Similarity Class 10 ICSE ML Aggarwal Chapter Test

ML Aggarwal Class 10 Solutions Similarity Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test.

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Question 1.
In the given figure, ∠1 = ∠2 and ∠3 = ∠4. Show that PT x QR = PR x ST.
Solution:
Given : In the given figure,
∠1 = ∠1 and ∠3 = ∠4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q1.1
To prove : PT x QR = PR x ST
Proof: ∠1 = ∠2
Adding ∠6 to both sides
∠1 + ∠6 = ∠2 + ∠6
∠SPT = ∠QPR
In ∆PQR and ∆PST
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q1.2

Question 2.
In the adjoining figure, AB = AC. If PM ⊥ AB and PN ⊥ AC, show that PM x PC = PN x PB.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q2.1
Solution:
Given : In the given figure,
AB = AC, PM ⊥ AB and PN ⊥ AC
To prove : PM x PC = PN x PB
Proof: In ∆ABC, AB = AC
∠B = ∠C
Now in ∆CPN and ∆BPM,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q2.2

Question 3.
(a) In the figure (1) given below. ∠AED = ∠ABC. Find the values of x and y.
(b) In the fig. (2) given below, CD = \(\\ \frac { 1 }{ 2 } \) AC, B is mid-point of AC and E is mid-point of DF. If BF || AG, prove that :
(i) CE || AG
(ii) 3 ED = GD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q3.1
Solution:
(a) Given : In following figure, ∠AED = ∠ABC
Required : The values of x and y.
Now, in ∆ABC and ∆ADE
∠AED = ∠ABC (given)
∠A = ∠A (common)
∴ ∆ABC ~ ∆ADE
(By A.A. axiom of similarity)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q3.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q3.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q3.4

Question 4.
In the given figure, 2 AD = BD, E is mid-point of BD and F is mid-point of AC and EC || BH. Prove that:
(i) DF || BH
(ii) AH = 3 AF.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q4.1
Solution:
Given : E is the mid-point of BD and F is mid-point of AC also 2 AD = BD and EC || BH
To Prove : (i) DF || BH
(ii) AH = 3 AF
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q4.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q4.3

Question 5.
In a ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE.
Solution:
Given : In ∆ABC, D and E are the points on the sides AB and AC respectively
DE || BC
AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm, BC = 5 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q5.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q5.2

Question 6.
In a ∆ABC, D and E are points on the sides AB and AC respectively such that AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm. Is DE || BC? Justify your answer.
Solution:
In ∆ABC, D and E are points on the sides AB and AC respectively
AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and AC = 8.8 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q6.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q6.2

Question 7.
In a ∆ABC, DE is parallel to the base BC, with D on AB and E on AC. If \(\frac { AD }{ DB } =\frac { 2 }{ 3 } ,\frac { BC }{ DE } \)
Solution:
In ∆ABC, DE || BC
D is on AB and E is on AC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q7.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q7.2

Question 8.
If the area of two similar triangles are 360 cm² and 250 cm² and if one side of the first triangle is 8 cm, find the length of the corresponding side of the second triangle.
Solution:
Let ∆ABC and ∆DEF are similar and area of
∆ABC = 360 cm²
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q8.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q8.2

Question 9.
In the adjoining figure, D is a point on BC such that ∠ABD = ∠CAD. If AB = 5 cm, AC = 3 cm and AD = 4 cm, find
(i) BC
(ii) DC
(iii) area of ∆ACD : area of ∆BCA.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q9.1
Solution:
In ∆ABC and ∆ACD
∠C = ∠C (Common)
∠ABC = ∠CAD (given)
∴ ∆ABC ~ ∆ACD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q9.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q9.3

Question 10.
In the adjoining figure the diagonals of a parallelogram intersect at O. OE is drawn parallel to CB to meet AB at E, find area of DAOE : area of ||gm ABCD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q10.1
Solution:
(a) In the figure
Diagonals of parallelogram ABCD are AC and BD which intersect each other at O.
OE is drawn parallel to CB to meet AB in E.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q10.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q10.3

Question 11.
In the given figure, ABCD is a trapezium in which AB || DC. If 2AB = 3DC, find the ratio of the areas of ∆AOB and ∆COD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q11.1
Solution:
In the given figure, ABCD is trapezium in
which AB || DC, 2AB = 3DC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q11.2

Question 12.
In the adjoining figure, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O and meet AB (produced) at F. Prove that .
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q12.1
(i) DO : OE = 2 : 1
(ii) area of ∆OEC : area of ∆OAD = 1 : 4.
Solution:
Given : In || gm ABCD,
E is mid point of BC and DE meets the diagonal AC at O and meet AB produced at F.
To prove : (i) DO : OE = 2 : 1
(ii) area of ∆OEC : area of ∆OAD = 1 : 4
Proof: In ∆AOD and ∆EDC
∠AOD = ∠EOC (vertically opposite angle)
∠OAD = ∠OCB (alt. angles)
∆AOD ~ ∆EOC (AA postulate)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q12.2

Question 13.
A model of a ship is made to a scale of 1 : 250. Calculate :
(i) the length of the ship, if the length of model is 1.6 m.
(ii) the area of the deck of the ship, if the area of the deck of model is 2.4 m².
(iii) the volume of the model, if the volume of the ship is 1 km³.
Solution:
Scale factor (k) of the model of the ship = \(\\ \frac { 1 }{ 250 } \)
(i) Length of model = 1.6 m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q13.1

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