Triangles | Grade 10th - Mathematics Chapter Summary & NCERT CBSE Study Resources

Study Materials

10th

10th - Mathematics

Triangles

Overview of the Chapter: Triangles

This chapter introduces students to the fundamental concepts related to triangles, their properties, and the various theorems associated with them. The focus is on understanding similarity and congruence of triangles, along with the application of the Pythagorean theorem. The chapter is essential for building a strong foundation in geometry.

Key Topics Covered

Similarity of Triangles
Criteria for Similarity of Triangles
Areas of Similar Triangles
Pythagoras Theorem

Detailed Summary

1. Similarity of Triangles

Two triangles are said to be similar if their corresponding angles are equal and their corresponding sides are proportional. The symbol used for similarity is '∼'.

Definition: Two triangles are similar if their corresponding angles are equal and corresponding sides are in the same ratio.

2. Criteria for Similarity of Triangles

There are three main criteria for the similarity of triangles:

AAA (Angle-Angle-Angle) Criterion: If the corresponding angles of two triangles are equal, then the triangles are similar.
SSS (Side-Side-Side) Criterion: If the corresponding sides of two triangles are proportional, then the triangles are similar.
SAS (Side-Angle-Side) Criterion: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the triangles are similar.

3. Areas of Similar Triangles

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

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

4. Pythagoras Theorem

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

Theorem: In a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Important Formulas

Pythagoras Theorem: \( a^2 + b^2 = c^2 \) (where \( c \) is the hypotenuse)
Area Ratio of Similar Triangles: \( \frac{Area_1}{Area_2} = \left( \frac{side_1}{side_2} \right)^2 \)

Applications

The concepts learned in this chapter are widely used in various fields such as architecture, engineering, and physics. Understanding the properties and theorems related to triangles helps in solving real-world problems involving shapes and measurements.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:

In ΔABC, if ∠A = 50° and ∠B = 70°, find ∠C.

Answer:

60°

Question 2:

State the Basic Proportionality Theorem (BPT).

Answer:

A line parallel to one side divides the other two proportionally.

Question 3:

If two triangles are similar, their corresponding sides are _________.

Answer:

Proportional

Question 4:

In ΔPQR, PQ = 6 cm, PR = 8 cm, and QR = 10 cm. Is it a right triangle?

Answer:

Yes (6² + 8² = 10²)

Question 5:

What is the ratio of areas of two similar triangles with sides in ratio 2:3?

Answer:

4:9

Question 6:

If ΔABC ~ ΔDEF and AB/DE = 3/5, find ar(ΔABC)/ar(ΔDEF).

Answer:

9:25

Question 7:

In ΔXYZ, XY || BC. If AX/XB = 2/3, find AY/YC.

Answer:

2:3

Question 8:

The sides of a triangle are 7 cm, 24 cm, and 25 cm. What is its area?

Answer:

84 cm²

Question 9:

If two triangles have equal areas, are they always congruent?

Answer:

Question 10:

In ΔABC, D and E are midpoints of AB and AC. What is DE/BC?

Answer:

1:2

Question 11:

If two triangles are similar, what can you say about their corresponding angles?

Answer:

The corresponding angles of two similar triangles are equal.

Question 12:

In ΔABC and ΔDEF, if ∠A = ∠D and ∠B = ∠E, what is the similarity criterion satisfied here?

Answer:

The triangles satisfy the AA (Angle-Angle) similarity criterion because two angles of one triangle are equal to two angles of the other triangle.

Question 13:

The sides of two similar triangles are in the ratio 3:5. What is the ratio of their areas?

Answer:

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Thus, the ratio of their areas = (3/5)² = 9/25.

Question 14:

In a right-angled triangle ABC, right-angled at B, if AB = 6 cm and BC = 8 cm, find the length of AC.

Answer:

Using the Pythagoras theorem:
AC² = AB² + BC²
AC² = 6² + 8² = 36 + 64 = 100
AC = √100 = 10 cm

Question 15:

State the Basic Proportionality Theorem (Thales Theorem).

Answer:

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.

Question 16:

In ΔPQR, if PS is the angle bisector of ∠P, meeting QR at S, what relation does it establish?

Answer:

By the Angle Bisector Theorem:
PQ/PR = QS/SR

Question 17:

If the ratio of the corresponding altitudes of two similar triangles is 4:9, what is the ratio of their corresponding sides?

Answer:

The ratio of the corresponding sides of two similar triangles is equal to the ratio of their corresponding altitudes.
Thus, the ratio of their sides = 4:9.

Question 18:

In ΔABC, D and E are points on AB and AC respectively such that DE || BC. If AD = 4 cm, DB = 6 cm, and AE = 5 cm, find EC.

Answer:

By the Basic Proportionality Theorem:
AD/DB = AE/EC
4/6 = 5/EC
EC = (5 × 6)/4 = 7.5 cm

Question 19:

What is the Pythagorean triplet that includes the number 5?

Answer:

A Pythagorean triplet including 5 is (3, 4, 5) because 3² + 4² = 5² (9 + 16 = 25).

Question 20:

If two triangles have their corresponding sides proportional, what can you conclude about them?

Answer:

The triangles are similar by the SSS (Side-Side-Side) similarity criterion.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:

State the Basic Proportionality Theorem (Thales Theorem) with a diagram.

Answer:

The Basic Proportionality Theorem states that 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.

Diagram: Draw triangle ABC with DE || BC, intersecting AB at D and AC at E. Then, AD/DB = AE/EC.

Question 2:

In a right-angled triangle, the lengths of the legs are 6 cm and 8 cm. Find the length of the hypotenuse.

Answer:

Using the Pythagoras theorem:
Hypotenuse² = Perpendicular² + Base²
Hypotenuse² = 6² + 8² = 36 + 64 = 100
Hypotenuse = √100 = 10 cm.

Question 3:

If two triangles are similar, what can you say about their corresponding angles and sides?

Answer:

For two similar triangles:

Corresponding angles are equal.
Corresponding sides are in the same ratio (proportional).

Question 4:

In ΔABC and ΔDEF, ∠A = ∠D and ∠B = ∠E. What additional information is needed to prove ΔABC ~ ΔDEF by AA similarity?

Answer:

No additional information is needed. If two angles of one triangle are equal to two angles of another triangle, the triangles are similar by AA (Angle-Angle) criterion.

Question 5:

The sides of a triangle are 7 cm, 24 cm, and 25 cm. Is it a right-angled triangle? Justify.

Answer:

Check using Pythagoras theorem:
25² = 625
7² + 24² = 49 + 576 = 625
Since 25² = 7² + 24², the triangle is right-angled.

Question 6:

If the ratio of corresponding sides of two similar triangles is 3:5, what is the ratio of their areas?

Answer:

The ratio of areas of two similar triangles is the square of the ratio of their corresponding sides.
Ratio of areas = (3/5)² = 9:25.

Question 7:

In ΔPQR, DE is parallel to QR and intersects PQ and PR at D and E respectively. If PD = 4 cm, DQ = 6 cm, and PE = 5 cm, find ER.

Answer:

By Basic Proportionality Theorem:
PD/DQ = PE/ER
4/6 = 5/ER
ER = (5 × 6)/4 = 7.5 cm.

Question 8:

State the SSS similarity criterion for triangles with an example.

Answer:

The SSS similarity criterion states that if the corresponding sides of two triangles are in the same ratio, then the triangles are similar.
Example: If ΔABC has sides 2, 3, 4 and ΔDEF has sides 4, 6, 8, then ΔABC ~ ΔDEF (ratio 1:2).

Question 9:

The areas of two similar triangles are 81 cm² and 121 cm². Find the ratio of their corresponding sides.

Answer:

Ratio of areas = (Ratio of sides)²
81/121 = (Ratio of sides)²
Ratio of sides = √(81/121) = 9:11.

Question 10:

In ΔABC, AD is the bisector of ∠A. If AB = 6 cm, AC = 8 cm, and BD = 3 cm, find DC.

Answer:

By Angle Bisector Theorem:
AB/AC = BD/DC
6/8 = 3/DC
DC = (3 × 8)/6 = 4 cm.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:

State and prove the Basic Proportionality Theorem (Thales Theorem).

Answer:

The Basic Proportionality Theorem states that if a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides those sides proportionally.

Given: A triangle ABC with DE parallel to BC, intersecting AB at D and ACE.

To Prove: AD/DB = AE/EC

Proof:
1. Join BE and CD.
2. Triangles BDE and CDE have the same base DE and lie between the same parallels DE and BC, so their areas are equal.
3. Now, ADE and BDE have the same height from E, so their area ratio is equal to AD/DB.
4. Similarly, ADE and CDE have the same height from D, so their area ratio is equal to AE/EC.
5. Since areas of BDE and CDE are equal, AD/DB = AE/EC.

Question 2:

In a right-angled triangle ABC, right-angled at B, if AB = 6 cm and BC = 8 cm, find the length of AC.

Answer:

Given a right-angled triangle ABC with ∠B = 90°, AB = 6 cm, and BC = 8 cm.

Using the Pythagoras Theorem:

AC² = AB² + BC²
AC² = 6² + 8²
AC² = 36 + 64
AC² = 100
AC = √100
AC = 10 cm

Question 3:

Explain the AAA similarity criterion for triangles with an example.

Answer:

The AAA (Angle-Angle-Angle) similarity criterion states that if the corresponding angles of two triangles are equal, then the triangles are similar.

Example: Consider two triangles ABC and DEF where ∠A = ∠D, ∠B = ∠E, and ∠C = ∠F.

Since all corresponding angles are equal, the triangles are similar by AAA criterion. This means their corresponding sides are proportional, i.e., AB/DE = BC/EF = AC/DF.

Question 4:

If the areas of two similar triangles are in the ratio 25:36, find the ratio of their corresponding sides.

Answer:

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given area ratio = 25:36.

Let the ratio of corresponding sides be x:y.

Then, x²/y² = 25/36
x/y = √(25/36)
x/y = 5/6

Thus, the ratio of corresponding sides is 5:6.

Question 5:

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Answer:

Let ABCD be a rhombus with sides of length a and diagonals AC = d₁ and BD = d₂ intersecting at O.

In a rhombus, diagonals bisect each other at right angles.

So, AO = OC = d₁/2 and BO = OD = d₂/2.

Applying Pythagoras theorem in △AOB:
AB² = AO² + BO²
a² = (d₁/2)² + (d₂/2)²
4a² = d₁² + d₂²

Since all sides are equal, sum of squares of all sides = 4a².

Thus, sum of squares of sides = sum of squares of diagonals.

Question 6:

In a triangle ABC, if AB = AC and ∠B = 50°, find the measure of ∠A.

Answer:

Given: Triangle ABC is isosceles with AB = AC and ∠B = 50°.
Since AB = AC, angles opposite to equal sides are equal.
Thus, ∠C = ∠B = 50°.
Using the angle sum property of a triangle:
∠A + ∠B + ∠C = 180°
∠A + 50° + 50° = 180°
∠A = 180° - 100° = 80°.
Therefore, ∠A = 80°.

Question 7:

Prove that the angles opposite to equal sides of an isosceles triangle are equal.

Answer:

Consider an isosceles triangle ABC with AB = AC.
Draw the angle bisector of ∠A, meeting BC at D.
In triangles ABD and ACD:
AB = AC (given)
∠BAD = ∠CAD (AD is angle bisector)
AD = AD (common side)
By SAS congruence rule, ΔABD ≅ ΔACD.
Thus, ∠B = ∠C (CPCT).
Hence, angles opposite to equal sides are equal.

Question 8:

In a right-angled triangle PQR, right-angled at Q, PR = 13 cm and PQ = 5 cm. Find the length of QR.

Answer:

Given: Right-angled triangle PQR with ∠Q = 90°, PR = 13 cm, and PQ = 5 cm.
Using the Pythagoras theorem:
PR² = PQ² + QR²
13² = 5² + QR²
169 = 25 + QR²
QR² = 169 - 25 = 144
QR = √144 = 12 cm.
Therefore, the length of QR is 12 cm.

Question 9:

If the areas of two similar triangles are in the ratio 16:25, find the ratio of their corresponding sides.

Answer:

Given: The ratio of the areas of two similar triangles is 16:25.
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Let the ratio of corresponding sides be a:b.
Thus, (a/b)² = 16/25.
Taking square root on both sides:
a/b = √(16/25) = 4/5.
Therefore, the ratio of their corresponding sides is 4:5.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Use an example from NCERT to illustrate.

Answer:

Introduction

We studied that similar triangles have equal corresponding angles and proportional sides. The area ratio depends on their side ratio.

Argument 1

Let ΔABC ~ ΔPQR with sides AB/PQ = BC/QR = AC/PR = k. Areas are (1/2)×base×height. Since heights are also proportional, Area(ABC)/Area(PQR) = (AB/PQ)² = k².

Argument 2

NCERT Example: If ΔABC ~ ΔDEF with AB/DE = 2/3, then Area(ABC)/Area(DEF) = (2/3)² = 4/9.

Conclusion

Thus, the area ratio is the square of the side ratio, proven mathematically and with NCERT examples.

Question 2:

In a right-angled triangle ABC, right-angled at B, prove that AC² = AB² + BC² using Pythagoras Theorem. Also, mention one real-life application.

Answer:

Introduction

Pythagoras Theorem states that in a right-angled triangle, the square of the hypotenuse equals the sum of squares of the other two sides.

Argument 1

Given ΔABC with ∠B = 90°, we draw a perpendicular BD to AC. Using similarity, ΔABD ~ ΔABC ⇒ AB² = AD×AC. Similarly, ΔBCD ~ ΔABC ⇒ BC² = CD×AC. Adding both, AB² + BC² = AC(AD+CD) = AC².

Argument 2

Real-life application: It is used in construction to ensure walls are perpendicular (e.g., checking 3-4-5 triangle rule).

Conclusion

Hence, AC² = AB² + BC² is verified, and the theorem has practical uses.

Question 3:

Using the Basic Proportionality Theorem (BPT), prove that the line drawn parallel to one side of a triangle divides the other two sides proportionally. Include a diagram and real-life application.

Answer:

Introduction

We studied in NCERT that the Basic Proportionality Theorem (BPT) states a line parallel to one side divides the other two sides proportionally.

Argument 1

Let’s take ∆ABC with DE ∥ BC. By BPT, AD/DB = AE/EC. We can prove this using similarity of ∆ADE and ∆ABC (AA criterion).

Argument 2

Real-life application: Architects use BPT to scale down structures proportionally. [Diagram: Triangle ABC with DE ∥ BC]

Conclusion

Thus, BPT helps in dividing sides proportionally, useful in scaling designs.

Question 4:

In a right-angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides. Prove this statement (Pythagoras Theorem) using similarity of triangles.

Answer:

Introduction

Our textbook shows the Pythagoras Theorem as a fundamental property of right-angled triangles.

Argument 1

Let ∆ABC be right-angled at B. Draw BD ⊥ AC. Now, ∆ADB ~ ∆ABC and ∆BDC ~ ∆ABC by AA similarity.

Argument 2

Using similarity ratios: AB² = AD × AC and BC² = CD × AC. Adding these, AB² + BC² = AC(AD + CD) = AC².

Conclusion

Hence, Pythagoras Theorem is verified, essential in construction and navigation.

Question 5:

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Use an example from NCERT to illustrate.

Answer:

Introduction

We studied that similar triangles have equal corresponding angles and proportional sides. The areas of such triangles follow a specific ratio.

Argument 1

Let ΔABC ~ ΔPQR with a scale factor k. If AB = k·PQ, then by the area formula (½×base×height), Area(ΔABC) = k²·Area(ΔPQR).

Argument 2

Our textbook shows ΔABC (sides 6cm, 8cm, 10cm) and ΔPQR (sides 3cm, 4cm, 5cm). Here, k = 2, and Area(ΔABC)/Area(ΔPQR) = 4 (which is 2²).

Conclusion

Thus, the ratio of areas is the square of the ratio of corresponding sides, as proved.

Question 6:

In a right-angled triangle, prove the Pythagoras Theorem using similarity of triangles. Also, mention one real-life application.

Answer:

Introduction

We learned that Pythagoras Theorem states: In a right-angled ΔABC, AB² + BC² = AC², where AC is the hypotenuse.

Argument 1

Draw altitude BD to hypotenuse AC. Now, ΔADB ~ ΔABC and ΔBDC ~ ΔABC by AA similarity. Using proportions, AB² = AD·AC and BC² = DC·AC. Adding both, AB² + BC² = AC².

Argument 2

Real-life application: This theorem helps carpenters ensure corners are right-angled. For example, a 3m, 4m, 5m triangle confirms a 90° angle.

Conclusion

Hence, Pythagoras Theorem is proved using similarity, and it has practical uses.

Question 7:

In ∆PQR, right-angled at Q, PR = 25 cm and PQ = 24 cm. Using the Pythagoras theorem, find QR. Also, relate this to an NCERT example.

Answer:

Introduction

Our textbook shows Pythagoras theorem: Hypotenuse² = Base² + Height². Here, PR is the hypotenuse.

Argument 1

Given: PQ = 24 cm, PR = 25 cm. Using Pythagoras, QR² = PR² - PQ² = 25² - 24² = 49.

Argument 2

Thus, QR = √49 = 7 cm. This resembles NCERT Example 6.2 where sides of a right triangle are calculated.

Conclusion

QR measures 7 cm, verifying the theorem practically.

Question 8:

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Use an example from NCERT.

Answer:

Introduction

We studied that similar triangles have proportional sides and equal angles. Our textbook shows this property helps compare their areas.

Argument 1

Let ΔABC ~ ΔPQR with ratio of sides = k.
Area formula: (1/2) × base × height.

Argument 2

If AB/PQ = k, then height ratio is also k (by similarity). Thus, Area(ABC)/Area(PQR) = (k × base × k × height)/(base × height) = k².

Conclusion

NCERT Example 9.25 confirms this for ΔABC ~ ΔDEF with sides 2:1, giving area ratio 4:1.

Question 9:

In a right-angled triangle, prove the Pythagoras Theorem using similarity of triangles. Include a diagram.

Answer:

Introduction

Pythagoras Theorem states: In ΔABC, right-angled at B, AB² + BC² = AC². We prove it using similarity.

Argument 1

[Diagram: Right ΔABC with altitude BD to hypotenuse AC]. ΔADB ~ ΔABC (AA similarity). So, AD/AB = AB/AC ⇒ AB² = AD × AC.

Argument 2

Similarly, ΔBDC ~ ΔABC ⇒ DC/BC = BC/AC ⇒ BC² = DC × AC. Adding both: AB² + BC² = AC(AD + DC) = AC².

Conclusion

Our textbook (Theorem 6.8) derives this step-by-step, confirming the proof.

Question 10:

A ladder 10m long reaches a window 8m high. How far is the foot of the ladder from the wall? Apply Pythagoras Theorem and include units.

Answer:

Introduction

This is a real-life application of Pythagoras Theorem. The ladder, wall, and ground form a right triangle.

Argument 1

Let distance from wall = x. Given: Hypotenuse (ladder) = 10m, height = 8m. Using Pythagoras: x² + 8² = 10².

Argument 2

Solving: x² = 100 - 64 = 36 ⇒ x = 6m. NCERT Example 6.25 solves a similar problem.

Conclusion

Thus, the foot of the ladder is 6 meters from the wall, matching our calculation.

Question 11:

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Answer:

To prove this, let's consider two similar triangles ΔABC and ΔDEF such that ΔABC ~ ΔDEF.

Given: ΔABC ~ ΔDEF, so their corresponding sides are proportional, i.e., AB/DE = BC/EF = AC/DF = k (where k is the scale factor).

We know that the area of a triangle is given by (1/2) × base × height.

Let the heights of ΔABC and ΔDEF be h₁ and h₂, respectively.

Since the triangles are similar, their corresponding heights are also proportional:
h₁/h₂ = AB/DE = k.

Now, calculate the areas:
Area of ΔABC = (1/2) × BC × h₁
Area of ΔDEF = (1/2) × EF × h₂

Divide the areas:
(Area of ΔABC)/(Area of ΔDEF) = [(1/2) × BC × h₁] / [(1/2) × EF × h₂]
Simplify:
= (BC × h₁) / (EF × h₂)

Substitute the proportional relationships:
= (k × EF × k × h₂) / (EF × h₂)
= k²

Thus, (Area of ΔABC)/(Area of ΔDEF) = (AB/DE)², proving the statement.

Question 12:

In a right-angled triangle, prove that the square of the hypotenuse is equal to the sum of the squares of the other two sides (Pythagoras Theorem).

Answer:

Let's prove Pythagoras Theorem for a right-angled triangle ΔABC, where ∠B = 90°.

Given: ΔABC is right-angled at B, with AB and BC as the legs and AC as the hypotenuse.

To prove: AC² = AB² + BC².

Construction: Draw BD ⊥ AC.

Proof:
1. In ΔABC and ΔADB:
- ∠A is common.
- ∠ABC = ∠ADB = 90°.
Thus, ΔABC ~ ΔADB by AA similarity.

2. Therefore, AB/AC = AD/AB (corresponding sides of similar triangles).
Cross-multiplying: AB² = AD × AC — (1).

3. Similarly, in ΔABC and ΔBDC:
- ∠C is common.
- ∠ABC = ∠BDC = 90°.
Thus, ΔABC ~ ΔBDC by AA similarity.

4. Therefore, BC/AC = DC/BC.
Cross-multiplying: BC² = DC × AC — (2).

5. Add equations (1) and (2):
AB² + BC² = AD × AC + DC × AC
Factorize: = AC (AD + DC)
Since AD + DC = AC,
AB² + BC² = AC × AC = AC².

Hence, AC² = AB² + BC², proving the theorem.

Question 13:

State and prove the Basic Proportionality Theorem (Thales Theorem).

Answer:

The Basic Proportionality Theorem (Thales Theorem) states: 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.

Given: In ΔABC, DE || BC, intersecting AB at D and AC at E.

To prove: AD/DB = AE/EC.

Proof:
1. Since DE || BC, by AA similarity, ΔADE ~ ΔABC.

2. Therefore, the ratios of corresponding sides are equal:
AD/AB = AE/AC — (1).

3. Rewrite AB and AC as:
AB = AD + DB
AC = AE + EC

4. Substitute into equation (1):
AD/(AD + DB) = AE/(AE + EC)

5. Cross-multiply:
AD × (AE + EC) = AE × (AD + DB)
Simplify:
AD × AE + AD × EC = AE × AD + AE × DB
Cancel AD × AE from both sides:
AD × EC = AE × DB

6. Divide both sides by DB × EC:
AD/DB = AE/EC.

Hence, the theorem is proved.

Question 14:

In a triangle ABC, D and E are points on AB and AC respectively such that DE || BC. If AD = 4 cm, DB = 6 cm, and AE = 5 cm, find the length of EC. Justify your answer using the Basic Proportionality Theorem (Thales Theorem).

Answer:

Given: In triangle ABC, DE || BC with AD = 4 cm, DB = 6 cm, and AE = 5 cm.

To find: Length of EC.

Solution:

According to the Basic Proportionality Theorem (Thales 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.

Thus, in triangle ABC with DE || BC, we have:

AD/DB = AE/EC

Substituting the given values:

4/6 = 5/EC

Cross-multiplying:

4 × EC = 6 × 5

4 × EC = 30

Dividing both sides by 4:

EC = 30/4

EC = 7.5 cm

Conclusion: The length of EC is 7.5 cm.

Value-added information: The Basic Proportionality Theorem is foundational for understanding similarity in triangles and is widely used in geometry problems involving parallel lines and proportional divisions.

Question 15:

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Use an example to illustrate the proof.

Answer:

To prove: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Let us consider two similar triangles, ABC and DEF, such that ABC ~ DEF.

Let the ratio of their corresponding sides be k, i.e., AB/DE = BC/EF = CA/FD = k.

We know that the area of a triangle is given by (1/2) × base × height.
For triangle ABC:
Area = (1/2) × AB × h1.
For triangle DEF:
Area = (1/2) × DE × h2.

Since the triangles are similar, their corresponding heights are also proportional to their corresponding sides.

Thus, h1/h2 = AB/DE = k.
Now, the ratio of their areas is:
Area(ABC)/Area(DEF) = [(1/2) × AB × h1] / [(1/2) × DE × h2].
Simplifying:
Area(ABC)/Area(DEF) = (AB/DE) × (h1/h2).
Substituting AB/DE = k and h1/h2 = k:
Area(ABC)/Area(DEF) = k × k = k².

Hence, the ratio of the areas is equal to the square of the ratio of their corresponding sides.

Example: If two triangles have corresponding sides in the ratio 3:5, then the ratio of their areas will be 9:25.

Question 16:

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Explain with a suitable diagram.

Answer:

To prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides, let us consider two triangles ΔABC and ΔDEF such that ΔABC ~ ΔDEF.

Given: ΔABC ~ ΔDEF, which means their corresponding angles are equal, and their corresponding sides are proportional.

To Prove: (Area of ΔABC)/(Area of ΔDEF) = (AB/DE)² = (BC/EF)² = (AC/DF)²

Proof:

1. Since ΔABC ~ ΔDEF, the ratio of their corresponding sides is equal. Let this ratio be k.

∴ AB/DE = BC/EF = AC/DF = k

2. Draw the altitudes AP and DQ from vertices A and D to sides BC and EF, respectively.

3. Since the triangles are similar, their corresponding altitudes are also in the same ratio as their corresponding sides.

∴ AP/DQ = k

4. Now, the area of ΔABC = (1/2) × BC × AP

The area of ΔDEF = (1/2) × EF × DQ

5. Taking the ratio of the areas:

(Area of ΔABC)/(Area of ΔDEF) = [(1/2) × BC × AP] / [(1/2) × EF × DQ] = (BC/EF) × (AP/DQ)

6. Substituting the values from step 1 and step 3:

(Area of ΔABC)/(Area of ΔDEF) = k × k = k² = (AB/DE)² = (BC/EF)² = (AC/DF)²

Conclusion: Hence, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Diagram: (Draw ΔABC and ΔDEF with corresponding sides and altitudes labeled as per the proof.)

Question 17:

In a right-angled triangle, prove that the square of the hypotenuse is equal to the sum of the squares of the other two sides (Pythagoras Theorem). Explain with a suitable diagram and example.

Answer:

To prove the Pythagoras Theorem, which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides, let us consider a right-angled triangle ΔABC with ∠B = 90°.

Given: ΔABC is right-angled at B, where AB and BC are the legs, and AC is the hypotenuse.

To Prove: AC² = AB² + BC²

Proof:

1. Draw a perpendicular BD from B to the hypotenuse AC.

2. Now, ΔABD ~ ΔABC (by AA similarity criterion, as both have ∠A common and ∠ADB = ∠ABC = 90°).

∴ AB/AC = AD/AB ⇒ AB² = AD × AC ...(i)

3. Similarly, ΔBDC ~ ΔABC (by AA similarity criterion, as both have ∠C common and ∠BDC = ∠ABC = 90°).

∴ BC/AC = DC/BC ⇒ BC² = DC × AC ...(ii)

4. Adding equations (i) and (ii):

AB² + BC² = AD × AC + DC × AC = AC (AD + DC) = AC × AC = AC²

Conclusion: Hence, AC² = AB² + BC², proving the Pythagoras Theorem.

Example: Consider a right-angled triangle with sides AB = 3 cm, BC = 4 cm, and AC = 5 cm.

AB² + BC² = 3² + 4² = 9 + 16 = 25

AC² = 5² = 25

Thus, AB² + BC² = AC², verifying the theorem.

Diagram: (Draw ΔABC right-angled at B, with perpendicular BD to AC, labeling all sides and angles as per the proof.)

Question 18:

In a triangle ABC, D and E are points on sides AB and AC respectively such that DE is parallel to BC. If AD = 4 cm, DB = 6 cm, and AE = 5 cm, find the length of EC. Justify your answer using the Basic Proportionality Theorem (Thales Theorem).

Answer:

Given: In triangle ABC, DE is parallel to BC, with AD = 4 cm, DB = 6 cm, and AE = 5 cm.

To find: The length of EC.

Solution:

Mathematically, if DE is parallel to BC, then:
AD/DB = AE/EC

Substituting the given values:
4/6 = 5/EC

Cross-multiplying:
4 × EC = 6 × 5
4 × EC = 30

Solving for EC:
EC = 30/4
EC = 7.5 cm

Thus, the length of EC is 7.5 cm.

Verification: The ratio AD/DB = 4/6 = 2/3 and AE/EC = 5/7.5 = 2/3, confirming the theorem's application.

Question 19:

In a triangle ABC, D and E are points on sides AB and AC respectively such that DE || BC. If AD = 4 cm, DB = 6 cm, and AE = 5 cm, find the length of EC. Justify your answer using the Basic Proportionality Theorem (Thales Theorem).

Answer:

Given: In triangle ABC, DE || BC, AD = 4 cm, DB = 6 cm, and AE = 5 cm.

To find: Length of EC.

Solution:

Mathematically, if DE || BC, then:

AD / DB = AE / EC

Substituting the given values:

4 / 6 = 5 / EC

Cross-multiplying:

4 × EC = 6 × 5

4 × EC = 30

EC = 30 / 4

EC = 7.5 cm

Conclusion: The length of EC is 7.5 cm.

Value-added information: The Basic Proportionality Theorem is fundamental in proving similarity of triangles and is widely used in geometric constructions and real-world applications like scaling maps or designing structures.

Question 20:

In a triangle ABC, D and E are points on sides AB and AC respectively such that DE || BC. If AD = 4 cm, DB = 6 cm, and AE = 5 cm, find the length of EC. Prove your answer using the Basic Proportionality Theorem (Thales Theorem).

Answer:

Given: In triangle ABC, DE || BC, AD = 4 cm, DB = 6 cm, and AE = 5 cm.

To find: Length of EC.

Proof using Basic Proportionality Theorem:

According to the Basic Proportionality Theorem (Thales 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.

Thus, in triangle ABC, since DE || BC, we have:
AD/DB = AE/EC

Substituting the given values:
4/6 = 5/EC

Cross-multiplying:
4 × EC = 6 × 5
4 × EC = 30

Solving for EC:
EC = 30/4
EC = 7.5 cm

Therefore, the length of EC is 7.5 cm.

Key takeaway: The Basic Proportionality Theorem is a fundamental concept in the study of similar triangles and helps in solving problems involving parallel lines and proportional sides.

Question 21:

In a triangle ABC, D and E are points on AB and AC respectively such that DE is parallel to BC. If AD = 4 cm, DB = 6 cm, and AE = 5 cm, find the length of EC. Justify your answer using the Basic Proportionality Theorem (Thales Theorem).

Answer:

Given:

AD = 4 cm
DB = 6 cm
AE = 5 cm
DE is parallel to BC

Using the theorem:

AD/DB = AE/EC

Substituting the given values:

4/6 = 5/EC

Cross-multiplying:

4 × EC = 6 × 5

4 × EC = 30

EC = 30/4

EC = 7.5 cm

Therefore, the length of EC is 7.5 cm.

Question 22:

In a triangle ABC, D and E are points on sides AB and AC respectively such that DE || BC. If AD = 4 cm, DB = 6 cm, and AE = 5 cm, find the length of EC. Prove your answer using the Basic Proportionality Theorem (Thales Theorem).

Answer:

According to the Basic Proportionality Theorem (or Thales 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.

Given:
In triangle ABC, DE || BC, AD = 4 cm, DB = 6 cm, and AE = 5 cm.

To find: Length of EC.

Step 1: Apply the Basic Proportionality Theorem.
Since DE || BC, by the theorem, we have:
AD/DB = AE/EC

Step 2: Substitute the given values.
4/6 = 5/EC

Step 3: Simplify the ratio.
2/3 = 5/EC

Step 4: Cross-multiply to solve for EC.
2 × EC = 3 × 5
2 × EC = 15
EC = 15/2
EC = 7.5 cm

Conclusion: The length of EC is 7.5 cm.

Additional Insight: This theorem is foundational in geometry and helps establish similarity between triangles, which is crucial for solving problems involving proportional sides and angles.

Question 23:

In a triangle ABC, D and E are points on sides AB and AC respectively such that DE || BC. If AD = 4 cm, DB = 6 cm, and AE = 5 cm, find the length of EC. Justify your answer using the Basic Proportionality Theorem (Thales Theorem).

Answer:

According to the Basic Proportionality 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.

Given: DE || BC in triangle ABC.
AD = 4 cm, DB = 6 cm, AE = 5 cm.
We need to find EC.

By the theorem, AD/DB = AE/EC.
Substituting the given values: 4/6 = 5/EC.
Cross-multiplying: 4 × EC = 6 × 5.
4 × EC = 30.
EC = 30 / 4 = 7.5 cm.

Thus, the length of EC is 7.5 cm.

Question 24:

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Use an example with two triangles having sides in the ratio 2:3 to illustrate this property.

Answer:

To prove: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Let’s consider two similar triangles, ΔABC and ΔDEF, such that AB/DE = BC/EF = AC/DF = 2/3 (given).

Step 1: Since the triangles are similar, their corresponding angles are equal, and their sides are proportional.
Step 2: Let the area of ΔABC be A1 and the area of ΔDEF be A2.
Step 3: The formula for the area of a triangle is (1/2) × base × height.
For ΔABC: A1 = (1/2) × AB × h1.
For ΔDEF: A2 = (1/2) × DE × h2.
Step 4: Since the triangles are similar, the ratio of their heights is equal to the ratio of their corresponding sides, i.e., h1/h2 = AB/DE = 2/3.
Step 5: Now, A1/A2 = [(1/2) × AB × h1] / [(1/2) × DE × h2] = (AB/DE) × (h1/h2).
Substituting the values: A1/A2 = (2/3) × (2/3) = (2/3)² = 4/9.

Thus, the ratio of the areas is 4:9, which is the square of the ratio of their corresponding sides (2:3). This proves the theorem.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:

A ladder 10 m long leans against a wall, making a 60° angle with the ground. Using properties of right-angled triangles, find how far the ladder's base is from the wall.

Answer:

Problem Interpretation

We need to find the horizontal distance (base) between the ladder's base and the wall, given its length and angle.

Mathematical Modeling

Using the cosine ratio in right-angled ΔABC (where ∠B = 90°), cos 60° = base/hypotenuse.

Solution

cos 60° = 1/2 = base/10 m
Base = 10 × (1/2) = 5 m (units included)

Question 2:

In ΔPQR, PQ = 6 cm, PR = 8 cm, and QR = 10 cm. Verify if it is a right-angled triangle using the Pythagoras theorem.

Answer:

Problem Interpretation

We must check if ΔPQR satisfies Pythagoras' theorem (hypotenuse² = sum of squares of other sides).

Mathematical Modeling

Our textbook shows that for right-angled triangles, the longest side (QR) should be the hypotenuse.

Solution

PQ² + PR² = 6² + 8² = 36 + 64 = 100
QR² = 10² = 100
Since PQ² + PR² = QR², ΔPQR is right-angled at P.

Question 3:

In a right-angled triangle ABC, right-angled at B, AB = 6 cm and BC = 8 cm. Using the Pythagoras theorem, find the length of AC. Also, state whether triangle ABC is scalene, isosceles, or equilateral.

Answer:

Problem Interpretation

We are given a right-angled triangle ABC with sides AB = 6 cm and BC = 8 cm. We need to find AC and classify the triangle.

Mathematical Modeling

Using Pythagoras theorem: AC² = AB² + BC².

Solution

AC² = 6² + 8² = 36 + 64 = 100
AC = √100 = 10 cm
Since all sides (6 cm, 8 cm, 10 cm) are unequal, ABC is scalene.

Question 4:

Two triangles, PQR and XYZ, are such that PQ/XY = QR/YZ = PR/XZ = 3/2. Are the two triangles similar? If yes, state the criterion and find the ratio of their areas.

Answer:

Problem Interpretation

We need to check if triangles PQR and XYZ are similar given the ratio of their corresponding sides.

Mathematical Modeling

Our textbook shows that if corresponding sides are proportional, triangles are similar by SSS criterion.

Solution

Since PQ/XY = QR/YZ = PR/XZ, PQR ~ XYZ by SSS.
Ratio of areas = (3/2)² = 9/4.

Question 5:

A ladder 10 m long leans against a wall such that its foot is 6 m from the wall. Using the Pythagoras theorem, find the height the ladder reaches on the wall.

Answer:

Problem Interpretation

We studied that a ladder, wall, and ground form a right-angled triangle. The ladder is the hypotenuse.

Mathematical Modeling

Let the height be h. Using Pythagoras theorem: h² + 6² = 10².

Solution

Solving, h² = 100 - 36 = 64. Thus, h = 8 m. The ladder reaches 8 m high.

Question 6:

In ΔABC, DE || BC such that AD = 4 cm, DB = 6 cm, and AE = 5 cm. Using Basic Proportionality Theorem (BPT), find EC.

Answer:

Problem Interpretation

Our textbook shows that when DE || BC, ΔABC ~ ΔADE by BPT.

Mathematical Modeling

By BPT, AD/DB = AE/EC. Substituting values: 4/6 = 5/EC.

Solution

Cross-multiplying: 4 × EC = 30. Thus, EC = 7.5 cm.

Question 7:

A ladder 10 m long leans against a wall such that its foot is 6 m away from the wall. Using the Pythagoras theorem, find the height the ladder reaches on the wall.

Answer:

Problem Interpretation

We studied that a ladder, wall, and ground form a right-angled triangle. The ladder acts as the hypotenuse.

Mathematical Modeling

Let height be h. Using Pythagoras theorem: h² + 6² = 10².

Solution

Solving, h² = 100 - 36 = 64. Thus, h = 8 m. The ladder reaches 8 m high.

Question 8:

In ΔABC and ΔPQR, AB/PQ = BC/QR = AC/PR = 1/2. If area of ΔABC is 12 cm², find the area of ΔPQR using the Area Ratio Theorem.

Answer:

Problem Interpretation

Our textbook shows that if sides of two triangles are proportional, their areas are in the ratio of the squares of corresponding sides.

Mathematical Modeling

Given ratio of sides = 1/2. Thus, area ratio = (1/2)² = 1/4.

Solution

Area of ΔPQR = 12 × 4 = 48 cm², since ΔABC's area is 12 cm².

Question 9:

A ladder 10 m long leans against a wall such that its foot is 6 m away from the wall. Using the Pythagoras theorem, find the height the ladder reaches on the wall. Also, state the similarity criterion if another ladder forms a similar triangle.

Answer:

Problem Interpretation

We studied that a ladder, wall, and ground form a right triangle. Here, the ladder is the hypotenuse.

Mathematical Modeling

Using Pythagoras theorem: Height² + Base² = Hypotenuse². Given base = 6 m, hypotenuse = 10 m.

Solution

Height² + 6² = 10² → Height = √(100 - 36) = 8 m
Similarity criterion: AA (Angle-Angle) as both triangles are right-angled and share an acute angle.

Question 10:

In ΔABC and ΔPQR, AB/PQ = BC/QR = 3/2 and ∠B = ∠Q. Prove that ΔABC ~ ΔPQR using the SAS similarity criterion. Also, if AB = 9 cm, find PQ.

Answer:

Problem Interpretation

Our textbook shows that two triangles are similar if their sides are proportional and included angles are equal.

Mathematical Modeling

Given AB/PQ = 3/2, ∠B = ∠Q, and BC/QR = 3/2, ΔABC ~ ΔPQR by SAS similarity.

Solution

Proof: AB/PQ = BC/QR and ∠B = ∠Q → ΔABC ~ ΔPQR (SAS).
AB = 9 cm → PQ = (2/3) × AB = 6 cm.

Question 11:

In a park, two triangular flower beds ABC and PQR are designed such that AB/PQ = BC/QR = CA/RP = 3/2. If the area of ABC is 36 cm², find the area of PQR. Justify your answer using the Basic Proportionality Theorem.

Answer:

Given: ABC and PQR are two triangles with sides in the ratio AB/PQ = BC/QR = CA/RP = 3/2.
This implies the triangles are similar by the SSS similarity criterion.

The ratio of their areas is equal to the square of the ratio of their corresponding sides.
So, Area(ABC)/Area(PQR) = (3/2)² = 9/4.

Given Area(ABC) = 36 cm²,
36/Area(PQR) = 9/4.

Cross-multiplying:
9 × Area(PQR) = 36 × 4
Area(PQR) = (36 × 4)/9 = 16 cm².

Thus, the area of PQR is 16 cm².

Question 12:

A ladder 10 m long reaches a window 8 m above the ground. The ladder is tilted such that its foot is moved 2 m away from the wall. Determine whether the new position forms a right-angled triangle and justify using the Pythagoras Theorem.

Answer:

Initial position:
The ladder, wall, and ground form a right-angled triangle.
Using the Pythagoras Theorem:
AC² = AB² + BC²
10² = 8² + BC²
100 = 64 + BC²
BC² = 36
BC = 6 m (initial distance from the wall).

After moving the ladder:
New distance from the wall = 6 m + 2 m = 8 m.
Now, check if 10² = 8² + 8²:
100 = 64 + 64
100 ≠ 128.

Since the equation does not hold, the new position does not form a right-angled triangle.

Question 13:

In a park, two triangular flower beds ABC and PQR are designed such that AB/PQ = BC/QR = AC/PR = 3/2. The area of ABC is 108 sq. meters. A gardener claims that the two triangles are congruent because their corresponding sides are proportional. Evaluate his claim and find the area of PQR.

Answer:

The gardener's claim is incorrect because congruent triangles must have equal corresponding sides and angles, not just proportional sides. Here, the triangles are similar (not congruent) as their sides are proportional (AB/PQ = BC/QR = AC/PR = 3/2).

Given: ABC ~ PQR and ar(ABC) = 108 sq. meters.

Since the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides:

ar(ABC)/ar(PQR) = (AB/PQ)² = (3/2)² = 9/4.

Substitute the known area:

108/ar(PQR) = 9/4.

Cross-multiply to solve for ar(PQR):

ar(PQR) = (108 × 4)/9 = 48 sq. meters.

Thus, the area of PQR is 48 sq. meters.

Question 14:

A ladder 10 meters long leans against a wall, touching it at a height of 8 meters. The foot of the ladder is moved 2 meters away from the wall, causing the top to slide down. Using properties of triangles, find:

The original distance of the ladder's foot from the wall.
The new height at which the ladder touches the wall.

Answer:

Step 1: Original distance calculation

Let the original distance of the ladder's foot from the wall be x meters. The ladder, wall, and ground form a right-angled triangle.

Using the Pythagoras theorem:

10² = 8² + x².

100 = 64 + x².

x² = 36.

x = 6 meters (distance cannot be negative).

Step 2: New height calculation

After moving the foot 2 meters away, the new distance from the wall is 6 + 2 = 8 meters.

Let the new height be h meters. Again, apply the Pythagoras theorem:

10² = 8² + h².

100 = 64 + h².

h² = 36.

h = 6 meters.

Final answers:

Original distance: 6 meters.
New height: 6 meters.

Question 15:

In a park, two triangular flower beds ABC and PQR are designed such that AB/PQ = BC/QR = CA/RP = 3/2. A gardener claims that the two flower beds are similar because their corresponding sides are proportional. Is the gardener correct? Justify your answer using the Basic Proportionality Theorem (BPT) or any other relevant theorem.

Answer:

The gardener is correct. According to the SSS (Side-Side-Side) similarity criterion, two triangles are similar if their corresponding sides are in proportion.

Given: AB/PQ = BC/QR = CA/RP = 3/2.

Since all three corresponding sides are proportional, by the SSS similarity rule, ΔABC ~ ΔPQR.

Additionally, the Basic Proportionality Theorem (BPT) states that if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. However, BPT is not directly applicable here since we are comparing two separate triangles, not a line within a triangle.

Thus, the gardener's claim is valid because the sides satisfy the SSS similarity condition.

Question 16:

In a right-angled triangle XYZ, right-angled at Y, a perpendicular YM is drawn from Y to the hypotenuse XZ. Prove that ΔXYM ~ ΔXYZ using the AA (Angle-Angle) similarity criterion.

Answer:

To prove: ΔXYM ~ ΔXYZ using the AA similarity criterion.

Step 1: Identify the common angle.
Both triangles ΔXYM and ΔXYZ share the angle at X (i.e., ∠YXM = ∠YXZ).

Step 2: Identify the right angles.
ΔXYZ is right-angled at Y, so ∠XYZ = 90°.
YM is perpendicular to XZ, so ∠XMY = 90°.

Step 3: Apply the AA criterion.
Since both triangles have two pairs of equal angles (∠YXM = ∠YXZ and ∠XMY = ∠XYZ = 90°), by the AA similarity rule, ΔXYM ~ ΔXYZ.

This proves the similarity between the two triangles.

Question 17:

In a park, two triangular flower beds ABC and PQR are designed such that AB/PQ = BC/QR = CA/RP = 1/2. If the area of ABC is 25 sq. units, find the area of PQR. Justify your answer using the appropriate theorem.

Answer:

Given the sides of ABC and PQR are in the ratio 1:2, the triangles are similar by the SSS (Side-Side-Side) similarity criterion.

According to the Area Ratio Theorem, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Here, the ratio of corresponding sides is 1:2, so the ratio of their areas will be (1)²:(2)² = 1:4.

Given the area of ABC is 25 sq. units, the area of PQR is calculated as:

Area of PQR = 4 × Area of ABC
= 4 × 25
= 100 sq. units.

Question 18:

A ladder 10 m long reaches a window 8 m above the ground. The ladder is turned to the other side of the street and now reaches a window 6 m high. Using the properties of right-angled triangles, find the width of the street.

Answer:

Let the width of the street be x meters. The ladder forms two right-angled triangles on either side of the street.

First Position:
Using the Pythagoras theorem, the distance from the foot of the ladder to the building is:
√(10² - 8²) = √(100 - 64) = √36 = 6 m.

Second Position:
Similarly, the distance from the foot of the ladder to the other building is:
√(10² - 6²) = √(100 - 36) = √64 = 8 m.

Total Width:
The width of the street is the sum of these distances:
x = 6 m + 8 m = 14 m.

Thus, the street is 14 meters wide.

Question 19:

Answer:

The given triangles ABC and PQR are similar because their corresponding sides are in proportion (AB/PQ = BC/QR = CA/RP = 3/2).

According to the Area Ratio Theorem of Similar Triangles, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Here, the ratio of corresponding sides is 3/2, so the ratio of their areas will be (3/2)² = 9/4.

Given the area of ABC is 36 cm², let the area of PQR be x cm².

Thus, 36/x = 9/4.

Solving for x, we get x = (36 × 4)/9 = 16 cm².

Therefore, the area of PQR is 16 cm².

Question 20:

A ladder 10 m long reaches a window 8 m above the ground. The ladder is moved such that its foot remains at the same point, but it now reaches a window 6 m high on the opposite side of the street. Using the properties of triangles, find the width of the street.

Answer:

This problem involves two right-angled triangles formed by the ladder in its two positions.

First Position: The ladder reaches 8 m high.

Using the Pythagoras theorem, the distance of the foot of the ladder from the wall is:

√(10² - 8²) = √(100 - 64) = √36 = 6 m.

Second Position: The ladder reaches 6 m high on the opposite side.

Similarly, the distance of the foot of the ladder from the opposite wall is:

√(10² - 6²) = √(100 - 36) = √64 = 8 m.

Since the foot of the ladder remains at the same point, the total width of the street is the sum of these two distances.

Thus, the width of the street = 6 m + 8 m = 14 m.

Therefore, the width of the street is 14 m.

Triangles – CBSE NCERT Study Resources

Overview of the Chapter: Triangles

Key Topics Covered

Detailed Summary

1. Similarity of Triangles

2. Criteria for Similarity of Triangles

3. Areas of Similar Triangles

4. Pythagoras Theorem

Important Formulas

Applications

All Question Types with Solutions – CBSE Exam Pattern

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)