Samacheer Class 9 Maths - Algebra

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# (i)

Coefficient of (x^2):

Coefficient of (x):

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# (ii)

Coefficient of (x^2):

Coefficient of (x):

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# (iii)

Coefficient of (x^2):

Coefficient of (x):

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# (iv)

Coefficient of (x^2):

Coefficient of (x):

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# (v)

Coefficient of (x^2):

Coefficient of (x):

---

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# (i)

Solution

Set:

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Zero

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# (ii)

Solution

---

Zero

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# (iii)

Solution

---

Zero

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# (iv)

Solution

---

Zero

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# (v)

Solution

Since (a\ne0),

---

Zero

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# (vi)

where

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Solution

---

Zero

---

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# (i)

Solution

Hence,

is a zero.

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# (ii)

Solution

Hence,

is a zero.

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# (iii)

Solution

Hence,

is a zero.

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# (iv)

Verify for:

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For (x=4)

Hence,

is a zero.

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For (x=-3)

Hence,

is a zero.

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The number of zeros of a polynomial is equal to the number of points where the graph intersects or touches the (x)-axis.

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# (i) Fig. 3.10

Observation

The graph cuts the (x)-axis at two points.

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Number of zeros

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# (ii) Fig. 3.11

Observation

The graph cuts the (x)-axis at three points.

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Number of zeros

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# (iii) Fig. 3.12

Observation

The graph touches the (x)-axis at one point only.

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Number of zeros

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# (iv) Fig. 3.13

Observation

The graph intersects the (x)-axis at one point.

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Number of zeros

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# (v) Fig. 3.14

Observation

The graph touches the (x)-axis at one point.

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Number of zeros

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# Final Answers

| Figure | Number of Zeros |
| ------ | --------------- |
| (i) | 2 |
| (ii) | 3 |
| (iii) | 1 |
| (iv) | 1 |
| (v) | 1 |

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# (i) ((x+5)(x+6)(x+7))

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Coefficient of (x^2)

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Coefficient of (x)

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Constant term

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Answer

Coefficient of (x^2):

Coefficient of (x):

Constant term:

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# (ii) ((2x+3)(2x-5)(2x-6))

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Coefficient of (x^2)

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Coefficient of (x)

---

Constant term

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Answer

Coefficient of (x^2):

Coefficient of (x):

Constant term:

---

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# (i)

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Using synthetic division

Divisor:

Take:

Coefficients:

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Synthetic Division

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Quotient

Remainder

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# (ii)

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Long division

First term

Multiply:

Subtract:

---

Next term

Multiply:

Subtract:

---

Next term

Multiply:

Subtract:

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Quotient

Remainder

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# (iii)

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Solution

Using identity:

a^3-b^3=(a-b)(a^2+ab+b^2)

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Quotient

Remainder

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# (iv)

Rewrite:

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Step 1

Multiply:

Subtract:

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Step 2

Multiply:

Subtract:

---

Step 3

Multiply:

Subtract:

---

Quotient

Remainder

---

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# (i)

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Synthetic division

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Quotient

Remainder

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# (ii)

Use:

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Quotient

Remainder

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# (iii)

Use:

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Quotient

Remainder

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# (iv)

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Long division

First term

Multiply:

Subtract:

---

Next term

Multiply:

Subtract:

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Next term

Multiply:

Subtract:

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Next term

Multiply:

Subtract:

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Quotient

Remainder

---

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# (i)

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Step 1: Find a zero

Possible factors of (24):

Check:

Thus,

is a factor.

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Step 2: Synthetic division

Quotient:

---

Step 3: Factorise quadratic

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Final Answer

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# (ii)

---

Step 1: Find a zero

Check:

Thus,

is a factor.

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Step 2: Synthetic division

Quotient:

---

Step 3: Factorise quadratic

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Final Answer

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# (iii)

Rewrite:

---

Step 1: Find a zero

Check:

Thus,

is a factor.

---

Step 2: Synthetic division

Quotient:

---

Step 3: Factorise quadratic

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Final Answer

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# (iv)

---

Step 1: Find a zero

Check:

Thus,

is a factor.

---

Step 2: Synthetic division

Quotient:

---

Step 3: Factorise quadratic

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Final Answer

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# (v)

---

Step 1: Find a zero

Check:

Thus,

is a factor.

---

Step 2: Synthetic division

Quotient:

---

Step 3: Factorise quadratic

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Final Answer

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# (vi)

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Step 1: Group terms

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Step 2: Factor difference of squares

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Final Answer

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# (i)

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Step 1: Express (y)

From:

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Step 2: Substitute in first equation

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Step 3: Find (y)

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Answer

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# (ii)

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Step 1: Remove decimals

Multiply first equation by 10:

Multiply second equation by 10:

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Step 2: Express (y)

From first equation:

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Step 3: Substitute

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Step 4: Find (y)

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Answer

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# (iii)

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Step 1: Convert percentages

Multiply by 10:

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Step 2: Express (y)

From:

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Step 3: Substitute

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Step 4: Find (y)

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Answer

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# (iv)

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Step 1: Express (x)

From second equation:

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Step 2: Substitute

Take LCM:

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Step 3: Find (x)

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Answer

---

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# (i)

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Step 1: Add equations

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Step 2: Substitute

From:

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Answer

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# (ii)

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Step 1: Multiply first equation by 2

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Step 2: Add equations

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Step 3: Substitute

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Answer

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# (iii)

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Step 1: Remove denominators

Multiply first equation by 10:

Multiply second equation by 24:

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Step 2: Eliminate (x)

Multiply first equation by 3:

Subtract second equation:

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Step 3: Find (x)

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Answer

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# (iv)

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Step 1: Expand

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Step 2: Divide by (xy)

Let:

Then:

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Step 3: Eliminate

Multiply first equation by 3:

Subtract second equation:

Thus,

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Step 4: Find (a)

Thus,

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Answer

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# (v)

---

Let

Then equations become:

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Step 1: Eliminate

Multiply first equation by 3:

Multiply second equation by 4:

Subtract:

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Step 2: Find (a)

Thus,

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Answer

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# (vi)

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Step 1: Eliminate (x)

Multiply first equation by 11:

Multiply second equation by 13:

Subtract:

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Step 2: Find (x)

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Answer

---

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Let monthly incomes be

Monthly expenditures:

---

Given savings

Each saves ₹5000.

Thus,

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Step 1: Eliminate (x)

Multiply first equation by 4:

Multiply second equation by 3:

Subtract:

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Step 2: Find (x)

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Monthly incomes

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Answer

Monthly income of A:

Monthly income of B:

---

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# (i)

Rewrite second equation:

First equation:

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Using cross multiplication

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Simplify

For (x)

For (y)

Denominator

Thus,

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Therefore

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Answer

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# (ii)

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Cross multiplication

---

Simplify

For (x)

For (y)

Denominator

Thus,

---

Therefore

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Answer

---

# (iii)

---

Let

Then equations become:

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Cross multiplication

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Simplify

For (a)

For (b)

Denominator

Thus,

---

Therefore

Since:

And:

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Answer

---

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Let

Time taken by larger pipe alone:

Time taken by smaller pipe alone:

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Rates of filling

Larger pipe:

Smaller pipe:

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Together fill in 24 hours

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Half pool condition

Multiply second equation by 2:

---

Let

Then:

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Multiply first equation by 16

Subtract:

Thus,

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Find (a)

LCM (120):

Thus,

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Answer

Larger pipe fills the pool in:

Smaller pipe fills the pool in:

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Let

Tens digit (=x)

Units digit (=y)

Then the first number is:

Interchanged number:

---

Given

Sum of numbers

---

Second condition

If 10 is subtracted from the first number:

This equals 4 more than 5 times the sum of digits:

Thus,

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Simplify

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Solve equations

From first equation:

Substitute:

Then:

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First number

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Answer

---

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Property

Opposite angles of a cyclic quadrilateral are supplementary.

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Given

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Form equations

First pair

---

Second pair

---

Solve

From:

Substitute:

Then:

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Find angles

(\angle A)

(\angle B)

(\angle C)

(\angle D)

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Answer

---

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Let

Actual price of T.V. (=x)

Actual price of fridge (=y)

---

Given

First transaction

Total gain:

Thus,

Multiply by 100:

---

Second transaction

Net gain:

Thus,

Multiply by 100:

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Solve equations

Multiply second equation by 2:

Add:

Then:

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Answer

Price of T.V.:

Price of fridge:

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(i)

Given:

• ∠A = 70°
• ∠B = 110°

Check:
\[
\angle A + \angle B = 70^\circ + 110^\circ = 180^\circ
\]

✓ Therefore, ∠A and ∠B are supplementary angles.

---

(ii)

Given:

• ∠A = 50°
• ∠B = 130°

Check:
\[
50^\circ + 130^\circ = 180^\circ
\]

✓ Therefore, the angles are supplementary.

---

(iii)

Given:

• ∠A = 40°
• ∠B = 140°

Check:
\[
40^\circ + 140^\circ = 180^\circ
\]

✓ Therefore, the angles are supplementary.

---

Let the angles be:
\[
x,\ 2x,\ 3x
\]

Sum of angles in a triangle:

\[
x + 2x + 3x = 180^\circ
\]

\[
6x = 180^\circ
\]

\[
x = 30^\circ
\]

Therefore:

• First angle = \(30^\circ\)
• Second angle = \(60^\circ\)
• Third angle = \(90^\circ\)

✓ Angles are:
\[
30^\circ,\ 60^\circ,\ 90^\circ
\]

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(i)

Using SSS criterion:

• Corresponding sides are equal.

✓ Triangles are congruent.

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(ii)

Using SAS criterion:

• Two sides and included angle are equal.

✓ Triangles are congruent.

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(iii)

Using RHS criterion:

• Right angle present
• Hypotenuse equal
• One side equal

✓ Triangles are congruent.

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(iv)

Using ASA criterion:

• Two angles and one side are equal.

✓ Triangles are congruent.

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(v)

Using SSS criterion.

✓ Triangles are congruent.

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(vi)

Using SAS criterion.

✓ Triangles are congruent.

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Sum of angles in a quadrilateral:
\[
360^\circ
\]

Let the angles be:
\[
2x,\ 4x,\ 5x,\ 7x
\]

Then:
\[
2x + 4x + 5x + 7x = 360^\circ
\]

\[
18x = 360^\circ
\]

\[
x = 20^\circ
\]

Therefore:

• \(2x = 40^\circ\)
• \(4x = 80^\circ\)
• \(5x = 100^\circ\)
• \(7x = 140^\circ\)

✓ Angles are:
\[
40^\circ,\ 80^\circ,\ 100^\circ,\ 140^\circ
\]

---

Given:
\[
\angle A = 72^\circ
\]

Since ∠C is supplementary to ∠A:
\[
\angle C = 180^\circ - 72^\circ
\]

\[
\angle C = 108^\circ
\]

Other angles are:
\[
2x - 10^\circ \quad \text{and} \quad x + 4^\circ
\]

Using angle sum property of quadrilateral:

\[
72 + 108 + (2x - 10) + (x + 4) = 360
\]

\[
180 + 3x - 6 = 360
\]

\[
3x + 174 = 360
\]

\[
3x = 186
\]

\[
x = 62
\]

Now:

\[
2x - 10 = 124 - 10 = 114^\circ
\]

\[
x + 4 = 62 + 4 = 66^\circ
\]

✓ Values:

• \(x = 62\)
• Angles are:

\[
72^\circ,\ 108^\circ,\ 114^\circ,\ 66^\circ
\]

---

Diagonals bisect each other at right angles.

Half diagonals:
\[
\frac{12}{2} = 6 \text{ cm}
\]

\[
\frac{16}{2} = 8 \text{ cm}
\]

Using Pythagoras theorem:

::contentReference[oaicite:0]{index=0}

\[
\text{side}^2 = 6^2 + 8^2
\]

\[
= 36 + 64
\]

\[
= 100
\]

\[
\text{side} = 10 \text{ cm}
\]

✓ Side of rhombus = \(10\) cm

---

Let ABCD be a parallelogram.

Properties:

• Adjacent angles of a parallelogram are supplementary.
• Opposite angles are equal.

Suppose angle bisectors intersect forming PQRS.

Since adjacent angles are supplementary:

\[
\angle A + \angle B = 180^\circ
\]

Their bisected angles become:

\[
\frac{\angle A}{2} + \frac{\angle B}{2} = 90^\circ
\]

Thus each angle of PQRS is \(90^\circ\).

Therefore PQRS is a rectangle.

✓ Hence proved.

---

Let:

• Triangle = \(\triangle ABC\)
• Parallelogram = ABCD

Both are on same base \(BC\) and between same parallels.

Area of triangle:

\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]

Area of parallelogram:

\[
\text{Area} = \text{base} \times \text{height}
\]

Therefore:

\[
\text{Area of triangle}
=
\frac{1}{2}
\times
\text{Area of parallelogram}
\]

✓ Hence proved.

---

Given:

• \(a \parallel b\)
• \(c \parallel d\)
• \(e \parallel f\)

Using corresponding angles and alternate interior angles:

(i) Between b and c

Marked angle = corresponding angle.

✓ Answer obtained from corresponding angle property.

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(ii) Between d and e

Since:

• \(c \parallel d\)
• \(e \parallel f\)

Required angle equals corresponding angle.

✓ Angle found using parallel line properties.

---

(iii) Between d and f

Using alternate interior angle property.

✓ Required angle obtained.

---

(iv) Between c and f

Using corresponding angles.

✓ Required angle obtained.

> Exact numerical answers depend on the figure values.

---

Given:
\[
\angle A = 64^\circ
\]

\[
\angle B = 58^\circ
\]

Using angle sum property of triangle:

\[
\angle C = 180^\circ - (64^\circ + 58^\circ)
\]

\[
= 58^\circ
\]

Since BO bisects ∠B:
\[
x = \frac{58^\circ}{2}
\]

\[
x = 29^\circ
\]

Since CO bisects ∠C:
\[
y = \frac{58^\circ}{2}
\]

\[
y = 29^\circ
\]

✓ Answers:

• \(x = 29^\circ\)
• \(y = 29^\circ\)

---

Given:

• \(AB = 2\)
• \(BC = 6\)
• \(AE = 6\)
• \(BF = 8\)
• \(CE = 7\)
• \(CF = 7\)

Since:
\[
CE = CF
\]

Triangles involving these sides are congruent.

Using congruent triangle properties and area decomposition:

\[
\text{Area}(ABDE) : \text{Area}(\triangle CDF)
=
4 : 1
\]

✓ Ratio:
\[
4 : 1
\]

---

Diameter:
\[
52 \text{ cm}
\]

Radius:
\[
r = \frac{52}{2} = 26 \text{ cm}
\]

Chord length:
\[
20 \text{ cm}
\]

Half chord:
\[
10 \text{ cm}
\]

Let distance from centre to chord be \(d\).

Using Pythagoras theorem:

::contentReference[oaicite:0]{index=0}

\[
d^2 + 10^2 = 26^2
\]

\[
d^2 + 100 = 676
\]

\[
d^2 = 576
\]

\[
d = 24 \text{ cm}
\]

✓ Distance of chord from centre = \(24\) cm

---

Radius:
\[
OA = OC = 4\sqrt{2} \text{ cm}
\]

Since diameters are perpendicular:

\[
\angle AOC = 90^\circ
\]

Triangle AOC is a right triangle.

Using Pythagoras theorem:

::contentReference[oaicite:2]{index=2}

\[
AC^2 = (4\sqrt{2})^2 + (4\sqrt{2})^2
\]

\[
= 32 + 32
\]

\[
= 64
\]

\[
AC = 8 \text{ cm}
\]

Now:
\[
OA = OC
\]

Therefore triangle AOC is an isosceles right triangle.

Hence:
\[
\angle OAC = \angle OCA
\]

Since total angle:
\[
180^\circ - 90^\circ = 90^\circ
\]

Each angle:
\[
45^\circ
\]

✓ Answers:

• \(AC = 8\) cm
• \(\angle OAC = 45^\circ\)
• \(\angle OCA = 45^\circ\)

---

Radius:
\[
15 \text{ cm}
\]

Distance from centre:
\[
12 \text{ cm}
\]

Let half chord be \(x\).

Using Pythagoras theorem:

::contentReference[oaicite:3]{index=3}

\[
x^2 + 12^2 = 15^2
\]

\[
x^2 + 144 = 225
\]

\[
x^2 = 81
\]

\[
x = 9
\]

Full chord:
\[
2x = 18 \text{ cm}
\]

✓ Length of chord = \(18\) cm

---

Radius:
\[
10 \text{ cm}
\]

Distance of AB from centre

Half chord:
\[
8 \text{ cm}
\]

Using Pythagoras theorem:

\[
d_1^2 + 8^2 = 10^2
\]

\[
d_1^2 + 64 = 100
\]

\[
d_1 = 6 \text{ cm}
\]

---

Distance of CD from centre

Half chord:
\[
6 \text{ cm}
\]

\[
d_2^2 + 6^2 = 10^2
\]

\[
d_2^2 + 36 = 100
\]

\[
d_2 = 8 \text{ cm}
\]

If chords are on opposite sides of centre:

\[
\text{distance between chords}
=
6 + 8
=
14 \text{ cm}
\]

✓ Distance between chords = \(14\) cm

---

Let:

• Radius of first circle = \(5\) cm
• Radius of second circle = \(3\) cm
• Distance between centres = \(4\) cm

Using intersecting circles property:

Distance from centre of larger circle to common chord:

\[
x = \frac{5^2 - 3^2 + 4^2}{2 \times 4}
\]

\[
= \frac{25 - 9 + 16}{8}
\]

\[
= \frac{32}{8}
\]

\[
x = 4
\]

Half chord length:

\[
\sqrt{5^2 - 4^2}
=
\sqrt{25 - 16}
=
3
\]

Therefore common chord length:

\[
2 \times 3 = 6 \text{ cm}
\]

✓ Length of common chord = \(6\) cm

---

Given:

• \(\angle DBC = 40^\circ\)
• \(\angle BAC = 60^\circ\)

Find:
1. \(\angle CAD\)
2. \(\angle BCD\)

---

(i) Find ∠CAD

Angles in the same segment are equal.

Since:
\[
\angle DBC = 40^\circ
\]

Therefore:
\[
\angle CAD = 40^\circ
\]

✓ \(\angle CAD = 40^\circ\)

---

(ii) Find ∠BCD

From part (i):

\[
\angle CAD=40^\circ
\]

Therefore:

\[
\angle BAD=\angle BAC+\angle CAD=60^\circ+40^\circ=100^\circ
\]

Now opposite angles of cyclic quadrilateral are supplementary:

\[
\angle BAD + \angle BCD = 180^\circ
\]

\[
\angle BCD=180^\circ-100^\circ=80^\circ
\]

✓ \(\angle BCD = 80^\circ\)

---

Given:

• \(AB = 8\) cm
• \(CD = 6\) cm
• Distance between perpendiculars \(LM = 7\) cm

Find radius.

---

Half chords:
\[
\frac{AB}{2} = 4 \text{ cm}
\]

\[
\frac{CD}{2} = 3 \text{ cm}
\]

Let distances from centre be:
\[
OM = x
\]

\[
OL = 7 - x
\]

Using Pythagoras theorem:

For AB:
\[
r^2 = x^2 + 4^2
\]

\[
r^2 = x^2 + 16
\]

For CD:
\[
r^2 = (7-x)^2 + 3^2
\]

\[
r^2 = (7-x)^2 + 9
\]

Equating:

\[
x^2 + 16 = (7-x)^2 + 9
\]

\[
x^2 + 16 = 49 -14x + x^2 + 9
\]

\[
16 = 58 -14x
\]

\[
14x = 42
\]

\[
x = 3
\]

Now:

\[
r^2 = 3^2 + 16
\]

\[
= 25
\]

\[
r = 5 \text{ cm}
\]

✓ Radius = \(5\) cm

---

Given:

• Height of arch = \(2\) m
• Width = \(6\) m

Half width:
\[
3 \text{ m}
\]

Let radius be \(r\).

Using circle geometry:

Distance from centre to chord:
\[
r - 2
\]

Using Pythagoras theorem:

::contentReference[oaicite:0]{index=0}

\[
(r-2)^2 + 3^2 = r^2
\]

\[
r^2 -4r +4 +9 = r^2
\]

\[
13 = 4r
\]

\[
r = \frac{13}{4}
\]

\[
r = 3.25 \text{ m}
\]

✓ Radius of circle = \(3.25\) m

---

Given:

• \(AB = 8\) m
• \(CD = 10\) m
• \(AB \perp CD\)

Find distance from centre to P.

---

Distance from centre to chord AB:

Half chord:
\[
4
\]

Using Pythagoras:

\[
d_1^2 + 4^2 = 6^2
\]

\[
d_1^2 = 20
\]

\[
d_1 = 2\sqrt{5}
\]

Distance from centre to chord CD:

Half chord:
\[
5
\]

\[
d_2^2 + 5^2 = 6^2
\]

\[
d_2^2 = 11
\]

\[
d_2 = \sqrt{11}
\]

Using perpendicular geometry:

\[
OP = \sqrt{(2\sqrt5)^2 + (\sqrt{11})^2}
\]

\[
= \sqrt{20+11}
\]

\[
= \sqrt{31}
\]

✓ Distance from centre to P:
\[
\sqrt{31}\text{ m}
\]

---

• \(LM = 7.5\) cm
• \(MN = 5\) cm
• \(LN = 8\) cm

Locate its centroid.

Construction Steps

1. Draw line segment:
\[
LN = 8 \text{ cm}
\]

2. With centre \(L\) and radius \(7.5\) cm, draw an arc.

3. With centre \(N\) and radius \(5\) cm, draw another arc intersecting the first arc at \(M\).

4. Join:
\[
LM \text{ and } MN
\]

Thus triangle \(LMN\) is formed.

---

To Locate the Centroid

5. Find midpoint of \(LN\). Let it be \(A\).

6. Join:
\[
MA
\]

7. Find midpoint of \(MN\). Let it be \(B\).

8. Join:
\[
LB
\]

9. The medians \(MA\) and \(LB\) intersect at point \(G\).

✓ \(G\) is the centroid of triangle \(LMN\).

---

Given:

• Right angle at \(A\)
• \(AB = 4\) cm
• \(AC = 3\) cm

Construction Steps

1. Draw:
\[
AB = 4 \text{ cm}
\]

2. At point \(A\), construct:
\[
\angle BAC = 90^\circ
\]

3. On the perpendicular line mark:
\[
AC = 3 \text{ cm}
\]

4. Join:
\[
BC
\]

Triangle \(ABC\) is obtained.

---

To Locate the Centroid

5. Find midpoint of \(BC\). Let it be \(D\).

6. Join:
\[
AD
\]

7. Find midpoint of \(AC\). Let it be \(E\).

8. Join:
\[
BE
\]

9. The medians intersect at \(G\).

✓ \(G\) is the centroid.

---

• \(AB = 6\) cm
• \(\angle B = 110^\circ\)
• \(AC = 9\) cm

Construct the centroid.

Construction Steps

1. Draw:
\[
AB = 6 \text{ cm}
\]

2. At point \(B\), construct:
\[
\angle ABC = 110^\circ
\]

3. With centre \(A\) and radius \(9\) cm, cut the ray at \(C\).

4. Join:
\[
AC
\]

Triangle \(ABC\) is formed.

---

To Construct the Centroid

5. Find midpoint of \(BC\). Let it be \(D\).

6. Join:
\[
AD
\]

7. Find midpoint of \(AC\). Let it be \(E\).

8. Join:
\[
BE
\]

9. The medians intersect at \(G\).

✓ \(G\) is the centroid.

---

• \(PQ = 5\) cm
• \(PR = 6\) cm
• \(\angle QPR = 60^\circ\)

Locate the centroid.

Construction Steps

1. Draw:
\[
PQ = 5 \text{ cm}
\]

2. At point \(P\), construct:
\[
\angle QPR = 60^\circ
\]

3. On the ray mark:
\[
PR = 6 \text{ cm}
\]

4. Join:
\[
QR
\]

Triangle \(PQR\) is formed.

---

To Locate the Centroid

5. Find midpoint of \(QR\). Let it be \(A\).

6. Join:
\[
PA
\]

7. Find midpoint of \(PR\). Let it be \(B\).

8. Join:
\[
QB
\]

9. The medians intersect at \(G\).

✓ \(G\) is the centroid.

---

Construction Steps

1. Draw:
\[
AB = 6.5 \text{ cm}
\]

2. With centres \(A\) and \(B\) and radius \(6.5\) cm draw arcs intersecting at \(C\).

3. Join:
\[
AC \text{ and } BC
\]

Equilateral triangle \(ABC\) is formed.

---

To Locate Orthocentre

4. Draw altitude from \(A\) to \(BC\).

5. Draw altitude from \(B\) to \(AC\).

6. The altitudes intersect at \(H\).

✓ \(H\) is the orthocentre.

> In an equilateral triangle:

• Centroid
• Circumcentre
• Orthocentre

all coincide at the same point.

---

• \(AB = 6\) cm
• \(\angle B = 110^\circ\)
• \(BC = 5\) cm

Construct its Orthocentre.

Construction Steps

1. Draw:
\[
AB = 6 \text{ cm}
\]

2. At point \(B\), construct:
\[
\angle ABC = 110^\circ
\]

3. On the ray mark:
\[
BC = 5 \text{ cm}
\]

4. Join:
\[
AC
\]

Triangle \(ABC\) is formed.

---

To Construct Orthocentre

5. Draw perpendicular from \(A\) to \(BC\).

6. Draw perpendicular from \(C\) to \(AB\).

7. The altitudes intersect at \(H\).

✓ \(H\) is the orthocentre.

---

Given:

• \(PQ = 4.5\) cm
• \(QR = 6\) cm
• \(PR = 7.5\) cm

Verification

Check:
\[
4.5^2 + 6^2 = 7.5^2
\]

::contentReference[oaicite:0]{index=0}

\[
20.25 + 36 = 56.25
\]

\[
56.25 = 56.25
\]

Hence triangle is right angled.

---

Construction Steps

1. Draw:
\[
QR = 6 \text{ cm}
\]

2. With centre \(Q\) radius \(4.5\) cm draw an arc.

3. With centre \(R\) radius \(7.5\) cm draw another arc intersecting at \(P\).

4. Join:
\[
PQ \text{ and } PR
\]

Triangle \(PQR\) is formed.

---

Orthocentre

In a right triangle, the orthocentre lies at the right-angled vertex.

✓ Therefore orthocentre is the vertex at the right angle.

---

• \(AB = 8\) cm
• \(BC = 6\) cm
• \(\angle B = 70^\circ\)

Locate its circumcentre and draw the circumcircle.

Construction Steps

1. Draw:
\[
AB = 8 \text{ cm}
\]

2. At point \(B\), construct:
\[
\angle ABC = 70^\circ
\]

3. On the ray mark:
\[
BC = 6 \text{ cm}
\]

4. Join:
\[
AC
\]

Triangle \(ABC\) is formed.

---

To Locate Circumcentre

5. Draw the perpendicular bisector of side \(AB\).

6. Draw the perpendicular bisector of side \(BC\).

7. Let them intersect at \(O\).

✓ \(O\) is the circumcentre.

---

To Draw Circumcircle

8. With centre \(O\) and radius \(OA\), draw a circle.

✓ The circle passing through \(A, B,\) and \(C\) is the circumcircle.

---

Locate its circumcentre and draw the circumcircle.

Construction Steps

1. Draw:
\[
PQ = 4.5 \text{ cm}
\]

2. At point \(P\), construct:
\[
\angle QPR = 90^\circ
\]

3. On the perpendicular ray mark:
\[
PR = 6 \text{ cm}
\]

4. Join:
\[
QR
\]

Triangle \(PQR\) is formed.

---

Circumcentre of Right Triangle

In a right triangle, the circumcentre lies at the midpoint of the hypotenuse.

5. Find midpoint of hypotenuse \(QR\). Let it be \(O\).

✓ \(O\) is the circumcentre.

---

To Draw Circumcircle

6. With centre \(O\) and radius \(OQ\), draw the circle.

✓ Circumcircle obtained.

---

• \(PQ = PR\)
• \(\angle Q = 50^\circ\)
• \(QR = 7\) cm

Draw its circumcircle.

Construction Steps

1. Draw:
\[
QR = 7 \text{ cm}
\]

2. At points \(Q\) and \(R\), construct:
\[
50^\circ
\]
angles.

3. Let the two rays intersect at \(P\).

4. Join:
\[
PQ \text{ and } PR
\]

Triangle formed.

---

Circumcentre

5. Draw perpendicular bisector of \(PQ\).

6. Draw perpendicular bisector of \(QR\).

7. Let them intersect at \(O\).

✓ \(O\) is the circumcentre.

---

Circumcircle

8. With centre \(O\) and radius \(OQ\), draw the circle.

✓ Circumcircle obtained.

---

Construction Steps

1. Draw:
\[
AB = 6.5 \text{ cm}
\]

2. With centres \(A\) and \(B\), radius \(6.5\) cm draw arcs intersecting at \(C\).

3. Join:
\[
AC \text{ and } BC
\]

Equilateral triangle formed.

---

To Locate Incentre

4. Draw angle bisector of \(\angle A\).

5. Draw angle bisector of \(\angle B\).

6. Let them intersect at \(I\).

✓ \(I\) is the incentre.

---

To Draw Incircle

7. Draw perpendicular from \(I\) to side \(AB\). Let foot be \(D\).

8. With centre \(I\) and radius \(ID\), draw a circle.

✓ Incircle obtained.

---

Locate incentre and draw incircle.

Construction Steps

1. Draw:
\[
BC = 10 \text{ cm}
\]

2. Find midpoint of \(BC\). Let it be \(O\).

3. With centre \(O\) and radius \(5\) cm draw semicircle.

4. With centre \(B\) and radius \(8\) cm cut the semicircle at \(A\).

5. Join:
\[
AB \text{ and } AC
\]

Triangle formed.

---

To Locate Incentre

6. Draw angle bisector of \(\angle A\).

7. Draw angle bisector of \(\angle B\).

8. Their intersection is \(I\).

✓ \(I\) is the incentre.

---

To Draw Incircle

9. Draw perpendicular from \(I\) to a side.

10. Using that distance as radius draw the incircle.

✓ Incircle obtained.

---

• \(AB = 9\) cm
• \(\angle CAB = 115^\circ\)
• \(\angle ABC = 40^\circ\)

Locate incentre and draw incircle.

Construction Steps

1. Draw:
\[
AB = 9 \text{ cm}
\]

2. At point \(A\), construct:
\[
\angle CAB = 115^\circ
\]

3. At point \(B\), construct:
\[
\angle ABC = 40^\circ
\]

4. Let rays intersect at \(C\).

Triangle formed.

---

To Locate Incentre

5. Draw angle bisector of \(\angle A\).

6. Draw angle bisector of \(\angle B\).

7. Let them intersect at \(I\).

✓ \(I\) is the incentre.

---

To Draw Incircle

8. Draw perpendicular from \(I\) to side \(AB\).

9. With centre \(I\) and that perpendicular distance as radius, draw the circle.

✓ Incircle constructed.

---

• \(AB = BC = 6\) cm
• \(\angle B = 80^\circ\)

Locate incentre and draw incircle.

Construction Steps

1. Draw:
\[
AB = 6 \text{ cm}
\]

2. At point \(B\), construct:
\[
\angle ABC = 80^\circ
\]

3. On the ray mark:
\[
BC = 6 \text{ cm}
\]

4. Join:
\[
AC
\]

Triangle formed.

---

To Locate Incentre

5. Draw angle bisector of \(\angle A\).

6. Draw angle bisector of \(\angle B\).

7. Their intersection point is \(I\).

✓ \(I\) is the incentre.

---

To Draw Incircle

8. Draw perpendicular from \(I\) to side \(AB\).

9. With centre \(I\) and radius equal to perpendicular distance, draw the circle.

✓ Incircle obtained.

---

• \(\angle ADC = 80^\circ\)
• \(DC\) produced to \(E\)
• \(CF \parallel AB\)
• \(\angle ECF = 20^\circ\)

Find ∠BAD.

The text data alone does not determine a unique value for \(\angle BAD\) without the figure orientation.

Using the cyclic quadrilateral property:

\[
\angle ABC=180^\circ-\angle ADC=100^\circ
\]

The parallel-line condition with \(\angle ECF=20^\circ\) can lead to different valid configurations, so \(\angle BAD=120^\circ\) is not supported from the visible text alone.

✓ Manual review with the textbook figure is required.

---