Problem Set 2 Geometry 10th Std Maths Part 2 Answers Chapter 2 Pythagoras Theorem
Question 1
Maharashtra Board Solution
Some questions and their alternative answers are given. Select the correct alternative. [1 Mark each]
i. Out of the following which is the Pythagorean triplet?
(A) (1,5,10)
(B) (3,4,5)
(C) (2,2,2)
(D) (5,5,2)
Solution & Step-by-Step Answer:
(B)
Hint: Refer Practice set 2.1 Q.1 (i)
Hint: Refer Practice set 2.1 Q.1 (i)
ii. In a right angled triangle, if sum of the squares of the sides making right angle is 169, then what is the length of the hypotenuse?
(A) 15
(B) 13
(C) 5
(D) 12
Answer: (B)
Hint:
iii. Out of the dates given below which date constitutes a Pythagorean triplet?
(A) 15/08/17
(B) 16/08/16
(C) 3/5/17
(D) 4/9/15
Answer: (A)
Hint:
iv. If a, b, c are sides of a triangle and a2+ b2= c2, name the type of the triangle.
(A) Obtuse angled triangle
(B) Acute angled triangle
(C) Right angled triangle
(D) Equilateral triangle
Answer: (C)
v. Find perimeter of a square if its diagonal is 10 cm.
(A) 10 cm
(B) 40 cm
(C) 20 cm
(D) 40 cm
Answer: (D)
Hint:
vi. Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude.
(A) 9 cm
(B) 4 cm
(C) 6 cm
(D) 2
Answer: (C)
Hint:
vii. Height and base of a right angled triangle are 24 cm and 18 cm find the length of its hypotenuse.
(A) 24 cm
(B) 30 cm
(C) 15 cm
(D) 18 cm
Answer: (B)
Hint:
viii. In ∆ABC, AB = 6 cm, AC = 12 cm, BC = 6 cm. Find measure of ∠A.
(A) 30°
(B) 60°
(C) 90°
(D) 45°
Answer: (A)
Hint:
Question 2
Maharashtra Board Solution
Solve the following examples. i. Find the height of an equilateral triangle having side 2a. ii. Do sides 7 cm, 24 cm, 25 cm form a right angled triangle? Give reason. iii. Find the length of a diagonal of a rectangle having sides 11 cm and 60 cm. iv. Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm. v. A side of an isosceles right angled triangle is x. Find its hypotenuse. vi. In ∆PQR, PQ = , QR = , PR = . Is ∆PQR a right angled triangle? If yes, which angle is of 90°?
Solution & Step-by-Step Answer:
i. Let ∆ABC be the given equilateral triangle. ∴ ∠B = 60° [Angle of an equilateral triangle] Let AD ⊥BC, B – D – C. In ∆ABD, ∠B = 60°, ∠ADB = 90° ∴ ∠BAD = 30° [Remaining angle of a triangle] ∴ ∆ABD is a 30° – 60° – 90° triangle. ∴ AD = AB [Side opposite to 60°] = × 2a = a units The height of the equilateral triangle is a units.
ii. The sides of the triangle are 7 cm, 24 cm and 25 cm. iii. Let ꠸ABCD be the given rectangle. iv. Let ∆PQR be the given right angled triangle. v. Let ∆PQR be the given right angled isosceles triangle. vi. Longest side of ∆PQR = PQ =
Question 3
Maharashtra Board Solution
In ∆RST, ∠S = 90°, ∠T = 30°, RT = 12 cm, then find RS and ST.
Solution & Step-by-Step Answer:
in ∆RST, ∠S = 900, ∠T = 30° [Given] ∴ ∠R = 60° [Remaining angle of a triangle] ∴ ∆RST is a 30° – 60° – 90° triangle. ∴ RS = RT [Side opposite to 30°] = × 12 = 6cm Also, ST = RT [Side opposite to 60°] = × 12 = 6 cm ∴ RS = 6 cm and ST = 6 cm
Question 4
Maharashtra Board Solution
Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq. cm.
Solution & Step-by-Step Answer:
Let ꠸ABCD be the given rectangle. BC = 16cm Area of rectangle = length × breadth Area of ꠸ABCD = BC × AB ∴ 192 = I6 × AB ∴ AB = = 12cm Now, in ∆ABC, ∠B = 90° [Angle of a rectangle] ∴ AC2 = AB2 + BC2 [Pythagoras theorem] = 122 + 162 = 144 + 256 =400 ∴ AC = [Taking square root of both sides] = 20cm ∴ The diagonal of the rectangle is 20 cm.
Question 5
Maharashtra Board Solution
Find the length of the side and perimeter of an equilateral triangle whose height is cm.
Solution & Step-by-Step Answer:
Let ∆ABC be the given equilateral triangle. ∴ ∠B = 60° [Angle of an equilateral triangle] AD ⊥ BC, B – D – C. In ∆ABD, ∠B =60°, ∠ADB = 90° ∴ ∠BAD = 30° [Remaining angle of a triangle] ∴ ∆ABD is a 30° – 60° – 90° triangle. ∴ AD = AB [Side opposite to 600] ∴ = AB ∴ AB = ∴ AB = 2cm ∴ Side of equilateral triangle = 2cm Perimeter of ∆ABC = 3 × side = 3 × AB = 3 × 2 = 6cm ∴ The length of the side and perimeter of the equilateral triangle are 2 cm and 6 cm respectively.
Question 6
Maharashtra Board Solution
In ∆ABC, seg AP is a median. If BC = 18, AB2 + AC2 = 260, find AP.
Solution & Step-by-Step Answer:
PC = BC [P is the midpoint of side BC] = × 18 = 9cm in ∆ABC, seg AP is the median, Now, AB2 + AC2 = 2 A2 + 2 PC2 [Apollonius theorem] ∴ 260 = 2 AP2 + 2 (9)2 ∴ 130 = AP2 + 81 [Dividing both sides by 2] ∴ AP2 = 130 – 81 ∴ AP2 = 49 ∴ AP = [Taking square root of both sides] ∴ AP = 7 units
Question 7
Maharashtra Board Solution
∆ABC is an equilateral triangle. Point P is on base BC such that PC = BC, if AB = 6 cm find AP. Given: ∆ABC is an equilateral triangle. PC = BC, AB = 6cm. To find: AP Consttuction: Draw seg AD ± seg BC, B – D – C.
Solution & Step-by-Step Answer:
∆ABC is an equilateral triangle. ∴ AB = BC = AC = 6cm [Sides of an equilateral triangle] pc = BC [Given] = (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 = AC [Side opposite to 60°] ∴ AD = (6) ∴ AD = 3 cm CD = AC [Side opposite to 30°] ∴ CD = (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)2 + (1)2 ∴ AP2 = 9 × 3 + 1 = 27 + 1 ∴ AP2 = 28 ∴ AP = ∴ AP = ∴ AP = 2 cm
Question 8
Maharashtra Board Solution
From the information given in the adjoining figure, prove that PM = PN = × a
Solution & Step-by-Step Answer:
Proof: In ∆PMR, QM = QR = a [Given] ∴ Q is the midpoint of side MR. ∴ seg PQ is the median. ∴ PM2 + PR2 = 2PQ2 + 2QM2 [Apollonius theorem] ∴ PM2 + a2 = 2a2 + 2a2 ∴ PM2 + a2 = 4a2 ∴ PM2 = 3a2 ∴ PM,= a (i) [Taking square root of both sides] SimlarIy, in ∆PNQ, R is the midpoint of side QN. ∴ seg PR is the median. ∴ PN2 + PQ2 = 2 PR2 + 2 RN2 [Apollonius theorem] ∴ PN2 + a2 = 2a2 + 2a2 ∴ PN2 + a2 = 4a2 ∴ PN2 = 3a2 ∴ PN = a (ii) [Taking square root of both sides] ∴ PM = PN = a [From (i) and (ii)]
Question 9
Maharashtra Board Solution
Prove that the sum of the squares of the diagonals of a parallelogram ¡s equal to the sum of the squares of its sides. Given: ꠸ABCD is a parallelogram, diagonals AC and BD intersect at point M. To prove: AC2 + BD2 = AB2 + BC2 + CD2 + AD2
Solution & Step-by-Step Answer:
Proof: ꠸ABCD is a parallelogram. ∴ AB = CD and BC = AD (i) [Opposite sides of a parallelogram] AM = AC and BM = BD (ii) [Diagonals of a parallelogram bisect each other] ∴ M is the midpoint of diagonals AC and BD. (iii) In ∆ABC. seg BM is the median. [From (iii)] AB2 + BC2 = 2AM2 + 2BM2 (iv) [Apollonius theorem] ∴ AB2 + BC2 = 2( AC)2 + 2( BD)2 [From (ii) and (iv)] ∴ AB2 + BC2 = 2 × ∴ AB2 + BC2 = ∴ 2AB2 + 2BC2 = BD2 + AC2 [Multiplying both sides by 2] ∴ AB2 + AB2 + BC2 + BC2 = BD2 + AC2 ∴ AB2 + CD2 + BC2 + AD2 = BD2 + AC2 [From(i)] i.e. AC2 + BD2 = AB2 + BC2 + CD2 + AD2
Question 10
Maharashtra Board Solution
Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was 15 km. Find their speed per hour.
Solution & Step-by-Step Answer:
Suppose Pranali and Prasad started walking from point A, and reached points B and C respectively after 2 hours. Distance between them = BC = 15 km Since, their speed is same, both travel the same distance in the given time. ∴ AB = AC Let AB = AC = x km (i) Now, in ∆ ABC, ∠A = 90° ∴ BC2 = AB2 + AC2 [Pythagoras theorem] ∴ (15)2 = x2 + x2 [From (i)] ∴ 225 × 2 = 2 x2 ∴ x2 = 225 ∴ x = [Taking square root of both sides] ∴ x = 15 km ∴ AB = AC = 15km = 7.5 km/hr ∴ The speed of Pranali and Prasad is 7.5 km/hour.
Question 11
Maharashtra Board Solution
In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that 4 (BL2 + CM2) = 5 BC2. Given : ∠BAC = 90° seg BL and seg CM are the medians. To prove: 4(BL2 + CM2) = 5BC2
Solution & Step-by-Step Answer:
Proof: In ∆BAL, ∠BAL 90° [Given] ∴ BL2 = AB2 + AL2 (i) [Pythagoras theorem] In ∆CAM, ∠CAM = 90° [Given] ∴ CM2 = AC2 + AM2 (ii) [Pythagoras theorem] ∴ BL2 + CM2 = AB2 + AC2 + AL2 + AM2 (iii) [Adding (i) and (ii)] Now, AL = AC and AM = AB (iv) [seg BL and seg CM are the medians] ∴ BL2 + CM2 = AB2 + AC2 + ( AC)2 + ( AB)2 [From (iii) and (iv)] ∴ BL2 + CM2 = (AB2 + AC2) ∴ 4(BL2 + CM2) = 5(AB2 + AC2) (v) In ∆BAC, ∠BAC = 90° [Given] ∴ BC2 = AB2 + AC2 (vi) [Pythagoras theorem] ∴ 4(BL2 + CM2) = 5BC2 [From (v) and (vi)]
Question 12
Maharashtra Board Solution
Sum of the squares of adjacent sides of a parallelogram is 130 cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal.
Solution & Step-by-Step Answer:
Let ꠸ABCD be the given parallelogram and its diagonals AC and BD intersect at point M. ∴ AB2 + AD2 = 130cm, BD = 14cm MD = BD (i) [Diagonals of a parallelogram bisect each other] = × 14 = 7 cm In ∆ABD, seg AM is the median. [From (i)] ∴ AB2 + = 2AM2 + 2MD2 [Apollonius theorem] ∴ 130 = 2 AM2 + 2(7)2 ∴ 65 = AM2 +49 [Dividing both sides by 2] ∴ AM2 = 65 – 49 ∴ AM2 = 16 [Taking square root of both sides] ∴ AM = = 4cm Now, AC =2 AM [Diagonals of a parallelogram bisect each other] 2 × 4 = 8 cm ∴ The length of the other diagonal of the parallelogram is 8 cm.
Question 13
Maharashtra Board Solution
In ∆ABC, seg AD ⊥ seg BC and DB = 3 CD. Prove that: 2 AB2 = 2 AC2 + BC2. Given: seg AD ⊥ seg BC DB = 3CD To prove: 2AB2 = 2AC2 + BC2
Solution & Step-by-Step Answer:
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 = (v) AB2 = AC2 + 8()2 [From (iv) and (v)] ∴ AB2 = AC2 + 8 × ∴ AB2 = AC2 + ∴ 2AB2 = 2AC2 + BC2 [Multiplying both sides by 2]
Question 14
Maharashtra Board Solution
In an isosceles triangle, length of the congruent sides is 13 em and its.base is 10 cm. Find the distance between the vertex opposite to the base and the centroid. 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.
Solution & Step-by-Step Answer:
Centroid G of ∆ABC lies on AD ∴ seg AD is the median. (i) ∴ D is the midpoint of side BC. ∴ DC = BC = × 10 = 5 In ∆ABC, seg AD is the median. [From (i)] ∴ AB2 + AC2 = 2 AD2 + 2 DC2 [Apollonius theorem] ∴ 132 + 132 = 2 AD2 + 2 (5)2 ∴ 2 × 132 = 2 AD2 + 2 × 25 ∴ 169 = AD2 + 25 [Dividing both sides by 2] ∴ AD2 = 169 – 25 ∴ AD2 = 144 ∴ AD = [Taking square root of both sides] = 12 cm We know that, the centroid divides the median in the ratio 2 : 1. ∴ = ∴ = [By invertendo] ∴ = [By componendo] ∴ = [A – G – D] ∴ = ∴ AG = = 8cm ∴ The distance between the vertex opposite to the base and the centroid is 8 cm.
Question 15
Maharashtra Board Solution
In a trapezium ABCD, seg AB || seg DC, seg BD ⊥ seg AD, seg AC ⊥ seg BC. If AD = 15, BC = 15 and AB = 25, find A (꠸ABCD). Construction: Draw seg DE ⊥ seg AB, A – E – B and seg CF ⊥ seg AB, A – F- B.
Solution & Step-by-Step Answer:
In ∆ ACB, ∠ACB = 90° [Given] ∴ AB2 = AC2 + BC2 [Pythagoras theorem] ∴ 252 = AC2 + 152 ∴ AC2 = 625 – 225 = 400 ∴ AC = [Taking square root of both sides] = 20 units Now, A(∆ABC) = × BC × AC Also, A(∆ABC) = × AB × CF ∴ × BC × AC = × AB × CF ∴ BC × AC = AB × CF ∴ 15 × 20 = 25 × CF ∴ CF = = 12 units In ∆CFB, ∠CFB 90° [Construction] ∴ BC2 = CF2 + FB2 [Pythagoras theorem] ∴ 152 = 122 + FB2 ∴ FB2 = 225 – 144 ∴ FB2 = 81 ∴ FB = [Taking square root of both sides] = 9 units Similarly, we can show that, AE = 9 units Now, AB = AE + EF + FB [A – E – F, E – F – B] ∴ 25 = 9 + EF + 9 ∴ EF = 25 – 18 = 7 units In ꠸CDEF, seg EF || seg DC [Given, A – E – F, E – F – B] seg ED || seg FC [Perpendiculars to same line are parallel] ∴ ꠸CDEF is a parallelogram. ∴ DC = EF 7 units [Opposite sides of a parallelogram] A(꠸ABCD) = × CF × (AB + CD) = × 12 × (25 + 7) = × 12 × 32 ∴ A(꠸ABCD) = 192 sq. units
Question 16
Maharashtra Board Solution
In the adjoining figure, ∆PQR is an equilateral triangle. Point S is on seg QR such that QS = QR. Prove that: 9 PS2 = 7 PQ2. Given: ∆PQR is an equilateral triangle. QS = QR To prove: 9PS2 = 7PQ2
Solution & Step-by-Step Answer:
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 = PQ (iv) [Side opposite to 60°] QT = PQ (v) [Side opposite to 30°] QS + ST = QT [Q – S – T] ∴ QR + ST = PQ [Given and from (v)] ∴ PQ + ST = PQ [From (ii)] ∴ ST = – ∴ ST = ∴ ST = (vi) [From (iii), (iv) and (vi)] ∴ ∴ ∴ ∴PS2 = PQ2 ∴ 9PS2 = 7 PQ2
Question 17
Maharashtra Board Solution
Seg PM is a median of APQR. If PQ = 40, PR = 42 and PM = 29, find QR.
Solution & Step-by-Step Answer:
In ∆PQR, seg PM is the median. [Given] ∴ M is the midpoint of side QR. ∴ PQ2 + PR2 = 2 PM2 + 2 MR2 [Apollonius theorem] ∴ 402 + 422 = 2 (29)2 + 2 MR2 ∴ 1600 + 1764 = 2 (841) + 2 MR2 ∴ 3364 = 2 (841) + 2 MR2 ∴ 1682 = 841 +MR2 [Dividing both sides by 2] ∴ MR2 = 1682 – 841 ∴ MR2 = 841 ∴ MR = [Taking square root of both sides] = 29 units Now, QR = 2 MR [M is the midpoint of QR] = 2 × 29 ∴ QR = 58 units
Question 18
Maharashtra Board Solution
Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM.
Solution & Step-by-Step Answer:
In ∆ABC, seg AM is the median. [Given] ∴ M is the midpoint of side BC. ∴ MC = BC = × 24 = 12 units Now, AB2 + AC2 = 2 AM2 + 2 MC2 [Apollonius theorem] ∴ 222 + 342 = 2 AM2 + 2 (12)2 ∴ 484 + 1156 = 2 AM2 + 2 (144) ∴ 1640 = 2 AM2 + 2 (144) ∴ 820 = AM2 + 144 [Dividing both sides by 2] ∴ AM2 = 820 – 144 ∴ AM2 = 676 ∴ AM = [Taking square root of both sides] ∴ AM = 26 units
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