Class 9 Mathematics Exercise-10.1 NCERT Solutions Olympiad Board Exam

Chapter 10

CIRCLES

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

6 Questions

15–20 min Ideal time

Q1 Now at

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📘 Concept & Theory Concept Used ›

An equilateral triangle is a triangle in which all three sides are equal.

Heron’s Formula is used to find the area of a triangle when the lengths of all three sides are known.

If the sides of a triangle are \(\small a\), \(\small b\), and \(\small c\), then:

\[\small s = \frac{a+b+c}{2} \]

where \(\small s\) is the semi-perimeter.

Area of triangle:

\[\small \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]

Since this is an equilateral triangle:

\[\small a=b=c \]

🗺️ Solution Roadmap Step-by-step Plan ›

Find the length of each side using the perimeter of the equilateral triangle.
Calculate the semi-perimeter of the triangle.
Apply Heron’s Formula step by step.
Simplify the square root carefully to obtain the final area.

📊 Graph / Figure Graph / Figure ›

Equilateral Signal Board

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

The signal board is an equilateral triangle.
Perimeter of the triangle \(\small = 180~\text{cm}\)
Given
Find the length of each side
Since all sides of an equilateral triangle are equal: \[\small \begin{aligned}\text{Side} &= \frac{\text{Perimeter}}{3}\\ a &= \frac{180}{3}\\ &= 60~\text{cm}\end{aligned}\]
Therefore,\[\small a=b=c=60~\text{cm}\]
Calculate the semi-perimeter
Semi-perimeter is: \[\small s = \frac{a+b+c}{2}\]
Substituting the values: \[\small \begin{aligned}s &= \frac{60+60+60}{2}\\ &= \frac{180}{2}\\ &= 90~\text{cm}\end{aligned}\]
Apply Heron’s Formula
\[\small \text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\]
Substituting the values: \[\small \begin{aligned} \text{Area} &= \sqrt{90(90-60)(90-60)(90-60)}\\ &=\sqrt{90 \times 30 \times 30 \times 30}\\ &=\sqrt{3\times \overline{30 \times 30}\; \times\; \overline{30 \times 30}}\\ &=30\times30\sqrt{3}\\ &=900\sqrt{3}\;\text{cm}^2 \end{aligned}\]

🎯 Exam Significance Exam Significance ›

This problem strengthens understanding of Heron’s Formula and its practical applications.
Board examinations frequently ask numerical problems based on perimeter, semi-perimeter, and area calculations.
Competitive entrance exams often test simplification of square roots and formula-based geometry problems.
Students learn how to convert word problems into mathematical expressions systematically.
This question also develops accuracy in algebraic manipulation and arithmetic calculations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

In an equilateral triangle, all sides are equal.
Semi-perimeter is half of the perimeter.
Heron’s Formula is useful when all three sides of a triangle are known.
Always simplify square roots carefully in the final step.
Area of an equilateral triangle with side \(\small 60~\text{cm}\) is: \[\small 900\sqrt{3}~\text{cm}^2 \]

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Q2 →

📘 Concept & Theory Concept Used ›

Heron’s Formula is used to calculate the area of a triangle when all three sides are known.

If the sides of a triangle are \(\small a\), \(\small b\), and \(\small c\), then:

\[\small s = \frac{a+b+c}{2} \]

where \(\small s\) is the semi-perimeter of the triangle.

Area of the triangle:

\[\small \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]

After finding the area, the advertising rent is calculated using:

\[\small \text{Rent} = \text{Area} \times \text{Rate} \times \text{Time Fraction} \]

🗺️ Solution Roadmap Step-by-step Plan ›

Write the given sides of the triangular wall.
Find the semi-perimeter of the triangle.
Compute the values of \(\small s-a\), \(\small s-b\), and \(\small s-c\).
Apply Heron’s Formula step by step to calculate the area.
Use the given yearly advertisement rate and calculate the rent for 3 months.

📊 Graph / Figure Graph / Figure ›

Flyover Side Wall

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps

Given
The sides of the triangular wall are:

\[\small a = 122~\text{m} \]

\[\small b = 22~\text{m} \]

\[\small c = 120~\text{m} \]

Advertisement earning:

\[\small \text{₹}5,000 \text{ per m}^2 \text{ per year} \]

Time period:

\[\small 3 \text{ months} \]
Given
Calculate the semi-perimeter
Semi-perimeter of triangle:\[\small s = \frac{a+b+c}{2}\]
Substituting the values: \[\small \begin{aligned} s &= \frac{122+22+120}{2}\\ &= \frac{264}{2}\\ &= 132~\text{m} \end{aligned} \]
Calculate \(\small s-a\), \(\small s-b\), and \(\small s-c\)
\[\small \begin{aligned} s-a &= 132-122= 10~\text{m}\\ s-b &= 132-22=110~\text{m}\\ s-c &= 132-120=12~\text{m} \end{aligned} \]
Apply Heron’s Formula
\[\small \text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\]
Substituting the values: \[\small \begin{aligned} \text{Area} &= \sqrt{132 \times 10 \times 110 \times 12}\\ &=\sqrt{1320 \times 1320}\\ &=1320 \text{ m}^2 \end{aligned} \]
Calculate advertisement rent for 3 months
Yearly earning per square metre:\[\small \text{₹ }5000 \text{ per m}^2\]
Area of wall:=\(\small 1320~\text{m}^2\)
Time fraction for 3 months:\[\small \frac{3}{12} = \frac{1}{4}\]
Therefore, \[\small \begin{aligned} \text{Rent}&=1320 \times 5000 \times \frac{1}{4}\\ &=\frac{66,00,000}{4}\\ &=16,50,000 \end{aligned} \]

💡 Answer Final Answer ›

The company paid \(\small \text{₹ }16,50,000\) as rent for 3 months.

🎯 Exam Significance Exam Significance ›

This question combines geometry with real-life commercial applications.
Students learn how Heron’s Formula is used in practical area calculations.
Board exams frequently ask application-based questions involving area and cost calculations.
Competitive entrance examinations test arithmetic accuracy and formula application in multi-step problems.
This problem improves numerical simplification and unit interpretation skills.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 4 points

Semi-perimeter is essential for applying Heron’s Formula.
Always calculate \(\small s-a\), \(\small s-b\), and \(\small s-c\) carefully before substitution
Area obtained using Heron’s Formula can be directly applied in commercial calculations.
Convert months into fraction of year while calculating yearly rent.

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Q3 →

📘 Concept & Theory Concept Used ›

Heron’s Formula helps us calculate the area of a triangle when all three sides are known.

For a triangle with sides \(\small a\), \(\small b\), and \(\small c\):

\[\small s = \frac{a+b+c}{2} \]

where \(\small s\) is the semi-perimeter.

Area of the triangle:

\[\small \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]

This formula is especially useful when the height of the triangle is not given.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the lengths of the three sides of the triangular wall.
Calculate the semi-perimeter of the triangle.
Find the values of \(\small s-a\), \(\small s-b\), and \(\small s-c\).
Substitute all values into Heron’s Formula.
Simplify the square root carefully to obtain the final area.

📊 Graph / Figure Graph / Figure ›

Fig. 1 — Free body diagram

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

Given
The sides of the triangular wall are:

\[\small a = 15~\text{m} \]

\[\small b = 11~\text{m} \]

\[\small c = 6~\text{m} \]
Calculate the semi-perimeter
Semi-perimeter of a triangle:\[\small s = \frac{a+b+c}{2}\]
Substituting the values: \[\small \begin{aligned} s &= \frac{15+11+6}{2}\\ &= \frac{32}{2}\\ &=16~\text{m} \end{aligned} \]
Calculate \(\small s-a\), \(\small s-b\), and \(\small s-c\)
\[\small \begin{aligned} s-a &= 16-15=1\\ s-b &= 16-11=5\\ s-c &= 16-6=10 \end{aligned} \]
Apply Heron’s Formula
\[\small \text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\]
Substituting the values: \[\small \begin{aligned} \text{Area}&=\sqrt{16 \times 1 \times 5 \times 10}\\ &=\sqrt{16 \times 50}\\ &=\sqrt{4\times 4 \times 25\times 2}\\ &=\sqrt{\overline{4\times 4}\; \times\; \overline{5\times 5}\;\times 2}\\ &=4\times5\times\sqrt{2}\\ &=20\sqrt{2}\text{ m}^2 \end{aligned} \]

💡 Answer Final Answer ›

The area painted in colour is \(\small 20\sqrt{2}\text{ m}^2\)

🎯 Exam Significance Exam Significance ›

This question tests direct application of Heron’s Formula in practical situations.
Students learn simplification of irrational numbers and square roots.
Board examinations frequently include area problems involving real-life structures.
Competitive entrance exams often test numerical accuracy and simplification techniques.
This problem improves understanding of geometry applications in engineering and design.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 4 points

Heron’s Formula is useful when only side lengths are given.
Semi-perimeter must always be calculated before applying the formula.
Perfect square factors should be separated for easier simplification.
Real-life area problems can often be solved using triangle formulas.

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Q4 →

📘 Concept & Theory Concept Used ›

When two sides and the perimeter of a triangle are known, the third side can be calculated first.

After finding all three sides, Heron’s Formula is used to calculate the area.

Semi-perimeter of a triangle:

\[\small s = \frac{a+b+c}{2} \]

Area of a triangle using Heron’s Formula:

\[\small \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]

This method is useful when height of the triangle is not given.

🗺️ Solution Roadmap Step-by-step Plan ›

Use the perimeter to calculate the third side of the triangle.
Find the semi-perimeter of the triangle.
Calculate \(\small s-a\), \(\small s-b\), and \(\small s-c\).
Apply Heron’s Formula carefully.
Simplify the square root to obtain the final area.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps

Find the third side
We know:\[\small \text{Perimeter} = a+b+c\]
Therefore\[\small c = P-(a+b)\]
Substituting the values:\[\small \begin{aligned}c &= 42-(18+10)\\c &= 42-28\\ &= 14~\text{cm}\end{aligned}\]
Thus, the three sides of the triangle are:\[\small 18~\text{cm},\ 10~\text{cm},\ 14~\text{cm}\]
Calculate the semi-perimeter
Semi-perimeter:\[\small \begin{aligned}s &= \frac{P}{2}\\&= \frac{42}{2}\\&=21\text{ cm}\end{aligned}\]
Calculate \(\small s-a\), \(\small s-b\), and \(\small s-c\)
\[\small \begin{aligned} s-a = 21-18=3\text{ cm}\\ s-b = 21-10=11~\text{cm}\\ s-c = 21-14=7~\text{cm} \end{aligned} \]
Apply Heron’s Formula
\[\small \text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\]
Substituting the values:\[\small \text{Area}=\sqrt{21 \times 3 \times 11 \times 7}\]
Rearanging the terms:\[\small \begin{aligned}\text{Area}&=\sqrt{21 \times 7 \times 3 \times 11}\\&=\sqrt{\overline{21 \times 21}\; \times 11}\\&=21\sqrt{11}\end{aligned}\]

💡 Answer Final Answer ›

The area of the triangle is: \(\small \text{Area} = 21\sqrt{11}~\text{cm}^2\)

🎯 Exam Significance Exam Significance ›

This problem combines perimeter concepts with Heron’s Formula.
Students learn how to derive a missing side before calculating area.
Board examinations frequently include multi-step geometry problems like this.
Competitive exams test simplification of surds and algebraic accuracy.
This question improves logical sequencing of mathematical steps.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 4 points

he third side of a triangle can be found using perimeter.
Semi-perimeter is always half of the perimeter.
Heron’s Formula is applied only after all three sides are known.
Perfect square factors help simplify irrational expressions.

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Q5 →

📘 Concept & Theory Concept Used ›

When the sides of a triangle are given in a ratio, we assume the sides as multiples of a common variable.

If the ratio of sides is:

\[\small 12 : 17 : 25 \]

then the sides can be written as:

\[\small 12x,\ 17x,\ 25x \]

Using the perimeter, we first find the value of \(\small x\).

After finding all three sides, Heron’s Formula is applied:

\[\small \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]

where:

\[\small s = \frac{a+b+c}{2} \]

🗺️ Solution Roadmap Step-by-step Plan ›

Assume the sides using the given ratio.
Use the perimeter to calculate the common factor \(\small x\).
Determine the actual lengths of all three sides.
Find the semi-perimeter of the triangle.
Apply Heron’s Formula step by step.
Simplify the square root carefully to obtain the final area.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

Given
Ratio of sides:

\[\small 12 : 17 : 25 \]

Perimeter of the triangle:

\[\small 540~\text{cm} \]
Assume the sides
Let the sides of the triangle be:\[\small 12x,\ 17x,\ 25x\]
Sum of the ratio terms: 12+17+25=54
Therefore,\[\small 54x = 540\]
Find the value of \(\small x\)
\[\small \begin{aligned} x &= \frac{540}{54}\\x&=10\end{aligned}\]
Find the actual sides of the triangle
sides=\[\small \begin{aligned} a &= 12 \times 10=120\\ b &= 17\times 10=170\\ c &= 25 \times 10=250 \end{aligned}\]
Calculate the semi-perimeter
\[\small s = \frac{a+b+c}{2}\]
Substituting the values: \[\small \begin{aligned} s &= \frac{120+170+250}{2}\\ &= \frac{540}{2}\\ &=270~\text{cm} \end{aligned}\]
Calculate \(\small s-a\), \(\small s-b\), and \(\small s-c\)
\[\small \begin{aligned} s-a &= 270-120=150~\text{cm}\\ s-b &= 270-170=100~\text{cm}\\ s-c &= 270-250= 20~\text{cm} \end{aligned}\]
Apply Heron’s Formula
\[\small \text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\]
Substituting the values: \[\small \begin{aligned} \text{Area}&=\sqrt{270 \times 150 \times 100 \times 20}\\ &=\sqrt{3\times9\times10 \times 3\times5\times10 \times 10\times10 \times 2\times10}\\ &=\sqrt{2\times2\times3\times3\times3\times3\times5\times5\times10 \times 10\times10 \times 10}\\ &=\sqrt{\overline{2\times2}\;\times\overline{3\times3}\;\times\overline{3\times3}\;\times\overline{5\times5}\;\times\overline{10 \times 10}\;\times\overline{10 \times 10}}\\ &=2\times3\times3\times5\times10\times10\\ &=9000\text{ cm}^2 \end{aligned}\]

💡 Answer Final Answer ›

Answer: Area of triangle is \(\small 9000\text{ cm}^2\)

🎯 Exam Significance Exam Significance ›

This question develops understanding of ratio-based geometry problems.
Students learn how to convert ratios into actual side lengths.
Board examinations frequently include application-based Heron’s Formula questions.
Competitive entrance exams test arithmetic simplification and multi-step calculations.
The problem improves algebraic reasoning and formula application skills.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 4 points

Ratio problems are solved by assuming a common multiplying factor.
Perimeter helps determine the actual side lengths.
Heron’s Formula requires all three side lengths.
Perfect square decomposition helps simplify square roots easily.

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Q6 →

📘 Concept & Theory Concept Used ›

An isosceles triangle is a triangle in which two sides are equal.

When the perimeter and two equal sides are given, the third side can be found using:

\[\small \text{Third Side} = \text{Perimeter} - (\text{Sum of Equal Sides}) \]

After finding all three sides, Heron’s Formula is used to calculate the area.

Semi-perimeter:

\[\small s = \frac{a+b+c}{2} \]

Area of triangle:

\[\small \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]

🗺️ Solution Roadmap Step-by-step Plan ›

Use the perimeter to calculate the third side of the triangle.
List all three sides of the triangle.
Find the semi-perimeter.
Calculate \(\small s-a\), \(\small s-b\), and \(\small s-c\).
Apply Heron’s Formula carefully.
Simplify the square root to obtain the final area.

📊 Graph / Figure Graph / Figure ›

Isosceles Triangle

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps

Given
Perimeter of the isosceles triangle:

\[\small 30~\text{cm} \]

Equal sides:

\[\small 12~\text{cm},\ 12~\text{cm} \]
Find the third side
Perimeter of a triangle:\[\small P = a+b+c\]
Let the third side be \(\small c\).
Therefore:\[\small \begin{aligned} c &= 30-(12+12)\\&= 30-24\\&= 6~\text{cm}\end{aligned}\]
List all side lengths
\[\small \begin{aligned} a &= 12~\text{cm}\\ b &= 12~\text{cm}\\ c &= 6~\text{cm} \end{aligned}\]
Calculate the semi-perimeter
Semi-perimeter:\[\small s = \frac{a+b+c}{2}\]
Substituting the values: \[\small \begin{aligned} s &= \frac{12+12+6}{2}\\ &= \frac{30}{2}\\ &= 15~\text{cm} \end{aligned}\]
Calculate \(\small s-a\), \(\small s-b\), and \(\small s-c\)
\[\small \begin{aligned} s-a &= 15-12= 3~\text{cm}\\ s-b &= 15-12= 3~\text{cm}\\ s-c &= 15-6 = 9~\text{cm} \end{aligned}\]
Apply Heron’s Formula
\[\small \begin{aligned} \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}\\ \end{aligned}\]
Substituting the values: \[\small \begin{aligned} \text{Area}&=\sqrt{15 \times 3 \times 3 \times 9}\\ &=\sqrt{15 \times 3 \times 3 \times 3 \times 3}\\ &=\sqrt{3 \times 3 \times 3 \times 3\times 15} \\ &=\sqrt{\overline{3 \times 3}\; \times \overline{3 \times 3}\;\times 15} \\ &=3\times3\sqrt{15}\\ &=9\sqrt{15} \end{aligned}\]

💡 Answer Final Answer ›

Answer: The area of the triangle is \(\small 9\sqrt{15}~\text{cm}^2\)

🎯 Exam Significance Exam Significance ›

This question combines properties of isosceles triangles with Heron’s Formula.
Students learn how to derive a missing side from the perimeter.
Board examinations frequently include geometry problems involving special triangles.
Competitive entrance exams test square root simplification and formula application.
The problem strengthens logical sequencing and algebraic manipulation skills.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 4 points

In an isosceles triangle, two sides are always equal.
The third side can be found using the perimeter.
Heron’s Formula requires all three sides of the triangle.
Perfect square factors should be separated for easier simplification.

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