RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4e

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 4 Triangles EX 4e 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 EX 4e Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 4 Triangles EX 4e Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 4
Chapter NameTriangles
Exercise4e
CategoryRS Aggarwal Solutions

RS Aggarwal Solutions Class 10 Chapter 4 Triangles EX 4e

Question 1.
Solution:
Two properties for similarity of two triangles are:
(i) Angle-Angle-Angle (AAA) property.
(ii) Angle-Side-Angle (ASA) property.

Question 2.
Solution:
In a triangle, if a line parallel to one side is drawn, it will divide the other two sides proportionally.

Question 3.
Solution:
If a line divides any two sides of a triangle in the same ratio. Then, the line must be parallel to the third side.

Question 4.
Solution:
The line joining the midpoints of two sides of a triangle, is parallel to the third side.

Question 5.
Solution:
In two triangles, if three angles of the one triangle are equal to the three angles of the other, the triangles are similar.

Question 6.
Solution:
In two triangles, if two angles of the one triangle are equal to the corresponding angles of the other triangle, then the triangles are similar.

Question 7.
Solution:
In two triangles, if three sides of the one are proportional to the corresponding sides of the other, the triangles are similar.

Question 8.
Solution:
In two triangles, if two sides of the one are proportional to the corresponding sides of the other and their included angles are equal, the two triangles are similar.

Question 9.
Solution:
In a right angled triangle, the square on the hypotenuse is equal to the sum of squares on the other two sides.

Question 10.
Solution:
In a triangle, if the square on the longest side is equal to the sum of the squares on the other two sides, then the angle opposite to the hypotenuse is a right angle.

Question 11.
Solution:
The ratio of their areas will be 1 : 4.

Question 12.
Solution:
In two triangles ∆ABC and ∆PQR,
AB = 3 cm, AC = 6 cm, ∠A = 70°
PR = 9 cm, ∠P = 70° and PQ= 4.5 cm
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 12.1

Question 13.
Solution:
∆ABC ~ ∆DEF
2AB = DE, BC = 6 cm
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 13.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 13.2

Question 14.
Solution:
In the given figure,
DE || BC
AD = x cm, DB = (3x + 4) cm
AE = (x + 3) cm and EC = (3x + 19) cm
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 14.1

Question 15.
Solution:
AB is the ladder and A is window.
AB =10 m, AC = 8 m
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 15.1
We have to find the distance of BC
Let BC = x m
In right ∆ABC,
AB² = AC² + BC² (Pythagoras Theorem)
(10)² = 8² + x²
⇒ 100 = 64 + x²
⇒ x² = 100 – 64 = 36 = (6)²
x = 6
Distance between foot of ladder and base of the wall = 6 m.

Question 16.
Solution:
∆ABC is an equilateral triangle with side = 2a cm
AD ⊥ BC
and AD bisects BC at D
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 16.1
Now, in right ∆ABD,
AB² = AD² + BD² (Pythagoras Theorem)
⇒ (2a)² = AD² + (a)²
⇒ 4a² = AD² + a²
⇒ AD² = 4a² – a² = 3a²
AD = √3 a² = √3 a cm

Question 17.
Solution:
Given : ∆ABC ~ ∆DEF
and ar (∆ABC) = 64 cm²
and ar (∆DEF) = 169 cm², BC = 4 cm.
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 17.1

Question 18.
Solution:
In trapezium ABCD,
AB || CD
AB = 2CD
Diagonals AC and BD intersect each other at O
and area(∆AOB) = 84 cm².
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 18.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 18.2

Question 19.
Solution:
Let ∆ABC ~ ∆DEF
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 19.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 19.2

Question 20.
Solution:
In an equiangular ∆ABC,
AB = BC = CA = a cm.
Draw AD ⊥ BC which bisects BC at D.
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 20.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 20.2

Question 21.
Solution:
ABCD is a rhombus whose sides are equal and diagonals AC and BD bisect each other at right angles.
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 21.1
∠AOB = 90° and AO = OC, BO = OD
AO = 242 = 12 cm
and BO = 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 22.
Solution:
∆DEF ~ ∆GHK
∠D = ∠G = 48°
∠E = ∠H = 57°
∠F = ∠K
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 22.1
Now, in ∆DEF,
∠D + ∠E + ∠F = 180° (Angles of a triangle)
⇒ 48° + 57° + ∠F = 180°
⇒ 105° + ∠F= 180°
⇒ ∠F= 180°- 105° = 75°

Question 23.
Solution:
In the given figure,
In ∆ABC,
MN || BC
AM : MB = 1 : 2
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 23.1

Question 24.
Solution:
In ∆BMP,
PB = 5 cm, MP = 6 cm and BM = 9 cm
and in ∆CNR, NR = 9 cm
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 24.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 24.2

Question 25.
Solution:
In ∆ABC,
AB = AC = 25 cm
BC = 14 cm
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 25.1
AD ⊥ BC which bisects the base BC at D.
BD = DC = 142 = 7 cm
Now, in right ∆ABD,
AB² = AD² + BD² (Pythagoras Theorem)
(25)² = AD² + 7²
625 = AD² + 49
⇒ AD² = 625 – 49 = 576
⇒ AD² = 576 = (24)²
AD = 24 cm
Length of altitude = 24 cm

Question 26.
Solution:
A man goes 12 m due north of point O reaching A and then 37 m due west reaching B.
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 26.1
Join OB,
In right ∆OAB,
OB² = OA² + AB² (Pythagoras Theorem)
= (12)² + (35)² = 144 + 1225 = 1369 = (37)²
OB = 37 m
The man is 37 m away from his starting point.

Question 27.
Solution:
In ∆ABC, AD is the bisector of ∠A which meets BC at D.
AB = c, BC = a, AC = b
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 27.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 27.2

Question 28.
Solution:
In the given figure, ∠AMN = ∠MBC = 76°
p, q and r are the lengths of AM, MB and BC Express the length of MN in terms of p, q and r.
In ∆ABC,
∠AMN = ∠MBC = 76°
But there are corresponding angles
MN || BC
∆AMN ~ ∆ABC
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 28.1

Question 29.
Solution:
In rhombus ABCD,
Diagonals AC and BD bisect each other at O at right angles.
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 29.1
AO = OC = 402 = 20 cm
BO = OD = 422 = 21 cm
Now, in right ∆AOB,
AB2 = AO2 + BO2 = (20)2 + (21)2 = 400 + 441 = 841
AB = √841 cm = 29 cm
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 29.2
Each side of rhombus = 29 cm

For each of the following statements state whether true (T) or false (F):
Question 30.
Solution:
(i) True.
(ii) False, as sides will not be proportion in each case.
(iii) False, as corresponding sides are proportional, not necessarily equal.
(iv) True.
(v) False, in ∆ABC,
AB = 6 cm, ∠A = 45° and AC = 8 cm
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 30.1
(vi) False, the polygon joining the midpoints of a quadrilateral is not a rhombus but it is a parallelogram.
(vii) True.
(viii) True.
(ix) True, O is any point in rectangle ABCD then
OA² + OC² = OB² + OD² is true.
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4E 30.2
(x) True.

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