END - OF - CHAPTER EXERCISES Chapter 6 Class 9 Maths

 

End-of-Chapter Exercises

Note: In the problems below, unless stated otherwise, use \(\pi = \frac{22}{7}\).

Question 1

Identities in algebra can sometimes be shown as area relationships. For example, the figure shown corresponds to the identity \((a + b)^2 = a^2 + 2ab + b^2\). Do you see how?

Draw figures corresponding to the identities \((a + b)(a - b) = a^2 - b^2\) and \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\).

Solution:

Part 1: Explanation of \((a + b)^2 = a^2 + 2ab + b^2\)

A square with side length \((a + b)\) has a total area of \((a + b)^2\). It is divided into four sub-regions:

• One square of side \(a\) with area \(a^2\)
• Two rectangles of sides \(a\) and \(b\) with area \(ab\) each (total area = \(2ab\))
• One square of side \(b\) with area \(b^2\)

Adding all sub-regions gives: \(\text{Total Area} = a^2 + 2ab + b^2\).

Part 2: Visual Explanation for \((a + b)(a - b) = a^2 - b^2\)

1. Start with a square of side \(a\). Its area is \(a^2\).

2. Remove a small square of side \(b\) from one corner. Its area is \(b^2\).

3. The remaining area is \(a^2 - b^2\).

4. Rearrange the remaining L-shaped region into a single rectangle. The new rectangle has length \((a + b)\) and width \((a - b)\).

\[\text{Area of Rectangle} = \text{Length} \times \text{Width} = (a + b)(a - b) = a^2 - b^2\]

Part 3: Area Model for \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\)

A square with side \((a + b + c)\) is divided into 9 sub-regions. Summing all sub-regions:

\[(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\]

Question 2

An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.

Solution:

Let equal sides be \(a = 15\text{ cm}\) and \(b = 15\text{ cm}\). Let the third side (base) be \(c\).

\[\text{Perimeter} = a + b + c = 40\]

\[15 + 15 + c = 40 \implies 30 + c = 40 \implies c = 10\text{ cm}\]

Heron's Formula:

\[\text{Semi-perimeter } (s) = \frac{a + b + c}{2} = \frac{15 + 15 + 10}{2} = 20\text{ cm}\] \[\text{Area of Triangle} = \sqrt{s(s - a)(s - b)(s - c)}\]

Substituting the values into Heron's Formula:

\[\text{Area} = \sqrt{20 \cdot (20 - 15) \cdot (20 - 15) \cdot (20 - 10)}\] \[\text{Area} = \sqrt{20 \cdot 5 \cdot 5 \cdot 10} = \sqrt{5000} = \sqrt{2500 \cdot 2} = 50\sqrt{2}\text{ cm}^2 \approx 70.71\text{ cm}^2\]

Question 3

An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?

Solution:

1. Find height (\(h\)) using the area formula:

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

\[60 = \frac{1}{2} \times 10 \times h \implies 5h = 60 \implies h = 12\text{ cm}\]

2. The altitude bisects the base of 10 cm into two equal parts of 5 cm each.

3. Apply Pythagoras' theorem to find equal side length (\(x\)):

\[x^2 = h^2 + 5^2 \implies x^2 = 12^2 + 5^2 = 144 + 25 = 169\] \[x = \sqrt{169} = 13\text{ cm}\]

The length of each equal side is 13 cm.

Question 4

The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.

Solution:

Let height (\(h\)) = 12 cm, and base = \(b\).

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

\[54 = \frac{1}{2} \times b \times 12 \implies 6b = 54 \implies b = 9\text{ cm}\]

Find hypotenuse (\(c\)) using Pythagoras' theorem:

\[c = \sqrt{\text{base}^2 + \text{height}^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15\text{ cm}\]

\[\text{Perimeter} = \text{side}_1 + \text{side}_2 + \text{hypotenuse} = 12 + 9 + 15 = \mathbf{36\text{ cm}}\]

Question 5

The sides of a triangle are in the ratio 2 : 3 : 4, and its perimeter is 45 cm. Find its area.

Solution:

Let the side lengths be \(2x\), \(3x\), and \(4x\).

\[\text{Perimeter} = 2x + 3x + 4x = 45 \implies 9x = 45 \implies x = 5\text{ cm}\]

The side lengths are:

• \(a = 2 \times 5 = 10\text{ cm}\)
• \(b = 3 \times 5 = 15\text{ cm}\)
• \(c = 4 \times 5 = 20\text{ cm}\)

Heron's Formula Calculation:

\[\text{Semi-perimeter } (s) = \frac{\text{Perimeter}}{2} = \frac{45}{2} = 22.5\text{ cm}\] \[\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}\]

\[\text{Area} = \sqrt{22.5 \cdot (22.5 - 10) \cdot (22.5 - 15) \cdot (22.5 - 20)}\] \[\text{Area} = \sqrt{22.5 \cdot 12.5 \cdot 7.5 \cdot 2.5} = \sqrt{5273.4375} \approx \mathbf{72.62\text{ cm}^2}\]

Question 6

The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.

Solution:

Method 1: Right Triangle Formula

Check if it is a right triangle: \(7^2 + 24^2 = 49 + 576 = 625 = 25^2\). Thus, it is a right triangle with base = 24 cm and height = 7 cm.

\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 24 \times 7 = \mathbf{84\text{ cm}^2}\]

Method 2: Heron's Formula

\[s = \frac{7 + 24 + 25}{2} = 28\text{ cm}\] \[\text{Area} = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{28 \cdot (28 - 7) \cdot (28 - 24) \cdot (28 - 25)}\]

\[\text{Area} = \sqrt{28 \cdot 21 \cdot 4 \cdot 3} = \sqrt{7056} = \mathbf{84\text{ cm}^2}\]

Question 7

If the wheel of a bicycle has a diameter of 60 cm, find how far a cyclist will have travelled after the wheel has rotated 100 times.

Solution:

Diameter (\(d\)) = 60 cm \(\implies\) Radius (\(r\)) = 30 cm.

\[\text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 30 = \frac{1320}{7}\text{ cm}\]

Calculating distance in 100 rotations:

\[\text{Distance} = 100 \times \frac{1320}{7} = \frac{132000}{7}\text{ cm} \approx \mathbf{18857.14\text{ cm}} = \mathbf{188.57\text{ meters}}\]

Question 8

Find the area of a quadrant of a circle whose circumference is 66 cm.

Solution:

1. Find radius (\(r\)) from circumference = 66 cm:

\[2\pi r = 66 \implies 2 \times \frac{22}{7} \times r = 66 \implies r = \frac{66 \times 7}{44} = 10.5\text{ cm}\]

2. Area of Quadrant:

\[\text{Area of Quadrant} = \frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times (10.5)^2 = \mathbf{86.625\text{ cm}^2}\]

Question 9

The wheel of a car has an outer radius of 28 cm. Calculate how far the car travels after one complete turn of the wheel, and how many times the wheel turns during a journey of 1 km.

Solution:

Radius (\(r\)) = 28 cm.

Distance per turn:

\[\text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 28 = \mathbf{176\text{ cm}} \quad (1.76\text{ m})\]

Number of turns in 1 km (100,000 cm):

\[\text{Number of turns} = \frac{\text{Total Distance}}{\text{Circumference}} = \frac{100000}{176} \approx \mathbf{568.18\text{ turns}}\]

Question 10

*Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?

Solution:

Yes, two rectangles with the same area and same perimeter MUST be congruent.

Detailed Algebraic Step-by-Step Proof:

Let Rectangle 1 have length \(x_1\) and width \(y_1\). Let Rectangle 2 have length \(x_2\) and width \(y_2\).

1. Equal Perimeter Condition:

\[2(x_1 + y_1) = 2(x_2 + y_2) = P \implies x_1 + y_1 = x_2 + y_2 = S \quad \text{(where } S = \text{semi-perimeter)}\] \[y_1 = S - x_1 \quad \text{and} \quad y_2 = S - x_2\]

2. Equal Area Condition:

\[x_1 \cdot y_1 = x_2 \cdot y_2 = A \quad \text{(where } A = \text{area)}\]

3. Substitute \(y = S - x\) into the Area equation:

\[x(S - x) = A \implies S x - x^2 = A \implies x^2 - S x + A = 0\]

4. Conclusion:

The side lengths of both rectangles are solutions (roots) to the exact same quadratic equation \(t^2 - S t + A = 0\).

By the quadratic formula, the two roots are:

\[t = \frac{S \pm \sqrt{S^2 - 4A}}{2}\]

Therefore, one root gives the length and the other gives the width. Since both rectangles get the exact same dimensions \(\{x_1 = x_2, y_1 = y_2\}\), they are congruent.

Question 11

You know that the area of a parallelogram is \(\text{base} \times \text{height}\). Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides \(\times \text{height}\), i.e., \(\frac{1}{2}(a + b)h\).

Solution:

The trapezium is split into a parallelogram (base = \(a\), height = \(h\)) and a triangle (base = \(b - a\), height = \(h\)):

\[\text{Area of Parallelogram} = a \times h\] \[\text{Area of Triangle} = \frac{1}{2}(b - a)h\]

\[\text{Total Area} = a \cdot h + \frac{1}{2}(b - a)h = h \left(a + \frac{1}{2}b - \frac{1}{2}a\right) = \mathbf{\frac{1}{2}(a + b)h}\]

Question 12

By dividing a trapezium into two triangles, show that its area is half the sum of the parallel sides multiplied by the height.

Solution:

A diagonal divides the trapezium into two triangles:

• Triangle 1: Base = \(b\), Height = \(h \implies \text{Area} = \frac{1}{2} b h\)
• Triangle 2: Base = \(a\), Height = \(h \implies \text{Area} = \frac{1}{2} a h\)

\[\text{Total Area} = \frac{1}{2} a h + \frac{1}{2} b h = \mathbf{\frac{1}{2}(a + b)h}\]

Question 13

Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?

Solution:

1. Take two identical trapeziums, each having parallel sides \(a\) and \(b\), and height \(h\).
2. Rotate the second trapezium by 180° and place it adjacent to the first trapezium.
3. The combined figure forms a single large Parallelogram with base = \((a + b)\) and height = \(h\).

\[\text{Area of Parallelogram} = \text{base} \times \text{height} = (a + b)h\]

\[\text{Area of 1 Trapezium} = \frac{1}{2} \times \text{Area of Parallelogram} = \mathbf{\frac{1}{2}(a + b)h}\]

Question 14

Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra, and (ii) using geometry.

Solution:

(i) Using Geometry:

The vertical diagonal \(d_1\) splits the kite into two triangles, each having base = \(d_1\). The horizontal diagonal \(d_2\) is split into two heights \(h_1\) and \(h_2\) such that \(h_1 + h_2 = d_2\).

\[\text{Area} = \frac{1}{2} \cdot d_1 \cdot h_1 + \frac{1}{2} \cdot d_1 \cdot h_2 = \frac{1}{2} \cdot d_1 \cdot (h_1 + h_2) = \mathbf{\frac{1}{2} d_1 d_2}\]

(ii) Using Algebra:

The perpendicular diagonals intersect at origin to form 4 right-angled triangles with leg pairs \((a, b)\), \((b, c)\), \((c, d)\), and \((d, a)\), where \(d_1 = a + c\) and \(d_2 = b + d\).

\[\text{Total Area} = \frac{1}{2}ab + \frac{1}{2}bc + \frac{1}{2}cd + \frac{1}{2}da = \frac{1}{2}[b(a + c) + d(a + c)]\] \[\text{Area} = \frac{1}{2}(a + c)(b + d) = \mathbf{\frac{1}{2} d_1 d_2}\]