AIOU 248 Solved Assignment Autumn 2025

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ALLAMA IQBAL OPEN UNIVERSITY

(Department of Mathematics)

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WARNING

1. Plagiarism or hiring of ghost writer(s) for solving the assignment(s) will debar the student from award of degree/certificate if found at any stage.

2. Submitting assignment(s) borrowed or stolen from other(s) as one's own will be penalized as defined in the "Aiou Plagiarism Policy".

Assignment Submission Schedule
6 Credit Hours	Due Date	3 Credit Hours	Due Date
Assignment 1	15-12-2025	Assignment 1	08-01-2026
Assignment 2	08-01-2026
Assignment 3	30-01-2026	Assignment 2	20-02-2026
Assignment 4	20-02-2026

Course: Mathematics-II (248)	Semester: Autumn,2025
Level: Matric / SSC

Please read the following instructions for writing your assignments. (SSC, HSSC & BA Programmes)
1. All questions are compulsory and carry equal marks but within a question the marks are distributed according to its requirements.
2. Read the question carefully and then answer it according to the requirements of the questions.
3. Late submission of assignments will not be accepted.
4. Your own analysis and synthesis will be appreciated.
5. Avoid irrelevant discussion/information and reproducing from books, study guide of allied material.

Total Marks: 100	Pass Marks: 40

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ASSIGNMENT No. 1

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Q1(a). If \( x = \sqrt{3} - \sqrt{2} \), then find the values of: (i) Find \( x - \frac{1}{x} \) and (ii) Find \( x^2 + \frac{1}{x^2} \). ▶

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Given:

\( x = \sqrt{3} - \sqrt{2} \)

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(i) Find \( x - \frac{1}{x} \)

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\(\frac{1}{x} = \frac{1}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \sqrt{3} + \sqrt{2}\)

So, \( x - \frac{1}{x} = (\sqrt{3} - \sqrt{2}) - (\sqrt{3} + \sqrt{2}) = -2\sqrt{2} \)

\( x - \frac{1}{x} = -2\sqrt{2} \)

\( x - \frac{1}{x} = -2.828 \)

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(ii) Find \( x^2 + \frac{1}{x^2} \)

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Using the identity: \( x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2 \)

\( \left(x - \frac{1}{x}\right)^2 = (-2\sqrt{2})^2 = 8 \)

So, \( x^2 + \frac{1}{x^2} = 8 + 2 = 10 \)

\( x^2 + \frac{1}{x^2} = 10 \)

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Q1(b). Find the value of x³ + y³ when x + y = 11 and xy = -8. ▶

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\( x + y = 11, \quad xy = -8 \)

To Find:

\( x^3 + y^3 \)

Step 1: Identity

\( x^3 + y^3 = (x + y)^3 - 3xy(x + y) \)

Step 2: Substitute Values

\( x^3 + y^3 = (11)^3 - 3(-8)(11) \)

Step 3: Simplify

\( 11^3 = 1331 \)

\( -3(-8)(11) = +264 \)

\( x^3 + y^3 = 1331 + 264 = 1595 \)

\( x^3 + y^3 = 1595 \)

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Q2(a). Factorize 8x³ - 6x - 9y + 27y³. ▶

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\( 8x^3 - 6x - 9y + 27y^3 \)

Step 1: Group Terms

\( (8x^3 + 27y^3) - (6x + 9y) \)

Step 2: Factorize Cubes

\( 8x^3 + 27y^3 = (2x)^3 + (3y)^3 = (2x + 3y)(4x^2 - 6xy + 9y^2) \)

Step 3: Factor Out Common Term

\( - (6x + 9y) = -3(2x + 3y) \)

So, \( 8x^3 - 6x - 9y + 27y^3 = (2x + 3y)(4x^2 - 6xy + 9y^2) - 3(2x + 3y) \)

Step 4: Final Factorization

\( = (2x + 3y)\big(4x^2 - 6xy + 9y^2 - 3\big) \)

\( 8x^3 - 6x - 9y + 27y^3 = (2x + 3y)(4x^2 - 6xy + 9y^2 - 3) \)

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Q2(b). Resolve into factors. 3u² - 10u + 8. ▶

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\( 3u^2 - 10u + 8 \)

Step 1: Multiply Coefficients

\( 3 \times 8 = 24 \)

We need two numbers whose product is \(24\) and sum is \(-10\). The numbers are \(-6\) and \(-4\).

Step 2: Split the Middle Term

\( 3u^2 - 6u - 4u + 8 \)

Step 3: Group and Factor

\( (3u^2 - 6u) - (4u - 8) \)

\( = 3u(u - 2) - 4(u - 2) \)

Step 4: Factorize

\( = (3u - 4)(u - 2) \)

\( 3u^2 - 10u + 8 = (3u - 4)(u - 2) \)

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Q3(a). The product of two polynomials and their H.C.F are
x⁴ + 6x³ - 3x² - 56x - 48 and x³ + 2x² - 11x - 12 respectively. Find their L.C.M. ▶

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\( \text{Product} = x^{4} + 6x^{3} - 3x^{2} - 56x - 48 \)

\( \text{H.C.F.} = x^{3} + 2x^{2} - 11x - 12 \)

To Find:

\( \text{L.C.M.} \)

Use the relation

\( \text{Product} = \text{H.C.F.} \times \text{L.C.M.} \)

So

\( \text{L.C.M.} = \dfrac{\text{Product}}{\text{H.C.F.}} \)

Divide

\( \dfrac{x^{4} + 6x^{3} - 3x^{2} - 56x - 48}{x^{3} + 2x^{2} - 11x - 12} = x + 4 \)

Verification

\( (x^{3} + 2x^{2} - 11x - 12)(x + 4) = x^{4} + 6x^{3} - 3x^{2} - 56x - 48 \)

\( \text{L.C.M.} = x + 4 \)

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Q3(b). Find the square root of the following: \( (x - \tfrac{1}{x})^2 - 4\left(x + \tfrac{1}{x}\right) + 8, \quad x \neq 0 \) ▶

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Step 1: Simplify inside

\( (x - \tfrac{1}{x})^2 - 4(x + \tfrac{1}{x}) + 8 \)

\( = x^2 + \tfrac{1}{x^2} - 4(x + \tfrac{1}{x}) + 6 \)

Step 2: Check if it is a perfect square

Assume it can be written as \( (x + \tfrac{1}{x} - k)^2 \).

\( (x + \tfrac{1}{x} - k)^2 = (x + \tfrac{1}{x})^2 - 2k(x + \tfrac{1}{x}) + k^2 \)

\( = x^2 + \tfrac{1}{x^2} + 2 - 2k(x + \tfrac{1}{x}) + k^2 \)

Step 3: Match coefficients

Comparing with \( x^2 + \tfrac{1}{x^2} - 4(x + \tfrac{1}{x}) + 6 \):

\(-2k = -4 \Rightarrow k = 2 \)

\( 2 + k^2 = 2 + 4 = 6 \)

Step 4: Final Result

\( (x - \tfrac{1}{x})^2 - 4(x + \tfrac{1}{x}) + 8 = (x + \tfrac{1}{x} - 2)^2 \)

So,

\( \sqrt{(x - \tfrac{1}{x})^2 - 4(x + \tfrac{1}{x}) + 8} = \big|x + \tfrac{1}{x} - 2\big| \)

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Q4(a). Solve \( x = \sqrt{5n + 9} = n - 1 \) ▶

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\( \sqrt{5n + 9} = n - 1 \)

Step 1: Condition

Since square root is always nonnegative, we require

\( n - 1 \geq 0 \quad \Rightarrow \quad n \geq 1 \)

Step 2: Square Both Sides

\( (\sqrt{5n + 9})^2 = (n - 1)^2 \)

\( 5n + 9 = n^2 - 2n + 1 \)

Step 3: Rearrange

\( 0 = n^2 - 2n + 1 - 5n - 9 \)

\( 0 = n^2 - 7n - 8 \)

Step 4: Factorize

\( n^2 - 7n - 8 = (n - 8)(n + 1) = 0 \)

So, \( n = 8 \) or \( n = -1 \)

Step 5: Check Solutions

For \( n = 8 \): \( \sqrt{5(8)+9} = \sqrt{49} = 7 \), and \( n - 1 = 7 \).

For \( n = -1 \): \( \sqrt{5(-1)+9} = \sqrt{4} = 2 \), but \( n - 1 = -2 \).

\( n = 8 \)

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Q4(b). Simplify \( \dfrac{x^4 - y^4}{x^2 - 2xy + y^2}\times\dfrac{x - y}{x(x + y)}\div\dfrac{x^2 + y^2}{x} \) ▶

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Step 1: Factor \(x^4 - y^4\)

\( x^4 - y^4 = (x^2 - y^2)(x^2 + y^2) = (x - y)(x + y)(x^2 + y^2) \)

Step 2: Factor the denominator

\( x^2 - 2xy + y^2 = (x - y)^2 \)

Step 3: Simplify the first fraction

\( \dfrac{x^4 - y^4}{x^2 - 2xy + y^2} = \dfrac{(x - y)(x + y)(x^2 + y^2)}{(x - y)^2} = \dfrac{(x + y)(x^2 + y^2)}{x - y} \)

Step 4: Multiply by the second factor

\( \dfrac{(x + y)(x^2 + y^2)}{x - y}\times\dfrac{x - y}{x(x + y)} = \dfrac{x^2 + y^2}{x} \)

Step 5: Divide by the third factor

\( \dfrac{x^2 + y^2}{x}\div\dfrac{x^2 + y^2}{x} = 1 \)

Final Answer:

\( 1 \)

Restrictions:

\( x \neq 0,\; x \neq y,\; x \neq -y,\; x^2 + y^2 \neq 0 \)

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Q5(a). Solve \( (x + 4)(x - 1) + (x + 5)(x + 2) = 6 \) by using the quadratic formula. ▶

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Step 1: Expand

\( (x + 4)(x - 1) = x^2 + 3x - 4 \)

\( (x + 5)(x + 2) = x^2 + 7x + 10 \)

Step 2: Add and simplify

\( x^2 + 3x - 4 + x^2 + 7x + 10 = 2x^2 + 10x + 6 \)

So the equation becomes

\( 2x^2 + 10x + 6 = 6 \)

\( 2x^2 + 10x = 0 \)

\( 2x(x + 5) = 0 \)

Step 3: Quadratic formula check

Write the quadratic as \( ax^2 + bx + c = 0 \) with \( a = 2 \), \( b = 10 \), \( c = 0 \).

\( x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( x = \dfrac{-10 \pm \sqrt{10^2 - 4(2)(0)}}{2(2)} = \dfrac{-10 \pm \sqrt{100}}{4} \)

\( x = \dfrac{-10 \pm 10}{4} \)

Hence

\( x = \dfrac{-10 + 10}{4} = 0 \)

or

\( x = \dfrac{-10 - 10}{4} = -5 \)

\( x = 0 \) or \( x = -5 \)

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Q5(b). Find two consecutive positive odd numbers. ▶

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Let the first positive odd number be \(n\). Then the next consecutive odd number is \(n+2\).

General form:

Two consecutive positive odd numbers are \(n\) and \(n+2\), where \(n\) is an odd positive integer.

Alternative using an integer parameter

Write \(n = 2k + 1\) with \(k \ge 0\) an integer. Then the two numbers are

\(2k + 1\) and \(2k + 3\).

Examples:

If \(k = 0\) then the numbers are \(1\) and \(3\).

If \(k = 1\) then the numbers are \(3\) and \(5\).

If \(k = 2\) then the numbers are \(5\) and \(7\).

\(n\) and \(n+2\) or equivalently \(2k+1\) and \(2k+3\)

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ASSIGNMENT No. 2

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Q1(a). If \( \begin{vmatrix} 2 & 7 \\[6pt] 3 & 2a \end{vmatrix} \begin{vmatrix} b \\[6pt] 7 \end{vmatrix} = \begin{vmatrix} 45 \\[6pt] 12 \end{vmatrix} \), find the values of \(a\) and \(b\). ▶

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We are given:

\[ \begin{vmatrix} 2 & 7 \\[6pt] 3 & 2a \end{vmatrix} \begin{vmatrix} b \\[6pt] 7 \end{vmatrix} = \begin{vmatrix} 45 \\[6pt] 12 \end{vmatrix} \]

First, simplify the determinant on the left:

\[ \begin{vmatrix} 2 & 7 \\[6pt] 3 & 2a \end{vmatrix} = (2)(2a) - (7)(3) = 4a - 21 \]

So the equation becomes:

\[ (4a - 21) \begin{vmatrix} b \\[6pt] 7 \end{vmatrix} = \begin{vmatrix} 45 \\[6pt] 12 \end{vmatrix} \]

Multiplying scalar with a column vector:

\[ \begin{vmatrix} (4a - 21)b \\[6pt] (4a - 21)7 \end{vmatrix} = \begin{vmatrix} 45 \\[6pt] 12 \end{vmatrix} \]

Now equate corresponding entries:

\[ (4a - 21)b = 45 \quad \text{and} \quad (4a - 21)7 = 12 \]

From the second equation:

\[ 28a - 147 = 12 \]

\[ 28a = 159 \quad \Rightarrow \quad a = \tfrac{159}{28} \]

Now substitute this into the first equation:

\[ (4a - 21)b = 45 \]

\[ \left(4 \times \tfrac{159}{28} - 21\right)b = 45 \]

Simplify inside the bracket:

\[ \tfrac{636}{28} - 21 = \tfrac{159}{7} - 21 = \tfrac{159}{7} - \tfrac{147}{7} = \tfrac{12}{7} \]

So:

\[ \left(\tfrac{12}{7}\right)b = 45 \quad \Rightarrow \quad b = \tfrac{45 \times 7}{12} = \tfrac{315}{12} = \tfrac{105}{4} \]

Final Answer:

\[ a = \tfrac{159}{28}, \quad b = \tfrac{105}{4} \]

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Q1(b). If \( N = \begin{vmatrix} 23 & 4 \\[6pt] 7 & 5 \end{vmatrix}\), then (i) find \(N^{-1}\) (ii) Verify that \(N^{-1}N = NN^{-1}\). ▶

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We are given:

\[ N = \begin{bmatrix} 23 & 4 \\[6pt] 7 & 5 \end{bmatrix} \]

Step 1: Find determinant of \(N\)

\[ |N| = (23)(5) - (7)(4) = 115 - 28 = 87 \]

Step 2: Formula for inverse

For a \(2 \times 2\) matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\):

\[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

Step 3: Apply formula

\[ N^{-1} = \frac{1}{87} \begin{bmatrix} 5 & -4 \\ -7 & 23 \end{bmatrix} \]

\[ N^{-1} = \frac{1}{87} \begin{bmatrix} 5 & -4 \\ -7 & 23 \end{bmatrix} \]

Step 4: Verification

\[ N^{-1}N = \frac{1}{87} \begin{bmatrix} 5 & -4 \\ -7 & 23 \end{bmatrix} \begin{bmatrix} 23 & 4 \\ 7 & 5 \end{bmatrix} \]

Multiply:

\[ = \frac{1}{87} \begin{bmatrix} (5)(23)+(-4)(7) & (5)(4)+(-4)(5) \\ (-7)(23)+(23)(7) & (-7)(4)+(23)(5) \end{bmatrix} \]

\[ = \frac{1}{87} \begin{bmatrix} 115 - 28 & 20 - 20 \\ -161 + 161 & -28 + 115 \end{bmatrix} \]

\[ = \frac{1}{87} \begin{bmatrix} 87 & 0 \\ 0 & 87 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

Similarly,

\[ NN^{-1} = \begin{bmatrix} 23 & 4 \\ 7 & 5 \end{bmatrix} \cdot \frac{1}{87}\begin{bmatrix} 5 & -4 \\ -7 & 23 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

\[ N^{-1}N = NN^{-1} = I \]

Hence verified.

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Q2(a). The sum of the two angles is \(100^\circ\), the supplement of the first angle exceeds the supplement of the second angle by \(40^\circ\) Find the angles. ▶

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Step 1: Let the angles

Let the first angle be \(x\). Then the second angle is \(100 - x\).

Step 2: Write supplements

Supplement of first angle = \(180 - x\)

Supplement of second angle = \(180 - (100 - x) = 80 + x\)

Step 3: Apply condition

\((180 - x) - (80 + x) = 40\)

Step 4: Simplify

\(180 - x - 80 - x = 40\)

\(100 - 2x = 40\)

\(2x = 60\)

\(x = 30\)

Step 5: Find other angle

Second angle = \(100 - 30 = 70\)

Answer:

The two angles are \(30^\circ\) and \(70^\circ\).

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Q2(b). (b) Use the Cramer’s rule to solve the simultaneous equations. Give the reason why the solution is not possible.
\(x - 3y = 5\)
\(2x - 5y = 9\) ▶

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Step 1: Write in standard form

\(1x - 3y = 5\)

\(2x - 5y = 9\)

Step 2: Determinant of coefficients

\[ D = \begin{vmatrix} 1 & -3 \\ 2 & -5 \end{vmatrix} = (1)(-5) - (2)(-3) = -5 + 6 = 1 \]

Step 3: Determinant for \(x\)

\[ D_x = \begin{vmatrix} 5 & -3 \\ 9 & -5 \end{vmatrix} = (5)(-5) - (9)(-3) = -25 + 27 = 2 \]

Step 4: Determinant for \(y\)

\[ D_y = \begin{vmatrix} 1 & 5 \\ 2 & 9 \end{vmatrix} = (1)(9) - (2)(5) = 9 - 10 = -1 \]

Step 5: Solve

\( x = \dfrac{D_x}{D} = \dfrac{2}{1} = 2 \)

\( y = \dfrac{D_y}{D} = \dfrac{-1}{1} = -1 \)

Final Answer:

\( x = 2, \; y = -1 \)

Reason solution is possible:

Since \( D = 1 \neq 0 \), the system has a unique solution. The two equations are independent.

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Q3(a). Draw an equilateral triangle, each of whose side is 8cm. ▶

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Q3(b). Define triangle. Explain different types of triangles. ▶

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Triangle

A triangle is one of the simplest and most important shapes in geometry. It is a closed figure that is formed by joining three non-collinear points with straight lines. These three line segments are called the sides of the triangle, and the three meeting points are called its vertices. The angles formed inside the triangle are known as its interior angles. A very important property of every triangle is that the sum of its interior angles is always equal to 180°.

Triangles are found everywhere around us. They are commonly used in construction, art, design, and engineering because of their stability and strength. For example, bridges, towers, and roof trusses often use triangular shapes because a triangle cannot be deformed easily, making it a very strong structure.

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Classification of Triangles

Triangles can be classified in two main ways: (i) according to the lengths of their sides, and (ii) according to the size of their angles.

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1. Based on Sides

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(a) Equilateral Triangle: An equilateral triangle has all three sides equal in length. Since the sides are equal, all three angles are also equal. Each angle in an equilateral triangle measures exactly 60°. This makes it a very symmetrical and regular type of triangle. Equilateral triangles are often used in tiling and decorative designs because of their neat appearance.

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(b) Isosceles Triangle: An isosceles triangle has two sides of equal length and the third side different. The angles opposite the equal sides are also equal. This property makes it easier to study and calculate various measurements in an isosceles triangle. Isosceles triangles are often used in architectural patterns and artistic designs where balance and proportion are needed.

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(c) Scalene Triangle: A scalene triangle has all three sides of different lengths. As a result, all three angles are also different. Scalene triangles are the most general form of triangles. They do not have any equal sides or angles. Many irregular shapes found in real life can be broken down into scalene triangles, making them very useful in geometry and trigonometry.

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2. Based on Angles

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(a) Acute Triangle: An acute triangle is one in which all three angles are less than 90°. Since all angles are sharp or acute, the triangle looks narrow and pointed. Acute triangles are common in art and design and can be seen in many decorative structures.

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(b) Right Triangle: A right triangle has one angle equal to exactly 90°. The side opposite this right angle is called the hypotenuse, which is the longest side of the triangle. Right triangles are extremely important in mathematics, especially in trigonometry and the Pythagoras theorem. They are also widely used in construction, measurement, and navigation.

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(c) Obtuse Triangle: An obtuse triangle has one angle greater than 90°. Because of this large angle, the triangle appears wider and more open. The other two angles in an obtuse triangle are always acute. Obtuse triangles are less common in structures but they play an important role in geometry and design.

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Conclusion

Triangles are fundamental shapes that help us understand geometry and the world around us. They are not only simple figures but also very strong and useful in practical applications. Whether it is the perfect symmetry of an equilateral triangle, the balance of an isosceles triangle, the generality of a scalene triangle, or the practical use of right triangles in construction and measurement, each type of triangle has its own importance. By studying triangles and their properties, we develop a deeper understanding of mathematics, symmetry, and real-world design.

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Q4(a). Find the volume of a sphere, with a radius of 7.5cm. ▶

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Formula: \(V = \frac{4}{3}\pi r^3\)

Given radius \(r = 7.5\ \text{cm}\)

Step 1. Cube the radius: \(7.5^3 = 421.875\)

Step 2. Multiply by \(\pi\): \(421.875 \times \pi = 1325.36\)

Step 3. Multiply by four thirds: \(\dfrac{4}{3} \times 1325.36 = 1767.15\)

\(V = 1767.2\ \text{cm}^3\)

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Q4(b). Find the hypotenuse of the right isosceles triangle, each of whose leg is 7cm. ▶

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Hypotenuse of a right isosceles triangle

Each leg = 7 cm

Formula (Pythagoras theorem): \(\text{Hypotenuse}^2 = (\text{leg})^2 + (\text{leg})^2\)

Step 1. \((7)^2 + (7)^2 = 49 + 49 = 98\)

Step 2. \(\text{Hypotenuse} = \sqrt{98} = \sqrt{49 \times 2}\)

Step 3. \(\text{Hypotenuse} = 7\sqrt{2} = 9.9 \, \text{cm}\)

\(7\sqrt{2} \, \text{cm}\) or about 9.9 cm

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Q5(a). Show that the points A(6,1), B(2,7) and C(-6,7) are of a scalene triangle. ▶

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Distance formula: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Step 1. Find AB

\(AB = \sqrt{(2 - 6)^2 + (7 - 1)^2} = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} = 7.21\)

Step 2. Find BC

\(BC = \sqrt{(-6 - 2)^2 + (7 - 7)^2} = \sqrt{(-8)^2 + 0^2} = \sqrt{64} = 8\)

Step 3. Find AC

\(AC = \sqrt{(-6 - 6)^2 + (7 - 1)^2} = \sqrt{(-12)^2 + 6^2} = \sqrt{144 + 36} = \sqrt{180} = 6\sqrt{5} = 13.42\)

Step 4. Compare lengths

\(AB = 7.21,\ BC = 8,\ AC = 13.42\)

All three sides are unequal, so △ABC is a scalene triangle.

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Q5(b). Find the area of the rectangle 2m long and 20cm wide. ▶

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Area of a rectangle

Length = 2 m = 200 cm

Width = 20 cm

Formula: \(\text{Area} = \text{Length} \times \text{Width}\)

Step 1. \(200 \times 20 = 4000 \, \text{cm}^2\)

Step 2. Convert to square meters: \(4000 \, \text{cm}^2 = 0.4 \, \text{m}^2\)

4000 cm\(^2\) or 0.4 m\(^2\)

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