Similar Triangles Class 10 OP Malhotra Exe-12D ICSE Maths Solutions Ch-12. We Provide Step by Step Solutions / Answer of Exe-12D Questions of S Chand OP Malhotra Maths . In this article you would learn solving problems on Basic Theorem of Proportionality. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Similar Triangles Class 10 OP Malhotra Exe-12D ICSE Maths Solutions Ch-12
Board | ICSE |
Publications | S Chand |
Subject | Maths |
Class | 10th |
Chapter-12 | Similar Triangles |
Writer | OP Malhotra |
Exe-12E | Basic Theorem of Proportionality |
Edition | 2024-2025 |
Basic Theorem of Proportionality
- If a line is drawn parallel to one side of a triangle and intersects the other two sides, then it divides those two sides in the same ratio.
- The area of similar triangles are are proportional to the squire of corresponding sides.
Converse of Basic Proportionality Theorem
According to this theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side
Exercise- 12D (Basic Theorem of Proportionality)
Similar Triangles Class 10 OP Malhotra ICSE Maths Solutions Ch-12
Que-1: In the figure, in ∆PQR if XY || QR, PX = 1 cm, XQ = 3 cm, YR = 4.5 cm and QR = 9 cm, find (i) PY and (ii) XY. Further, if the area of the ∆PXY is A cm², find in terms of A, (iii) the area of the triangle PQR (iv) the area of the figure XYRQ.
Sol: In ∆PQR,
XY || QR
PX = 1 cm, XQ = 3 cm, YR = 4.5 cm and QR = 9 cm
Let PY = x cm and XY = y cm
∵ XY || QR
∴ ∆PXY ~ ∆PQR
∴ PX/PQ = PY/PR = XY/QR
⇒ 1/(1+3) = x/(x+4.5) = y/9
⇒ 1/4 = x/(x+4.5) = y/9
x/(x+4.5) = 1/4
⇒ 4x = x + 4.5
⇒ 4x – x = 4.5 ⇒ 3x = 4.5
(i) ⇒ x = 4.5/3 = 1.5 cm
and y/9 = 1/4 = 4y = 9
⇒ 9/4 = 2.25 cm
(ii) ∴ PY = 1.5 cm and XY = 2.25 cm
(iii) ∵ ∆PXY ~ ∆PQR (proved)
∴ 1/2
(Areas of the similar triangles are proportional to the square of their corresponding sides)
⇒ 1/2
(iv) Now area of figure XYRQ = area of ∆PQR – area of ∆PXY
= 16A – A = 15A cm²
Que-2: The figure shows two isosceles similar triangles. If PQ and BC are not parallel and PC = 4cm, AQ = 3cm, QB = 12cm and BC = 15cm, calculate : (i) the length of AP. (ii) the ratio of the areas of triangle APQ
Sol: In the figure two triangles are similar
i.e. ∆ABC ~ ∆APQ, PQ is not parallel
To BC, PC = 4cm, AQ = 3cm, QB = 12cm, BC = 15cm
Let AP = x, and PQ = y
Then AQ/AC = AP/AB = PQ/BC
⇒ 3/(x+4) = x/(3+12) = y/15
⇒ 3/(x+4) = x/15
⇒ x² + 4x = 45 (By cross multiplication)
⇒ x² + 4x – 45 = 0
⇒ x² + 9x – 5x – 45 = 0
⇒ x (x + 9) – 5 (x + 9) = 0
⇒ (x + 9) (x – 5) = 0
Either x + 9 = 0, then x = – 9 which is not possible being negative
or x – 5 = 0, then x = 5
∴ AP = 5
and 3/(x+4) = y/15
⇒3/(5+4) = y/15
⇒ y = (3×15)/2 = 5
∵ ∆APQ ~ ∆ABC
∴ {area △APQ}/{area △ABC} = PQ²/BC²
{Areas of two similar triangles are proportional to the square of their corresponding sides}
(5)²/(15)² = 25/225 = 1/9
Ratio between the areas of ∆APQ and ∆ABC = 1 : 9
Que-3: Given that ∆s ABC, DEF are similar. Find (a) the ratio of the area of ∆ABC to the area of ∆DEF if AB = 2 and DE = 4. (b) AB/DE, if ∆ABC : ADEF = 16 : 25.
Sol: ∆ABC ~ ∆DEF
AB = 2, DE = 4
(a) ∆s are similar
{area ∆ABC}/{area ∆DEF} = AB²/DE² = 2²/4² = 4/16 = 1/4
Ratio between their area is 1 : 4.
(b) ∆ABC : ∆DEF = 16 : 25
∆ABC/∆DEF = 16/25
∆ABC/∆DEF = AB²/DE²
AB²/DE² = 16/25 = (4/5)²
AB/DE = 4/5.
Que-4: In a ∆ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point on AC such that PQ || BC. Find the ratio of the areas of ∆APQ and trapezium BPQC.
Sol: In ∆ABC,
P is a point on AB such that
AP : PB = 1 : 2
and Q is a point on AC such that PQ || BC
∵ PQ || BC
∴ ∆APQ ~ ∆ABC
∵ {area of △APQ}/{area of △ABC} = AP²/AB²
(Areas of similar triangles are proportional to the squares of their corresponding sides)
= (1)²/{(1+2)²} = (1)²/(3)² = 1/9
⇒ 9 areas of ∆APQ = area of ∆ABC
Subtracting area of ∆APQ from both sides area of ∆ABC – area of ∆APQ = 9 area of ∆APQ – area of ∆APQ
⇒ area of trap. BPQC = 8 area of ∆APQ
⇒ {area of △APQ}/{area of trap. BPQC} = 1/8
∴ area of ∆APQ : area of trap. BPQC = 1 : 8
Que-5: The areas of two similar triangles ABC and DEF are 36 cm² and 81 cm² respectively. If EF = 6.9 cm, determine BC.
Sol: Area of ∆ABC = 36 cm²
and area of ADEF = 81 cm²
EF = 6.9 cm
∵ ∆ABC ~ ADEF
∴ {area of △ABC}/{area of △DEF} = BC²/EF²
(Areas of two similar triangles are in the ratio of the squares on their corresponding sides)
⇒ 36/81 = (BC)²/(6.9)²
⇒ (6)²/(9)² = (BC)²/(6.9)²
⇒ BC/6.9 = 6/9
⇒BC = (6.9×6)/9
⇒ BC = 4.6 cm
Que-6: In the figure, DE || BC. AD = 3 cm, DB = 2 cm, area of ∆ABC = 10 cm². Find the area of ∆ADE.

Sol: In ∆ADE, BC || DE
AD = 3 cm, DB = 2 cm
∴ AB = AD – DB = 3 – 2 = 1 cm
Area of ∆ABC = 10 cm
∵ In ∆ADE; BC || DE
∴ ∆ABC ~ ∆ADE
∴ {area △ABC}/{area △ADE} = AB²/AD²
{Areas of two similar triangles are proportional to the square of their corresponding sides}
⇒ 10/{area △ADE} = (1)²/(3)² = 1/9
⇒ area ∆ADE =10 x 9 = 90 cm²
Que-7: In the figure, the angles PRS and PQR are similar PS = 2 cm and PR = 3 cm. If the area of the triangle PRS is 2 cm², calculate the area of ∆PQR.

Sol: In ∆PRS and ∆PQR,
∠PRS – ∠PQR (given)
∠RPS – ∠RPQ (common)
∆PRS – ∆PQR (AA axiom)
∴ {area of △PRS}/{area of △PQR} = P²/P²
{Areas of two similar triangles are proportional to the squares of their corresponding sides}
⇒ 2/{area of △PQR} = (2)²/(3)² = 4/9
⇒ area of ∆PQR = (2×9)/4 = 9/2 = 4.5 cm²
Que-8: In the figure, DE || BC and AD : DB = 5 : 4. Find {Area (△ADE)}/{Area (△ABC)}
Sol: In △ABC, we have
DE || BC
⇒ ∠ADE = ∠ABC and ∠AED=∠ACB [Corresponding angles]
Thus, in triangles ADE and ABC, we have
∠A = ∠A [Common]
∠ADE = ∠ABC
and, ∠AED = ∠ACB
∴ △AED∼△ABC [By AAA similarity]
⇒ AD/AB = DE/BC
We have,
AD/DB = 5/4
⇒ DB/AD = 4/5
⇒ (DB/AD) + 1 = (4/5) + 1
⇒ (DB+AD)/AD = 9/5
⇒ AB/AD = 9/5
⇒ AD/AB = 5/9
∴ DE/BC = 5/9
In △DFE and △CFB, we have
∠1=∠3 [Alternate interior angles]
∠2=∠4 [Vertically opposite angles]
Therefore, by AA-similarity criterion, we have
△DFE∼△CFB
⇒ {Area(△DFE)}/{Area(△CFB)} = DE²/BC²
⇒ {Area(△DFE)}/{Area(△CFB)} = (5/9)² = 25/81.
Que-9: In the figure, in ∆PQR, PQ = 8 cm, PR = 10 cm and ∠Q = 90°. A and B are points on sides PQ and PR respectively such that AB = 2 cm and ∠ABP = 90°. Find : (i) the area of APAB (ii) the area of quad. AQRB : area of ∆PQR.

Sol: In ∆PAB and ∆PQR,
∠ABP = ∠Q (each 90°)
∠P = ∠P (common)
∴ ∆PAB ~ ∆PQR (AA axiom)
In right ∆PQR, PQ = 8 cm and PR = 10 cm But PR² = PQ² + QR² (Pythagoras theorem)
⇒ (10)² = (8)² + (QR)²
⇒ 100 = 64 + (QR)²
⇒ QR² = 100 – 64 = 36 = (6)²
∴ QR = 6 cm
∵ ∆PAB ~ ∆PQR
∴ AB/QR = PB/PQ
⇒ 2/6 = PB/8
⇒ PB = (2 x 8)/6 = 8/3 cm
∴ Area of right ∆PAB = (1/2) AB x PB
= (1/2) x 2 x (8/3) = 8/3 cm²
Again ∆PAB ~ ∆PQR
∴ {area(ΔPAB)}/{area(ΔPQR)} = AB²/QR²
{Areas of similar triangles are proportional to the squares of their corresponding sides}
= (2)²/(6)² = 4/36 = 1/9
∴ area (∆PQR) = 9 area (∆PAB) … (i)
Subtracting area (∆PAB) from both sides, area (∆PQR) – are (∆PAB) = 9 area (∆PAB) – area (∆PAB)
⇒ area quad. AQRB = 8 area (∆PAB)
⇒ area (quad. AQRB) 8 area (∆PAB)
{area ( quad. AQRB)}/{area (△PQR)} = {8 area (△PAB)}/{9 area (△PAB)} = 8/9 (from (i))
∴ Area quad. AQRB : area (∆PQR) = 8 : 9
Que-10: The areas of two similar triangles ABC and PQR are 64 sq. cm and 121 sq. cm respectively. If QR = 15.4 cm, find BC.
Sol: ∆ABC ~ ∆PQR
Area (∆ABC) = 64 cm
and area (APQR) = 121 cm
QR = 15.4 cm
∵ Triangles are similar
∴ {area(ΔABC)}/{area(ΔPQR)} = BC²/QR²
{Areas of similar triangles are similar to the squares of their corresponding sides}
⇒ 64/121 = (BC)²/(15.4)²
⇒ (8)²/(11)² = (BC)²/(15.4)²
⇒ BC/15.4 = 8/11
⇒ BC = (8/11) x 15.4
⇒ BC = 11.2 cm
Que-11: D, E, F are the mid-points of the sides BC, CA and AB respectively of a triangle ABC. Determine the ratio of the areas of the triangles DEF and ABC.

Sol: D, E and F are the mid-points of the sides
BC, CA and AB of ∆ABC
DE, EF and FD are joined
∵ E and F are mid-points of AC and AB
∴ EF || BC and = (1/2) BC
⇒ EF/BC = 1/2 … (i)
Similarly D and E are the mid-point of BC and AC
∵ DE || AB and = (1/2) AB
⇒ DE/AB = 1/2 … (ii)
and D and F are mid-points of BC and AB
∴ DF || AC and = (1/2) AC
⇒ DF/AC = 1/2 … (iii)
From (i), (ii) and (iii)
EF/BC = DE/AB = DF/AC ( each =1/2)
∆DEF ~ ∆ABC
∴ {area △DEF}/{area △ABC} = EF²/BC² = (1/2)² = 1/4
∴ area ADEF : area ∆ABC = 1 : 4
Que-12: ABCD is a trapezium in which AB = 2 DC and AB || DC. If AC and BD intersect at O, prove that area (∆AQB) = 4 area (ACOD).
Sol: ABCD is a trapezium in which AB || DC and
AB = 2DC
AC and BD intersect each other at O
In ∆AOB and ∆COD,
∠AOB = ∠COD
(vertically opposite anlges)
∠ABO = ∠ODC (alternate angles)
∴ ∆AOB ~ ∆COD (AAA axiom)
∵ ∵ {area(ΔAOB)}/{area(ΔCOD)} = AB²/DC² = (2DC)²/DC²
= 4DC²/DC² = 4/1
∴ area (∆AOQ) = 4 area (ACOD)
Hence proved.
Que-13: If the areas of two similar triangles are equal, prove that they are congruent.
Sol: Let ∆ABC and ADEF are similar
and area (∆ABC) = area (ADEF)
Area of ΔABC / Area of ΔDEF = (AB)2 / (DE)2 = (BC)2 / (EF)2 = (CA)2 / (FD)2 (According to
theorem 6.6) …………… (2)
From equation (1) and (2),
(AB)2 / (DE)2 = 1
⇒ (AB)2 = (DE)2
⇒ AB = DE …………… (3)
Similarly,
⇒ BC = EF ….(4)
⇒ CA = FD …..(5)
Now,
In ΔABC and ΔDEF
⇒ AB = DE [from equation(3)]
⇒ BC = EF [from equation (4)]
⇒ CA = FD [from equation (5)]
Thus, ΔABC ≅ ΔDEF (SSS axiom)
Que-14: Prove that the ratio of corresponding altitudes of two similar triangles is equal to the ratio of their corresponding sides.
Sol: In ∆ABC and ∆DEF,
AP ⊥ BC and DQ ⊥ EF
∵ ∆ABC ~ ADEF
∴ AB/DE = BC/EF = AC/DF and ∠B = ∠E
Now in ∆ABP and ∆DEQ,
∠B = ∠E (proved)
∠P = ∠Q (each 90°)
∴ ∆ABP ~ ∆DEQ (AA axiom)
∴ AB/DE = AP/DQ
But AB/DE = BC/EF (proved)
∴ AP/DQ = BC/EF
Hence proved.
Que-15: ABC is a triangle, PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides ∆ABC into two of equal parts equal area. Find BP/AB
Sol: In ∆ABC, P and Q are the points on AB and AC such that
and area (∆APQ) = area (quad, PBCQ)
⇒ area (∆APQ) + area (∆APQ) = area (quad. PBCQ) + area (∆APQ)
⇒ 2 area (∆PQ) = (1/2) area (∆ABC)
⇒ {area(ΔAPQ)}/{area(ΔABC)} = 1/2
∵ PQ || BC
∴ ∆APQ ~ ∆ABC
∴ {area(ΔAPQ)}/{area(ΔABC)} = AP²/AB² = 1/2
∴ AP²/AB² = 12
⇒ AP/AB = 1/√2
(AB−BP)/AB = 1/√2
⇒ (AB/AB)−(BP/AB) = 1/√2
1 – (BP/AB) = 1/√2
⇒ BP/AB = 1 − (1√2) = (√2−1)/√2
Hence BP/AB = (√2−1)/√2.
–: End of Similar Triangles Class 10 OP Malhotra Exe-12D ICSE Maths Solutions of Ch-12. :–
Return to :- OP Malhotra S Chand Solutions for ICSE Class-10 Maths
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