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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 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 13 Similarity Chapter Test

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test

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

To prove : PT × QR = PR × ST
Proof: ∠1 = ∠2
Adding ∠6 to both sides
∠1 + ∠6 = ∠2 + ∠6
∠SPT = ∠QPR
In ∆PQR and ∆PST

Question 2.
In the adjoining figure, AB = AC. If PM ⊥ AB and PN ⊥ AC, show that PM x PC = PN x PB.

Solution:
Given : In the given figure,
AB = AC, PM ⊥ AB and PN ⊥ AC
To prove : PM × PC = PN × PB
Proof: In ∆ABC, AB = AC
∠B = ∠C
Now in ∆CPN and ∆BPM,

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.

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)

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.

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

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

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

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

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

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.

Solution:
In ∆ABC and ∆ACD
∠C = ∠C (Common)
∠ABC = ∠CAD (given)
∴ ∆ABC ~ ∆ACD

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.

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.

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.

Solution:
In the given figure, ABCD is trapezium in
which AB || DC, 2AB = 3DC

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 .

(i) DO : OE = 2 : 1
(ii) area of ∆OEC : area of ∆OAD = 1 : 4.
Solution:
Given : In || gm ABCD,
E is the midpoint 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)

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

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS

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 13 Similarity MCQS

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test

Choose the correct answer from the given four options (1 to 22):

Question 1.
In the given figure, ∆ABC ~ ∆QPR. Then ∠R is
(a) 60°
(b) 50°
(c) 70°
(d) 80°

Solution:
In the given figure,
∆ABC ~ ∆QPR
∴ ∠A = ∠Q, ∠B = ∠P and ∠C = ∠R
But ∠A = 70°, ∠B = 50°
∴ ∠C = 180° – (70° + 500) = 180° – 120° = 60°
∠C = ∠R
∴ ∠R = ∠C = 60°

Question 2.
In the given figure, ∆ABC ~ ∆QPR. The value of x is
(a) 2.25 cm
(b) 4 cm
(c) 4.5 cm
(d) 5.2 cm

Solution:
In the given figure
∆ABC ~ ∆QPR

Question 3.
In the given figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to
(a) 50°
(b) 30°
(c) 60°
(d) 100°

Solution:
In the given figure two line segments AC and BD intersect each other at P
and PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°
In ∆APB and ∆CPD

Question 4.
If in two triangles ABC and PQR,
\(\frac { AB }{ QR } =\frac { BC }{ PR } =\frac { CA }{ PQ } \)
then
(a) ∆PQR ~ ∆CAB
(b) ∆PQR ~ ∆ABC
(c) ∆CBA ~ ∆PQR
(d) ∆BCA ~ ∆PQR
Solution:
In two ∆ABC and ∆PQR

Question 5.
In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE, then the two triangles are
(a) congruent but not similar
(b) similar but not congruent
(c) neither congruent nor similar
(d) congruent as well as similar
Solution:
In ∆ABC and ∆DEF
∠B = ∠E, ∠F = ∠C
AB = 3DE
∵ Two angles of the one triangles are equal
to corresponding two angles of the other
But sides are not equal
∵ Triangles are similar but not congruent, (b)

Question 6.
In the given figure, if D, E and F are midpoints of the sides BC, CA and AB respectively, then the two triangles ABC and DEF are
(a) similar
(b) congruent
(c) both similar and congruent
(d) neither similar nor congruent

Solution:
D, E and F are the mid-points of the sides BC, CA and AB of ∆ABC
then two triangles ABC and DEF are similar. (a)

Question 7.
The given figure, AB || DE. The length of CD is
(a) 2.5 cm
(b) 2.7 cm
(c) \(\\ \frac { 10 }{ 3 } \) cm
(d) 3.5 cm

Solution:
In the given figure, AB || DE
∆ABC ~ ∆DCE

Question 8.
If in triangles ABC and DEF,\(\frac { AB }{ DE } =\frac { BC }{ FD } \) , then they will be similar when
(a) ∠B = ∠E
(b) ∠A = ∠D
(c) ∠B = ∠D
(d) ∠A = ∠F
Solution:
In two triangles ABC and DEF
\(\frac { AB }{ DE } =\frac { BC }{ FD } \)
They will be similar if their included angles are equal
∠B = ∠D (c)

Question 9.
If ∆PQR ~ ∆ABC, PQ = 6 cm, AB = 8 cm and perimeter of ∆ABC is 36 cm, then perimeter of ∆PQR is
(a) 20.25 cm
(b) 27 cm
(c) 48 cm
(d) 64 cm
Solution:
∆PQR ~ ∆ABC
PQ = 6 cm, AB = 8 cm
Perimeter of ∆ABC = 36 cm
Let perimeter of ∆PQR be x cm

Question 10.
In the given figure, DE || BC and all measurements are in centimetres. The length of AE is
(a) 2 cm
(b) 2.25 cm
(c) 3.5 cm
(d) 4 cm

Solution:
In the given figure,
DE || BC

Question 11.
In the given figure, PQ || CA and all lengths are given in centimetres. The length of BC is
(a) 6.4 cm
(b) 7.5 cm
(c) 8 cm
(d) 9 cm

Solution:
In the given figure,
PQ || CA
Let BC = x

Question 12.
In the given figure, MN || QR. If PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm, then PM is
(a) 4 cm
(b) 3.6 cm
(c) 2 cm
(d) 3 cm

Solution:
In the given figure, MN || QR
PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm
Let PM = x cm

Question 13.
D and E are respectively the points on the sides AB and AC of a ∆ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then the length of DE is
(a) 2.5 cm
(b) 3 cm
(c) 5 cm
(d) 6 cm
Solution:
D and E are the points on sides AB and AC of ∆ABC,
AD = 2 cm, BD = 3 cm,
BC = 7.5 cm

Question 14.
It is given that ∆ABC ~ ∆PQR with \(\frac { BC }{ QR } =\frac { 1 }{ 3 } \) then \(\frac { area\quad of\quad \Delta PQR }{ area\quad of\quad \Delta ABC } \) equal to
(a) 9
(b) 3
(c) \(\\ \frac { 1 }{ 3 } \)
(d) \(\\ \frac { 1 }{ 9 } \)
Solution:
∆ABC ~ ∆PQR
\(\frac { BC }{ QR } =\frac { 1 }{ 3 } \)

Question 15.
If the areas of two similar triangles are in the ratio 4 : 9, then their corresponding sides are in the ratio
(a) 9 : 4
(b) 3 : 2
(c) 2 : 3
(d) 16 : 81
Solution:
Ratio in the areas of two similar triangles = 4 : 9
Ratio in their corresponding sides
= √4 : √9
= 2 : 3 (c)

Question 16.
If ∆ABC ~ ∆PQR, BC = 8 cm and QR = 6 cm, then the ratio of the areas of ∆ABC and ∆PQR is
(a) 8 : 6
(b) 3 : 4
(c) 9 : 16
(d) 16 : 9
Solution:
∆ABC ~ ∆PQR, BC = 8 cm, QR = 6 cm
\(\frac { area\quad of\quad \Delta ABC }{ area\quad of\quad \Delta PQR } =\frac { { BC }^{ 2 } }{ { QR }^{ 2 } } \)
= \(\frac { { 8 }^{ 2 } }{ { 6 }^{ 2 } } =\frac { 64 }{ 36 } =\frac { 16 }{ 9 } \) (d)

Question 17.
If ∆ABC ~ ∆QRP \(\frac { area\quad of\quad \Delta ABC }{ area\quad of\quad \Delta PQR } \) AB = 18 cm and BC = 15 cm, then the length of PR is equal to
(a) 10 cm
(b) 12 cm
(c) \(\\ \frac { 20 }{ 3 } \)
(d) 8 cm
Solution:
∆ABC ~ ∆QRP
\(\frac { area\quad of\quad \Delta ABC }{ area\quad of\quad \Delta PQR } \)

Question 18.
If ∆ABC ~ ∆PQR, area of ∆ABC = 81 cm², area of ∆PQR = 144 cm² and QR = 6 cm, then length of BC is
(a) 4 cm
(b) 4.5 cm
(c) 9 cm
(d) 12 cm
Solution:
∆ABC ~ ∆PQR,
area of ∆ABC = 81 cm² and area of ∆PQR = 144 cm²,
QR = 6 cm, BC = ?

Question 19.
In the given figure, DE || CA and D is a point on BD such that BD : DC = 2 : 1. The ratio of area of ∆ABC to area of ∆BDE is
(a) 4 : 1
(b) 9 : 1
(c) 9 : 4
(d) 3 : 2

Solution:
In the given figure, DE || CA ,
D is a point on BC and BD : DC = 2 : 1
BD : BC = 2 : (2 + 1) = 2 : 3
In ∆ABC, DE || CA
∆ABC ~ ∆BDE

Question 20.
If ABC and BDE are two equilateral triangles such that D is mid-point of BC, then the ratio of the areas of triangles ABC and BDE is
(a) 2 : 1
(b) 1 : 2
(c) 1 : 4
(d) 4 : 1
Solution:
∆ABC and ∆BDE are an equilateral triangles

Question 21.
The areas of two similar triangles are 81 cm² and 49 cm² respectively. If an altitude of the smaller triangle is 3.5 cm, then the corresponding altitude of the bigger triangle is
(a) 9 cm
(b) 7 cm
(c) 6 cm
(d) 4.5 cm
Solution:
Areas of two similar triangles are 81 cm² and 49 cm²
The altitude of the smaller triangle is 3.5 cm
Let the altitude of the bigger triangle is x, then

Question 22.
Given ∆ABC ~ ∆PQR, area of ∆ABC = 54 cm² and area of ∆PQR = 24 cm². If AD and PM are medians of ∆’s ABC and PQR respectively, and length of PM is 10 cm, then length of AD is
(a) \(\\ \frac { 49 }{ 3 } \) cm
(b) \(\\ \frac { 20 }{ 3 } \) cm
(c) 15 cm
(d) 22.5 cm
Solution:
∆ABC ~ ∆PQR
area of ∆ABC = 54 cm²
and of ∆PQR = 24 cm²
AD and PM are their median respectively
PM = 10 cm
Let AD = x cm, then

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3

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 13 Similarity Ex 13.3

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test

Question 1.
Given that ∆s ABC and PQR are similar.
Find:
(i) The ratio of the area of ∆ABC to the area of ∆PQR if their corresponding sides are in the ratio 1 : 3.
(ii) the ratio of their corresponding sides if area of ∆ABC : area of ∆PQR = 25 : 36.
Solution:
(i) ∴ ∆ABC ~ ∆PQR

(By theorem 15.1)
But BC = QR =1 : 3

Question 2.
∆ABC ~ DEF. If area of ∆ABC = 9 sq. cm., area of ∆DEF =16 sq. cm and BC = 2.1 cm., find the length of EF.
Solution:
Let EF = x
Given that
∆ABC ~ ∆DEF,

Question 3.
∆ABC ~ ∆DEF. If BC = 3 cm, EF = 4 cm and area of ∆ABC = 54 sq. cm. Determine the area of ∆DEF.
Solution:
∆ABC ~ ∆DEF

Question 4.
The area of two similar triangles are 36 cm² and 25 cm². If an altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other triangle.
Solution:
Let ABC ~ ∆DEF, AL and DM are their altitudes
then area of ∆ABC = 36 cm²
area of ∆DEF = 25 cm² and AL = 2.4 cm.
Let DM = x
Now ∆ABC ~ ∆DEF

Question 5.
(a) In the figure, (i) given below, PB and QA are perpendiculars to the line segment AB. If PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm², find the area of ∆QOA. (2006)

(b) In the figure (ii) given below, AB || DC. AO = 10 cm, OC = 5cm, AB = 6.5 cm and OD = 2.8 cm.
(i) Prove that ∆OAB ~ ∆OCD.
(ii) Find CD and OB.
(iii) Find the ratio of areas of ∆OAB and ∆OCD.

Solution:
In ∆AOQ and ∆BOP, we have
∠ OAQ = ∠ OBP [Each = 90°]
∠ AOQ= ∠BOP
[Vertically opposite angles]
∆AOQ ~ ∆BOP [A.A. similarity]

Question 6.
(a) In the figure (i) given below, DE || BC. If DE = 6 cm, BC = 9 cm and area of ∆ADE = 28 sq. cm, find the area of ∆ABC.

(b) In the figure (iii) given below, DE || BC and AD : DB = 1 : 2, find the ratio of the areas of ∆ADE and trapezium DBCE.

Solution:
(a) In the figure,
DE || BC
∠D = ∠B and ∠E = ∠C
(Corresponding angles)

Question 7.
In the given figure, DE || BC.
(i) Prove that ∆ADE and ∆ABC are similar.
(ii) Given that AD = \(\\ \frac { 1 }{ 2 } \) BD, calculate DE if BC = 4.5 cm.

(iii) If area of ∆ABC = 18cm², find the area of trapezium DBCE
Solution:
(i) Given : In ∆ABC, DE || BC.
To prove : ∆ADE ~ ∆ABC
Proof: In ∆ADE and ∆ABC,
∠A = ∠A (common)
∠ADE = ∠ABC (corresponding angles)
∴ ∆ADE ~ ∆ABC. (AA axiom)
(ii) ∴ ∆ADE ~ ∆ABC

Question 8.
In the given figure, AB and DE are perpendicular to BC.
(i) Prove that ∆ABC ~ ∆DEC
(ii) If AB = 6 cm: DE = 4 cm and AC = 15 cm, calculate CD.
(iii) Find the ratio of the area of ∆ABC : area of ∆DEC.

Solution:
(i) To prove : ∆ABC ~ ∆DEC
In ∆ABC and ∆DEC

Question 9.
In the adjoining figure, ABC is a triangle.

(ii) Prove that ∆DEF is similar to ∆CBF.
Hence, find \(\\ \frac { EF }{ FB } \).
(iii) What is the ratio of the areas of ∆DEF and ∆CBF ? (2007)

Solution:

Question 10.
In ∆ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find:
(i) area ∆APO : area ∆ABC.
(ii) area ∆APO : area ∆CQO. (2008)

Solution:
In the figure,
PQ || BC and PO is produced to Q such that CQ || BA
and AP : PB = 2 : 3.

Question 11.
(a) In the figure (i) given below, ABCD is a trapezium in which AB || DC and AB = 2 CD. Determine the ratio of the areas of ∆AOB and ∆COD.
(b) In the figure (ii) given below, ABCD is a parallelogram. AM ⊥ DC and AN ⊥ CB. If AM = 6 cm, AN = 10 cm and the area of parallelogram ABCD is 45 cm², find
(i) AB
(ii) BC
(iii) area of ∆ADM : area of ∆ANB.
(c) In the figure (iii) given below, ABCD is a parallelogram. E is a point on AB, CE intersects the diagonal BD at O and EF || BC. If AE : EB = 2 : 3, find
(i) EF : AD
(ii) area of ∆BEF : area of ∆ABD
(iii) area of ∆ABD : area of trap. AFED
(iv) area of ∆FEO : area of ∆OBC.

Solution:
(a) In trapezium ABCD, AB || DC.
∠OAB = ∠OCD [alternate angles]
∠OBA = ∠ODC
∆AOB ~ ∆COD

Question 12.
In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2 and DP produced meets AB produced at Q. If area of ∆CPQ = 20 cm², find

(i) area of ∆BPQ.
(ii) area ∆CDP.
(iii) area of || gm ABCD.
Solution:
In the figure, ABCD is a parallelogram.
P is a point on BC such that BP : PC = 1 : 2
and DP is produced to meet ABC produced at Q.
Area ∆CPQ = 20 cm²

Question 13.
(a) In the figure (i) given below, DE || BC and the ratio of the areas of ∆ADE and trapezium DBCE is 4 : 5. Find the ratio of DE : BC.

(b) In the figure (ii) given below, AB || DC and AB = 2 DC. If AD = 3 cm, BC = 4 cm and AD, BC produced meet at E, find (i) ED (ii) BE (iii) area of ∆EDC : area of trapezium ABCD.

Solution:
(a) In ∆ABC, DE || BC
Now in ∆ABC and ∆ADE
∠A = ∠A (common)
∠D = ∠B and ∠E = ∠C
(Corresponding angles)
∆ADE ~ ∆ABC

Question 14.
(a) In the figure given below, ABCD is a trapezium in which DC is parallel to AB. If AB = 9 cm, DC = 6 cm and BB = 12 cm., find
(i) BP
(ii) the ratio of areas of ∆APB and ∆DPC.

(b) In the figure given below, ∠ABC = ∠DAC and AB = 8 cm, AC = 4 cm, AD = 5 cm.
(i) Prove that ∆ACD is similar to ∆BCA
(ii) Find BC and CD
(iii) Find the area of ∆ACD : area of ∆ABC.

Solution:
(a) In trapezium ABCD, DC || AB

Question 15.
ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:
(i) ∆ADE ~ ∆ACB.
(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.
(iii) Find, area of ∆ADE : area of quadrilateral BCED.

Solution:
(i) Consider DADE and DACB
∠A = ∠A (Common)
m∠B = m∠E = 90°
Thus by angle-angle similarity, triangles,
∆ACB ~ ∆ADE
(ii) Consider ∆ADE and ∆ACB
Since they are similar triangles,
the sides are proportional Thus, we have,

Question 16.
Two isosceles triangles have equal vertical angles and their areas are in the ratio 7 : 16. Find the ratio of their corresponding height.
Solution:
In two isosceles ∆s ABC and DEF

Question 17.
On a map drawn to a scale of 1 : 250000, a triangular plot of land has the following measurements :
AB = 3 cm, BC = 4 cm and ∠ABC = 90°. Calculate
(i) the actual length of AB in km.
(ii) the area of the plot in sq. km:
Solution:
Scale factor k = 1 : 250000 = \(\\ \frac { 1 }{ 250000 } \)
Length on map,
AB = 3 cm, BC = 4 cm

Question 18.
On a map drawn to a scale of 1 : 25000, a rectangular plot of land, ABCD has the following measurements AB = 12 cm and BG = 16 cm.
Calculate:
(i) the distance of a diagonal of the plot in km.
(ii) the area of the plot in sq. km.
Solution:
Scale factor (k) = \(\\ \frac { 1 }{ 25000 } \)
Measurements of plot ABCD on the map are
AB = 12 cm and BC = 16 cm.

Question 19.
The model of a building is constructed with the scale factor 1 : 30.
(i) If the height of the model is 80 cm, find the actual height of the building in metres.
(ii) If the actual volume of a tank at the top of the building is 27 m³, find the volume of the tank on the top of the model. (2009)
Solution:
(i)

Question 20.
A model of a ship is made to a scale of 1 : 200.
(i) If the length of the model is 4 m, find the length of the ship.
(ii) If the area of the deck of the ship is 160000 m², find the area of the deck of the model.
(iii) If the volume of the model is 200 litres, find the volume of the ship in m³.
(100 litres = m³)
Solution:
Scale = 1 : 200
(i) Length of a model of ship = 4 m

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.2

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 13 Similarity Ex 13.2

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test

Question 1.
(a) In the figure (i) given below if DE || BG, AD = 3 cm, BD = 4 cm and BC = 5 cm. Find (i) AE : EC (ii) DE.
(b) In the figure (ii) given below, PQ || AC, AP = 4 cm, PB = 6 cm and BC = 8 cm. Find CQ and BQ.
(c) In the figure (iii) given below, if XY || QR, PX = 1 cm, QX = 3 cm, YR = 4.5 cm and QR = 9 cm, find PY and XY.

Solution:
(a) In the figure (i)
Given : DE || BC, AD = 3 cm, BD = 4 cm and BC = 5 cm.
To find (i) AE : EC and
(ii) DE Since DE || BC of ∆ABC

Question 2.
In the given figure, DE || BC.
(i) If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x.
(ii) If DB = x – 3, AB = 2x, EC = x – 2 and AC = 2x + 3, find the value of x.

Solution:
In the given figure, DE || BC
(i) AD = x, DB = x – 2, AE = x + 2, EC = x – 1
In ∆ABC,
∵DE || BC

Question 3.
E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 8 cm and RF = 9 cm.
(ii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.
Solution:
(i) In ∆PQR, E and F are the points on the sides PQ and PR respectively
PE = 3.9 cm, EQ = 3 cm, PF = 8 cm,
RF = 9 cm
Is EF || QR ?

Question 4.
A and B are respectively the points on the sides PQ and PR of a triangle PQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm. Is AB || QR? Give reasons for your answer.
Solution:
In ∆PQR, A and B are points on the sides PQ and PR such that
PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm

Question 5.
(a) In figure (i) given below, DE || BC and BD = CE. Prove that ABC is an isosceles triangle.
(b) In figure (ii) given below, AB || DE and BD || EF. Prove that DC² = CF x AC.
Solution:
(a) Given: In the figure,
DE || BC and BD = CE
To prove: ∆ABC is an isosceles triangle
Proof: In ∆ABC, DE || BC

Question 6.
(a) In the figure (i) given below, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and EC = 2 cm.
(b) In the given figure, ∠D = ∠E and \(\frac { AD }{ BD } =\frac { AE }{ EC } \). Prove that BAC is an isosceles triangle.

Solution:
(a) Given : CD || LA and DE || AC
Length of BE = 4 cm
Length of EC = 2 cm
Now, in ∆BCA
DE || AC
\(\frac { BE }{ BC } =\frac { BD }{ BA } \)
(Corallary of basic proportionality theorem)

Question 7.
In the figure given below, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. show that BC || QR.

Solution:
In the given figure, A, B, C are points on
OP, OQ and OR respectively
and AB || PQ and AC || PR
To prove: BC || QR
Proof: In POQ,
AB || PQ

Question 8.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O. Using Basic Proportionality theorem, prove that \(\frac { AO }{ BO } =\frac { CO }{ DO } \)
Solution:
Given : ABCD is a trapezium in which AB || DC
Its diagonals AC and BD intersect each other at O

Question 9.
(a) In the figure (1) given below, AB || CR and LM || QR.
(i) Prove that \(\frac { BM }{ MC } =\frac { AL }{ LQ } \)
(ii) Calculate LM : QR, given that BM : MC = 1 : 2.
(b) In the figure (2) given below AD is bisector of ∠BAC. If AB = 6 cm, AC = 4 cm and BD = 3cm, find BC

Solution:
(a) Given: AB || CR and LM II QR.
Also BM : MC = 1:2
To Prove:
(i) \(\frac { BM }{ MC } =\frac { AL }{ LQ } \)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.2 are helpful to complete your math homework.

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