Model Test Paper 3

Rationalising the denominator by multiplying the numerator and denominator by $\sqrt{3}$, we get :

$$\begin{aligned}\Rightarrow \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\[1em] = \frac{6\sqrt{3}}{3} \\[1em] = 2\sqrt{3}.\end{aligned}$$

Hence, option 2 is the correct option.

We know that,

⇒ 4ab = (a + b)² - (a - b)²

Substituting values we get :

⇒ 4ab = (16)² - (2)²

⇒ 4ab = 256 - 4

⇒ 4ab = 252

⇒ ab = 63.

Hence, option 3 is the correct option.

We know that, the standard identity for the difference of cubes is :

⇒ x³ - y³ = (x - y)(x² + xy + y²)

∴ x³ - y³ = (x - y)(x² + y² + xy).

Hence, option 2 is the correct option.

Solving the given expression :

$$\begin{aligned}\Rightarrow (-27)^{2/3} \\[1em] = \left((-3)^3\right)^{2/3} \\[1em] = (-3)^{3 \times \frac{2}{3}} \\[1em] = (-3)^2 \\[1em] = 9.\end{aligned}$$

Hence, option 1 is the correct option.

Given,

⇒ log₈ x = $\frac{2}{3}$

Writing in exponential form, we get :

$$\begin{aligned}\Rightarrow x = 8^{2/3} \\[1em] \Rightarrow x = (2^3)^{2/3} \\[1em] \Rightarrow x = 2^{3 \times \frac{2}{3}} \\[1em] \Rightarrow x = 4.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

In △ABC, ∠A = 60° and ∠B = 40°.

We know that, the side opposite to the greater angle is longer.

Side BC is opposite to ∠A and side AC is opposite to ∠B.

Since ∠A > ∠B (60° > 40°), the side opposite to ∠A is longer than the side opposite to ∠B.

⇒ BC > AC.

Hence, option 1 is the correct option.

Given,

In △ABC, DE ∥ BC and D is the mid-point of AB. DE = 2.5 cm.

Since D is the mid-point of AB and DE ∥ BC, by the converse of mid-point theorem, E is the mid-point of AC and DE = $\frac{1}{2}$ BC.

⇒ BC = 2 × DE

⇒ BC = 2 × 2.5

⇒ BC = 5 cm.

Hence, option 2 is the correct option.

We know that, the area of a parallelogram is given by the product of its base and the corresponding height.

∴ Area of parallelogram = base × height.

Hence, option 4 is the correct option.

The second class interval is 11 - 20.

By formula,

Class mark = $\frac{\text{lower limit} + \text{upper limit}}{2}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Class mark} = \frac{11 + 20}{2} \\[1em] = \frac{31}{2} \\[1em] = 15.5.\end{aligned}$$

Hence, option 3 is the correct option.

The first 9 odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15 and 17.

Here, the number of observations n = 9, which is odd.

By formula,

Median = $\left(\frac{n + 1}{2}\right)^{th}$ observation

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Median} = \left(\frac{9 + 1}{2}\right)^{th} \text{observation} \\[1em] = 5^{th} \text{observation} \\[1em] = 9.\end{aligned}$$

Hence, option 1 is the correct option.

Let the side of the equilateral triangle be a cm.

By formula,

Area of equilateral triangle = $\frac{\sqrt{3}}{4}a^2$

Substituting values we get :

$$\begin{aligned}\Rightarrow 12\sqrt{3} = \frac{\sqrt{3}}{4}a^2 \\[1em] \Rightarrow a^2 = \frac{12\sqrt{3} \times 4}{\sqrt{3}} \\[1em] \Rightarrow a^2 = 48 \\[1em] \Rightarrow a = 4\sqrt{3} \text{ cm}.\end{aligned}$$

By formula,

Height of equilateral triangle = $\frac{\sqrt{3}}{2}a$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Height} = \frac{\sqrt{3}}{2} \times 4\sqrt{3} \\[1em] = \frac{\sqrt{3} \times 4\sqrt{3}}{2} \\[1em] = \frac{4 \times 3}{2} \\[1em] = 6 \text{ cm}.\end{aligned}$$

Hence, option 4 is the correct option.

The wire forms a square of side 11 cm, so its total length equals the perimeter of the square.

Length of wire = 4 × 11 = 44 cm

When the same wire is bent into a circle, this length becomes the circumference of the circle.

⇒ 2πr = 44

⇒ r = $\frac{44}{2π} = \frac{44}{2 \times \frac{22}{7}} = \frac{44 \times 7}{2 \times 22}$ = 7 cm.

Now, the area of the circle is:

Area = πr² = $\frac{22}{7}$ × 7² = 22 × 7 = 154 cm²

Hence, option 1 is the correct option.

Given, the length of the diagonal of the cube is $3\sqrt{6}$ cm.

Let the edge of cube is a cm, the length of the diagonal of a cube = $a\sqrt{3}$.

$$\begin{aligned}\Rightarrow a\sqrt{3} = 3\sqrt{6} \\[1em] \Rightarrow a = \frac{3\sqrt{6}}{\sqrt{3}} \\[1em] \Rightarrow a = 3\sqrt{2} \text{ cm}.\end{aligned}$$

By formula,

Volume of cube = a³

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Volume} = (3\sqrt{2})^3 \\[1em] = 27 \times 2\sqrt{2} \\[1em] = 54\sqrt{2} \text{ cm}^3.\end{aligned}$$

Hence, option 3 is the correct option.

We know that,

⇒ sin²30° + cos²30° = 1 and tan² 45° = 1.

Substituting these values, we get :

$\Rightarrow \sin^2 30° + \cos^2 30° + \tan^2 45°$

= 1 + (1)²

= 1 + 1

= 2.

Hence, option 2 is the correct option.

The correct distance formula between points P(x₁, y₁) and Q(x₂, y₂) is :

⇒ PQ = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

The formula given in Reason (R) is incorrect, so Reason (R) is false.

Using the correct formula, the distance between A(-4, 2) and B(1, 6) :

$$\begin{aligned}\Rightarrow AB = \sqrt{(1 - (-4))^2 + (6 - 2)^2} \\[1em] = \sqrt{(5)^2 + (4)^2} \\[1em] = \sqrt{25 + 16} \\[1em] = \sqrt{41} \text{ units}.\end{aligned}$$

The Assertion uses the coordinates incorrectly and gives $\sqrt{61}$, whereas the correct distance is $\sqrt{41}$. So, Assertion (A) is false.

∴ Both A and R are false.

Hence, option 4 is the correct option.

Given equations,

⇒ 2x - 3y + 8 = 0 ......... (1)

⇒ 3x + y + 1 = 0 ......... (2)

Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get :

a₁ = 2, b₁ = -3, c₁ = 8 and a₂ = 3, b₂ = 1, c₂ = 1.

By cross multiplication method,

⇒ $\frac{x}{b_1 c_2 - b_2 c_1} = \frac{y}{c_1 a_2 - c_2 a_1} = \frac{1}{a_1 b_2 - a_2 b_1}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \frac{x}{(-3)(1) - (1)(8)} = \frac{y}{(8)(3) - (1)(2)} = \frac{1}{(2)(1) - (3)(-3)} \\[1em] \Rightarrow \frac{x}{-3 - 8} = \frac{y}{24 - 2} = \frac{1}{2 + 9} \\[1em] \Rightarrow \frac{x}{-11} = \frac{y}{22} = \frac{1}{11}\end{aligned}$$

Taking $\frac{x}{-11} = \frac{1}{11}$, we get :

⇒ x = $\frac{-11}{11}$ = -1.

Taking $\frac{y}{22} = \frac{1}{11}$, we get :

⇒ y = $\frac{22}{11}$ = 2.

Hence, x = -1 and y = 2.

Given,

ABCD is a square in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively.

In a square, the diagonals are equal, so AC = BD.

Let AC = BD = x

By mid-point theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it.

In △ABC, P and Q are the mid-points of AB and BC respectively.

By mid-point theorem,

⇒ PQ ∥ AC and PQ = $\frac{1}{2}\text{AC} = \frac{1}{2}x$ ......... (1)

In △ADC, S and R are the mid-points of DA and CD respectively.

⇒ SR ∥ AC and SR = $\frac{1}{2}\text{AC} = \frac{1}{2}x$ ......... (2)

In △ABD, S and P are the mid-points of AD and AB respectively.

⇒ SP ∥ BD and PS = $\frac{1}{2}\text{BD} = \frac{1}{2}x$ .......(3)

In △BCD, Q and R are the mid-points of BC and DC respectively.

⇒ QR ∥ BD and QR = $\frac{1}{2}\text{BD} = \frac{1}{2}x$ ......... (4)

From equations (1), (2), (3) and (4), we get :

⇒ PQ = SR = PS = QR.

So, all four sides of PQRS are equal.

Also, the diagonals of a square are perpendicular to each other, so AC ⊥ BD. Since PQ ∥ AC and PS ∥ BD, we have PQ ⊥ PS, i.e. ∠SPQ = 90°.

Since all sides of PQRS are equal and one angle is 90°, PQRS is a square.

Hence, proved that the quadrilateral formed is a square.

Given,

Rate of depreciation (r) = $16\frac{2}{3}% = \frac{50}{3}$% per annum.

Time = 3 years and present value = ₹62500.

Let the purchase price be ₹P.

By formula,

Present value = $P\left(1 - \frac{r}{100}\right)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 62500 = P\left(1 - \frac{50/3}{100}\right)^3 \\[1em] \Rightarrow 62500 = P\left(1 - \frac{50}{300}\right)^3 \\[1em] \Rightarrow 62500 = P\left(1 - \frac{1}{6}\right)^3 \\[1em] \Rightarrow 62500 = P\left(\frac{5}{6}\right)^3 \\[1em] \Rightarrow 62500 = P \times \frac{125}{216} \\[1em] \Rightarrow P = \frac{62500 \times 216}{125} \\[1em] \Rightarrow P = ₹108000.\end{aligned}$$

Hence, the purchase price of the machine = ₹1,08,000.

We have to find three rational numbers between $\frac{1}{2}$ and 1.

We can write $\frac{1}{2}$ and 1 with a common denominator. Since we need 3 numbers, we multiply numerator and denominator by a number greater than (3 + 1) = 4.

Writing both with denominator 8 :

⇒ $\frac{1}{2} = \frac{4}{8} \text { and } 1 = \frac{8}{8}$.

The rational numbers between $\frac{4}{8}\text { and } \frac{8}{8}\text { are } \frac{5}{8}, \frac{6}{8} \text { and } \frac{7}{8}$.

∴ Three rational numbers between $\frac{1}{2}\text { and } 1 \text { are } \frac{5}{8}, \frac{3}{4}\text { and } \frac{7}{8}$.

Hence, three rational numbers between $\frac{1}{2}\text { and }1 \text { are } \frac{5}{8}, \frac{3}{4} \text { and } \frac{7}{8}$.

Given,

(a + b + c) = 15 and (ab + bc + ca) = 74.

We know that,

⇒ (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

Rearranging, we get :

⇒ a² + b² + c² = (a + b + c)² - 2(ab + bc + ca)

Substituting values we get :

⇒ a² + b² + c² = (15)² - 2(74)

= 225 - 148

= 77.

Hence, the value of (a² + b² + c²) = 77.

Adjustment factor

$$\begin{aligned}= \frac{\text{Lower limit of a class - Upper limit of previous class}}{2} \\[1em] = \frac{35 - 34}{2} \\[1em] = \frac{1}{2} \\[1em] = 0.5\end{aligned}$$

Since the class-intervals are in the inclusive form, we first convert them to the continuous (exclusive) form by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit. We then find the class mark (mid-value) of each class.

Class-interval	Frequency
24.5 – 29.5	0
29.5 – 34.5	12
34.5 – 39.5	16
39.5 – 44.5	20
44.5 – 49.5	8
49.5 – 54.5	10
54.5 – 59.5	4
59.5 - 64.5	0

By formula,

Class mark = $\frac{\text{lower limit} + \text{upper limit}}{2}$

Class-interval	Class mark	Frequency
24.5 – 29.5	27	0
29.5 – 34.5	32	12
34.5 – 39.5	37	16
39.5 – 44.5	42	20
44.5 – 49.5	47	8
49.5 – 54.5	52	10
54.5 – 59.5	57	4
59.5 - 64.5	62	0

Steps of Construction:

1. On a graph paper, mark class-marks along x-axis and frequencies along y-axis.

2. Since, the scale on x-axis starts at 20, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 20.

3. Take 1 cm along the x-axis = 5 units.

4. Take 1 cm along the y-axis = 2 units.

5. On the graph, mark points taking values of class-marks along x-axis and the values of their corresponding frequencies along y-axis.

6. Plot the points (27, 0), (32, 12), (37, 16), (42, 20), (47, 8), (52, 10), (57, 4) and (62, 0), taking the class marks along the x-axis and the corresponding frequencies along the y-axis. The classes 24.5 - 29.5 and 59.5 - 64.5 with frequency 0 are taken at both the ends to complete the polygon.

7. Join the consecutive points in order by straight line segments to obtain the required frequency polygon.

Given,

Principal (P) = ₹4000, Compound Interest (C.I.) = ₹1324 and Time (n) = 3 years.

By formula,

Amount = Principal + Compound Interest

⇒ A = 4000 + 1324 = ₹5324.

By formula,

Amount = $P\left(1 + \frac{r}{100}\right)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 5324 = 4000\left(1 + \frac{r}{100}\right)^3 \\[1em] \Rightarrow \left(1 + \frac{r}{100}\right)^3 = \frac{5324}{4000} \\[1em] \Rightarrow \left(1 + \frac{r}{100}\right)^3 = \frac{1331}{1000} \\[1em] \Rightarrow \left(1 + \frac{r}{100}\right)^3 = \left(\frac{11}{10}\right)^3 \\[1em] \Rightarrow 1 + \frac{r}{100} = \frac{11}{10} \\[1em] \Rightarrow \frac{r}{100} = \frac{11}{10} - 1 \\[1em] \Rightarrow \frac{r}{100} = \frac{1}{10} \\[1em] \Rightarrow r = 10%.\end{aligned}$$

Hence, the required rate of interest = 10% per annum.

Given,

AD = AE and ∠BAD = ∠CAE.

Since AD = AE, in △ADE the angles opposite to equal sides are equal.

⇒ ∠ADE = ∠AED .........(1)

Now, ∠ADB and ∠ADE are angles on a straight line, and ∠AEC and ∠AED are angles on a straight line.

⇒ ∠ADB = 180° - ∠ADE ......... (2)

⇒ ∠AEC = 180° - ∠AED ......... (3)

From equations (1), (2) and (3), we get :

⇒ ∠ADB = ∠AEC ......... (4)

Given, ∠BAD = ∠CAE.

Adding ∠DAE to both sides, we get :

⇒ ∠BAD + ∠DAE = ∠CAE + ∠DAE

⇒ ∠BAE = ∠CAD ......... (5)

Now, in △ABD and △ACE :

⇒ ∠BAD = ∠CAE [Given]

⇒ AD = AE [Given]

⇒ ∠ADB = ∠AEC [From equation (4)]

∴ △ABD ≅ △ACE [By ASA congruence criterion]

⇒ AB = AC [By C.P.C.T.]

Hence, proved that AB = AC.

We know that,

⇒ cos 60° = $\frac{1}{2}$, sin 30° = $\frac{1}{2}$ and cos 30° = $\frac{\sqrt{3}}{2}$.

Solving the first expression (1 - 2 sin²30°) :

$$\begin{aligned}\Rightarrow 1 - 2\sin^2 30° = 1 - 2\left(\frac{1}{2}\right)^2 \\[1em] = 1 - 2 \times \frac{1}{4} \\[1em] = 1 - \frac{1}{2} \\[1em] = \frac{1}{2}.\end{aligned}$$

Solving the second expression (2 cos²30° - 1) :

$$\begin{aligned}\Rightarrow 2\cos^2 30° - 1 = 2\left(\frac{\sqrt{3}}{2}\right)^2 - 1 \\[1em] = 2 \times \frac{3}{4} - 1 \\[1em] = \frac{3}{2} - 1 \\[1em] = \frac{1}{2}.\end{aligned}$$

Since cos 60° = $\frac{1}{2}, (1 - 2 \sin^2 30°) = \frac{1}{2} \text { and } (2 \cos^2 30° - 1) = \frac{1}{2}$, all three are equal.

Hence, verified that cos 60° = (1 - 2 sin²30°) = (2 cos²30° - 1).

Solving the given expression by adding and subtracting 4x² :

$$\begin{aligned}\Rightarrow x^4 + 4 \\[1em] \Rightarrow x^4 + 4x^2 + 4 - 4x^2 \\[1em] = (x^2 + 2)^2 - (2x)^2\end{aligned}$$

Using the identity a² - b² = (a + b)(a - b), we get :

$$\begin{aligned}\Rightarrow (x^2 + 2)^2 - (2x)^2 \\[1em] \Rightarrow (x^2 + 2 + 2x)(x^2 + 2 - 2x) \\[1em] = (x^2 + 2x + 2)(x^2 - 2x + 2).\end{aligned}$$

Hence, x⁴ + 4 = (x² + 2x + 2)(x² - 2x + 2).

Solving the given expression :

$$\begin{aligned}\Rightarrow \log\frac{75}{16} - 2\log\frac{5}{9} + \log\frac{32}{243} \\[1em] = \log\frac{75}{16} - \log\left(\frac{5}{9}\right)^2 + \log\frac{32}{243} \\[1em] = \log\frac{75}{16} - \log\frac{25}{81} + \log\frac{32}{243}\end{aligned}$$

Using the quotient and product rules of logarithms, we get :

$$\begin{aligned}\Rightarrow \log\left(\frac{75}{16} \times \frac{81}{25} \times \frac{32}{243}\right) \\[1em] = \log\left(\frac{75 \times 81 \times 32}{16 \times 25 \times 243}\right) \\[1em] = \log\left(\frac{194400}{97200}\right) \\[1em] = \log 2 \\[1em] = 0.3010.\end{aligned}$$

Hence, the value of the given expression = 0.3010.

Given,

Diagonal of the cube = $16\sqrt{3}$ cm.

Let the edge of the cube be a cm.

By formula,

Diagonal of cube = $a\sqrt{3}$

Substituting values we get :

$$\begin{aligned}\Rightarrow a\sqrt{3} = 16\sqrt{3} \\[1em] \Rightarrow a = 16 \text{ cm}.\end{aligned}$$

By formula,

Surface area of cube = 6a²

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Surface area} = 6 \times 16^2 \\[1em] = 6 \times 256 \\[1em] = 1536 \text{ cm}^2.\end{aligned}$$

By formula,

Volume of cube = a³

Substituting values we get :

$\Rightarrow \text{Volume} = 16^3$

= 4096 cm³.

Hence, the surface area = 1536 cm² and the volume = 4096 cm³.

Given,

Area of the equilateral triangle = $121\sqrt{3}$ cm².

Let the side of the equilateral triangle be a cm.

By formula,

Area of equilateral triangle = $\frac{\sqrt{3}}{4}a^2$

Substituting values we get :

$$\begin{aligned}\Rightarrow 121\sqrt{3} = \frac{\sqrt{3}}{4}a^2 \\[1em] \Rightarrow a^2 = \frac{121\sqrt{3} \times 4}{\sqrt{3}} \\[1em] \Rightarrow a^2 = 484 \\[1em] \Rightarrow a = 22 \text{ cm}.\end{aligned}$$

∴ Length of the wire = perimeter of the triangle = 3a = 3 × 22 = 66 cm.

When the same wire is bent into a circle, the circumference of the circle = 66 cm.

By formula,

Circumference = 2πr

Substituting values we get :

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r = 66 \\[1em] \Rightarrow r = \frac{66 \times 7}{2 \times 22} \\[1em] \Rightarrow r = 10.5 \text{ cm}.\end{aligned}$$

By formula,

Area of circle = πr²

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Area} = \frac{22}{7} \times (10.5)^2 \\[1em] = \frac{22}{7} \times 110.25 \\[1em] = 346.5 \text{ cm}^2.\end{aligned}$$

Hence, the area enclosed by the wire when bent into a circle = 346.5 cm².

Let a = $x^{1/3} \text { and } \text { b } = x^{-1/3}$.

Then $a^2 = x^{2/3}, b^2 = x^{-2/3} \text { and } \text { ab } = x^{1/3} \times x^{-1/3} = x^0 = 1$.

So the given expression becomes :

⇒ (a - b)(a² + ab + b²) [since the middle term 1 = ab]

We know that, the identity for the difference of cubes is :

⇒ (a - b)(a² + ab + b²) = a³ - b³

Substituting back the values of a and b, we get :

$$\begin{aligned}\Rightarrow a^3 - b^3 = \left(x^{1/3}\right)^3 - \left(x^{-1/3}\right)^3 \\[1em] = x - x^{-1} \\[1em] = x - \frac{1}{x}.\end{aligned}$$

Hence, the simplified value = $x - \frac{1}{x}$.

Given,

ABCD is a trapezium in which AB ∥ DC, E is the mid-point of AD and EF ∥ AB.

To prove : F is the mid-point of BC.

Construction : Join the diagonal AC, which meets EF at point G.

Since AB ∥ DC and EF ∥ AB, we have EF ∥ DC as well.

In △ADC, E is the mid-point of AD and EG ∥ DC.

By converse of mid-point theorem, which states that if a line passes through the midpoint of one side of a triangle and is parallel to another side, it will bisect the third side.

G is the mid-point of AC.

⇒ AG = GC

In △CAB, G is the mid-point of AC and GF ∥ AB.

By converse of mid-point theorem, F is the mid-point of BC.

Hence, proved that F is the mid-point of BC.

Given,

ABCD is a parallelogram in which E and F are the mid-points of the opposite sides AB and DC respectively. EF is joined.

Since ABCD is a parallelogram, AB ∥ DC and AB = DC.

So, AE ∥ DF. Also, E and F are mid-points, so AE = $\frac{1}{2} \text { AB } \text { and } \text { DF } = \frac{1}{2}$ DC.

Since AB = DC,

⇒ AE = DF.

Since AE ∥ DF and AE = DF, AEFD is a parallelogram.

As, ABCD is a parallelogram, AB ∥ DC and AB = DC.

So, EB ∥ FC. Also, E and F are mid-points, so EB = $\frac{1}{2} \text { AB } \text { and } \text { FC } = \frac{1}{2}$ DC.

Since AB = DC,

⇒ EB = FC

Since EB ∥ FC and EB = FC, EBCF is a parallelogram.

We know that, parallelograms on equal bases and between the same parallels are equal in area.

Now, parallelograms AEFD and EBCF have equal bases (AE = EB) and lie between the same parallels AB and DC.

⇒ ar(AEFD) = ar(EBCF).

Hence, proved that the line segment joining the mid-points of a pair of opposite sides of a parallelogram, divides it into two equal parallelograms.

First Equation : x + y = 1

Step 1 :

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 1, then y = 0

Let x = 0, then y = 1

Let x = 4, then y = -3

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
1	0
0	1
4	-3

Step 3 :

Plot the points (1, 0), (0, 1), (4, -3) on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second Equation : 3x + 2y = 6

Step 1 :

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then y = 3

Let x = 2, then y = 0

Let x = 4, then y = -3

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
0	3
2	0
4	-3

Step 3 :

Plot the points (0, 3), (2, 0), (4, -3) on a graph paper and then draw a straight line passing through the points plotted on the graph.

On the same graph paper, draw the graph for each given equation.

Both the straight lines drawn meet at a point. As it is clear from the graph, co-ordinates of the common point are (4, -3).

Hence, x = 4, y = -3.

We have to find 9 rational numbers between 0 and 0.1.

We can write 0 and 0.1 as 0.00 and 0.10 respectively.

The rational numbers lying between 0.00 and 0.10 are :

⇒ 0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08 and 0.09.

Hence, 9 rational numbers between 0 and 0.1 are 0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08 and 0.09.

Given,

ABCD is a rectangle whose diagonals intersect at O. AC is produced to E and ∠ECD = 140°.

Since ∠DCA and ∠ECD form a linear pair,

⇒ ∠DCA = 180° - ∠ECD

⇒ ∠DCA = 180° - 140°

⇒ ∠DCA = 40°.

We know that, the diagonals of a rectangle are equal and bisect each other.

∴ OC = OD, which makes △OCD isosceles.

⇒ ∠ODC = ∠OCD = 40°.

In △OCD, by angle sum property,

⇒ ∠COD = 180° - ∠OCD - ∠ODC

⇒ ∠COD = 180° - 40° - 40°

⇒ ∠COD = 100°.

Since ∠AOB and ∠COD are vertically opposite angles,

⇒ ∠AOB = ∠COD = 100°.

In △OAB, OA = OB (halves of equal diagonals), so △OAB is isosceles.

⇒ ∠OAB = ∠OBA.

By angle sum property,

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ 2∠OAB + 100° = 180°

⇒ 2∠OAB = 80°

⇒ ∠OAB = 40°.

∴ ∠OAB = ∠OBA = 40° and ∠AOB = 100°.

Hence, the angles of △OAB are ∠OAB = 40°, ∠OBA = 40° and ∠AOB = 100°.

Given,

Side of the rhombus = 20 cm and one diagonal = 24 cm.

We know that, the diagonals of a rhombus bisect each other at right angles.

Let the diagonals be d₁ = 24 cm and d₂.

Half of the first diagonal = $\frac{24}{2}$ = 12 cm.

In the right-angled triangle formed by half the diagonals and a side, by Pythagoras theorem :

$$\begin{aligned}\Rightarrow \text{half of } d_2 = \sqrt{20^2 - 12^2} \\[1em] = \sqrt{400 - 144} \\[1em] = \sqrt{256} \\[1em] = 16 \text{ cm}.\end{aligned}$$

∴ d₂ = 2 × 16 = 32 cm.

By formula,

Area of rhombus = $\frac{1}{2} \times d_1 \times d_2$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Area} = \frac{1}{2} \times 24 \times 32 \\[1em] = 12 \times 32 \\[1em] = 384 \text{ cm}^2.\end{aligned}$$

Hence, the area of the rhombus = 384 cm².

Given,

In the isosceles △ABC inscribed in a circle, AB = AC = $12\sqrt{5}$ cm and BC = 24 cm.

Let O be the centre of the circle and let AM be the perpendicular from A to BC, meeting BC at M. Since the triangle is isosceles, AM passes through the centre O and bisects BC.

∴ BM = $\frac{BC}{2} = \frac{24}{2}$ = 12 cm.

In right-angled △AMB,

By Pythagoras theorem,

⇒ (AB)² = (AM)² + (BM)²

⇒ (AM)² = (AB)² - (BM)²

$$\begin{aligned}\Rightarrow AM = \sqrt{(AB)^2 - (BM)^2} \\[1em] = \sqrt{(12\sqrt{5})^2 - 12^2} \\[1em] = \sqrt{720 - 144} \\[1em] = \sqrt{576} \\[1em] = 24 \text{ cm}.\end{aligned}$$

Let the radius of the circle be R cm. Then OA = OB = R, and OM = AM - OA = (24 - R) cm.

In right-angled △OMB, by Pythagoras theorem,

$$\begin{aligned}\Rightarrow OB^2 = OM^2 + BM^2 \\[1em] \Rightarrow R^2 = (24 - R)^2 + 12^2 \\[1em] \Rightarrow R^2 = 576 - 48R + R^2 + 144 \\[1em] \Rightarrow 48R = 720 \\[1em] \Rightarrow R = 15 \text{ cm}.\end{aligned}$$

Hence, the radius of the circle = 15 cm.

Given,

O is any point inside △ABC.

To prove : ∠BOC > ∠A.

Join BO and produce it to meet AC at point D.

We know that, an exterior angle of a triangle is greater than each of its interior opposite angles.

In △ABD, ∠BDC is an exterior angle.

⇒ ∠BDC > ∠A ......... (1)

In △ODC, ∠BOC is an exterior angle.

⇒ ∠BOC > ∠BDC ......... (2)

From equations (1) and (2), we get :

⇒ ∠BOC > ∠BDC > ∠A

⇒ ∠BOC > ∠A.

Hence, proved that ∠BOC > ∠A.

From the given figure, ABCD is a trapezium in which AB and DC are the two parallel sides and the perpendicular distance between them is the height.

As, AED is a right angled triangle, right angled at E;

By Pythagoras theorem;

⇒ (AD)² = (AE)² + (ED)²

⇒ (ED)² = (AD)² - (AE)²

⇒ (ED)² = 5² - 4²

⇒ (ED)² = 25 - 16

⇒ (ED)² = 9

⇒ ED = $\sqrt{9}$

⇒ ED = 3 cm

Now, DC = EC - ED

DC = 5 - 3 = 2 cm

By formula,

$$\begin{aligned}\text{Area of trapezium = }\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\\[1em] = \frac{1}{2}\times \text{(AB + DC)}\times AE\\[1em] = \frac{1}{2} \times ( 5 + 2) \times 4\\[1em] = \frac{1}{2} \times 7 \times 4 = 14\end{aligned}$$

⇒ Area of trapezium ABCD = 14 cm².

Hence, the area of trapezium ABCD = 14 cm².

Given,

Two chords AB and CD of a circle intersect at point P and AB = CD.

To prove : arc AD = arc CB.

We know that, equal chords of a circle subtend equal arcs.

Since AB = CD,

⇒ arc AB = arc CD ......... (1)

The arc AB can be written as the sum of arc AD and arc DB.

⇒ arc AB = arc AD + arc DB ......... (2)

The arc CD can be written as the sum of arc CB and arc BD.

⇒ arc CD = arc CB + arc BD ......... (3)

Substituting equations (2) and (3) in equation (1), we get :

⇒ arc AD + arc DB = arc CB + arc BD

Since arc DB = arc BD, subtracting it from both sides, we get :

⇒ arc AD = arc CB.

Hence, proved that arc AD = arc CB.

Counting the number of times each value (number of children) occurs in the data, we get the following frequency distribution :

Number of children	Tally marks	Frequency (Number of families)
0	||||	4
1	|||| ||	7
2	|||| |||| ||	12
3	||||	5
4	|||| ||	7
5	||	2
6	|||	3
Total		40

Hence, the required frequency distribution is represented as shown above.

Given,

In rectangle ABCD, AB = 12 cm and ∠BAC = 30°.

Since ABCD is a rectangle, ∠ABC = 90°. So △ABC is right-angled at B, with AC as the diagonal (hypotenuse).

In right-angled △ABC,

By formula,

tan(∠BAC) = $\frac{\text{BC}}{\text{AB}}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \tan 30° = \frac{BC}{12} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{BC}{12} \\[1em] \Rightarrow BC = \frac{12}{\sqrt{3}} \\[1em] \Rightarrow BC = 4\sqrt{3} \text{ cm}.\end{aligned}$$

By formula,

cos(∠BAC) = $\frac{\text{AB}}{\text{AC}}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \cos 30° = \frac{12}{AC} \\[1em] \Rightarrow \frac{\sqrt{3}}{2} = \frac{12}{AC} \\[1em] \Rightarrow AC = \frac{12 \times 2}{\sqrt{3}} \\[1em] \Rightarrow AC = \frac{24}{\sqrt{3}} \\[1em] \Rightarrow AC = 8\sqrt{3} \text{ cm}.\end{aligned}$$

Hence, the length of side BC = $4\sqrt{3}$ cm and the length of diagonal AC = $8\sqrt{3}$ cm.