Triangles RD Sharma Class 10 Solutions Exercise 4.6

Triangles RD Sharma Class 10 Solutions Exercise 4.6

RD Sharma Class 10 Solutions Triangles Exercise 4.6

Question 1.
Triangles ABC and DEF are similar.
(i) If area (∆ABC) = 16 cm², area (∆DEF) = 25 cm² and BC = 2.3 cm, find EF. (C.B.S.E. 1992)
(ii) If area (∆ABC) = 9 cm², area (∆DEF) = 64 cm² and DE = 5.1 cm, find AB.
(iii) If AC = 19 cm and DF = 8 cm, find the ratio of the area of two triangles. (C.B.S.E. 1992C)
(iv) If area (∆ABC) = 36 cm², area (∆DEF) = 64 cm² and DE = 6.2 cm, find AB. (C.B.S.E. 1992)
(v) If AB = 1.2 cm and DE = 1.4 cm, find the ratio of the areas of ∆ABC and ∆DEF. (C.B.S.E. 1991C)
Solution:

Question 2.
In the figure, ∆ACB ~ ∆APQ. If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. Also, find the area (∆ACB) : area (∆APQ).

Solution:

Question 3.
The areas of two similar triangles are 81 cm² and 49 cm² respectively. Find the ratio of their corresponding heights, what is the ratio of their corresponding medians ?
Solution:
Areas of two similar triangles are 81 cm² and 49 cm²
The ratio of the areas of two similar triangles are proportion to the square of their corresponding altitudes and also squares of their corresponding medians
Ratio in their altitudes = √81 : √49 = 9 : 7
Similarly, the ratio in their medians = √81 : √49 = 9 : 7

Question 4.
The areas of two similar triangles are 169 cm² and 121 cm² respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.
Solution:
Triangles are similar Area of larger triangle = 169 cm²
and area of the smaller triangle =121 cm²
Length of longest sides of the larger triangles = 26 cm
Let the length of longest side of the smaller triangle = x

Question 5.
The areas of two similar triangles are 25 cm² and 36 cm² respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.
Solution:
Area of first triangle = 25 cm²
Area of second = 36 cm²
Altitude of the first triangle = 2.4 cm
Let altitude of the second triangle = x
The triangles are similar

Question 6.
The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.
Solution:
Length of the corresponding altitude of two triangles are 6 cm and 9 cm
triangles are similar

Question 7.
ABC is a triangle in which ∠A = 90°, AN ⊥ BC, BC = 12 cm and AC = 5 cm. Find the ratio of the areas of the ∆ANC and ∆ABC.
Solution:
In ∆ABC, ∠A = 90°
AN ⊥ BC
BC = 12 cm, AC = 5 cm

Question 8.
In the figure, DE || BC
(i) If DE = 4 cm, BC = 6 cm and area (∆ADE) = 16 cm², find the area of ∆ABC.
(ii) If DE = 4 cm, BC = 8 cm and area of (∆ADE) = 25 cm², find the area of ∆ABC. (C.B.S.E. 1991)
(iii) If DE : BC = 3 : 5, calculate the ratio of the areas of ∆ADE and the trapezium BCED

Solution:

Question 9.
In ∆ABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ∆ADE and ∆ABC.
Solution:
In ∆ABC, D and E are the mid points of AB and AC respectively

Question 10.
The areas of two similar triangles are 100 cm² and 49 cm² respectively. If the altitude of the bigger triangle is 5 cm, find the corresponding altitude of the other. (C.B.S.E. 2002)
Solution:
∆ABC ~ ∆DEF
area ∆ABC = 100 cm²
and area ∆DEF = 49 cm²

Question 11.
The areas of two similar triangles are 121 cm² and 64 cm² respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other. (C.B.S.E. 2001)
Solution:
∆ABC ~ ∆DEF
area of ∆ABC = 121 cm² area of ∆DEF = 64 cm²
AL and DM are the medians of ∆ABC and ∆DEF respectively
AL = 12.1 cm

Question 12.
In ∆ABC ~ ∆DEF such that AB = 5 cm and (∆ABC) = 20 cm² and area (∆DEF) = 45 cm², determine DE.
Solution:
∆ABC ~ ∆DEF
area (∆ABC) = 20 cm²
area (∆DEF) = 45 cm²

Question 13.
In ∆ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides ∆ABC into two parts equal in area. Find \(\frac { BP }{ AB }\).
Solution:
In ∆ABC, PQ || BC and PQ divides ∆ABC in two parts ∆APQ and trap. BPQC of equal in area
i.e., area ∆APQ = area BPQC

Question 14.
The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR. (C.B.S.E. 2004)
Solution:
∆ABC ~ ∆PQR
area (∆ABC) : area (∆PQR) = 9 : 16
and BC = 4.5 cm

Question 15.
ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of ∆APQ is one sixteenth of the area of ∆ABC. (C.B.S.E. 2005)
Solution:
In ∆ABC, P and Q are two points on AB and AC respectively such that
AP = 1 cm, PB = 3 cm, AQ = 1.5 cm and QC = 4.5 cm

Question 16.
If D is a point on the side AB of ∆ABC such that AD : DB = 3 : 2 and E is a point on BC such that DE || AC. Find the ratio of areas of ∆ABC and ∆BDE. (C.B.S.E. 2006C)
Solution:
In ∆ABC, D is a point on AB such that AD : DB = 3 : 2

Question 17.
If ∆ABC and ∆BDE are equilateral triangles, where D is the mid point of BC, find the ratio of areas of ∆ABC and ∆BDE. [CBSE 2010]
Solution:
∆ABC and ∆DBE are equilateral triangles Where D is mid point of BC

Question 18.
Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.
Solution:
Two isosceles triangles have equal vertical angles
So their base angles will also be the equal to each other
Triangles will be similar Now, ratio in their areas = 36 : 25

Question 19.
In the figure, ∆ABC and ∆DBC are on the same base BC. If AD and BC intersect

Solution:
Given : Two ∆ABC and ∆DBC are on the same base BC as shown in the figure
AC and BD intersect eachother at O

Question 20.
ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove that
(i) ∆AOB ~ ∆COD
(ii) If OA = 6 cm, OC = 8 cm, find

Solution:
Given : ABCD is a trapezium in which AB || CD
Diagonals AC and BD intersect each other at O

Question 21.
In ∆ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ∆APQ and trapezium BPQC.
Solution:
In ∆ABC, P is a point on AB such that AP : PQ = 1 : 2
PQ || BC
Now we have to find the ratio between area ∆APQ and area trap BPQC

Question 22.
AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Prove that Area (∆ADE) : Area (∆ABC) = 3 : 4. [CBSE 2010]
Solution:
Given: In equilateral ∆ABC, AD ⊥ BC and with base AD, another equilateral ∆ADE is constructed

Triangles RD Sharma Class 10 Solutions Exercise 4.6

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RD Sharma Class 10 Solutions

• Chapter 4 Triangles Exercise 4.1
• Chapter 4 Triangles Exercise 4.2
• Chapter 4 Triangles Exercise 4.3
• Chapter 4 Triangles Exercise 4.4
• Chapter 4 Triangles Exercise 4.5
• Chapter 4 Triangles Exercise 4.6
• Chapter 4 Triangles Exercise 4.7