∠MNP = 90° and [Given] 

seg NQ ⊥ seg MP

NQ² = MQ × QP [Theorem of geometric mean]

∴ NQ = \(\sqrt{MQ \times QP}\) [Taking square root of both sides]

\(=\sqrt{9 \times 4}\)

\(=3 \times 2\)

∴ NQ = 6 units

In ∆PSR, ∠S = 90° , ∠P = 30° [Given] 

∴ ∠R = 60° [Remaining angle of a triangle] 

∴ ∆PSR is a 30° – 60° – 90° triangle. 

RS = 1/2 RP [Side opposite to 30°] 

∴6 = 1/2 RP 

∴ RP = 6 × 2 = 12 units 

Also, PS = \( {\sqrt{3} \over 2}\) RP [Side opposite to 60°] 

= \( {\sqrt{3} \over 2}\) × 12 

= \( {6\sqrt{3} }\) units 

∴ RP = 12 units, PS = \( {6\sqrt{3} }\) units

In ∆PSR, ∠S = 90°, ∠P = 30° [Given] 

∴ ∠R = 60° [Remaining angle of a triangle]

∴ ∆PSR is a 30° – 60° – 90° triangle.

RS = 1/2 RP [Side opposite to 30°]

∴ 6 = 1/2 RP

∴ RP = 6 × 2 = 12 units

Also, PS = √3/2 RP [Side opposite to 60°]

= √3/2 × 12

= 6√3 units

∴ RP = 12 units, PS = 6√3 units.

Given: ∆PQR is an equilateral triangle. 

QS = 1/3 QR 

To prove: 9PS² = 7PQ²

Proof: 

∆PQR is an equilateral triangle [Given] 

∴ ∠P = ∠Q = ∠R = 60° (i) [Angles of an equilateral triangle] 

PQ = QR = PR (ii) [Sides of an equilateral triangle]

In ∆PTS, ∠PTS = 90° [Given] 

PS² = PT² + ST² (iii) [Pythagoras theorem] 

In ∆PTQ, 

∠PTQ = 90° [Given] 

∠PQT = 60° [From (i)] 

∴ ∠QPT = 30° [Remaining angle of a triangle]

∴ ∆PTQ is a 30° – 60° – 90° triangle 

∴ PT = √3/2 PQ (iv) [Side opposite to 60°] 

QT = 1/2 PQ (v) [Side opposite to 30°] 

QS + ST = QT [Q – S – T] 

∴ 1/3 QR + ST = 1/2 PQ [Given and from (v)] 

∴ PQ + ST = 1/2 PQ [From (ii)]

∴ ST = PQ/2 – PQ/2 

∴ ST = (3PQ - 2PQ)/6 

∴ ST = PQ/6 (vi) 

PS² = |(√3/2PQ)² + (PQ/6)² [From (iii), (iv) and (vi)]

∴ PS² = (3PQ²/4) + (PQ²/36) 

∴ PS² = (27PQ²/36) + (PQ²/36)

∴ PS² = (28PQ²)/36 

∴ PS² = 7/3PQ² 

∴PS² = 7/3 PQ² 

In ∆PQR, ∠QPR = 90° and [Given] 

seg PM ⊥ seg QR

∴ PM² = OM × MR [Theorem of geometric mean] 

∴ 10² = 8 × MR

∴ MR = \(\frac{100}{8}\)

= 12.5

Now, QR = QM + MR [Q – M – R] 

= 8 + 12.5 

∴ QR = 20.5 units

In ∆PQR, ∠QPR = 90° and [Given] 

seg PM ⊥ seg QR 

∴ PM² = OM × MR [Theorem of geometric mean] 

∴ 10² = 8 × MR 

∴ MR = 100/8 

= 12.5 

Now, QR = QM + MR [Q – M – R] 

= 8 + 12.5 

∴ QR = 20.5 units

∆ABC is an equilateral triangle. 

∴ AB = BC = AC = 6cm [Sides of an equilateral triangle] 

pc = 1/3 BC [Given]

= 1/3 (6) 

∴ PC = 2cm 

In ∆ADC, 

∠D = 90° [Construction] 

∠C = 60° [Angle of an equilateral triangle] 

∠DAC = 30° [Remaining angle of a triangle] 

∴ ∆ ADC is a 30° – 60° – 90° triangle.

∴ AD = √3/2 AC [Side opposite to 60°] 

∴ AD = √3/2  (6) 

∴ AD = 3√3 cm

∴ CD = 1/2 AC [Side opposite to 30°] 

∴ CD = 1/2 (6) 

∴ CD = 3cm 

Now DP + PC = CD [D – P – C] 

∴ DP + 2 = 3 

∴ DP = 1cm 

In ∆ADP, 

∠ADP = 900

AP² = AD² + DP² [Pythagoras theorem] 

∴ AP² = (3√3)² + (1)² 

∴ AP² = 9 × 3 + 1 = 27 + 1 

∴ AP² = 28 

∴ AP = √28 

∴ AP = (√4 x 7)

∴ AP = 2√7 cm 

The correct answer is : (c) 6 cm

The correct answer is : (B) 30 cm

The correct answer is : (A) 30°

Correct answer is

(A) 30°

∆ABC is an equilateral triangle.

∴ AB = BC = AC = 6cm [Sides of an equilateral triangle]

pc = 1/3 BC [Given]

= 1/3 (6)

∴ PC = 2cm

In ∆ADC,

∠D = 90° [Construction]

∠C = 60° [Angle of an equilateral triangle]

∠DAC = 30° [Remaining angle of a triangle]

∴ ∆ ADC is a 30° – 60° – 90° triangle.

∴ AD = √3/2 AC [Side opposite to 60°]

∴ AD = √3/2 (6)

∴ AD = 3 √3 cm

CD = 1/2 AC [Side opposite to 30°]

∴ CD = 1/2 (6)

∴ CD = 3 cm

Now DP + PC = CD [D – P – C]

∴ DP + 2 = 3

∴ DP = 1 cm

In ∆ADP,

∠ADP = 900

AP² = AD² + DP² [Pythagoras theorem]

∴ AP² = (3√3)² + (1) 

∴ AP² = 9 × 3 + 1 = 27 + 1

∴ AP² = 28

∴ AP = √28

∴ AP = \(\sqrt{4 \times 7}\)

∴ AP = 2√7 cm.

The sides of the triangle are 7 cm, 24 cm and 25 cm.

The longest side of the triangle is 25 cm.

∴ (25)² = 625

Now, sum of the squares of the remaining sides is,

(7)² + (24)² = 49 + 576

= 625

∴ (25)² = (7)² + (24)²

∴ Square of the longest side is equal to the sum of the squares of the remaining two sides. 

∴ The given sides will form a right angled triangle. 

[Converse of Pythagoras theorem]

Correct answer is

(D) 40 cm

Correct answer is

(A) 15/08/17