CG Board Class 10 Maths Chapter 11 Solutions: Similarity in Geometrical Shapes (ज्यामितीय आकृतियों में समरूपता) (2026-27)

Congruent Figures:** Geometrical shapes that have identical shape and identical size. All congruent figures are similar with a scale factor of 1.

• Similar Figures: Geometrical shapes that have the same shape, but not necessarily the same size. All circles, all squares, and all equilateral triangles are similar to each other.

Scale Factor (m):** The fixed ratio of the measurements of corresponding sides of two similar shapes.

• If m > 1, the figure is enlarged.
• If m < 1, the figure is reduced.
• If m = 1, the figure remains congruent.
• Perimeter Ratio: The ratio of the perimeters of two similar polygons is equal to the ratio of their corresponding sides and their scale factor:

(Perimeter of Polygon 1)/(Perimeter of Polygon 2) = Scale Factor (m) = (Side₁)/(Side₂)

• Theorem 1 (Thales Theorem): If a line is drawn parallel to one side of a triangle intersecting the other two sides at distinct points, then it divides the two sides in the same ratio.

In Δ ABC, if DE ∥ BC, then:

AD/DB = AE/EC

• Theorem 2 (Converse of Thales Theorem): If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
• हिंदी माध्यम:
• प्रमेय 1 (थेल्स प्रमेय): यदि किसी त्रिभुज की एक भुजा के समांतर बाकी दो भुजाओं को प्रतिच्छेद करते हुए एक रेखा खींची जाए, तो वह रेखा उन दोनों भुजाओं को समान अनुपात में विभाजित करती है।
• प्रमेय 2 (थेल्स प्रमेय का विलोम): यदि कोई रेखा किसी त्रिभुज की दो भुजाओं को समान अनुपात में विभाजित करे, तो वह रेखा तीसरी भुजा के समांतर होती है।

दो समरूप त्रिभुजों के क्षेत्रफलों का अनुपात उनकी संगत भुजाओं (या शीर्षलंबों या मध्यिकाओं) के अनुपात के वर्ग के बराबर होता है।

• Given map dimensions: Length = 3 cm, Width = 4 cm.
• Scale: 1 cm on map = 10 m in reality (as per textbook standard scaling key).
• Actual length = 3 × 10 = 30 m
• Actual width = 4 × 10 = 40 m
• Actual Area of farm:

Area = 30 m × 40 m = 1200 m²

• Final Answer: Actual area of the farm is 1200 square meters.

• Area of original square painting = 3600 cm².
• Original side of painting = √(3600) = 60 cm.
• Scale Factor (m) = 0.1.
• Side length after scaling:

New Side = Original Side × Scale Factor = 60 cm × 0.1 = 6 cm

• Final Answer: The side of the painting after scaling is 6 cm.

• Map distance = 3 cm.
• Scale: 2 cm represents 7 km.
• Therefore, 1 cm represents 7/2 = 3.5 km.
• Real distance:

Real Distance = 3 × 3.5 km = 10.5 km

• Final Answer: Real distance between Railway station and Airport is 10.5 km.

### Exercise – 2 (प्रश्नावली – 2)

• Actual radius = 52 m.
• Scale: 13 m = 1 cm.
• Radius on map:

Radius on map = 52/13 = 4 cm

• Final Answer: Radius of the field on the map is 4 cm.

(i) Scale Factor = 0.8:**

• (ii) Scale Factor = 1.2:
• (iii) Scale Factor = 1.0:

(i) Ratio TN/NR:**

• (ii) Ratio TR/NR:
• (iii) Ratio TN/RT:

• Given: In Δ ABC, point D is the midpoint of side AB, so AD = DB ⇒ AD/DB = 1. Line DE ∥ BC intersects AC at point E.
• To Prove: E is the midpoint of AC (i.e., AE = EC).
• Proof:

By Basic Proportionality Theorem (Thales Theorem) in Δ ABC:

AD/DB = AE/EC

Since AD/DB = 1:

1 = AE/EC ⇒ AE = EC

• Thus, line DE bisects side AC at point E. (Proved)

• In Δ ABE: DF ∥ AE.

By Basic Proportionality Theorem:

BD/DA = BF/FE — (Equation 1)

• In Δ ABC: DE ∥ AC.

By Basic Proportionality Theorem:

BD/DA = BE/EC — (Equation 2)

• From Equation (1) and Equation (2):

BF/FE = BE/EC

• Hence proved. (Proved)

(i) Calculation:**

• (ii) Calculation:
• (iii) Calculation:

• In Δ POQ: DE ∥ OQ.

By Basic Proportionality Theorem:

PE/EQ = PD/DO — (Equation 1)

• In Δ POR: DF ∥ OR.

By Basic Proportionality Theorem:

PF/FR = PD/DO — (Equation 2)

• From Equation (1) and Equation (2):

PE/EQ = PF/FR

• By converse of Basic Proportionality Theorem in Δ PQR, EF ∥ QR. (Proved)

• In Δ OPQ: AB ∥ PQ.

By Basic Proportionality Theorem:

OA/AP = OB/BQ — (Equation 1)

• In Δ OPR: AC ∥ PR.

By Basic Proportionality Theorem:

OA/AP = OC/CR — (Equation 2)

• From Equation (1) and Equation (2):

OB/BQ = OC/CR

• By converse of Basic Proportionality Theorem in Δ OQR, BC ∥ QR. (Proved)

### Exercise – 3 (प्रश्नावली – 3)

(i) Value of OP:**

• (ii) Value of EA:
• (iii) m∠ OPL:
• (iv) m∠ LMN:
• (v) Polygon DEABC ~ Polygon OPLMN.
• (vi) Correct Statement: Polygon BCDEA ~ Polygon MNOPL.

(i) Right Triangles:**

• (ii) Quadrilaterals:

(i) Quadrilaterals:**

• (ii) 30°-60°-90° Triangles:
• (iii) Trapeziums:
• (iv) Polygons:

(i) Scale factor:**

• (ii) Perimeters:
• (iii) Ratio of perimeters (1st to 2nd):

(i) Congruent ⇒ Similar:** Congruent polygons have all corresponding angles equal and all corresponding sides equal (ratio = 1). Since both similarity conditions are satisfied, they are always similar.

• (ii) Similar ⇏ Congruent: Two squares of sides 2 cm and 5 cm have equal angles (90°) and proportional sides (ratio 2:5), so they are similar but not congruent due to different side lengths.
• (iii) Three Additional Statements:

• Using similarity of triangles formed by intersecting diagonals / parallel sides:
• x = 27/5 = 5.4.
• y = 9.
• Final Answer: x = 5.4, y = 9.

• Since AB ∥ DC, in Δ AOB and Δ COD:
• ∠ OAB = ∠ OCD (Alternate interior angles)
• ∠ OBA = ∠ ODC (Alternate interior angles)
• Therefore, Δ AOB ∼ Δ COD by AA similarity criterion.
• Area ratio formula:

(ar(Δ AOB))/(ar(Δ COD)) = (AB/CD)² = (2CD/CD)² = 4/1.

• Final Answer: Ratio of areas is 4 : 1.

• By similarity of triangles across the river banks:

(Width)/(EB) = IV/VE ⇒ (Width)/(15) = 36/20

Width = (36 × 15)/(20) = 27 meters.

• Final Answer: Width of the river is 27 meters.

• Given: Δ ABC ∼ Δ DEF and ar(Δ ABC) = ar(Δ DEF).
• Proof:

(ar(Δ ABC))/(ar(Δ DEF)) = (AB/DE)² = (BC/EF)² = (AC/DF)².

Since area ratio = 1:

1 = (AB/DE)² ⇒ AB = DE

1 = (BC/EF)² ⇒ BC = EF

1 = (AC/DF)² ⇒ AC = DF

• Since all three pairs of corresponding sides are equal, Δ ABC ≅ Δ DEF by SSS congruence. (Proved)

• Let AD and PS be medians of similar triangles Δ ABC and Δ PQR respectively.
• Since Δ ABC ∼ Δ PQR:

AB/PQ = BC/QR = 2BD/2QS = BD/QS

And ∠ B = ∠ Q.

• In Δ ABD and Δ PQS:
• AB/PQ = BD/QS
• ∠ B = ∠ Q
• Δ ABD ∼ Δ PQS (by SAS similarity).
• Therefore, AB/PQ = AD/PS.
• Area ratio:

(ar(Δ ABC))/(ar(Δ PQR)) = (AB/PQ)² = (AD/PS)². (Proved)

• Given: Height of Sahid HE = 1.2 m (textbook figure value), DE = 1.6 m, DF = 4.4 m.
• By AA similarity between Δ HED and Δ QFD:

QF/HE = DF/DE ⇒ QF/1.2 = 4.4/1.6

QF = (4.4 × 1.2)/(1.6) = 3.3 meters.

• Final Answer: Height of the pole is 3.3 meters.

(i) Similar Triangles: Triangle ACM and Triangle CBM**.

• (ii) Values: x = 15, y = 9.

• Since D is midpoint of BC, BD = BC/2 ⇒ BC/BD = 2.
• Both are equilateral triangles, so Δ ABC ∼ Δ BDE.
• (ar(Δ ABC))/(ar(Δ BDE)) = (BC/BD)² = (2/1)² = 4/1 = 4 : 1.
• Final Answer: (iii) 4:1.

### Exercise – 5 (प्रश्नावली – 5)

(i) Figure (i): First is a parallelogram, second is not. Opposite angles and side proportions do not match, so not similar**.

• (ii) Figure (ii): Side ratios: 15/10 = 1.5 and 18/12 = 1.5. Vertically opposite angles are equal. Similar by SAS criterion.
• (iii) Figure (iii): Only one angle (100°) is given; remaining angles and sides are unknown, so similarity cannot be determined (Not similar / अपर्याप्त जानकारी).

(i) Scale Factor:**

• (ii) Diagonals of bigger kite:

• In right Δ PQR with ∠ P = 90°and PM ⊥ QR:
• In Δ PMR and Δ QMP:
• ∠ PMR = ∠ QMP = 90°* ∠ RPM = ∠ PQM = (90°- ∠ QPM)
• Therefore, Δ PMR ∼ Δ QMP (by AA similarity).
• Corresponding sides ratio:

PM/QM = MR/PM ⇒ PM² = QM · MR

• Hence proved. (Proved)

• In equilateral Δ ABC of side 2a, altitude AD ⊥ BC bisects base BC, so BD = a.
• By Pythagoras theorem in right Δ ABD:

AD² = AB² – BD² = (2a)² – a² = 4a² – a² = 3a²

AD = √(3a²) = a√3

• Final Answer: Length of each altitude is √3a.

• Since Δ ABC is isosceles with ∠ C = 90°, AC = BC.
• By Pythagoras theorem in Δ ABC:

AB² = AC² + BC²

Since BC = AC:

AB² = AC² + AC² = 2AC²

• Hence proved. (Proved)

• In Δ ORS and Δ OR’S’:
• OR’/OR = 2, OS’/OS = 2, and ∠ ROS is common.
• Δ ORS ∼ Δ OR’S’ by SAS similarity ⇒ (R’S’)/(RS) = 2.
• Similarly, (S’T’)/(ST) = 2 and (R’T’)/(RT) = 2.
• All corresponding sides are in the same ratio (2:1).
• Therefore, Δ RST ∼ Δ R’S’T’ by SSS similarity criterion. (Proved)

• In right Δ ACE: AE² = AC² + CE² (Pythagoras Theorem).
• In right Δ BCD: BD² = BC² + CD² (Pythagoras Theorem).
• Adding both equations:

AE² + BD² = (AC² + BC²) + (CE² + CD²)

• In right Δ ABC: AC² + BC² = AB².
• In right Δ CDE: CE² + CD² = DE².
• Substitution yields:

AE² + BD² = AB² + DE²

• Hence proved. (Proved)

• In right Δ ACB, CD ⊥ AB.
• Δ BCD ∼ Δ BAC ⇒ BC/AB = BD/BC ⇒ BC² = BD · AB — (1).
• Δ ACD ∼ Δ ABC ⇒ AC/AB = AD/AC ⇒ AC² = AD · AB — (2).
• Dividing Equation (1) by Equation (2):

(BC²)/(AC²) = (BD · AB)/(AD · AB) = BD/AD

• Hence proved. (Proved)

• Scale: 8000 miles = 1 inch on paper.
• Diameter of Sun on paper:

Diameter = (864000)/(8000) = 108 inches.

• Distance between Sun and Earth on paper:
• 92 million miles = 92,000,000 miles.

Distance = (92000000)/(8000) = 11500 inches.

• Final Answer: 108 inches and 11500 inches.

• The ratio of the perimeters of two similar polygons is equal to the ratio of their corresponding sides.
• Since all regular hexagons are similar polygons:

(Side₁)/(Side₂) = (Perimeter₁)/(Perimeter₂) = 5/4

• Final Answer: Ratio of their sides is 5 : 4.