Maharashtra Board Class 10 Math Part – 2 Solution Chapter 2 Problem Set 2 – Pythagoras Theorem
Balbharati Maharashtra Board Class 10 Math Part – 2 Solution Chapter 2: Pythagoras Theorem. Marathi or English Medium Students of Class 10 get here Pythagoras Theorem full Exercise Solution.
Std |
Maharashtra Class 10 |
Subject |
Math Part 2 Solution |
Chapter |
Pythagoras Theorem |
Problem Set |
2 |
Problem set – 2
(1) Some questions and their alternative answers are given. Select the correct alternative.
(1) 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:
(1) (B) (3, 4, 5)
32 = 9, 42 = 16, 52 = 25
∴ 52 = 32 + 42
∴ (3, 4, 5) are Pythagoras triplets.
Length of hypotenuse = √169 = 13
Ans – (B) 13.
(2) 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
Solution:
Length of hypotenuse = √169 = 13
Ans – (B) 13.
(3) 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
Solution:
Ans (A) 15/08/17 are Pythagoras triplet.
152 = 225, 82 = 64, 172 = 289
225+64 = 289
∴ 152 + 82 = 172
(4) If a, b, c are sides of a triangle and a2 + b2 = c2, name the type of triangle.
(A) Obtuse angled triangle
(B) Acute angled triangle
(C) Right angled triangle
(D) Equilateral triangle
Solution:
(C) Right angled triangle.
(5) Find perimeter of a square if its diagonal is 10√2 cm.
(A)10 cm
(B) 40√2 cm
(C) 20 cm
(D) 40 cm
Solution:
Solution:
Diagonal of sq = 10√2 cm.
Let, side be a.
∴ By Pythagoras theorem,
a2 + a2 = (10√2)2
Or, 2a2 = 100×2
Or, a = √10 = 10cm
Ans – (A) 10cm
(6) 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√6 cm
Solution:
Length of altitude = √Product of two parts of hypotenuse
= √4×9
= √36
= 6
Ans – (C) 6 cm.
(7) Height and base of a right angled triangle are 24 cm and 18 cm find the length of its hypotenus
(A) 24 cm
(B) 30 cm
(C) 15 cm
(D) 18 cm
Solution:
h2 = b2 + A2
h = √242 + 182
= √576+324
= 30
Ans – (B) 30 cm.
(8) In ∆ABC, AB = 6√3 cm, AC = 12cm, BC = 6cm. Find measure of ∠A.
(A) 30°
(B) 60°
(C) 90°
(D) 45°
Solution:
In ∆ABC,
AB2 = (6√3)2 = 108
AC2 = 122 = 144
BC2 = 62 = 36
∴ 108 + 31 = 144
∴ AB2 + BC2 = AC2
∴ ∆ABC is a right angled triangle at B.
Now, one side is half of hypotenuse AC. Which is BC.
∴ Angle opposite to it ∠A = 30°
Ans – (A) 30°
(2) Solve the following examples.
(1) Find the height of an equilateral triangle having side 2a
Solution:
Let, the triangle be ABC with height AM.
Side AB = BC = AC = 2a
∴ BM = C = 21/2 = a [∵ ∆ABC is equilateral ∆ So the median bisects the base]
∴ by Apollonius theorem,
AB2 + AC2 = 2AM2 + 2BM2
Or, (2a)2 + (2a)2 = 2AM2 + 2a2
Or, 4a2 + 4a2 = 2AM2 + 2a2
Or, AM2 = 36a2/2
Or, AM = √3a2
= a√3
∴ a√3 is the height of ∆ABC.
(2) Do sides 7 cm, 24 cm, 25 cm form a right angled triangle? Give reason.
Solution:
72 = 49, 242 = 576, 252 = 625
∴ 49+576 = 625
∴ 72 + 242 = 252
∴ The triangle is a right angled triangle since their side form Pythagoras triplets.
(3) Find the length a diagonal of a rectangle having sides 11 cm and 60cm.
Solution:
Length of ▭ = 60 cm
Breadth of ▭ = 11cm
∴ diagonal2 = lengths2 + breadth2
Or, diagonal = √602 + 112
= √3600 + 121
= √3721
= 61 cm
(4) Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm.
Solution:
We know, by Pythagoras theorem,
h2 = b2 + a2
Or, h = √92 + 122 [Given]
= √81 + 144
= √225
= 15 cm
(5) A side of an isosceles right angled triangle is x. Find its hypotenuse.
Solution:
Isosceles triangle have two equal sides one side is x.
∴ h2 = b2 + a2 [By Pythagoras theorem]
Or, h = √x2 + x2
= √2x2
= x√2
(6) In ∆PQR; PQ = √8, QR = √5, PR = √3. Is ∆PQR a right angled triangle? If yes, which angle is of 90°?
Solution:
Given, PQ = √8, QR = √5, PR = √3
∴ PQ2 = (√8)2 = 8, QR2 = (√5)2 = 5, PR2 = (√3)2 = 3
∴ QR2 + PR2 = PQ2 or, 5+3 = 8
∴ ∆PQR is a right angled triangle right angled at R since PQ is the hypotenuse.
(3) In ∆RST, ∠S = 90°, ∠T = 30°, RT = 12 cm then find RS and ST.
Solution:
Given, In ∆RST,
∠S = 90°, ∠T = 30°, RT = 12cm
∴ ∠R = 180 – (90 + 30) ∴ RT is hypotenuse
= 180 – 120
= 60°
∴ [RS = 1/2 RT Theorem of 30° – 60° – 90°]
= ½ × 12
= 6cm
ST = √3/2 × RT [30° – 60° – 40° theorem]
= √3/2 × 12 = 6√3 cm
(4) Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq. cm.
Solution:
Given, area of rectangle = 192cm2
Length = 16 cm
∴ area = l × b
Or, 192 = 16 × b
Or, b = 192/16 = 12 cm
∴ diagonal2 = length2 + breadth2
Or, Diagonal = √162 + 122
= √256 + 144
= √400
= 20 cm
(5) Find the length of the side and perimeter of an equilateral triangle whose height is √3 cm.
Solution:
Height of equilateral triangle = √3 cm. ABC is the equilateral triangle AM is the height.
We know, height of equilateral triangle bisects the base.
∴ BM = MC = 1/2 × 3C
Let, side of ∆ABC be x
∴ BM = x/2 [∵ AB = BC = AC sides of equilateral triangle]
∴ By Apollonius theorem,
AB2 + AC2 = 2AM2 + 2BM2
Or, x2 + x2 = 2(√3)2 + 2 (x/2)2 [∵ AB = BC = AC]
Or, 2x2 = 2×3 + 2× x/4
Or, 2x2 – x2/2 = 6
Or, 4x2 – x2 = 12
Or, x2 = 12/3 = 4 cm
∴ Length of side of ∆ABC – 4cm
(6) In ∆ABC seg AP is a median. If BC = 18, AB2 + AC2 = 260 Find AP.
Solution:
Given, AP is median, BC = 18
AB2 + AC2 = 200
∴ By Apollonius theorem,
AB2 + AC2 = 2AP2 + 2BP2
Or, 260 = 2AP2 + 2(BC/2)2 [∵ At median]
Or, 260 = 2AP2 + 2 (18/2)2
Or, AP2 = 260-162/2 Or, AP = √49 = 7
(7) ∆ABC is an equilateral triangle. Point P is on base BC such that PC 1/3 BC, if AB = 6 cm find AP.
Solution:
Given, ∆ABC is an equilateral triangle.
PC = 1/3 BC, AB = 6cm
Draw median AM to base BC.
AM ⊥ BC [∵ median drawn on equilateral triangle intersect the base at right angle]
Also, BM = MC = 1/2 BC [∵ AM median]
∴ In ∆, AMC
By Pythagoras theorem,
AC2 = AM2 + MC2
Or, AM2 = AB2 – MC2 [∵ AB = AC sides of equilateral triangle]
Or, AM2 = 62 – (BC/2)2
= 36 – (6/2)2
= 36 – 9
= 27
Now, in ∆AMP by Pythagoras theorem
AP2 = AM2 + MP2
Or, AP2 = 27 + (MC – PC)2 [∵ B – M – P – C]
Or, AP2 = 27 + (BC/2 – BC/3)2 [given]
Or, 27 + (6/2 – 6/3)2
Or, AP2 = 27 + 12
Or, AP = √28
= 2√7
(8) From the information given in the figure 2.31, prove that PM = PN = √3 × a
Solution:
From the given info in the adjacent fig. We can say that ∆PQR is an equilateral ∆ with side a.
PS ⊥ QR so, PS is the median
Since, PQR is equilateral.
QS = SR = 1/2 QR = a/2
MS = MQ + QS
= a + a/2 = 3a/2
SN = SR + RN = a + a/2 = 3a/2
Now, In ∆PQR, by Apollonius theorem,
PQ2 + PR2 = 2PS2 + 2QS2
Or, a2 + a2 = 2PS2 + 2(a/2)2
Or, 2PS2 = 2a2 – a2/4 × 2
Or 2PS2 = 4a2 – a2/2
Or, PS2 = 3a2/2×2 = 3a2/4 —- (i)
Now, In ∆PSM by Pythagoras theorem,
PM2 = PS2 + MS2
Or, PM2 = 3a2/4 + (3a/2)2 [From (i)]
Or, PM2 = 3a2/4 + 9a2/4
Or, PM2 = 12a2/4
Or, PM = √3a2
= a√3
In ∆PSN
By Pythagoras theorem,
PN2 = PS2 + SN2
Or, PN2 = 3a2/4 + (3a/2)2 [From (i)]
Or, PN2 = 3a2/4 + 9a2/4
Or, PN = √12a2/4
Or, PN = a√3
∴ PM = PN = a√3 Proved
(9) Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
Solution:
In the adjacent fig, ABCD is a parallelogram.
AC and BD are diagonals intersecting at O.
We know, diagonals of a parallelogram bisect each other.
∴ AO = OC & BO = OD
Now, AO = OC = 1/2 AC & BO = OD = 1/2 BO.
∴ In ∆BAD
By Apollonius theorem,
AB2 + AD2 = 2AO2 + 2BO2
= 2 (AC/2)2 + 2 (BD/2)2
= 2 × AC/4 + 2 × BD2/4
Or, AB2 + AD2 = AC2/2 + BD2/2 —- (i)
Also, In ∆BCD
BC2 + CD2 = 2OC2 + 2BO2
= 2 × (AC/2)2 + 2 (BD/2)2
= 2 × AC2/4 + 2 BD2/4
Or, BC2 + CD2 = AC2/2 + BD2/2 —- (ii)
Adding equation (i) & (ii) we get,
AB2 + AD2 + BC2 + CD2 = AC2/2 + BD2/2 + AC2/2 + BD2/2
= 2 × AC2/2 + 2 × BD2/2
= AC2 + BD2
Or, AB2 + AD2 + BC2 + CD2
Hence, we can say that sum of squares of all sides of a parallelogram is equal to sum of its squares of its diagonal.
(10) 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√2 km. Find their speed per hour.
Solution:
The adjacent triangle represents the path of Pranali and Prasad and their distance.
Let, their speed be x km/hr.
∴ Distance covered by them in 2 hrs
= x × 2
= 2x km
∴ OA = OB = 2x km [∵ they maintained same speed]
∴ By Pythagoras theorem, distance between them in two hours AB = 15√2 km
OA2 + OB2 = AB2
Or, (2x)2 + (2x)2 = (15√2)2
Or, 4x2 + 4x2 = 225×2
Or, 8x2 = 450
Or, x2 = √450/8 = √56.25 = 7.5
∴ Speed of pranali and prasad per hour = 7.5 km/hr.
(11) In ∆ABC, ∠BAC = 90°, seg BL and CM are medians of ∆ABC. Then prove that:
4 (BL2 + CM2) = 5BC2
Solution:
Given, ∠BAC = 90°
BL & CM are medians of ∆ABC.
∴In ∆ABC
By Apolloginous theorem,
BC2 + AB2 = 2BL2 + 2CL2
Or, BC2 = 2BL2 + 2CL2 – (2AM)2
= 2BL2 + 2CL2 – 4AM2 [∵ CM is median]
Or, 2BL2= BC2 = BC2 – 2CL2 + 4AM2
Or, 4BL2 = 2BC2 – 4CL2 + 8AM2 —- (i)
Also, by Apolloginous theorem
BC2 + AC2 = 2CM2 + AM2
BC2 = 2CM2 + AM2 – (2CL)2
Or, BC2 = 2CM2 + 2AM2 – 4CL2
Or, 2CM2 = BC2 – 2AM2 + 4CL2
Or, 4CM2 = 2BC2 – 4AM2 + 8CL2 —- (ii)
Now Adding equation (i) & (ii) we get,
4BL2 + 4CM2 = 2BC2 – 4CL2 + 8AM2 + 2BC2 – 4AM2 + 8CL
Or, 4BC+ 4CM2 = 4BC2 + 4AM2 + 4CL2 —- (iii)
Now, In ∆ABC also,
By Pythagoras theorem,
BC2 = AC2 + AB2
Or, BC2 = (2CL)2+(2AM)2 [∵ CM & BL are median of side AB and BC respectively]
Or, BC2 = 4CL2 + 4AM2
Or, 4CL2 + 4AM2 = BC2 —- (iv)
Now, Putting the value of 4CL2 + 4AM2 in equation (iii) we get –
4BL2 + 4CM2 = 4BC2 + 4AM2 + 4CL2
Or, 4(BL2 + CM2) = 4BC2 + BC2 [From (iv)]
Or, 4 (BL2 + CM2) = 5BC2 (Proved)
(12) Sum of the squares of adjacent sides of a parallelogram is 130 sq.cm and length of one of its diagonals is14 cm. Find the length of the other diagonal.
Solution:
Given, ABCD is a parallelogram AC & BD are diagonal intersecting at o
We know diagonal of a parallelogram bisect each other.
∴o is midpoint of AC and BD.
Also given,
AB2 + BC2 = 130 cm2, BD = 14cm.
∴In∆ABC, OB is median.
By Apollonious theorem.
AB2 + BC2 = 20B2 + 20A2
Or, 130 = 2 x (14/2) + 20A2
Or, 20A2 = 130 – 98
Or, OA = √32/2 = √16 = 4cm
[∵O midpoint of BD]
∴ AC = 20A
= 2×4 = 8 cm
∴ Length of other diagonal is 8cm.
(13) In ∆ABC, segAD⊥ seg BC
DB = 3CD. Prove that:
2AB2 = 2AC2 + BC2
Solution:
Given, AD⊥BC
DB = 3CD
In ∆ADB,
By Pythagoras theorem,
AD2 + DB2 = AB2
Or, AD2 = AB2 – (3CD)2
Or, AD2 = AB2 – 9CD2 —- (i)
Also, In ∆ADC
AD2 + CD2 = AC2
Or, AD2 = AC2 – CD2 —- (ii)
Comparing equation (i) & (ii) we get.
AB2 – 9CD2 = AC2 – CD2
Or, AB2 – AC2= 9CD2 – 8CD2
Or, AB2 – AC2 = 8CD2
Multiplying 2 on both sides we get.
Or, 2AB2 – 2AC2 = 16CD2
Or, 2AB2 – 2AC2 = (4CD)2
Or, 2AB2 – 2AC2 = BC2 [From equation (iii)]
Or, 2AB2 = 2AC2 + BC2 (Proved)
Also, BC = DB + CD
= 3CD + CD [∵ DB = 3CD](iii)
(14) In an isosceles triangle, length of the congruent sides is 13 cm and its base is 10 cm. Find the distance between the vertex opposite the base and the centroid.
Solution:
In the adjacent fig.
∆ABC is an isosceles triangle with AB = AC = 13 cm
BC = 10cm
AQ, BR & CP are medians of ∆ABC.
AQ is ⊥ to BC since it is the median that meets the base of the congruent sides so it makes 90° angle with the base.
0 is the centroid of ∆ABC.
Now, In ∆AQB
AQ2 + BQ2 = AB2 [By Pythagoras theorem]
Or, AQ2 + (BC/2)2 = 132 [∵ Q is midpoint of BC]
Or, AQ2 + (10/2)2 = 169
Or, AQ2 = 169-52
Or, AQ = √169-25
= √144 = 12 cm
Now, In isosceles triangles the median AQ to base BC is divided by the centroid o in 2:1 ratios it is a property of any isosceles triangle.
∴ AO:OQ = 2:1
Or, AO/OQ = 2/1 or, AO = 20Q
Also, AO + OQ = AQ [∵ A – O – Q]
Or, 20Q ≠ OQ = AQ
Or, OQ = 12/3 = 4
∴ AO = AQ – OQ = 12 – 4 = 8 cm
∴ Distance form vertex opposite to base to the centroid is 8cm
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 2 Pythagoras Theorem Problem Set 2
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