Ariyalur District Second Mid Term Exam 2024
Std: 10 Mathematics - Solutions
PART - A
I. Choose the best answer (7 x 1 = 7)
1. If number of columns and rows are not equal in a matrix then it is said to be a
Answer: b) rectangular matrix
Explanation: A matrix is called a rectangular matrix if the number of rows is not equal to the number of columns (m ≠ n).
Explanation: A matrix is called a rectangular matrix if the number of rows is not equal to the number of columns (m ≠ n).
2. If A is a 2x3 matrix and B is a 3x4 matrix how many columns does AB have.
Answer: b) 4
Explanation: When multiplying a matrix of order $m \times n$ by a matrix of order $n \times p$, the resulting matrix has the order $m \times p$. Here, A is $2 \times 3$ and B is $3 \times 4$. The resulting matrix AB will have the order $2 \times 4$. Therefore, it has 4 columns.
Explanation: When multiplying a matrix of order $m \times n$ by a matrix of order $n \times p$, the resulting matrix has the order $m \times p$. Here, A is $2 \times 3$ and B is $3 \times 4$. The resulting matrix AB will have the order $2 \times 4$. Therefore, it has 4 columns.
3. A tangent is perpendicular to the radius at the
Answer: b) point of contact
Explanation: A fundamental theorem in circle geometry states that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Explanation: A fundamental theorem in circle geometry states that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
4. In figure if PR and QR are tangents to the circle at P and Q and O is the centre of the circle, then $\angle POQ$ is
Note: The question is incomplete as the angle at R ($\angle PRQ$) is not given. Assuming a standard problem structure, we find the relationship. The sum of the angle at the center ($\angle POQ$) and the angle between the tangents ($\angle PRQ$) is 180°. Let's assume $\angle PRQ = 70^{\circ}$ to match one of the options.
Answer: c) $110^{\circ}$
Explanation: In the quadrilateral OPRQ, $\angle OPR = 90^{\circ}$ and $\angle OQR = 90^{\circ}$ (radius is perpendicular to tangent). The sum of angles in a quadrilateral is $360^{\circ}$. $$ \angle POQ + \angle OPR + \angle PRQ + \angle OQR = 360^{\circ} $$ $$ \angle POQ + 90^{\circ} + \angle PRQ + 90^{\circ} = 360^{\circ} $$ $$ \angle POQ + \angle PRQ = 180^{\circ} $$ If we assume $\angle PRQ = 70^{\circ}$, then $\angle POQ = 180^{\circ} - 70^{\circ} = 110^{\circ}$.
Explanation: In the quadrilateral OPRQ, $\angle OPR = 90^{\circ}$ and $\angle OQR = 90^{\circ}$ (radius is perpendicular to tangent). The sum of angles in a quadrilateral is $360^{\circ}$. $$ \angle POQ + \angle OPR + \angle PRQ + \angle OQR = 360^{\circ} $$ $$ \angle POQ + 90^{\circ} + \angle PRQ + 90^{\circ} = 360^{\circ} $$ $$ \angle POQ + \angle PRQ = 180^{\circ} $$ If we assume $\angle PRQ = 70^{\circ}$, then $\angle POQ = 180^{\circ} - 70^{\circ} = 110^{\circ}$.
5. A tower is 60m high. Its shadow reduces by x meters when the angle of elevation of the sun increases from 30° to 45° then x is equal to
Answer: b) 43.92m
Explanation: Let the height of the tower AB = 60 m. Let the initial and final positions of the shadow be C and D. When the angle is $30^{\circ}$, $\tan 30^{\circ} = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{60}{BC} \Rightarrow BC = 60\sqrt{3}$ m. When the angle is $45^{\circ}$, $\tan 45^{\circ} = \frac{AB}{BD} \Rightarrow 1 = \frac{60}{BD} \Rightarrow BD = 60$ m. The reduction in shadow length, $x = BC - BD = 60\sqrt{3} - 60 = 60(\sqrt{3} - 1)$. $x = 60(1.732 - 1) = 60(0.732) = 43.92$ m.
Explanation: Let the height of the tower AB = 60 m. Let the initial and final positions of the shadow be C and D. When the angle is $30^{\circ}$, $\tan 30^{\circ} = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{60}{BC} \Rightarrow BC = 60\sqrt{3}$ m. When the angle is $45^{\circ}$, $\tan 45^{\circ} = \frac{AB}{BD} \Rightarrow 1 = \frac{60}{BD} \Rightarrow BD = 60$ m. The reduction in shadow length, $x = BC - BD = 60\sqrt{3} - 60 = 60(\sqrt{3} - 1)$. $x = 60(1.732 - 1) = 60(0.732) = 43.92$ m.
6. If the radius of the base of a cone is tripled and the height is doubled then the volume is
Answer: b) made 18 times
Explanation: Original volume $V_1 = \frac{1}{3}\pi r^2 h$. New radius $r' = 3r$, new height $h' = 2h$. New volume $V_2 = \frac{1}{3}\pi (r')^2 h' = \frac{1}{3}\pi (3r)^2 (2h) = \frac{1}{3}\pi (9r^2)(2h) = 18 \left(\frac{1}{3}\pi r^2 h\right) = 18V_1$.
Explanation: Original volume $V_1 = \frac{1}{3}\pi r^2 h$. New radius $r' = 3r$, new height $h' = 2h$. New volume $V_2 = \frac{1}{3}\pi (r')^2 h' = \frac{1}{3}\pi (3r)^2 (2h) = \frac{1}{3}\pi (9r^2)(2h) = 18 \left(\frac{1}{3}\pi r^2 h\right) = 18V_1$.
7. The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is
Answer: d) 3 : 1 : 2
Explanation: Let the radius be $r$. Same diameter means same radius. Same height $h$. For a sphere, height = diameter, so $h = 2r$. $V_{cylinder} = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3$. $V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3$. $V_{sphere} = \frac{4}{3}\pi r^3$. Ratio: $V_{cylinder} : V_{cone} : V_{sphere} = 2\pi r^3 : \frac{2}{3}\pi r^3 : \frac{4}{3}\pi r^3$. Multiplying by $\frac{3}{2\pi r^3}$, we get the ratio: $3 : 1 : 2$.
Explanation: Let the radius be $r$. Same diameter means same radius. Same height $h$. For a sphere, height = diameter, so $h = 2r$. $V_{cylinder} = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3$. $V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3$. $V_{sphere} = \frac{4}{3}\pi r^3$. Ratio: $V_{cylinder} : V_{cone} : V_{sphere} = 2\pi r^3 : \frac{2}{3}\pi r^3 : \frac{4}{3}\pi r^3$. Multiplying by $\frac{3}{2\pi r^3}$, we get the ratio: $3 : 1 : 2$.
PART - B
II. Answer any five questions only [Q. No. 14 is compulsory] (5 x 2 = 10)
8. If $A = \begin{bmatrix} 0 & 4 & 9 \\ 8 & 3 & 7 \end{bmatrix}, B = \begin{bmatrix} 7 & 3 & 8 \\ 1 & 4 & 9 \end{bmatrix}$ find the value of B - 5A.
Solution:
First, calculate 5A:
$$ 5A = 5 \begin{bmatrix} 0 & 4 & 9 \\ 8 & 3 & 7 \end{bmatrix} = \begin{bmatrix} 5(0) & 5(4) & 5(9) \\ 5(8) & 5(3) & 5(7) \end{bmatrix} = \begin{bmatrix} 0 & 20 & 45 \\ 40 & 15 & 35 \end{bmatrix} $$
Now, calculate B - 5A:
$$ B - 5A = \begin{bmatrix} 7 & 3 & 8 \\ 1 & 4 & 9 \end{bmatrix} - \begin{bmatrix} 0 & 20 & 45 \\ 40 & 15 & 35 \end{bmatrix} $$
$$ = \begin{bmatrix} 7-0 & 3-20 & 8-45 \\ 1-40 & 4-15 & 9-35 \end{bmatrix} = \begin{bmatrix} 7 & -17 & -37 \\ -39 & -11 & -26 \end{bmatrix} $$
$B - 5A = \begin{bmatrix} 7 & -17 & -37 \\ -39 & -11 & -26 \end{bmatrix}$
9. If $A = \begin{bmatrix} 5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \frac{5}{2} \\ 8 & 3 & 1 \end{bmatrix}$ then verify $(A^T)^T = A$.
Solution:
Given matrix A:
$$ A = \begin{bmatrix} 5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \frac{5}{2} \\ 8 & 3 & 1 \end{bmatrix} $$
First, find the transpose of A, $A^T$:
$$ A^T = \begin{bmatrix} 5 & -\sqrt{17} & 8 \\ 2 & 0.7 & 3 \\ 2 & \frac{5}{2} & 1 \end{bmatrix} $$
Now, find the transpose of $A^T$, which is $(A^T)^T$:
$$ (A^T)^T = \begin{bmatrix} 5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \frac{5}{2} \\ 8 & 3 & 1 \end{bmatrix} $$
Comparing this with the original matrix A, we see that $(A^T)^T = A$.
Hence, verified.
10. Find the length of the tangent drawn from a point whose distance from the centre of a circle is 5cm and radius of the circle is 3 cm.
Solution:
Let O be the center of the circle, P be the external point, and T be the point of contact.
Given: Distance from the center, OP = 5 cm. Radius of the circle, OT = 3 cm.
The tangent is perpendicular to the radius at the point of contact, so $\triangle OTP$ is a right-angled triangle with the right angle at T.
By Pythagoras theorem:
$$ OP^2 = OT^2 + PT^2 $$
$$ 5^2 = 3^2 + PT^2 $$
$$ 25 = 9 + PT^2 $$
$$ PT^2 = 25 - 9 = 16 $$
$$ PT = \sqrt{16} = 4 $$
The length of the tangent is 4 cm.
11. A tower stands vertically on the ground. From a point on the ground, which is 48 m away from the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower.
The unit in the question paper is 'cm', which is likely a typo for a tower. The solution uses 'm'.
Solution:
Let the height of the tower be 'h' meters.
Let the distance from the foot of the tower be 'd' = 48 m.
The angle of elevation, $\theta = 30^{\circ}$.
In the right-angled triangle formed:
$$ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{d} $$
$$ \tan 30^{\circ} = \frac{h}{48} $$
$$ \frac{1}{\sqrt{3}} = \frac{h}{48} $$
$$ h = \frac{48}{\sqrt{3}} = \frac{48 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{48\sqrt{3}}{3} = 16\sqrt{3} $$
The height of the tower is $16\sqrt{3}$ m.
12. If the base area of a hemispherical solid is 1386 sq. meters then find its total surface area?
Solution:
The base of a hemispherical solid is a circle.
Given: Base area = $\pi r^2 = 1386$ sq. m.
The Total Surface Area (TSA) of a hemisphere is the sum of its curved surface area ($2\pi r^2$) and its base area ($\pi r^2$).
$$ \text{TSA of hemisphere} = 2\pi r^2 + \pi r^2 = 3\pi r^2 $$
We are given $\pi r^2 = 1386$.
$$ \text{TSA} = 3 \times (\pi r^2) = 3 \times 1386 $$
$$ \text{TSA} = 4158 $$
The total surface area of the hemisphere is 4158 sq. meters.
13. The volume of a solid right circular cone is 11088 cm³. If its height is 14cm then find the radius of the cone.
The numbers in this question might be unusual. A common textbook problem has a height of 24 cm for this volume. We will solve with the given value.
Solution:
Given: Volume of the cone, V = 11088 cm³. Height, h = 14 cm.
The formula for the volume of a cone is $V = \frac{1}{3}\pi r^2 h$.
$$ 11088 = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 14 $$
$$ 11088 = \frac{1}{3} \times 22 \times r^2 \times 2 $$
$$ 11088 = \frac{44}{3} r^2 $$
$$ r^2 = \frac{11088 \times 3}{44} $$
Dividing 11088 by 44: $11088 \div 44 = 252$.
$$ r^2 = 252 \times 3 = 756 $$
$$ r = \sqrt{756} = \sqrt{36 \times 21} = 6\sqrt{21} $$
The radius of the cone is $6\sqrt{21}$ cm.
14. (Compulsory) Find the angle of elevation of the top of a tower from a point on the ground, which is 30m away from the foot of a tower of height 10√3 m.
Solution:
Let the height of the tower, h = $10\sqrt{3}$ m.
Distance from the foot of the tower, d = 30 m.
Let the angle of elevation be $\theta$.
$$ \tan \theta = \frac{\text{Height of tower}}{\text{Distance from foot}} = \frac{10\sqrt{3}}{30} $$
$$ \tan \theta = \frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{1}{\sqrt{3}} $$
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$, the angle of elevation is $30^{\circ}$.
The angle of elevation is $30^{\circ}$.
PART - C
III. Answer any five questions [Q. No. 21 is compulsory] (5 x 5 = 25)
15. Find X and Y if $X + Y = \begin{bmatrix} 7 & 0 \\ 3 & 5 \end{bmatrix}$ and $X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}$.
Solution:
Let the given equations be:
$$ X + Y = \begin{bmatrix} 7 & 0 \\ 3 & 5 \end{bmatrix} \quad \cdots(1) $$
$$ X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} \quad \cdots(2) $$
Adding (1) and (2):
$$ (X+Y) + (X-Y) = \begin{bmatrix} 7 & 0 \\ 3 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} $$
$$ 2X = \begin{bmatrix} 7+3 & 0+0 \\ 3+0 & 5+4 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 3 & 9 \end{bmatrix} $$
$$ X = \frac{1}{2} \begin{bmatrix} 10 & 0 \\ 3 & 9 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ \frac{3}{2} & \frac{9}{2} \end{bmatrix} $$
Subtracting (2) from (1):
$$ (X+Y) - (X-Y) = \begin{bmatrix} 7 & 0 \\ 3 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} $$
$$ 2Y = \begin{bmatrix} 7-3 & 0-0 \\ 3-0 & 5-4 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 3 & 1 \end{bmatrix} $$
$$ Y = \frac{1}{2} \begin{bmatrix} 4 & 0 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ \frac{3}{2} & \frac{1}{2} \end{bmatrix} $$
$X = \begin{bmatrix} 5 & 0 \\ \frac{3}{2} & \frac{9}{2} \end{bmatrix}$, $Y = \begin{bmatrix} 2 & 0 \\ \frac{3}{2} & \frac{1}{2} \end{bmatrix}$
16. Given that $A=\begin{bmatrix} 1 & 3 \\ 5 & -1 \end{bmatrix}$, $B=\begin{bmatrix} 1 & -1 & 2 \\ 3 & 5 & 2 \end{bmatrix}$, $C=\begin{bmatrix} 1 & 3 & 2 \\ -4 & 1 & 3 \end{bmatrix}$ verify that $A(B+C) = AB+AC$.
Solution:
L.H.S = A(B+C)
First, find B+C:
$$ B+C = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 5 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 3 & 2 \\ -4 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 1+1 & -1+3 & 2+2 \\ 3-4 & 5+1 & 2+3 \end{bmatrix} = \begin{bmatrix} 2 & 2 & 4 \\ -1 & 6 & 5 \end{bmatrix} $$
Now, find A(B+C):
$$ A(B+C) = \begin{bmatrix} 1 & 3 \\ 5 & -1 \end{bmatrix} \begin{bmatrix} 2 & 2 & 4 \\ -1 & 6 & 5 \end{bmatrix} $$
$$ = \begin{bmatrix} (1)(2)+(3)(-1) & (1)(2)+(3)(6) & (1)(4)+(3)(5) \\ (5)(2)+(-1)(-1) & (5)(2)+(-1)(6) & (5)(4)+(-1)(5) \end{bmatrix} $$
$$ = \begin{bmatrix} 2-3 & 2+18 & 4+15 \\ 10+1 & 10-6 & 20-5 \end{bmatrix} = \begin{bmatrix} -1 & 20 & 19 \\ 11 & 4 & 15 \end{bmatrix} $$
R.H.S = AB + AC
First, find AB:
$$ AB = \begin{bmatrix} 1 & 3 \\ 5 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \\ 3 & 5 & 2 \end{bmatrix} = \begin{bmatrix} 1+9 & -1+15 & 2+6 \\ 5-3 & -5-5 & 10-2 \end{bmatrix} = \begin{bmatrix} 10 & 14 & 8 \\ 2 & -10 & 8 \end{bmatrix} $$
Next, find AC:
$$ AC = \begin{bmatrix} 1 & 3 \\ 5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ -4 & 1 & 3 \end{bmatrix} = \begin{bmatrix} 1-12 & 3+3 & 2+9 \\ 5+4 & 15-1 & 10-3 \end{bmatrix} = \begin{bmatrix} -11 & 6 & 11 \\ 9 & 14 & 7 \end{bmatrix} $$
Now, find AB + AC:
$$ AB+AC = \begin{bmatrix} 10 & 14 & 8 \\ 2 & -10 & 8 \end{bmatrix} + \begin{bmatrix} -11 & 6 & 11 \\ 9 & 14 & 7 \end{bmatrix} = \begin{bmatrix} 10-11 & 14+6 & 8+11 \\ 2+9 & -10+14 & 8+7 \end{bmatrix} = \begin{bmatrix} -1 & 20 & 19 \\ 11 & 4 & 15 \end{bmatrix} $$
Since L.H.S = R.H.S, the property is verified.
Hence, A(B+C) = AB + AC is verified.
17. Show that in a triangle, the medians are concurrent.
Proof using Coordinate Geometry:
Let the vertices of a triangle be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.
Let D, E, and F be the midpoints of the sides BC, CA, and AB respectively.
The coordinates of the midpoints are:
$$ D = \left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right) $$
$$ E = \left(\frac{x_3+x_1}{2}, \frac{y_3+y_1}{2}\right) $$
$$ F = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) $$
The medians are the line segments AD, BE, and CF.
The point that divides each median in the ratio 2:1 is called the centroid (G).
Let's find the point G that divides the median AD in the ratio 2:1 using the section formula:
$$ G = \left[\frac{1 \cdot x_1 + 2 \left(\frac{x_2+x_3}{2}\right)}{1+2}, \frac{1 \cdot y_1 + 2 \left(\frac{y_2+y_3}{2}\right)}{1+2}\right] $$
$$ G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) $$
Similarly, for median BE:
$$ G = \left[\frac{1 \cdot x_2 + 2 \left(\frac{x_3+x_1}{2}\right)}{1+2}, \frac{1 \cdot y_2 + 2 \left(\frac{y_3+y_1}{2}\right)}{1+2}\right] = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) $$
And for median CF:
$$ G = \left[\frac{1 \cdot x_3 + 2 \left(\frac{x_1+x_2}{2}\right)}{1+2}, \frac{1 \cdot y_3 + 2 \left(\frac{y_1+y_2}{2}\right)}{1+2}\right] = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) $$
Since all three medians pass through the same point G, they are concurrent.
Thus, the medians of a triangle are concurrent.
18. Two ships are sailing in the sea on either side of a lighthouse. The angle of elevation of the top of the lighthouse as observed from the ships are 30° and 45° respectively. If the lighthouse is 200m high. Find the distance between the two ships. (√3 = 1.732).
Solution:
Let AB be the lighthouse of height h = 200 m. Let the two ships be at positions C and D on either side of the lighthouse.
The angles of elevation are $\angle ACB = 30^{\circ}$ and $\angle ADB = 45^{\circ}$.
In the right-angled triangle $\triangle ABC$:
$$ \tan 30^{\circ} = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{200}{BC} $$
$$ BC = 200\sqrt{3} \text{ m} $$
In the right-angled triangle $\triangle ABD$:
$$ \tan 45^{\circ} = \frac{AB}{BD} \Rightarrow 1 = \frac{200}{BD} $$
$$ BD = 200 \text{ m} $$
The distance between the two ships is CD = BC + BD.
$$ CD = 200\sqrt{3} + 200 = 200(\sqrt{3} + 1) $$
Using $\sqrt{3} = 1.732$:
$$ CD = 200(1.732 + 1) = 200(2.732) = 546.4 $$
The distance between the two ships is 546.4 meters.
19. From a window (h meters high above the ground) of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are $\theta_1$ and $\theta_2$ respectively show that the height of the opposite house is $h\left(1+\frac{\cot \theta_2}{\cot \theta_1}\right)$.
There appears to be a typo in the question paper's formula. The correct formula is $h\left(1+\frac{\cot \theta_2}{\cot \theta_1}\right)$ or $h(1+\frac{\tan\theta_1}{\tan\theta_2})$. We will prove the correct formula as derived from the problem description.
Proof:
Let W be the window at height $h$ from the ground A, so $AW=h$. Let PQ be the opposite house, with P being the top and Q being the foot. Let the horizontal distance between the houses be $x$. Draw a horizontal line WR from W to the house PQ.
From the diagram, $WR = AQ = x$ and $RQ = AW = h$.
The angle of elevation to P from W is $\angle PWR = \theta_1$.
The angle of depression to Q from W is $\angle RWQ = \theta_2$.
In right-angled triangle $\triangle WRQ$:
$$ \tan \theta_2 = \frac{RQ}{WR} = \frac{h}{x} \Rightarrow x = \frac{h}{\tan \theta_2} = h \cot \theta_2 \quad \cdots(1)$$
In right-angled triangle $\triangle WRP$:
$$ \tan \theta_1 = \frac{PR}{WR} = \frac{PR}{x} \Rightarrow PR = x \tan \theta_1 \quad \cdots(2)$$
Substitute the value of $x$ from (1) into (2):
$$ PR = (h \cot \theta_2) \tan \theta_1 = h \frac{\cot \theta_2}{\cot \theta_1} $$
The total height of the opposite house is $H = PQ = PR + RQ$.
$$ H = h \frac{\cot \theta_2}{\cot \theta_1} + h $$
$$ H = h\left(1 + \frac{\cot \theta_2}{\cot \theta_1}\right) $$
Hence, the height of the opposite house is $h\left(1 + \frac{\cot \theta_2}{\cot \theta_1}\right)$. (The formula given in the question was slightly different, this is the correct derivation).
20. The internal and external diameter of a hollow hemispherical shell are 6cm and 10cm respectively. If it is melted and recast into solid cylinder of diameter 14cm, then find the height of the cylinder.
Solution:
For the hollow hemispherical shell:
External diameter = 10 cm $\Rightarrow$ External radius, $R = 5$ cm.
Internal diameter = 6 cm $\Rightarrow$ Internal radius, $r = 3$ cm.
Volume of the hollow hemisphere = $\frac{2}{3}\pi(R^3 - r^3)$
$$ V_{hemi} = \frac{2}{3}\pi(5^3 - 3^3) = \frac{2}{3}\pi(125 - 27) = \frac{2}{3}\pi(98) \text{ cm}^3 $$
For the solid cylinder:
Diameter = 14 cm $\Rightarrow$ Radius, $r_{cyl} = 7$ cm.
Let the height be $h$.
Volume of the cylinder = $\pi r_{cyl}^2 h = \pi (7)^2 h = 49\pi h \text{ cm}^3$.
Since the shell is melted and recast into the cylinder, their volumes are equal.
$$ V_{cyl} = V_{hemi} $$
$$ 49\pi h = \frac{2}{3}\pi(98) $$
$$ 49 h = \frac{2 \times 98}{3} $$
$$ h = \frac{2 \times 98}{3 \times 49} = \frac{2 \times 2}{3} = \frac{4}{3} $$
The height of the cylinder is $\frac{4}{3}$ cm or approximately 1.33 cm.
21. (Compulsory) If $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & -1 & 1 \end{bmatrix}, B = \begin{bmatrix} 2 & -1 \\ -1 & 4 \\ 0 & 2 \end{bmatrix}$ show that $(AB)^T = B^T A^T$.
Solution:
L.H.S = $(AB)^T$
First, calculate AB:
$$ AB = \begin{bmatrix} 1 & 2 & 1 \\ 2 & -1 & 1 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 4 \\ 0 & 2 \end{bmatrix} $$
$$ = \begin{bmatrix} (1)(2)+(2)(-1)+(1)(0) & (1)(-1)+(2)(4)+(1)(2) \\ (2)(2)+(-1)(-1)+(1)(0) & (2)(-1)+(-1)(4)+(1)(2) \end{bmatrix} $$
$$ = \begin{bmatrix} 2-2+0 & -1+8+2 \\ 4+1+0 & -2-4+2 \end{bmatrix} = \begin{bmatrix} 0 & 9 \\ 5 & -4 \end{bmatrix} $$
Now, find the transpose $(AB)^T$:
$$ (AB)^T = \begin{bmatrix} 0 & 5 \\ 9 & -4 \end{bmatrix} $$
R.H.S = $B^T A^T$
First, find the transposes $B^T$ and $A^T$:
$$ B^T = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 4 & 2 \end{bmatrix}, \quad A^T = \begin{bmatrix} 1 & 2 \\ 2 & -1 \\ 1 & 1 \end{bmatrix} $$
Now, calculate $B^T A^T$:
$$ B^T A^T = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 4 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & -1 \\ 1 & 1 \end{bmatrix} $$
$$ = \begin{bmatrix} (2)(1)+(-1)(2)+(0)(1) & (2)(2)+(-1)(-1)+(0)(1) \\ (-1)(1)+(4)(2)+(2)(1) & (-1)(2)+(4)(-1)+(2)(1) \end{bmatrix} $$
$$ = \begin{bmatrix} 2-2+0 & 4+1+0 \\ -1+8+2 & -2-4+2 \end{bmatrix} = \begin{bmatrix} 0 & 5 \\ 9 & -4 \end{bmatrix} $$
Since L.H.S = R.H.S, the property is shown.
Hence, $(AB)^T = B^T A^T$ is shown.
PART - D
IV. Answer any one: (1 x 8 = 8)
22. a) Draw the two tangents from a point which is 5cm away from the centre of a circle of diameter 6 cm. Also measure the lengths of the tangents.
Given: Diameter of the circle = 6 cm, so Radius (r) = 3 cm.
Distance of the external point from the centre (d) = 5 cm.
Steps of Construction:
Measurement:
On measuring the lengths of the tangents with a ruler, we find that PA = 4 cm and PB = 4 cm.
Verification by Calculation:
In the right-angled triangle $\triangle OAP$ (since radius OA is perpendicular to tangent PA):
By Pythagoras theorem, $OP^2 = OA^2 + PA^2$.
$$ 5^2 = 3^2 + PA^2 $$
$$ 25 = 9 + PA^2 $$
$$ PA^2 = 25 - 9 = 16 $$
$$ PA = \sqrt{16} = 4 \text{ cm} $$
The calculated length matches the measured length.
- Draw a circle with center O and radius 3 cm.
- Mark a point P, 5 cm away from the center O (i.e., OP = 5 cm).
- Draw the perpendicular bisector of the line segment OP. Let it intersect OP at M. M is the midpoint of OP.
- With M as the center and MO (or MP) as the radius, draw a new circle.
- This new circle will intersect the original circle at two points, say A and B.
- Join PA and PB. These are the required tangents to the circle from point P.
Measurement:
On measuring the lengths of the tangents with a ruler, we find that PA = 4 cm and PB = 4 cm.
Verification by Calculation:
In the right-angled triangle $\triangle OAP$ (since radius OA is perpendicular to tangent PA):
By Pythagoras theorem, $OP^2 = OA^2 + PA^2$.
$$ 5^2 = 3^2 + PA^2 $$
$$ 25 = 9 + PA^2 $$
$$ PA^2 = 25 - 9 = 16 $$
$$ PA = \sqrt{16} = 4 \text{ cm} $$
The calculated length matches the measured length.
The lengths of the tangents are PA = 4 cm and PB = 4 cm.
22. b) Discuss the nature of solutions of the following quadratic equation. $x^2 + x - 12 = 0$.
Solution:
The given quadratic equation is $x^2 + x - 12 = 0$.
To determine the nature of the roots (solutions), we use the discriminant, $\Delta = b^2 - 4ac$.
First, compare the given equation with the standard form $ax^2 + bx + c = 0$:
Here, $a = 1$, $b = 1$, and $c = -12$.
Now, calculate the discriminant:
$$ \Delta = b^2 - 4ac $$
$$ \Delta = (1)^2 - 4(1)(-12) $$
$$ \Delta = 1 + 48 $$
$$ \Delta = 49 $$
Analysis of the Discriminant:
The nature of the roots is determined by the value of $\Delta$:
- Since $\Delta = 49 > 0$, the roots are real and unequal (distinct).
- Since $\Delta = 49 = 7^2$, which is a perfect square, the roots are also rational.
The equation has two real, distinct, and rational roots.
