Triangles notes

6

Triangles

Detailed Theory • Easy English • Board Exam Focus

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Chapter Roadmap

1. Similar Figures and Similar Triangles 2. Similarity Criteria: AA, SSS and SAS 3. Basic Proportionality Theorem 4. Converse of BPT 5. Areas of Similar Triangles 6. Corresponding Sides and Perimeters 7. Pythagoras Theorem 8. Converse of Pythagoras Theorem 9. Board-Style Questions 10. Final 96% Target Revision

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1. Similar Figures and Similar Triangles

Meaning of Similar Figures

Two figures are called similar when they have the same shape, although their sizes may be different.

For similar triangles, corresponding angles are equal and corresponding sides are proportional.

\[\boxed{\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}}\]

Solved Example 1 — Identify similarity

Question: Two triangles have equal corresponding angles and proportional corresponding sides. Are they similar?

1. For similarity, corresponding angles must be equal and corresponding sides must be proportional.
2. The question gives both conditions.
3. Therefore, the two triangles have the same shape.
4. Answer: Yes, the triangles are \(\boxed{\text{similar}}\).

Solved Example 2 — Find a corresponding side

Question: If \(\triangle ABC\sim\triangle DEF\), \(AB=6\) cm, \(DE=9\) cm and \(EF=12\) cm, find \(BC\).

1. Since the triangles are similar, corresponding sides are proportional.
2. \(\frac{AB}{DE}=\frac{BC}{EF}\).
3. \(\frac{6}{9}=\frac{BC}{12}\).
4. \(BC=\frac{6\times12}{9}=8\) cm.
5. Answer: \(\boxed{8\text{ cm}}\).

Solved Example 3 — Scale factor

Question: Two similar triangles have corresponding sides \(5\) cm and \(15\) cm. Find the scale factor from the smaller triangle to the larger triangle.

1. Scale factor \(=\frac{\text{larger corresponding side}}{\text{smaller corresponding side}}\).
2. \(=\frac{15}{5}\).
3. \(=3\).
4. Therefore every corresponding side of the larger triangle is 3 times the smaller one.
5. Answer: \(\boxed{3}\).

2. Similarity Criteria: AA, SSS and SAS

AA Similarity Criterion

If two angles of one triangle are respectively equal to two angles of another triangle, the two triangles are similar.

\[\boxed{\angle A=\angle D,\quad \angle B=\angle E\Rightarrow\triangle ABC\sim\triangle DEF}\]

SSS Similarity Criterion

If the three corresponding sides of two triangles are proportional, the triangles are similar.

\[\boxed{\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}}\]

SAS Similarity Criterion

If one pair of corresponding angles is equal and the sides including those angles are proportional, the triangles are similar.

Solved Example 1 — AA criterion

Question: In two triangles, \(\angle A=\angle D=50^\circ\) and \(\angle B=\angle E=60^\circ\). Prove that the triangles are similar.

1. Given \(\angle A=\angle D=50^\circ\).
2. Given \(\angle B=\angle E=60^\circ\).
3. Thus two pairs of corresponding angles are equal.
4. By AA similarity criterion, the triangles are similar.
5. Answer: \(\boxed{\triangle ABC\sim\triangle DEF}\).

Solved Example 2 — SSS criterion

Question: The sides of two triangles are \(3,4,5\) cm and \(6,8,10\) cm. Check whether they are similar.

1. Compare corresponding sides.
2. \(\frac36=\frac12\).
3. \(\frac48=\frac12\).
4. \(\frac5{10}=\frac12\).
5. All three ratios are equal.
6. Therefore, by SSS similarity criterion, the triangles are similar.
7. Answer: \(\boxed{\text{Similar}}\).

Solved Example 3 — SAS criterion

Question: In two triangles, one included angle is equal and the two sides around it are in the ratio \(2:3\). Which criterion proves similarity?

1. There is one equal corresponding angle.
2. The two sides including that angle are proportional.
3. These are exactly the conditions of SAS similarity.
4. Answer: The triangles are similar by the \(\boxed{\text{SAS criterion}}\).

3. Basic Proportionality Theorem (BPT)

Statement of BPT

Basic Proportionality Theorem: 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.

\[\boxed{DE\parallel BC\Rightarrow\frac{AD}{DB}=\frac{AE}{EC}}\]

Solved Example 1 — Direct BPT

Question: In \(\triangle ABC\), \(DE\parallel BC\), \(AD=3\) cm, \(DB=2\) cm and \(AE=6\) cm. Find \(EC\).

1. By BPT, \(\frac{AD}{DB}=\frac{AE}{EC}\).
2. Substitute values: \(\frac32=\frac6{EC}\).
3. Cross multiply: \(3EC=12\).
4. Therefore \(EC=4\) cm.
5. Answer: \(\boxed{4\text{ cm}}\).

Solved Example 2 — Find a divided part

Question: If \(DE\parallel BC\), \(AD=4\) cm, \(AB=10\) cm and \(AE=6\) cm, find \(AC\).

1. First find \(DB=AB-AD=10-4=6\) cm.
2. By BPT, \(\frac{AD}{DB}=\frac{AE}{EC}\).
3. \(\frac46=\frac6{EC}\).
4. Cross multiply: \(4EC=36\).
5. \(EC=9\) cm.
6. Now \(AC=AE+EC=6+9=15\) cm.
7. Answer: \(\boxed{15\text{ cm}}\).

Solved Example 3 — Prove proportional division

Question: In \(\triangle ABC\), \(DE\parallel BC\), \(AD=5\), \(DB=10\), \(AE=4\), \(EC=8\). Verify BPT.

1. Calculate \(\frac{AD}{DB}=\frac5{10}=\frac12\).
2. Calculate \(\frac{AE}{EC}=\frac4{8}=\frac12\).
3. Both ratios are equal.
4. Hence \(\frac{AD}{DB}=\frac{AE}{EC}\).
5. Answer: BPT is \(\boxed{\text{verified}}\).

4. Converse of Basic Proportionality Theorem

Statement

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

\[\boxed{\frac{AD}{DB}=\frac{AE}{EC}\Rightarrow DE\parallel BC}\]

Solved Example 1 — Prove parallelism

Question: In \(\triangle ABC\), \(AD=3\), \(DB=2\), \(AE=6\), \(EC=4\). Show that \(DE\parallel BC\).

1. Calculate \(\frac{AD}{DB}=\frac32\).
2. Calculate \(\frac{AE}{EC}=\frac64=\frac32\).
3. Thus \(\frac{AD}{DB}=\frac{AE}{EC}\).
4. By the converse of BPT, \(DE\parallel BC\).
5. Answer: \(\boxed{DE\parallel BC}\).

Solved Example 2 — Check the condition

Question: \(AD=4,\ DB=6,\ AE=6,\ EC=9\). Is \(DE\parallel BC\)?

1. \(\frac{AD}{DB}=\frac46=\frac23\).
2. \(\frac{AE}{EC}=\frac69=\frac23\).
3. The ratios are equal.
4. Therefore, by the converse of BPT, \(DE\parallel BC\).
5. Answer: \(\boxed{\text{Yes}}\).

Solved Example 3 — Not parallel

Question: \(AD=2,\ DB=3,\ AE=4,\ EC=5\). Can we conclude \(DE\parallel BC\)?

1. \(\frac{AD}{DB}=\frac23\).
2. \(\frac{AE}{EC}=\frac45\).
3. Since \(\frac23\neq\frac45\), the ratios are not equal.
4. Therefore, the converse of BPT cannot be applied.
5. Answer: We cannot conclude that \(DE\parallel BC\).

5. Areas of Similar Triangles

Important Result

If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

\[\boxed{\frac{\operatorname{ar}(\triangle ABC)}{\operatorname{ar}(\triangle DEF)}=\left(\frac{AB}{DE}\right)^2}\]

Solved Example 1 — Find area ratio

Question: Two similar triangles have corresponding sides in the ratio \(2:3\). Find the ratio of their areas.

1. Side ratio \(=2:3\).
2. Area ratio is the square of side ratio.
3. \(=2^2:3^2\).
4. \(=4:9\).
5. Answer: \(\boxed{4:9}\).

Solved Example 2 — Find an area

Question: Two similar triangles have side ratio \(3:5\). If the area of the smaller triangle is \(27\text{ cm}^2\), find the area of the larger triangle.

1. Area ratio \(=3^2:5^2=9:25\).
2. So \(\frac{27}{A}=\frac9{25}\).
3. Cross multiply: \(9A=27\times25\).
4. \(A=75\text{ cm}^2\).
5. Answer: \(\boxed{75\text{ cm}^2}\).

Solved Example 3 — Reverse use

Question: The areas of two similar triangles are \(16\text{ cm}^2\) and \(64\text{ cm}^2\). Find the ratio of corresponding sides.

1. Area ratio \(=16:64=1:4\).
2. Side ratio is the square root of area ratio.
3. \(\sqrt{1:4}=1:2\).
4. Answer: Corresponding side ratio \(=\boxed{1:2}\).

6. Corresponding Sides and Perimeters

Perimeter Ratio

For similar triangles, the ratio of their perimeters is equal to the ratio of their corresponding sides.

\[\boxed{\frac{P_1}{P_2}=\frac{\text{corresponding side}_1}{\text{corresponding side}_2}}\]

Solved Example 1 — Find perimeter

Question: Two similar triangles have corresponding sides in the ratio \(2:5\). If the smaller perimeter is \(18\) cm, find the larger perimeter.

1. Perimeter ratio \(=2:5\).
2. Let larger perimeter be \(P\).
3. \(\frac{18}{P}=\frac25\).
4. \(2P=90\).
5. \(P=45\) cm.
6. Answer: \(\boxed{45\text{ cm}}\).

Solved Example 2 — Find a side

Question: Similar triangles have perimeters \(24\) cm and \(36\) cm. A corresponding side of the first triangle is \(8\) cm. Find the corresponding side of the second.

1. Perimeter ratio \(=24:36=2:3\).
2. Corresponding side ratio is also \(2:3\).
3. Let the required side be \(x\).
4. \(\frac8x=\frac23\).
5. \(2x=24\), so \(x=12\) cm.
6. Answer: \(\boxed{12\text{ cm}}\).

Solved Example 3 — Compare perimeters

Question: If two similar triangles have side ratio \(3:4\), find their perimeter ratio.

1. For similar triangles, corresponding side ratio equals perimeter ratio.
2. Given side ratio \(=3:4\).
3. Therefore perimeter ratio \(=3:4\).
4. Answer: \(\boxed{3:4}\).

7. Pythagoras Theorem

Statement

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

\[\boxed{c^2=a^2+b^2}\]

Here \(c\) is the hypotenuse, the side opposite the right angle.

Solved Example 1 — Find hypotenuse

Question: The perpendicular sides of a right triangle are \(6\) cm and \(8\) cm. Find the hypotenuse.

1. Let hypotenuse be \(c\).
2. Use \(c^2=a^2+b^2\).
3. \(c^2=6^2+8^2\).
4. \(c^2=36+64=100\).
5. \(c=\sqrt{100}=10\) cm.
6. Answer: \(\boxed{10\text{ cm}}\).

Solved Example 2 — Find a side

Question: A right triangle has hypotenuse \(13\) cm and one side \(5\) cm. Find the other side.

1. Let the unknown side be \(b\).
2. Use \(13^2=5^2+b^2\).
3. \(169=25+b^2\).
4. \(b^2=169-25=144\).
5. \(b=\sqrt{144}=12\) cm.
6. Answer: \(\boxed{12\text{ cm}}\).

Solved Example 3 — Verify a right triangle

Question: Check whether sides \(7,24,25\) form a right-angled triangle.

1. The largest side is \(25\), so take it as hypotenuse.
2. Calculate \(7^2+24^2\).
3. \(=49+576=625\).
4. Calculate \(25^2=625\).
5. Both values are equal.
6. Answer: Yes, the triangle is \(\boxed{\text{right-angled}}\).

8. Converse of Pythagoras Theorem

Statement

If the square of the largest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is right-angled.

\[\boxed{c^2=a^2+b^2\Rightarrow\text{right-angled triangle}}\]

Solved Example 1 — Prove right angle

Question: Show that a triangle with sides \(9,12,15\) cm is right-angled.

1. Largest side \(=15\) cm.
2. Calculate \(15^2=225\).
3. Calculate \(9^2+12^2=81+144=225\).
4. Thus \(15^2=9^2+12^2\).
5. By the converse of Pythagoras theorem, the triangle is right-angled.
6. Answer: \(\boxed{\text{Right-angled}}\).

Solved Example 2 — Not right-angled

Question: Check whether \(5,6,8\) form a right triangle.

1. Largest side \(=8\).
2. \(8^2=64\).
3. \(5^2+6^2=25+36=61\).
4. Since \(64\neq61\), the required condition is not satisfied.
5. Answer: \(\boxed{\text{Not right-angled}}\).

Solved Example 3 — Find the missing side for a right triangle

Question: A triangle has sides \(8\) cm and \(15\) cm. What third side would make it right-angled?

1. Assume \(8\) cm and \(15\) cm are the perpendicular sides.
2. By Pythagoras, \(c^2=8^2+15^2\).
3. \(c^2=64+225=289\).
4. \(c=\sqrt{289}=17\) cm.
5. Answer: The third side should be \(\boxed{17\text{ cm}}\).

9. Fully Solved Board-Style Questions

Question 1. If \(\triangle ABC\sim\triangle DEF\), \(AB=4\), \(DE=6\), \(BC=8\), find \(EF\).

Solution:

\[ \frac{AB}{DE}=\frac{BC}{EF} \]

\[ \frac46=\frac8{EF} \]

\[ 4EF=48\Rightarrow EF=12 \]

Answer: \(\boxed{12\text{ cm}}\).

Question 2. In \(\triangle ABC\), \(DE\parallel BC\), \(AD=2\), \(DB=3\), \(AE=4\). Find \(EC\).

Solution: By BPT,

\[ \frac{AD}{DB}=\frac{AE}{EC} \]

\[ \frac23=\frac4{EC} \]

\[ 2EC=12\Rightarrow EC=6 \]

Answer: \(\boxed{6\text{ cm}}\).

Question 3. The sides of a triangle are \(6,8,10\) cm. Prove that it is right-angled.

Solution:

Largest side \(=10\) cm.

\[ 10^2=100 \]

\[ 6^2+8^2=36+64=100 \]

Thus \(10^2=6^2+8^2\).

By the converse of Pythagoras theorem, the triangle is right-angled.

Answer: \(\boxed{\text{Right-angled}}\).

Question 4. Two similar triangles have areas \(25\text{ cm}^2\) and \(49\text{ cm}^2\). Find the ratio of their corresponding sides.

Solution:

\[ \frac{\text{Area}_1}{\text{Area}_2}=\frac{25}{49} \]

Side ratio is the square root of area ratio.

\[ \frac{\text{Side}_1}{\text{Side}_2}=\sqrt{\frac{25}{49}}=\frac57 \]

Answer: \(\boxed{5:7}\).

Question 5. If \(DE\parallel BC\), \(AD=4\), \(AB=12\), \(AC=15\), find \(AE\).

Solution:

Since \(DE\parallel BC\), by BPT,

\[ \frac{AD}{AB}=\frac{AE}{AC} \]

\[ \frac4{12}=\frac{AE}{15} \]

\[ AE=\frac{4\times15}{12}=5 \]

Answer: \(\boxed{5\text{ cm}}\).

10. Final 96% Target Revision

Must-Remember Results

\[\boxed{\triangle ABC\sim\triangle DEF\Rightarrow\text{corresponding sides are proportional}}\]

\[\boxed{DE\parallel BC\Rightarrow\frac{AD}{DB}=\frac{AE}{EC}}\]

\[\boxed{\frac{\operatorname{ar}(\triangle ABC)}{\operatorname{ar}(\triangle DEF)}=\left(\frac{AB}{DE}\right)^2}\]

\[\boxed{c^2=a^2+b^2}\]

Similarity criteria: AA, SSS and SAS.

BPT: Parallel line inside a triangle divides the other two sides proportionally.

Converse BPT: Equal division ratio implies parallelism.

One-Minute Checklist

• ✓ Know the meaning of similar triangles.
• ✓ Remember AA, SSS and SAS criteria.
• ✓ Learn the statement and application of BPT.
• ✓ Learn the converse of BPT.
• ✓ Remember that area ratio is the square of side ratio.
• ✓ Identify corresponding sides correctly.
• ✓ Apply Pythagoras theorem carefully.
• ✓ Use the converse of Pythagoras theorem to prove a triangle is right-angled.
• ✓ Write theorem names and complete steps in proof questions.

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Class 10 Maths • Chapter 6: Triangles

Designed as easy-to-read handwritten-style board exam notes.