RD Sharma Class 10 Solutions Chapter 4 VSAQs (Updated for 2021-22)

RD Sharma Class 10 Solutions Chapter 4 VSAQs

RD Sharma Class 10 Solutions Chapter 4 VSAQs: Students can download the RD Sharma Class 10 Solutions Chapter 4 VSAQs PDF to learn how to solve the questions in this exercise correctly. Students wishing to brush up on their concepts can check the RD Sharma Class 10 Solutions.

Access RD Sharma Class 10 Solutions Chapter 4 VSAQs PDF

Answer each of the following questions either in one word or one sentence or as per the requirement of the questions :
Question 1.
State basic proportionality theorem and its converse.
Solution:
Basic Proportionality Theorem: If a line is drawn parallel to one side of a triangle intersects the other two sides in distinct points, then the other two sides are divided in the same ratio.
Conversely: In a line that divides any two sides of a triangle in the same ratio, then the line is parallel to the third side of the triangle.

Question 2.
In the adjoining figure, find AC.
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 1
Solution:
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 2

Question 3.
In the adjoining figure, if AD is the bisector of ∠A, what is AC?
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 3
Solution:
In the figure, AD is the angle bisector of ∠A of ∆ABC
AB = 6 cm, BC = 3 cm, DC = 2 cm
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 4

Question 4.
State AAA similarity criterion.
Solution:
If in two triangles, corresponding angles are respectively equal then the triangles are similar.

Question 5.
State SSS similarity criterion.
Solution:
If in two triangles, corresponding sides are in the same ratio, then the two triangles are similar.

Question 6.
State SAS similarity criterion.
Solution:
If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, then the triangles are similar.

Question 7.
In the adjoining figure, DE is parallel to BC and AD = 1 cm, BD = 2 cm. What is the ratio of the area of ∆ABC to the area of ∆ADE?
Solution:
In ∆ABC, DE || BC
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 5

Question 8.
In the figure given below, DE || BC. If AD = 2.4 cm, DB = 3.6 cm and AC = 5 cm. Find AE.
Solution:
In ∆ABC, DE || BC
AD = 2.4 cm, DB = 3.6 cm, AC = 5 cm
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 6
x = 2
AE = 2 cm

Question 9.
If the areas of two similar triangles ABC and PQR are in the ratio 9: 16 and BC = 4.5 cm, what is the length of QR?
Solution:
∆ABC ~ ∆PQR
Area of ∆ABC : area of ∆PQR = 9 : 16
BC = 4.5 cm
Let QR = A
The area of two similar triangles are in the ratio of the squares of their corresponding sides
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 7

Question 10.
The areas of two similar triangles are 169 cm² and 121 cm² respectively. If the longest side of the larger triangle is 26 cm, what is the length of the longest side of the smaller triangle?
Solution:
Let ∆ABC be the larger triangle and ∆PQR
be the smaller triangle and their longest sides are BC and QR respectively
Area of ∆ABC = 169 cm²
area of ∆PQR = 121 cm²
BC = 26 cm
Let QR = x cm
∆ABC ~ ∆PQR
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 8
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 9

Question 11.
If ABC and DEF are similar triangles such that ∠A = 57° and ∠E = 73°, what is the measure of ∠C?
Solution:
∆ABC ~ ∆DEF
Their corresponding angles are equal ∠A = ∠D, ∠B = ∠E and ∠C = ∠F
But ∠A = 57°
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 10

Question 12.
If the altitude of two similar triangles are in the ratio 2 : 3, what is the ratio of their areas?
Solution:
Let ∆ABC ~ ∆PQR
and let AL ⊥ BC and PM ⊥ QR
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 11

Question 13.
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 12
Solution:
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 13

Question 14.
If ∆ABC and ∆DEF are similar triangles such that AB = 3 cm, BC = 2 cm, CA = 2.5 cm and EF = 4 cm, write the perimeter of ∆DEF.
Solution:
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 14
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 15
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 16

Question 15.
State Pythagoras Theorem and its converse.
Solution:
In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Conversely: If in a triangle, the square of one side is equal to the sum of squares of the remaining two sides, then the angle opposite to the first side is a right angle.

Question 16.
The lengths of the diagonals of a rhombus are 30 cm and 40 cm. Find the side of the rhombus. (C.B.S.E. 2008)
Solution:
In rhombus ABCD, BD = 30 cm, AC = 40 cm
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 17

Question 17.
In figure, PQ || BC and AP : PB = 1 : 2.
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 18
Solution:
In ∆ABC, PQ || BC
∆APQ ~ ∆ABC
But AP : PB = 1 : 2
The ratio of the areas of two similar triangles are proportional to the square of their corresponding sides
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 19

Question 18.
In the figure, S and T are points on the sides PQ and PR respectively of ∆PQR such that PT = 2 cm, TR = 4 cm, and ST is parallel to QR. Find the ratio of the areas of ∆PST and ∆PQR. [CBSE 2010]
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 20
Solution:
In ∆PQR, ST || QR and PT = 2 cm, TR = 4 cm
PR = PT + TR = 2 + 4 = 6 cm
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 21
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 22

Question 19.
In the figure, ∆AHK is similar to ∆ABC. If AK = 10 cm, BC = 3.5 cm and HK = 7 cm, find AC. [CBSE 2010]
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 23
Solution:
In the given figure, AK = 10 cm, BC = 3.5 cm, HK = 7 cm
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 24

Question 20.
In the figure, DE || BC in ∆ABC such that BC = 8 cm, AB = 6 cm and DA = 1.5 cm. Find DE.
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 25
Solution:
In the given figure,
DE || BC
BC = 8 cm, AB = 6 cm and DA = 1.5 cm
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 26

Question 21.
In the figure, DE || BC and AD = 12 BD. If BC = 4.5 cm, find DE. [CBSE 2010]
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 27
Solution:
In the given figure,
DE || BC
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 28

Question 22.
In the figure, ∠M = ∠N = 46°. Express x in terms of a, b, and c where a, b, c are lengths of LM, MN, and NK respectively.
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 29
Solution:
In the figure, ∠M = ∠N = 46°
∠M = a, PN = x, MN = b, NK = c
∠M = ∠N = 46°
But there are corresponding angle
PN || ML
∆PKN ~ ∆LKM
RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS 30

We have provided complete details of RD Sharma Class 10 Solutions Chapter 4 VSAQs. If you have any queries related to CBSE Class 10, feel free to ask us in the comment section below.

FAQs on RD Sharma Class 10 Solutions Chapter 4 VSAQs

Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 4 VSAQs?

Yes, learning all of the concepts included in RD Sharma Class 10 Solutions Chapter 4 VSAQs is required in order to achieve high scores on the Class 10 board exams. These solutions were created by subject matter specialists who compiled model questions that covered all of the textbook’s exercise questions.

Where can I download RD Sharma Class 10 Solutions Chapter 4 VSAQs free PDF?

You can download RD Sharma Class 10 Solutions Chapter 4 VSAQs free PDF from the above article.

What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 VSAQs?

1. Correct answers according to the latest CBSE guidelines and syllabus.
2. The RD Sharma Class 10 Solutions Chapter 4 VSAQs are written in simple language to assist students in their board examination, & competitive examination preparation.

Leave a Comment