Similar Triangles Class 10 OP Malhotra Exe-12C ICSE Maths Solutions

Similar Triangles Class 10 OP Malhotra Exe-12C ICSE Maths Solutions Ch-12. We Provide Step by Step Solutions / Answer of Exe-12C Questions of S Chand OP Malhotra Maths . In this article you would learn solving complex 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-12C ICSE Maths Solutions

Similar Triangles Class 10 OP Malhotra Exe-12C ICSE Maths Solutions Ch-12

Board ICSE
Publications  S Chand
Subject Maths
Class 10th
Chapter-12 Similar Triangles
Writer OP Malhotra
Exe-12C Complex Problems on Similarity of Triangles
Edition 2024-2025

Exercise- 12C

Similar Triangles Class 10 OP Malhotra Exe-12C ICSE Maths Solutions Ch-12

Que-1: In ∆ABC, AB = 6 cm and AC = 3 cm. If M is the mid-point of AB, and a straight line through M parallel to BC cuts AC in N, what is the length of AN?

Sol: In ∆ABC, AB = 6 cm, AC = 3 cm
M is mid-point of AB
∴ AM = 1/2 AB = 1/2 x 6 = 3 cm
MN || BC is drawn
∴ N is mid-point of AC
∴ AN = 1/2 AC = 1/2 x 3 = 32 = 1.5 cm

Que-2: Draw parallelogram ABCD with the following data: AB = 6 cm, AD = 5 cm and ∠DAB = 45° Let AC and DB meet in O and let E be the mid-point of BC. Join OE. Prove that (i) OE || AB, (ii) OE = 1/2 AB.

Sol: Given : In parallelogram ABCD AB = 6 cm, AD = 5 cm
Diagonals AC and BD intersect each other at O
E is mid-point of BC, OE is joined
Draw parallelogram ABCD with the following data: AB = 6 cm, AD = 5 cm and ∠DAB = 45° Let AC and DB meet in O and let E be the mid-point of BC. Join OE. Prove that (i) OE || AB, (ii) OE = 1/2 AB.
To prove :
(i) OE || AB, (ii) OE = 1/2 AB
Proof:
In ∆ABC,
O is mid-point of AC
(∵ Diagonals of a parallelogram bisect each other)
and E is the mid-point of BC (given)
∴ OE || AB and OE = 1/2 AB
Hence proved.

Que-3: In the ∆ABC, D is the mid-point of AB, E is the mid-point of AC. Calculate : (i) DE, if BC = 6 cm (ii) ∠ADE, if ∠DBC = 140°
In the ∆ABC, D is the mid-point of AB, E is the mid-point of AC. Calculate : (i) DE, if BC = 6 cm (ii) ∠ADE, if ∠DBC = 140°

Sol: In the figure, ABC is a triangle, D is mid point of AE and E is mid point of AC
∵ D and E are the mid points of ∆ABC

(i) ∴ DE || BC and DE = 1/2 BC
and ∆ABC ~ ∆ADE
∴ AD/AB = DE/BC
⇒ (1/2)(AB/AB) = DE/6
⇒ DE/6 = 1/2
⇒ DE = 6/2 = 3 cm

(ii) ∠DBC = 140°
But ∠ADE = ∠DBC (corresponding angles)
= 140° (∵ ∠DBC = 140° given)
∴ ∠ADE = 140°

Que-4: ABCD is a trapezium with AB parallel to DC. If AB = 10 cm, AD = BC = 4 cm and ∠DAB = ∠CBA = 60°, calculate (i) the length of CD, (ii) the distance between AB and CD.

Sol: In trapezium ABCD,
AB || DC
AD = BC = 4 cm
∠DAB = ∠CBA = 60°
ABCD is a trapezium with AB parallel to DC. If AB = 10 cm, AD = BC = 4 cm and ∠DAB = ∠CBA = 60°, calculate (i) the length of CD, (ii) the distance between AB and CD.

Draw DL and CM perpendiculars on AB In ∆ADL and ABCM
∠L = ∠M (each 90°)
∠A = ∠B (each 60°)
DA = CB (each 4 cm)
∴ ∆ADL ≅ ABCM
∴ AL = MB
Now in right ∆ADL
Since DL/AD ⇒ sin 60° = DL/4
⇒ √3/2 = DL/4
⇒ DL = (4×√3)/2 = 2√3 cm
and tan 60° = DL/AL = √3 = 2√3/AL
⇒ AL = (2√3)/√3 = 2
∴ MB = AL = 2 cm
But CD = LM = AB – AL – MB
= 10 – 2 – 2
= 10 – 4
= 6 cm

Que-5: Find the unknown marked lengths in centimeters in the following figures.
Que-5: Find the unknown marked lengths in centimetres in the following figures.

Sol: (i) In ||gm ABCD, AB || CD
Diagonals AC and BD bisects each other at K
Now in ∆ACD,
∵ E is mid-point of AD and K is mid-point of AD
(∵ DE = EA = 6 cm and AK = KC – 8 cm)
∴ EK or EKF || CD or AB
AE/AD = AK/AC = EK/CD
⇒ 6/(6+6) = 8/(8+8) = 5/y
⇒ 5y = 6/12
⇒ 5y = 1/2
⇒ y = 5 x 2 = 10
∴ y = 10 cm
Similarly in ∆ACB,
KE || AB
∴ CK/CA = KF/AB
⇒ 8/16 = x/10
⇒ x = (8×10)/16 = 5
Hence x = 5 cm, y= 10 cm

(ii) In the figure, PQRS is a parallelogram
OL = LT = 6, ST = 4, LS = 3
PQ = y and PL = x
We have to find the value of x and y
In ∆QTR,
∵ L is mid-point of QT (∵ QL = LT = 6 cm)
and PLS || QR
∴ S is mid-point of TR
∴ TS = SR ⇒ SR = 4
But PQ = SR (opposite sides of ||gm)
∴ PQ = 4 or y = 4
In APLQ and ASLT
LQ = LT (each = 6 cm)
PQ = TS (each = 4 cm)
∠PLQ = ∠SLT (vertically opposite angles)
∴ ∆PLQ = ∆SLT (SAS axiom)
∴ PL = LS ⇒ x = 3
Hence x = 3, y = 4

(iii) In ∆ABD,
AL = LD = 3
∴ L is mid-point of AD
∵ LM || BD
∴ M is mid-point of AB
∴ ∆ALM ~ ∆ADB
∴ AL/AD = LM/BD
⇒ 3/6 = q/p
⇒ q/p = 1/2
⇒ 2q = p … (i)
Similarly in ABCD,
CP = PD (each =11)
∴ P is mid-point of CD
and PQ || BD
∴ ∆CPQ ~ ∆CDB
∴ CP/CD = PQ/DB
⇒ 11/22 = 8/p
⇒ P = (22×8)/11 = 16
and from (i) 2q = p
2q = 16 ⇒ q = 162 = 8
∴ p = 16, q = 8.

Que-6: ABC is an isosceles triangle. AB = AC = 10 cm, BC = 12 cm. PQRS is a rectangle drawn inside the isosceles triangle. Given PQ = SR = y cm, prove that  x = 6 – {3y/4}.
Que-6: ABC is an isosceles triangle. AB = AC = 10 cm, BC = 12 cm. PQRS is a rectangle drawn inside the isosceles triangle. Given PQ = SR = y cm, prove that  x = 6 - {3y/4}.

Sol: L is mid-point of BC
∴ BL = LC = 12/2 = 6 cm
Now in right ∆ABL
AB² = AL² + BL²
⇒ (10)² = AL² + (6)²
⇒ AL² = (10)² – (6)²
⇒ AL² = 100 – 36 = 64 = (8)²
∴ AL = 8 cm
Now in ABPQ and ABAL
∠Q = ∠L (each 90°)
∠B = ∠B (common)
∴ ∆BPQ ~ ∆BAL
∴ PQ/AL = BQ/BC
⇒ y/8 = (6−x)/6
⇒ 6y = 48 – 8x
⇒ 8x = 48 – 6y
⇒ x = (48−6y)/8 = 6 − {3y/4}
Hence x = 6 – {3y/4}
Hence proved.

Que-7: In the figure, PQ || ST. Prove that (a) As PQR and STR are similar. (b) PR.RT = QR.RS.
Que-7: In the figure, PQ || ST. Prove that (a) As PQR and STR are similar. (b) PR.RT = QR.RS.

Sol: Given : In the figure, PQ || ST
(a) In APQR and ASTR
∠PRQ = ∠SRT (vertically opposite angles)
∠QPR = ∠RST (alternate angles)
∴ ∆PQR ~ ∆STR (AA axiom)

(b) PR/RS = RQ/RT (sides are proportional)
⇒ PR x RT = QR x RS
(By cross multiplication)
Hence proved.

Que-8: In the figure ABCD is a trapezium. AB || DC and the diagonals AC, BD intersect in E. Prove that AE.DE = BE.CE.
Que-8: In the figure ABCD is a trapezium. AB || DC and the diagonals AC, BD intersect in E. Prove that AE.DE = BE.CE.

Sol: In ∆AEB and ACED
∠AEB = ∠CED (vertically opposite angles)
∠EAB = ∠ECD (alternate angles)
and ∠EBA = ∠EDC
∴ ∆AEB ~ ∆CED (AAA axiom)
∴ AE/CE = BE/DE (sides are proportional)
∴ AE.DE = BE.CE (By cross multiplication)
Hence proved.

Que-9: Perpendiculars AL, BM are drawn from the vertices A, B of a triangle ABC to meet BC, AC at L, M. By proving the triangles ALC, BMC similar, or otherwise, prove that CM.CA = CL.CB.

Sol: In ∆ALC and ∆BMC
∠L = ∠M (each 90°)
∠C = ∠C (common)
∴ ∆ALC ~ ∆BMC (AA axiom)
∴ CM/CL = CB/CA (sides are proportional)
⇒ CM.CA = CB.CL (By cross multiplication)
⇒ CM.CA = CL.CB
Hence proved.

Que-10: If a perpendicular is drawn from the right angle of a right-angled triangle to the hypotenuse, prove that the triangles on each side of the perpendicular are similar to the whole triangle and to each other.

Sol: (i) In ∆ABD and ∆ABC
∠B = ∠B (common)
∠ADB = ∠BAC (each 90°)
∴ ∆ABD ~ ∆ABC (AA axiom)

(ii) Similarly in ∆ACD and ∆ABC
∠C = ∠C (common)
∠ADC = ∠BAC (each 90°)
∴ ∆ACD ~ ∆ABC (AA axiom)
Hence proved.

Que-11: In the figure, it is given that the angle BAD = the angle CAE and the angle B = the angle D. Prove that (i) AB/AD = AC/AE (ii) the angle ADB = the angle AEC. (SC)
In the figure, it is given that the angle BAD = the angle CAE and the angle B = the angle D. Prove that (i) AB/AD = AC/AE (ii) the angle ADB = the angle AEC. (SC)

Sol: Given : In the given figure, ∠BAD = ∠CAE
∠B = ∠D.
Adding ∠CAD both sides,
∠BAD + ∠CAD = ∠CAD + ∠CAE
⇒ ∠BAC = ∠DAE
Now in ∆ABC and ∆ADE
∠BAC = ∠DAE (proved)
∠B = ∠D (given)
(i) ∴ ∆ABC ~ ∆ADE (AA axiom)
∴ AB/AD = AC/AE
or AB/AC = AD/AE

(ii) Now in ∆ABD and ∆AEC
∠BAD = ∠CAE (given)
AB/AC = AD/AE ⇒ A (proved)
∴ ∆ABD ~ ∆AEC (SAS axiom)
∴ ∠ADB = ∠AEC
Hence proved.

Que-12: In the figure, ABCD and AEFG are squares. Prove that (i) AF : AG = AC : AD; (ii) triangles ACF and ADG are similar. (SC)
In the figure, ABCD and AEFG are squares. Prove that (i) AF : AG = AC : AD; (ii) triangles ACF and ADG are similar. (SC)

Sol: AC and AF are the diagonals of two squares ABCD and AEFG respectively
∴ ∆ADC ~ ∆AGF (A)
Now ∠BAF = ∠B AC – ∠FAC
∠GAC = ∠GAF – ∠FAC
But ∠BAC = ∠GAF = 45°
(Diagonals bisects the opposite angles of a square)
∠BAF = 45° – ∠FAC … (i)
and ∠GAC = 45° – ∠FAC … (ii)
From (i) and (ii)
∴ ∠BAF = ∠GAC
Similarly ∠DAG = ∠DAC – ∠GAC = 45° – ∠GAC
and ∠FAC = ∠BAC – ∠BAF = 45° -∠BAF
But ∠GAC = ∠BAF (proved)
∴ ∠DAG = ∠FAC
Now in ∆ADG and ∆ACF
∠DAG = ∠FAC (proved)
∴ From (A)
∆ADG ~ ∆ACF (SAS axiom)
∴ AG/AF = AD/AC (sides are proportional)
⇒ AF/AG = AC/AD
⇒ AF : AG = AC : AD
Hence proved.

Que-13: In a square ABCD, the bisector of the angle BAC cuts BD at X and BC at Y. Prove that the triangles ACY, ABX are similar. (SC)

Sol: In ∆ACY and ∆ABX
∠CAY =∠BAY
(∵ AXY is the bisector of ∠CAB)
∠ACY = ∠ABX (each 45°)
∴ ∆ACY ~ ∆ABX (∆A axiom)
Hence proved.

Que-14: On one of the longer sides PQ of a rectangle PQRS, a point X is taken such that SX² = PX.PQ. Prove that ∆s PXS, XSR are similar. (SC)

Sol: SX² = PX.PQ
⇒ SX/PQ = PX/SX
⇒ SX/SR = PX/SX
(∵ SR = PQ opposite sides of rectangle)
Now in APXS and AXSR
SX/SR = PX/SX (proved)
∠PXS = ∠RSX (alternate angles)
∴ ∆PXS ~ ∆XSR (SAS axiom)
Hence proved.

Que-15: In the figure, D is a point on the side BC of ∆ABC such that ∠ADC = ∠BAC. Prove that BC/CA = CA/CD.

Sol: In ∆ABC and ∆ADC
∠BAC = ∠ADC (given)
∠C = ∠C (common)
∴ ∆ABC ~ ∆ADC (AA axiom)
∴ BC/AC = AC/CD
⇒ BC/CA = CA/CD
Hence proved.

Que-16: In the figure, PQR is a triangle in which PQ = PR and Z is a point on the side PR such that QR² = PR x RZ. Prove that QZ = QR.
Que-16: In the figure, PQR is a triangle in which PQ = PR and Z is a point on the side PR such that QR² = PR x RZ. Prove that QZ = QR.

Sol: Given : In ∆PQR, PQ = PR
Z is a point on PR such that
QR² = PR x RZ
∴ ΔPQR ∼ ΔQ×R (SAS criterion)
∴ PQ/QZ = QR/ZR = PR/QR (corresponding parts)
⇒ PQ/QZ = QR/VR
or, PR/QZ = QR/ZR (∵PQ=PR)
⇒ QR = QZ proved.

Que-17: In the figure, segments AD and BE are perpendicular to the sides BC and AC respectively. Such that

(i) ∆ADC ~ ∆BEC;
(ii) CA.CE = CB.CD;
(iii) ∆ABC ~ ∆DEC;
(iv) CD.AB = CA.DE.
In the figure, segments AD and BE are perpendicular to the sides BC and AC respectively. Such that (i) ∆ADC ~ ∆BEC; (ii) CA.CE = CB.CD; (iii) ∆ABC ~ ∆DEC; (iv) CD.AB = CA.DE.

Sol: (i) In ∆ADC and ∆BEC
∠C = ∠C (common)
∠ADC = ∠BEC (each 90°)
∴ ∆ADC ~ ∆BEC (AA axiom)
∴ CA/CB = CD/CE  (sides are proportional)

(ii) ∴ CA.CE = CB.CD
Again in ∆ABC and ∆DEC
∠C = ∠C   (same)
CA/CD = CB/CE
∴ ∆ABC ~ ∆DEC (SAS axiom)
∴ CA/CD = AB/DE     (corresponding sides are proportional)

(iv) ⇒ CA.DE = CD.AB
or CD.AB = CA.DE
Hence proved.

Que-18: In the figure, ABCD is a quadrilateral P, Q, R, S are the points of trisection of the sides AB, BC, CD and DA respectively and are adjacent to A and C, prove that PQRS is a parallelogram.
Que-18: In the figure, ABCD is a quadrilateral P, Q, R, S are the points of trisection of the sides AB, BC, CD and DA respectively and are adjacent to A and C, prove that PQRS is a parallelogram.

Sol: Join diagonals AC
Proof: In ∆ABC
AP = 1/3 AB and CQ = 1/3 BC or AP/AB = CQ/BC
= 1/3
∴ PQ || AC and PQ = 2/3 AC … (i)
Similarly in ∆ADC,
AS = 1/3 AD and CR = 1/3 CD or AS/AD = CR/CD = 2/3
∴ RS || AC and RS = 2/3 AC … (ii)
From (i) and (ii)
PQRS is a parallelogram
Hence proved.

Que-19: In the figure, if AD ⊥ BC, BD = 4, AD = 6, CD = 9, prove that triangles ADB and CDA are similar. Also prove that ∠BAC is a right angle.
Que-19: In the figure, if AD ⊥ BC, BD = 4, AD = 6, CD = 9, prove that triangles ADB and CDA are similar. Also prove that ∠BAC is a right angle.

Sol: In ∆ADB and ∆CDA
∠ADB = ∠ADC (each 90°)
AD/BD = 6/4,  CD/AD = 9/6
⇒ AD/BD = 3/2 and CD/AD = 3/2
∴ AD/BD = CD/AD (each 2/3)
∴ ∆ADB ~ ACDA (SAS axiom)
∴ ∠ABD = ∠CAD and ∠BAD = ∠ACD
∴ ∠BAD + ∠CAB = ∠ABD + ∠ACD
⇒ ∠BAC = ∠ABD + ∠ACD
But ∠BAC + ∠ABD + ∠ACD = 180°
∴ ∠BAC = 90°
Hence proved.

Que-20: If two triangle are equiangular, prove that the ratio of the corresponding sides is same as the ratio of corresponding altitudes.

Sol: ∆ABC ~ ∆DEF
∠B = ∠E   and  AB/DE = BC/EF    ………… (i)
Now in ∆ABL and ∆DEM
∠B = ∠E   (proved)
∠L = ∠M    (each 90°)
∆ABL ~ ∆DEM   (AA axiom)
AB/DE = AL/DM   ……………… (ii)
From (i) and (ii)
AB/DE = BC/EF = AL/DM
Hence Proved.

Que-21: The diagonal BD of a parallelogram ABCD intersect AE at a point F, where E is any point on side BC. Prove that DF.EF = FB.FA.

Sol: In ∆AFD and ABFE
∠AFD = ∠BFE (vertically opposite angles)
∠ADF = ∠FBE (Alternate angles)
∴ ∆AFD ~ ∆BFE (AA axiom)
∴ DF/FB = FA/FE (Sides are proportional)
∴ DF.EF = FB.FA (By cross multiplication)
Hence proved.

Que-22: Any point X inside ADEF is joined to its vertices. From a point P in DX, PQ is drawn || DE meeting XE at Q and QR is drawn || EF meeting XF in R. Prove that PR || DF.
Que-22: Any point X inside ADEF is joined to its vertices. From a point P in DX, PQ is drawn || DE meeting XE at Q and QR is drawn || EF meeting XF in R. Prove that PR || DF.

Sol: P is any point on DX and PQ || DE
∴ XP/PD = XQ/QE
∵ QR || EF
∴ XQ/QE = XR/RF
From (i) and (ii)
XP/PD = XR/RF
∴ PR || DF
Hence proved.

–: End of Similar Triangles Class 10 OP Malhotra Exe-12C ICSE Maths Solutions Ch-12 :–

Return to :-  OP Malhotra S Chand Solutions for ICSE Class-10 Maths

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