CBSE X Maths — Triangles: Similarity: Notes, Practice Questions & Worked Solutions

CBSE Class X Mathematics (041/241) · Unit IV: Geometry

Triangles — Similarity

Similar triangles have the same shape but not necessarily the same size — their corresponding angles are equal, and their corresponding sides are all in the same ratio. This topic is built around the Basic Proportionality Theorem and the criteria used to prove two triangles similar.

1. The Basic Proportionality Theorem (BPT)

Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

The converse also holds: if a line divides two sides of a triangle in the same ratio, the line is parallel to the third side.

Worked example — applying BPT In triangle ABC, DE∥BC, with AD=1.5, DB=3, AE=1. Find EC.
By BPT: AD/DB = AE/EC → 1.5/3 = 1/EC → EC = 1/0.5 = 2

Worked example — using the converse of BPT D and E are on AB and AC with AD=4, DB=4.5, AE=8, EC=9. Is DE∥BC?
AD/DB = 4/4.5 ≈ 0.889. AE/EC = 8/9 ≈ 0.889. Since the ratios are equal, by the converse of BPT, DE is parallel to BC.

2. Criteria for Similarity of Triangles

Two triangles are similar if any one of the following holds:

Criterion	Condition
AA	Two pairs of corresponding angles are equal
SSS	Corresponding sides are all proportional
SAS	One pair of corresponding angles equal, and the sides including them proportional

Worked example — using similarity ratios ΔABC ~ ΔDEF. The perimeter of ABC is 25 cm, the perimeter of DEF is 15 cm, and DE = 9 cm. Find AB.
Since the triangles are similar, the ratio of corresponding sides equals the ratio of perimeters: 25/15 = 5/3.
AB/DE = 5/3 → AB = 9 × 5/3 = 15 cm

Worked example — a classic similarity proof In ΔABC, D is a point on BC such that AD⊥BC and AD² = BD×DC. Prove that ∠BAC = 90°.
From AD²=BD×DC: AD/BD = DC/AD. Since ∠ADB=∠ADC=90° (both right angles, as AD⊥BC), triangles ADB and CDA are similar by SAS (the included angle and the ratio of sides around it match).
So ∠BAD = ∠DCA and ∠ABD = ∠DAC. Adding: ∠BAD + ∠DAC = ∠DCA + ∠ABD, i.e. ∠BAC = ∠ABC + ∠ACB.
Since angles in ΔABC sum to 180°: ∠BAC + (∠ABC+∠ACB) = 180° → ∠BAC + ∠BAC = 180° → ∠BAC = 90°.

Common mistakes to watch for

Matching up the wrong vertices when writing a similarity statement — ΔABC~ΔDEF means A↔D, B↔E, C↔F specifically.
Confusing the ratio of sides with the ratio of perimeters — for similar triangles, they’re actually the same ratio, which is a useful shortcut.
Trying to use BPT on a line that doesn’t actually intersect both sides of the triangle in distinct points.

Key Vocabulary Recap

Similar triangles — same shape, corresponding angles equal, sides proportional

Basic Proportionality Theorem — a line parallel to one side divides the other two proportionally

Syllabus Reference

This note covers CBSE Class X Mathematics Unit IV, Triangles: definitions and examples of similar triangles, the Basic Proportionality Theorem and its converse, and the AA, SSS, and SAS criteria for similarity (stated without proof, except BPT which is proved).

Practice questions

CBSE Class X Mathematics · Unit IV: Triangles — Practice Question Bank (no answers)

Triangles — Practice

Covers: the Basic Proportionality Theorem, its converse, and similarity criteria. 10 questions.

In triangle ABC, DE∥BC, with AD=1.5, DB=3, AE=1. Find EC.

D and E are points on sides AB and AC such that AD=5.7, DB=9.5, AE=4.8, EC=8. State, with reasoning, whether DE∥BC.

D and E are points on sides AB and AC such that AD=4, DB=4.5, AE=8, EC=9. Is DE∥BC? Justify your answer.

Two triangles ABC and PQR are similar, with AB=4, BC=6, PQ=8. Find QR.

State the three criteria (AA, SSS, SAS) used to prove two triangles similar.

If ΔABC ~ ΔDEF, the perimeter of ABC is 25 cm, the perimeter of DEF is 15 cm, and DE=9 cm, find AB.

Prove the Basic Proportionality Theorem.

In triangle PQR, ST∥QR, with PS=3 cm, SQ=6 cm, PT=5 cm. Find TR.

State (without proof) the condition under which two triangles are similar if their corresponding sides are proportional.

Q10

In ΔABC, D is a point on BC such that AD⊥BC and AD² = BD×DC. Prove that ∠BAC = 90°.

More Practice

Q11

In triangle ABC, DE∥BC, with AD/DB = 3/5, and AC = 5.6 cm. Find AE.

Q12

ΔABC ~ ΔPQR, with ∠A = 32° and ∠R = 65°. Find ∠B.

Q13

Diagonals of a trapezium ABCD with AB∥CD intersect at point O. Show that AO/OC = BO/OD.

Q14

ΔABC ~ ΔEDF, with AB=5 cm, AC=7 cm, DF=15 cm, DE=12 cm. Find the remaining sides EF and BC.

Q15

E is a point on side AD produced of parallelogram ABCD, and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

Q16

Two poles of height 6 m and 11 m stand vertically on the ground. If the distance between their feet is 12 m, find the distance between their tops.

Q17

In ΔABC, X and Y are points on AB and AC such that AX=1 cm, XB=3 cm, AY=1.5 cm, YC=4.5 cm. State whether XY∥BC.

Q18

Sides AB and AC, and median AD of triangle ABC, are respectively proportional to sides PQ, PR, and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.

Q19

A vertical pole 6 m long casts a shadow 4 m long, and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Q20

Prove that if the square of one side of a triangle equals the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Worked solutions

Show worked solutions

CBSE Class X Mathematics · Unit IV: Triangles — Solutions

Triangles — Practice: Solutions

DE∥BC, AD=1.5,DB=3,AE=1. Find EC.

By BPT: AD/DB = AE/EC

1.5/3 = 1/EC

✓EC = 2

AD=5.7,DB=9.5,AE=4.8,EC=8. Is DE∥BC?

Compare the two ratios

AD/DB=5.7/9.5=0.6, AE/EC=4.8/8=0.6

✓The ratios are equal, so by the converse of BPT, DE∥BC

AD=4,DB=4.5,AE=8,EC=9. Is DE∥BC?

Compare the two ratios

AD/DB=4/4.5≈0.889, AE/EC=8/9≈0.889

✓The ratios are equal, so DE∥BC

ABC~PQR, AB=4,BC=6,PQ=8. Find QR.

Corresponding sides are proportional: AB/PQ=BC/QR

4/8 = 6/QR

✓QR = 12

State AA, SSS, SAS similarity criteria.

✓AA: two pairs of corresponding angles are equal. SSS: all three pairs of corresponding sides are proportional. SAS: one pair of corresponding angles is equal, and the sides including those angles are proportional.

Perimeters 25,15; DE=9. Find AB.

Ratio of sides = ratio of perimeters = 25/15

AB = 9 × (25/15)

✓AB = 15 cm

Prove the Basic Proportionality Theorem.

✓Given ΔABC with DE∥BC (D on AB, E on AC), join BE and CD, and draw DM⊥AC and EN⊥AB. Since triangles ADE and BDE share the same base DE and lie between the same parallels DE and BC, they have equal area, i.e. ar(ADE)=ar(BDE)… [and similarly ar(ADE)=ar(CDE)]. Using area = ½×base×height for triangles ADE & DEB (common vertex E, bases AD & DB on the same line), ar(ADE)/ar(DEB)=AD/DB. Similarly ar(ADE)/ar(DEC)=AE/EC. Since ar(DEB)=ar(DEC) (both share base DE, same parallels), it follows that AD/DB = AE/EC — proving the two sides are divided in the same ratio.

ST∥QR, PS=3,SQ=6,PT=5. Find TR.

By BPT: PS/SQ = PT/TR

3/6 = 5/TR

✓TR = 10

Condition for similarity from proportional sides.

✓If in two triangles, the corresponding sides are proportional, then their corresponding angles are equal, and the two triangles are similar (the SSS similarity criterion, stated without proof).

Q10

AD⊥BC, AD²=BD×DC. Prove ∠BAC=90°.

✓From AD²=BD×DC: AD/BD=DC/AD. Since ∠ADB=∠ADC=90°, triangles ADB and CDA are similar (SAS, using the matching ratio around the equal right angle). So ∠BAD=∠DCA and ∠ABD=∠DAC. Adding: ∠BAD+∠DAC=∠DCA+∠ABD, i.e. ∠BAC=∠ABC+∠ACB. Since the three angles of ΔABC sum to 180°: ∠BAC+∠BAC=180°, so ∠BAC=90°.

More Practice — Solutions

Q11

DE∥BC, AD/DB=3/5, AC=5.6. Find AE.

By BPT: AD/DB=AE/EC. Let AE=x, EC=5.6−x

3/5=x/(5.6−x) → 3(5.6−x)=5x

✓16.8=8x → AE = 2.1 cm

Q12

ABC~PQR, ∠A=32°, ∠R=65°. Find ∠B.

Correspondence: A↔P, B↔Q, C↔R, so ∠C=∠R=65°

∠B=180−32−65

✓∠B = 83°

Q13

Trapezium diagonals: show AO/OC=BO/OD.

✓In ΔAOB and ΔCOD: ∠AOB=∠COD (vertically opposite), and since AB∥CD, ∠OAB=∠OCD and ∠OBA=∠ODC (alternate angles). So ΔAOB~ΔCOD (AA). Hence AO/CO=BO/DO, i.e. AO/OC=BO/OD.

Q14

ABC~EDF, AB=5,AC=7,DF=15,DE=12. Find EF,BC.

Correspondence A↔E,B↔D,C↔F: AB/ED=AC/EF=BC/DF

5/12=7/EF → EF=16.8 cm

Find BC

5/12=BC/15 → BC=6.25 cm

✓EF = 16.8 cm, BC = 6.25 cm

Q15

Show ΔABE~ΔCFB in parallelogram with E,F construction.

✓Since AB∥CD (and hence AB∥CF), ∠AEB=∠CBF (alternate angles, with BE as transversal). Also AD∥BC gives ∠ABE=∠CFB (alternate angles, with BF as transversal, noting AE is along AD produced). With two pairs of angles equal, ΔABE~ΔCFB by AA.

Q16

Poles 6m,11m, 12m apart. Find distance between tops.

Difference in height=5m forms one leg of a right triangle with the 12m horizontal gap

distance=√(5²+12²)

✓13 m

Q17

AX=1,XB=3,AY=1.5,YC=4.5. Is XY∥BC?

Compare the two ratios

AX/XB=1/3, AY/YC=1.5/4.5=1/3

✓The ratios are equal, so by the converse of BPT, XY∥BC

Q18

Show ΔABC~ΔPQR from proportional sides and medians.

✓Produce AD to E and PM to N such that AD=DE and PM=MN, then join BE, CE, QN, RN. ABEC and PQNR become parallelograms (diagonals bisect each other), so BE=AC and QN=PR (opposite sides of a parallelogram). Since AB/PQ=AC/PR=AD/PM is given, and AC=BE, PR=QN, AD/PM=AE/PN (as AD=DE, AE=2AD etc.), triangle ABE and PQN become similar by SSS (all three ratios AB/PQ, BE/QN, AE/PN equal). This gives ∠BAE=∠QPN, and by symmetry ∠CAE=∠RPN, so ∠BAC=∠QPR. Combined with AB/PQ=AC/PR, ΔABC~ΔPQR by SAS.

Q19

Pole 6m/shadow4m; tower shadow28m. Find tower height.

Equal angle of elevation gives equal height:shadow ratio

6/4=h/28

✓h = 42 m

Q20

Prove the converse of Pythagoras’ theorem.

✓Given ΔABC with BC²=AB²+AC², construct a right triangle PQR with PQ=AB, PR=AC, and ∠P=90°. By Pythagoras, QR²=PQ²+PR²=AB²+AC²=BC², so QR=BC. Now ΔABC≅ΔPQR (SSS, since AB=PQ, AC=PR, BC=QR), so ∠A=∠P=90°. This proves the angle opposite the side satisfying the Pythagorean relation is a right angle.

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