RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQs

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 4 Triangles MCQs Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 4 Triangles MCQs Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQs Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 4
Chapter NameTriangles
ExerciseMCQs
CategoryRS Aggarwal Solutions

RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQs

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 Solutions Class 10 Chapter 4 Triangles MCQ 1.1
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 Solutions Class 10 Chapter 4 Triangles MCQ 2.1
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 Solutions Class 10 Chapter 4 Triangles MCQ 3.1
At the same time, let x cm be the shadow of 6 m long pole.
∆ABC ~ ∆DEF
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 3.2

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 Solutions Class 10 Chapter 4 Triangles MCQ 4.1

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 Solutions Class 10 Chapter 4 Triangles MCQ 5.1

Question 6.
Solution:
(a) Length of ladder AB = 25 m .
Height above the ground = 24 m
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 6.1
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 Solutions Class 10 Chapter 4 Triangles MCQ 9.1

Question 10.
Solution:
(d) In isosceles ∆ABC,
AB = AC = 13 cm
Length of altitude AB, (from A to BC) = 5 cm
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 10.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 10.2

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 Solutions Class 10 Chapter 4 Triangles MCQ 11.1

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 Solutions Class 10 Chapter 4 Triangles MCQ 12.1

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 Solutions Class 10 Chapter 4 Triangles MCQ 13.1

Question 14.
Solution:
(b) In a ∆ABC, AD ⊥ BC and BD = DC
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 14.1
In a ∆ABC, AD = 12 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 = 12 BC
Now, in right ∆ABD,
AB² = BD² + AD² (Pythagoras Theorem)
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 15.1

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 Solutions Class 10 Chapter 4 Triangles MCQ 16.1
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 Solutions Class 10 Chapter 4 Triangles MCQ 17.1
In rhombus ABCD
AO = OC, BO = OD
Let AO = OC = 242 = 12 cm
BO = OD = 102 = 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 Solutions Class 10 Chapter 4 Triangles MCQ 18.1
In quadrilateral ABCD, diagonals AC and BD intersect each-other at O and AOOC = BOOD
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 Solutions Class 10 Chapter 4 Triangles MCQ 19.1

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 Solutions Class 10 Chapter 4 Triangles MCQ 21.1

Question 22.
Solution:
(a) In ∆ABC,
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 22.1
∠B = 70° and ∠C = 50°
But ∠A + ∠B + ∠C = 180° (Angles of a triangle)
∠A = 180° – (∠B + ∠C)
= 180° – (70° + 50°)
= 180° – 120° = 60°
ABAC = BDDC
AD is the bisector of ∠A
∠BAD = 602 = 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 Solutions Class 10 Chapter 4 Triangles MCQ 23.1

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 Solutions Class 10 Chapter 4 Triangles MCQ 24.1

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 Solutions Class 10 Chapter 4 Triangles MCQ 25.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 25.2

Question 26.
Solution:
(d) In ∆ABC, DE || BC
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 26.1

Question 27.
Solution:
(b) ∆ABC ~ ∆DEF
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 27.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 27.2

Question 28.
Solution:
(a) ∆ABC ~ ∆DEF
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 28.1

Question 29.
Solution:
(d) ∆DEF ~ ∆ABC
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 29.1
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 Solutions Class 10 Chapter 4 Triangles MCQ 30.1

Question 31.
Solution:
(b) ∆ABC ~ ∆DFE.
∠A = 30°, ∠C = 50°, AB = 5cm, AC = 8 cm and DF = 7.5 cm
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 31.1

Question 32.
Solution:
(c) In ∆ABC, ∠A = 90°
AD ⊥ BC
In ∆ABD and ∆ADC
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 32.1

Question 33.
Solution:
(c) In ∆ABC, AB = 6 cm, AC = 12 cm and BC = 6 cm.
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 33.1
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, ABDE = BCFD
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 34.1
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 Solutions Class 10 Chapter 4 Triangles MCQ 35.1

Question 36.
Solution:
(c) ∆ABC ~ ∆EDF
∠A = ∠E, ∠B = ∠D, ∠C = ∠F
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 36.1

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 Solutions Class 10 Chapter 4 Triangles MCQ 38.1

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 Solutions Class 10 Chapter 4 Triangles MCQ 39.1

Question 40.
Solution:
(d) Corresponding sides of two similar triangles = 4:9
The areas of there triangle will be in the ratio
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 40.1

Question 41.
Solution:
(d)
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 41.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 41.2

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 Solutions Class 10 Chapter 4 Triangles MCQ 42.1

Question 43.
Solution:
(b)
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 43.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 43.2

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 Solutions Class 10 Chapter 4 Triangles MCQ 47.1

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 Solutions Class 10 Chapter 4 Triangles MCQ 49.1
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 Solutions Class 10 Chapter 4 Triangles MCQ 53.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 53.2
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 53.3
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 53.4
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 53.5

Question 54.
Solution:
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 54.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQ 54.2
correct answer is
(a) → (r)
(b) → (q)
(c) → (p)
(d) → (s)

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