RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQS.

Other Exercises

Choose the correct answer in each of the following questions.
Question 1.
Solution:
(c) A man goes from O to 24 m due west at A and then 10 m due north at B.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 1
Now, AB² = OA² + OB²
= (24)² + (10)² = 576 + 100 = 676 = (26)²
AB = 26 m

Question 2.
Solution:
(b) Two poles AB and CD are standing on the plane ground 8 m apart.
AB = 13 m and CD = 7 m, CE || DB
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 2
In right ∆ACE,
AC² = CE² + AE²
= (8)² + (6)² = 64 + 36 = 100 = (10)²
AC = 10 m
Distance between the tops of poles = 10 m

Question 3.
Solution:
(c) A vertical stick AB = 1.8 m
and its shadow = 45 cm = 0.45 m
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 3
At the same time, let x cm be the shadow of 6 m long pole.
∆ABC ~ ∆DEF
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 4

Question 4.
Solution:
(d) Shadow of a vertical pole 6 m long is 3.6 m on the ground and shadow of a tower at the same, is 18 m.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 5

Question 5.
Solution:
(d) Shadow of 5 m long stick = 2 m
Let shadow of 12.5 m high tree at the same time = x
∆ABC ~ ∆DEF
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 6

Question 6.
Solution:
(a) Length of ladder AB = 25 m .
Height above the ground = 24 m
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 7
Let its foot is x m away from the foot of building.
In right ∆ABC,
AB² = AC² + BC² (Pythagoras Theorem)
(25)² = (24)² + x²
⇒ 625 = 576 + x²
⇒ x² = 625 – 576 = 49 = (7)²
x = 7
Distance = 7 m

Question 7.
Solution:
(b) O is a point inside ∆MNP such that
MOP = 90°, OM = 16 cm, OP = 12 cm.
If MN = 21 cm ∠NMP = 90°, then NP = ?
Let MP = x Now, in right ∆MOP,
∠O = 90°
MP² = OM² + OP² (Pythagoras Theorem)
= (16)² + (12)² = 256 + 144 = 400 = (20)²
MP = 20 cm
Now, in right ∆MNP, ∠M = 90°
NP² = MN² + MP²
= (21)² + (20)² = 441 + 400 = 841 = (29)²
NP = 29 cm

Question 8.
Solution:
(b) Let ∆ABC is a right angled triangle with ∠B = 90°
AC = 25 cm
Let one side AB of the other two sides = x cm
then second side BC = (x + 5) cm
According to the Pythagoras Theorem,
AC² = AB² + BC²
(25)² = x² + (x + 5)²
625 = x² + x² + 10x + 25
⇒ 2x² + 10x + 25 – 625 = 0
⇒ 2x² + 10x – 600 = 0
⇒ x² + 5x – 300 = 0
⇒ x² + 20x – 15x – 300 = 0
⇒ x (x + 20) – 15 (x + 20) = 0
⇒ (x + 20)(x – 15) = 0
Either x + 20 = 0, then x = -20 which is not possible being negative,
or x – 15 = 0, then x = 15
First side = 15 cm
and second side = 15 + 5 = 20 cm

Question 9.
Solution:
(b) Side of an equilateral triangle = 12 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 8

Question 10.
Solution:
(d) In isosceles ∆ABC,
AB = AC = 13 cm
Length of altitude AB, (from A to BC) = 5 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 9
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 10

Question 11.
Solution:
(a) In the given figure,
AB = 6 cm, AC = 8 cm
AD is the bisector of ∠A which meets BC at D.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 11

Question 12.
Solution:
(d) In the given figure,
AD is the internal bisector of ∠A
BD = 4 cm, DC = 5 cm, AB = 6 cm
Let AC = x cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 12

Question 13.
Solution:
(b) In the given figure,
AD is the bisector of ∠A of ∆ABC.
AB = 10 cm, AC = 14 cm and BC = 6 cm
Let CD = x cm
Then BD = (6 – x) cm
Now, AD is the bisector of ∠A
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 13

Question 14.
Solution:
(b) In a ∆ABC, AD ⊥ BC and BD = DC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 14
In a ∆ABC, AD = \(\frac { 1 }{ 2 }\) BC and BD = DC.
In right ∆ABD and ∆ACD
AD = AD (common)
∠ABD = ∠ADC (each 90°)
BD = DC (given)
∆ABD = ∆ACD (SAS axiom)
AB = AC
∆ABC is an isosceles triangle.

Question 15.
Solution:
(c) In equilateral ∆ABC, AD ⊥ BC
Then BD = DC = \(\frac { 1 }{ 2 }\) BC
Now, in right ∆ABD,
AB² = BD² + AD² (Pythagoras Theorem)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 15

Question 16.
Solution:
(c) In a rhombus, each side = 10 cm and one diagonal = 12 cm
AB = BC = CD = DA = 10 cm BD = 12 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 16
The diagonals of a rhombus bisect each other at right angles.
In ∆AOB,
AB² = AO² + BO²
⇒ (10)² = AO² + (6)²
⇒ AO² = (10)² – (6)² = 100 – 36 = 64 = 8²
AO = 8 cm
Diagonals AC = 2 x AO = 2 x 8 = 16 cm

Question 17.
Solution:
(b) Length of diagonals of a rhombus are 24 cm and 10 cm.
The diagonals of a rhombus bisect each other at right angles.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 17
In rhombus ABCD
AO = OC, BO = OD
Let AO = OC = \(\frac { 24 }{ 2 }\) = 12 cm
BO = OD = \(\frac { 10 }{ 2 }\) = 5 cm
Now, in right ∆AOB,
AB² = AO² + BO² (Pythagoras Theorem)
= (12)² + (5)² = 144 + 25 = 169 = (13)²
AB = 13
Each side of rhombus = 13 cm

Question 18.
Solution:
(b) Diagonals of e. quadrilateral divides each other proportionally, then it is
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 18
In quadrilateral ABCD, diagonals AC and BD intersect each-other at O and \(\frac { AO }{ OC }\) = \(\frac { BO }{ OD }\)
Then, quadrilateral ABCD is a trapezium.

Question 19.
Solution:
(a) In the given figure,
ABCD is a trape∠ium and its diagonals AC
and BD intersect at O.
and OA = (3x – 1) cm OB = (2x + 1) cm, OC and OD = (6x – 5) cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 19

Question 20.
Solution:
(a) The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram

Question 21.
Solution:
(c) If the bisector of angle of a triangle bisects the opposite side of a triangle.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 20

Question 22.
Solution:
(a) In ∆ABC,
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 21
∠B = 70° and ∠C = 50°
But ∠A + ∠B + ∠C = 180° (Angles of a triangle)
∠A = 180° – (∠B + ∠C)
= 180° – (70° + 50°)
= 180° – 120° = 60°
\(\frac { AB }{ AC }\) = \(\frac { BD }{ DC }\)
AD is the bisector of ∠A
∠BAD = \(\frac { 60 }{ 2 }\) = 30°

Question 23.
Solution:
(b) In ∆ABC, DE || BC
AD = 2.4 cm, AE = 3.2 cm, EC = 4.8 cm
Let AD = x cm
DE || BC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 22

Question 24.
Solution:
(b) In ∆ABC, DE || BC
AB = 7.2 cm, AC = 6.4 cm, AD = 4.5 cm
Let AE = x cm
DE || BC
∆ADE ~ ∆ABC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 23

Question 25.
Solution:
(c) In ∆ABC, DE || BC
AD = (7x – 4) cm, AE = (5x – 2) cm DB = (3x + 4) cm and EC = 3x cm
In ∆ABC, DE || BC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 24
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 25

Question 26.
Solution:
(d) In ∆ABC, DE || BC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 26

Question 27.
Solution:
(b) ∆ABC ~ ∆DEF
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 27
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 28

Question 28.
Solution:
(a) ∆ABC ~ ∆DEF
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 29

Question 29.
Solution:
(d) ∆DEF ~ ∆ABC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 30
Perimeter of ∆DEF = DE + EF + DF
= 12 + 8 + 10 = 30 cm

Question 30.
Solution:
(d) ABC and BDE are two equilateral triangles such that D is the midpoint of BC.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 31

Question 31.
Solution:
(b) ∆ABC ~ ∆DFE.
∠A = 30°, ∠C = 50°, AB = 5cm, AC = 8 cm and DF = 7.5 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 32

Question 32.
Solution:
(c) In ∆ABC, ∠A = 90°
AD ⊥ BC
In ∆ABD and ∆ADC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 33

Question 33.
Solution:
(c) In ∆ABC, AB = 6 cm, AC = 12 cm and BC = 6 cm.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 34
Longest side (AC)2 = (12)2 = 144
AB2 + BC2 = (6√3)2 + (6)2 = 108 + 36 = 144
AC2 = AB2 + BC2 (Converse of Pythagoras Theorem)
∠B = 90°

Question 34.
Solution:
(c) In ∆ABC and ∆DEF, \(\frac { AB }{ DE }\) = \(\frac { BC }{ FD }\)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 34
For similarity,
Here, included angles must be equal and these
are ∠B = ∠D.

Question 35.
Solution:
(b) In ∆DEF and ∆PQR,
∠D = ∠Q and ∠R = ∠E
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 35

Question 36.
Solution:
(c) ∆ABC ~ ∆EDF
∠A = ∠E, ∠B = ∠D, ∠C = ∠F
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 36

Question 37.
Solution:
(b) In ∆ABC and ∆DEF,
∠B = ∠E, ∠F = ∠C and AB = 3DE
The triangles are similar as two angles are equal but including sides are not proportional.

Question 38.
Solution:
(a)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 37

Question 39.
Solution:
(d) In the given figure, two line segments AC and BD intersect each other at P such that
PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°
In ∆ABP and ∆CPD,
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 37

Question 40.
Solution:
(d) Corresponding sides of two similar triangles = 4:9
The areas of there triangle will be in the ratio
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 39

Question 41.
Solution:
(d)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 40
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 41

Question 42.
Solution:
(b) In the given figure,
∆ABC is an equilateral triangle.
D is midpoint of AB and E is the midpoint of AC.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 42

Question 43.
Solution:
(b)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 43
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 44

Question 44.
Solution:
(b) ∆ABC ~ ∆DEF
ar (∆ABC) = 36 cm² and ar (∆DEF) = 49 cm²
i.e. areas are in the ratio 36 : 49
Ratio in their corresponding sides = √36 : √49 = 6 : 7

Question 45.
Solution:
(c) Two isosceles triangles have their corresponding angles equal and ratio in their areas is 25 : 36.
The ratio in their corresponding altitude
(heights) = √25 : √36 = 5 : 6 (∆s are similar)

Question 46.
Solution:
(b) The line segments joining the midpoints of a triangle form 4 triangles which are similar to the given (original) triangle.

Question 47.
Solution:
(b) ∆ABC ~ ∆QRP
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 45

Question 48.
Solution:
(c) In the given figure, O is the point of intersection of two chords AB and CD.
OB = OD and ∠AOC = 45°
∠B = ∠D (Angles opposite to equal sides)
∠A = ∠D, ∠C = ∠B (Angles in the same segment)
and ∠AOC = ∠BOD = 45° each
∆OAC ~ ∆ODB (AAA axiom)
OA = OC (Sides opposite to equal angles)
∆OAC and ∆ODB are isosceles and similar.

Question 49.
Solution:
(d) In an isosceles ∆ABC,
AC = BC
⇒ AB² = 2 AC²
⇒ AB² = AC² + AC²
⇒ AB² = AC² + BC² (AC = BC)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 46
Converse of the Pythagoras Theorem,
∆ABC is a right triangle and angle opposite to AB = 90°
∠C = 90°

Question 50.
Solution:
(b) In ∆ABC,
AB = 16 cm, BC = 12 cm and AC = 20 cm
(Longest side)2 = 20² = 400
Sum of square on other sides = AB² + BC²
= 162 + 122 = 256 + 144 = 400
AC² = AB² + BC²
∆ABC is a right triangle.

True/False type
Question 51.
Solution:
(c) (a) False. Not always congruent.
(b) False. Two similar figures are similar if they have same shape, not size in every case.
(c) True.
(d) False. Not in each case.

Question 52.
Solution:
(a) True
(b) False, as ratio of the areas of the two similar triangles is equal to the ratio of the square of their corresponding sides.
(c) True
(d) True

Question 53.
Matching of columns : (2 marks)
Solution:
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 47
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 48
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 49
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 50
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 51

Question 54.
Solution:
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 52
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 53
correct answer is
(a) → (r)
(b) → (q)
(c) → (p)
(d) → (s)

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQS are helpful to complete your math homework.

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