Similar Triangles Class 10 OP Malhotra Exe-12B ICSE Maths Solutions Ch-12. We Provide Step by Step Solutions / Answer of Exe-12B Questions of S Chand OP Malhotra Maths . In this article you would learn problems on Similarity of Triangles. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Similar Triangles Class 10 OP Malhotra Exe-12B ICSE Maths Solutions Ch-12
Board | ICSE |
Publications | S Chand |
Subject | Maths |
Class | 10th |
Chapter-12 | Similar Triangles |
Writer | OP Malhotra |
Exe-12B | Similarity of Triangles |
Edition | 2024-2025 |
Similarity of Triangles
The three important criteria used in determining the similarity of triangles are
- AA criterion (Angle-Angle criterion)
- SSS criterion (Side-Side-Side criterion)
- SAS Criterion (Side-Angle-Side criterion)
Exercise- 12B
Similar Triangles Class 10 OP Malhotra Exe-12B ICSE Maths Solutions Ch-12
Que-1: In each question, of the three triangles, name the one that is different, i.e., not similar to the other two.
(i)
(ii)
(iii)
Sol: (i) In ∆ (a), third angle = 180° – (75° + 55°) = 180° – 130° = 50°
∴ Three angles are 75°, 55°, 50°
In ∆ (b), third angle is = 180° – (55° + 80°) = 180° – 135° = 45°
Three angles are 55°, 80°, 45°
In ∆ (c), third angle is = 180° – (45° + 80°) = 180°- 125° = 55°
Three angles are 45°, 80°, 55°
We see that triangle (a) is different
(ii) In ∆ (a), third angle is = 180° – (100°+ 45°) = 180° – 145° = 35°
∴ Three angles will be 100°, 45°, 35°
In ∆ (b), third angle is = 180° – (100° + 35°) = 180° + 35° = 45°
∴ Three angles will be = 100°, 35°, 45°
In ∆ (c), third angle is = 180° – (25° + 45°) = 180° – 70° = 110°
∴ Three angles will be 110°, 25°, 45°
We see that triangle (c) is different
(iii) In ∆ (a), two sides are equal
In ∆ (b), no two sides are equal
In ∆ (c), two sides are equal
∴ ∆ (b) is different
Que-2: Write down the ratio of the corresponding sides for each pair of triangles and check that it is the same.
(i)
(ii)
Sol: (i) In the first pair of triangle
The ratio in corresponding sides are
AB/YZ = BC/XY = AC/XZ
⇒ 6/9 = 4.5/6.75 = 7/10.5
⇒ 2/3 = 450/675 = 70/105
⇒ 2/3 = 2/3 = 2/3
Yes, it is the same ratio
AB/YZ = BC/XY = AC/XZ = 2/3
(ii) In the second pair of triangles
The ratio in the corresponding sides are
LM/RP = MN/RQ = LN/PQ
⇒ 16/2.4 = 8/1.2 = 14/2.1
⇒ {(16×10)/24} = {(8×10)/12} = {(14×10)/21}
⇒ 20/3 = 20/3 = 20/3 = 2/3 (Dividing by 10)
Yes, it is the same ratio
∴ LM/RP = MN/RQ = LN/PQ = 2/3
Que-3: In ∆ABCD and E are points on the sides AB and AC respectively such that DE || BC.
(i) If AD/BD = 4/5 and EC = 2.5 cm, find AE.
(ii) If AD = 4, AE = 8, DB = x – 4, and EC = 3x – 19, find x.
(in) If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the vlaue of x.
(iv) If AD = 6 cm, DB = 9 cm and AE = 8 cm, find AC.
(v) If AD/BD = 2/3 and AC = 18 cm, find AE.
(vi) If AD = 8 cm, AB = 12 cm and AE = 12 cm, find CE.
(vii) If AD = 2 cm, AB = 6 cm and AC = 9 cm, find AE.
Sol: In ∆ABC, D and E are points on AB and AC respectively and DE || BC
(i) AD/BD = 4/5, EC = 2.5 cm
In ∆ABC,
∵ DE || BC
∴ AD/BD = AE/EC
⇒ 4/5 = AE/2.5
⇒ AE = (4×2.5)/5 = 2 cm
∴ AE = 2 cm
(ii) In the figure, in ∆ABC, DE || BC
∴ AD/DB = AE/EC
AD = 4, AE = 8, DB = x – 4 and EC = 3x – 19
⇒ 4/(x−4) = 8/(3x−19)
⇒ 12x – 76 = 8x – 32 (By cross multiplication)
⇒ 12x – 8x = – 32 + 76
⇒ 4x = 44 ⇒ x = 44/4 = 11
∴ x = 11
(iii) In ∆ABC, DE || BC
∴ AD/DB = AE/EC
AD = x, DB = x – 2, AE = x + 2 and EC = x – 1
∴ x/(x−2) = (x+2)/(x−1)
⇒ x (x – 1) = (x + 2) (x – 2) (By cross multiplication)
⇒ x² – x = x² – 4
⇒ – x = – 4
⇒ x = 4
∴ x = 4
(iv) In ∆ABC, DE || BC
∴ AD/DB = AE/EC
AD = 6 cm, DB = 9 cm, AE = 8 cm
∴ 6/9 = 8/EC
⇒ EC = (9×8)/6
⇒ EC = 12 cm
∴ AC = AE + EC = 8 + 12 = 20 cm
(v) In ∆ABC, DE || BC
∴ AD/DB = AE/EC
AD/DB = 2/3, AC = 18 cm
Let AE = x cm
∴ EC = AC – AE = (18 -x) cm
∴ 2/3 = x/(18−x) ⇒ 3x = 36 – 2x (By cross multiplication)
⇒ 3x + 2x = 36 ⇒ 5x = 36
⇒ x = 23 = 7.2
∴ AE = 7.2
(vi) In ∆ABC, DE || BC
∴ AD/DB = AE/EC
AD = 8 cm, AB = 12 cm, AE = 12 cm DB = AB – AD = 12 – 8 = 4 cm
8/4 = 12/CE ⇒ CE = (4×12)/8 = 6
∴ CE = 6 cm
(vii) In ∆ABC, DE || BC
∴ AD/DB = AE/EC
AD = 2 cm, AB = 6 cm, AC = 9 cm
DB = AB – AD = 6 – 2 = 4 cm
Let AE = x, then EC = AC – AE = 9 – x
∴ 2/4 = x/(9−x) ⇒ 4x = 18 – 2x
⇒ 4x + 2x = 18
⇒ 6x = 18
⇒ x = 18/6 = 3
∴ AE = 3 cm
Que-4: In a ∆ABC, D and E are points on the sides AB and AC respectively, for each of the following cases, show that DE || BC.
(i) AD = 3, BD = 4.5, AE = 4, CE = 6.
(ii) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm.
(iii) AB = 5.6 cm, AD = 1.4 cm, AC 7.2 cm and AE = 1.8 cm.
Sol: (i) In ∆ABC, D and E are the points on the sides AB and AC respectively such that AD = 3, BD = 4.5, AE = 4 and CE = 6
Now, AD/DB = 3/4.5 = 2/3
and AE/EC = 4/6 = 2/3
AD/DB = AE/EC
DE || BC.
(ii) In ∆ABC, D and E are the points on the sides AB and AC respectively such that AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm
∴ DB = AB – AD = 12 – 8 = 12 – 8 = 4 cm
EC = AC – AE = 18 – 12 = 6 cm
Now AD/DB = 8/4 = 2/1
and AE/EC = 12/6 = 2/1
∵ AD/DB = AE/EC
∴ DE || BC
(iii) In ∆ABC, D and E are the points on the sides AB and AC respectively such that AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm
∴ DB = AB – AD = 5.6 – 1.4 = 4.2 cm
EC = AC – AE = 7.2 – 1.8 = 5.4 cm
Now AD/DB = 1.4/4.2 = 1/3
and AE/EC = 1.8/5.4 = 1/3
∵ AD/DB = AE/EC
∴ DE || BC
Que-5: In the figure, ABC and DEF and similar triangles, find the values of x and y, if the sides are as marked.
Sol: In the figure, ∆ABC and ∆DEF are similar
AB = x, AC = 7, BC = y
DE = 5, DF = 3 and EF = 8
∵ ∆s are similar
Corresponding sides are proportional
∴ AB/DE = AC/DF = BC/EF
x/5 = 7/3 = y/8
x/5 = 7/3
⇒ x = (5×7)/3 = 35/3 = 11*(2/3)
and y/8 = 7/3
⇒ y = (8×7)/3 = 56/3 = 18*(2/3)
Que-6: In the figure, ∆ABO ~ ADCO. If CD = 2 cm, AB = 3 cm, OC = 3.2 cm, OD = 2.4 cm, determine OA and OB.
Sol: ∵ ∆ABO ~ ADCO
∴ Their corresponding sides are proportional
∴ AB/CD = AO/OD = BO/OC
But AB = 3 cm, CD = 2 cm, OC = 3.2 cm and OD = 2.4 cm
Hence OA = 3.6 cm and OB = 4.8 cm
Que-7: In the figure, ∆ABC ~ ∆ADE. If AD : DB = 2:3 and DE = 5 cm. (i) find BC (ii) if x be the length of the perpendicular from A to DE, find the length of the perpendicular from A to BC in terms of x.
Sol: In the figure,
(i) ΔADE ≈ ΔABC
=> AD/AB = DE/BC
=> AD/(AD + DB) = DE/BC
=> 2/(2 + 3) = 5/BC
=> BC = 25/2
=> BC = 12.5 cm
(ii) ΔADE ≈ ΔABC
=> Their corresponding altitude will be in the same ratio as sides
hence
AD/AB = length of the perpendicular from a to de/length of the perpendicular from a to bc
=> 2/5 = x/length of the perpendicular from a to bc
=> length of the perpendicular from a to bc = 5x/2
Que-8: In the figure, ABCD is a trapezium with AB parallel to DC. Given that AB = 4 cm, BC = 3 cm and CD = 6 cm.
(i) Name two triangles in the figure which are similar.
(ii) Calculate the length of EB.
Sol: In trapezium ABCD, AB || DC
AB = 4 cm, BC = 3 cm and CD = 6 cm
(i) In the figure,
∆EDC and ∆EAB
∠E = ∠E (common)
∠EDC = ∠EAB (corresponding angles)
∠ECD = ∠EBA (corresponding angles)
∴ ∆EDC ~ ∆EAB
(ii) Let EB = x, then EC = x + 3
∵ ∆EAB ~ ∆EDC
∴ EB/EC = AB/DC
⇒ x/(x+3) = 4/6 ⇒ 6x = 4x + 12
⇒ 6x – 4x = 12 ⇒ 2x = 12
⇒ x = 12/2 = 6
∴ EB = 6 cm
Que-9: A man of height 1.8 m is standing 5 m away from a lamp post and observes that the length of his shadow is 1.5 m. Find the height of the lamp post.
Sol: The mean and the lamp post are perpendicular on the ground
Height of man PQ = 1.8 m
and his shadow QN = 1.5 m
Let LM be the lamp post and LM = x m
Now in ∆LMN
∵ PQ || LM
(both perpendiculars on the same line)
∴ ∆PNQ ~ ∆LNM
∵ PQ/LM = NQ/NM
⇒ 1.8/x = 1.5/(1.5+5)
⇒ 1.8/x = 1.5/6.5
⇒ x = (1.8×6.5)/1.5
∴ Height of the lamp post = 7.8 m
Que-10: In figure, ACE and BCD are two straight lines.

∠A = 40° and ∠B = 85°. Using the measurements given in the figure, complete the following true statements
(i) Triangles ABC and CDE are similar because ……… and ………
(ii) The size of ∠D is ………
(iii) If AB = x cm, then ED = ………
Sol: (i) Two lines ACE and BCD intersect each other at C
AC = 4 cm, CE = 6 cm
BC = 3 cm and CD = 8 cm
(i) In ∆ABC and ∆CDE
BC/AC = 3/4 and CE/CD = 6/8 = 3/4
∠ACB = ∠DCE (vertically opposite angles)
∴ ∆ABC ~ ACDE because BC/AC = CE/CD = 3/4 and ∠ACB = ∠DCE
(ii) ∠D = ∠A = 40° (corresponding angles)
(iii) AB = x cm
and AC/CD = AB/DE = 4/8 = 1/2
⇒ x/DE = 1/2
⇒ DE = 2x
∴ DE = 2x
Que-11: The triangle ABC is right-angled at C. From P, a point on the hypotenuse, PQ is drawn parallel to AC cutting BC at Q. If AC 2.5 cm, BC = 6 cm and PQ = 1 cm, find (i) BQ (ii) BP
Sol: In right angle ∆ABC, ∠C = 90
P is any point on AB
From P, PQ is drawn parallel to AC which meets BC at Q
AC = 2.5 cm, BC = 6 cm, PQ = 1 cm
In ∆ABC
∵ PQ || AC
(i) ∴ ∆ABC ~ APBQ
AC/PQ = BC/BQ
⇒ 2.5/1 = 6/BQ
⇒ BQ = (6×1)/2.5 = 6/2.5 = 2.4 cm
(ii) In right ∆PQB (∵ ∠Q = ∠C = 90°)
BP² = PQ² + BQ²
= (1)² + (2.4)²
= 1 + 5.76
= 6.75
= (2.6)²
BP = 2.6 cm
Que-12: State whether the following statement is true or false, briefly giving the reasons. If two angles of one triangle are 72° and 80° respectively and that of another triangle are 28° and 72° respectively, then the triangles are similar.
Sol: Let in one triangle ABC,
∠A = 72°, ∠B = 80°
Then ∠C =180°- (72° + 80°) = 180° – 152° = 28°
and in the other triangle PQB
∠P = 28°, ∠Q = 72°
∴ ZR = 180° – (28° + 72°) = 180° – 100° = 80°
∵ ∠A = ∠Q, ∠B = ZR and ∠C = ∠P
∴ ∆ABC ~ APQR (AAA axiom)
Que-13: In figure, ∠ADE = ∠B, show that ∆ABC ~ ∆ADE. If AD = 2.7 cm, AE = 2.5 cm, BE = 1.1 cm and BC = 5.2 cm, find DE.
Sol: (i) In the figure, AD = 2.7 cm, AE = 2.5 cm, BE = 1.1 cm and BC = 5.2 cm
(i) In ∆ABC and ∆ADE
∠A = ∠A (common)
∠ABC=∠ADE (given)
∴ ∆ABC ~ ∆ADE (AA axiom)
(ii) ∴ BC/DE = AB/AD = AC/AE
(corresponding sides are proportional)
⇒ 5.2/DE = (1.1+2.5)/2.7
⇒ 5.2/DE = 3.6/2.7
⇒ DE = (5.2×2.7)/3.6 = 3.9
Hence DE = 3.9 cm
Que-14: In figure, AB, EF and CD are parallel lines. Given that AB = 15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm. Calculate EF and AC.

Sol: In the figure, ABC is a triangle AB, EF and CD are parallel line
AB = 15 cm, EG = 5 cm, BC = 10 cm and DC =18 cm
We have to find EF and AC
(i) In ∆EFG and ∆CGD
∠EGF = ∠CGD
(vertically opposite angles)
∠EFG = ∠GDC (alternate angles)
∴ ∆EFG ~ ∆CGD (∆A axiom)
∴ EG/CG = EF/CD
(corresponding sides are proportional)
⇒ 5/10 = EF/18
⇒ EF = (5×18)/10 = 9
∴ EF = 9 cm
Similarly in ∆ABC and AECF
∠C = ∠C (common)
∠CAB = ∠CEF (corresponding angles)
∆ABC ~ AECF (AA axiom)
∴ AC/EC = AB/EF
⇒ AC/(10+5) = 15/9
⇒ AC/15 = 15/9
⇒ AC = (15×15)/9 = 225/9 = 25
∴ AC = 25 cm
Que-15: Calculate the other sides of a triangle whose shortest side is 6 cm and which is similar to a triangle whose sides are 4, 7 and 8 cm.
Sol: Let ∆ABC ~ ∆PQR and let
in ∆PQR, PQ = 4 cm, QR = 7 cm and RP = 8 cm
and let in ∆ABC, AB = 6 cm, BC = x and CA = y
Then PQ/AB = QR/BC = RP/CA (∵ As are similar)
⇒ 4/6 = 7/x = 8/y
Now 7/x = 4/6 ⇒ x = (7×6)/4 = 21/2 = 10.5
and 8/y = 4/6 ⇒ y = (6×8)/4 = 12
∴ Other two corresponding sides are 10.5 cm and 12 cm
Que-16: In ∆ABC, DE || BC
(a) If AD = 3 cm, DB = 4 cm, EC = 12 cm, find AE.
(b) If AE = 2.7 cm, EC = 4.5 cm, AD = 2.4 cm, find BD.
(c) If AD = 2.6 cm, DB = 6.5 cm and AE = 3 cm, find EC.
(d) If AB = 6 cm, AD = 2 cm and AC = 9 cm, calculate the length of CE.
Sol: (a) In ∆ABC, DE || CB
∴ ∆ABC ~ ∆ADE
Now AD = 3 cm, DB = 4 cm, EC = 12 cm,
Let AE = x cm
∵ In ∆ABC, DE || BC
AD/DB = AE/EC
⇒ 3/4 = AE/12
⇒ 3/4 = x/12
⇒ x = (3×12)/4 = 9
∴ AE = 9 cm
(b) In ∆ABC, DE || BC
∴ AD/DB = AE/EC
⇒ 2.4/BD = 2.7/4.5
⇒ BD = (2.4×4.5)/2.7
∴ BD = 4 cm
(c) In ∆ABC, DE || BC
∴ AD/DB = AE/EC
⇒ 2.6/6.5 = 3/EC
⇒ EC = (6.5×3)/2.6
⇒ EC = 15/2 = 7.5
∴ EC = 7.5 cm
(d) In ∆ABC, DE || BC
∴ ∆ABC ~ ∆ADE
∴ AD/DB = AE/AC
⇒ 2/6 = AE/9
⇒ AE = (2×9)/6 = 3
∴ CE = AC – AE = 9 – 3 = 6 cm
–: End of Similar Triangles Class 10 OP Malhotra Exe-12B ICSE Maths Solutions Ch-12 :–
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