Mathematics Part II (Geometry) - July 2024
Time: 2 Hours | Max. Marks: 40
Q.1 (A) Choose the correct alternative. (4 Marks)
1. In $\triangle ABC$ and $\triangle PQR$, in a one to one correspondence of vertices, if $\frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ}$, then which of the following statements is true?
Answer: (B)
Explanation: The sides are proportional as follows: $AB \leftrightarrow QR$, $BC \leftrightarrow PR$, $CA \leftrightarrow PQ$.
Ordering vertices: $A \leftrightarrow Q$, $B \leftrightarrow R$, $C \leftrightarrow P$.
Therefore, $\triangle ABC \sim \triangle QRP$ or $\triangle CAB \sim \triangle PQR$.
Explanation: The sides are proportional as follows: $AB \leftrightarrow QR$, $BC \leftrightarrow PR$, $CA \leftrightarrow PQ$.
Ordering vertices: $A \leftrightarrow Q$, $B \leftrightarrow R$, $C \leftrightarrow P$.
Therefore, $\triangle ABC \sim \triangle QRP$ or $\triangle CAB \sim \triangle PQR$.
2. Two circles of radii 5.5 cm and 3.3 cm respectively touch each other externally. Then the distance between their centres is ____.
Answer: (B)
Explanation: Distance between centres of externally touching circles = $r_1 + r_2 = 5.5 + 3.3 = 8.8$ cm.
Explanation: Distance between centres of externally touching circles = $r_1 + r_2 = 5.5 + 3.3 = 8.8$ cm.
3. Seg AB is parallel to Y-axis and co-ordinates of point A are (1, 3), then the co-ordinates of point B are ____.
Answer: (D)
Explanation: A line parallel to the Y-axis has a constant X-coordinate. Since A is (1, 3), the X-coordinate of B must be 1. Only option (D) matches.
Explanation: A line parallel to the Y-axis has a constant X-coordinate. Since A is (1, 3), the X-coordinate of B must be 1. Only option (D) matches.
4. The volume of a cube of side 10 cm is ____.
Answer: (A)
Explanation: Volume of cube = $(\text{side})^3 = 10^3 = 1000 \text{ cm}^3$.
Explanation: Volume of cube = $(\text{side})^3 = 10^3 = 1000 \text{ cm}^3$.
Q.1 (B) Solve the following subquestions. (4 Marks)
1. In $\triangle ABC$, $\angle B=90^{\circ}$, $\angle C=30^{\circ}$, $AC=12$ cm, then find AB.
By $30^{\circ}-60^{\circ}-90^{\circ}$ theorem, the side opposite to $30^{\circ}$ is half the hypotenuse.
$AB = \frac{1}{2} AC$
$AB = \frac{1}{2} \times 12$
$AB = 6$ cm
$AB = \frac{1}{2} AC$
$AB = \frac{1}{2} \times 12$
$AB = 6$ cm
2. In a circle, if $m(\text{arc } MN) = 70^{\circ}$, find $\angle MLN$.
By Inscribed Angle Theorem:
$\angle MLN = \frac{1}{2} m(\text{arc } MN)$
$\angle MLN = \frac{1}{2} \times 70^{\circ}$
$\angle MLN = 35^{\circ}$
$\angle MLN = \frac{1}{2} m(\text{arc } MN)$
$\angle MLN = \frac{1}{2} \times 70^{\circ}$
$\angle MLN = 35^{\circ}$
3. Find the value of $\sin \theta \times \csc \theta$.
We know that $\csc \theta = \frac{1}{\sin \theta}$
$\therefore \sin \theta \times \csc \theta = \sin \theta \times \frac{1}{\sin \theta}$
Value = 1
$\therefore \sin \theta \times \csc \theta = \sin \theta \times \frac{1}{\sin \theta}$
Value = 1
4. If radius of a circle is 4 cm and length of an arc is 10 cm, then find the area of the sector.
Given: $r = 4$ cm, length of arc ($l$) = 10 cm.
Area of Sector ($A$) = $\frac{l \times r}{2}$
$A = \frac{10 \times 4}{2} = \frac{40}{2}$
Area = 20 cm$^2$
Area of Sector ($A$) = $\frac{l \times r}{2}$
$A = \frac{10 \times 4}{2} = \frac{40}{2}$
Area = 20 cm$^2$
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Q.2 (A) Complete the activities (Any 2). (4 Marks)
1. Chord PQ and chord RS intersect at point T. Prove $\angle STQ = \frac{1}{2}[m(\text{arc } SQ) + m(\text{arc } PR)]$.
$\angle STQ = \angle SPQ + $ $\angle PSQ$ (Exterior angle theorem of a triangle)
$= \frac{1}{2} m(\text{arc } SQ) + $ $\frac{1}{2} m(\text{arc } PR)$ (Inscribed angle theorem)
$= \frac{1}{2} [$ $m(\text{arc } SQ)$ + $m(\text{arc } PR)$ $]$
$= \frac{1}{2} m(\text{arc } SQ) + $ $\frac{1}{2} m(\text{arc } PR)$ (Inscribed angle theorem)
$= \frac{1}{2} [$ $m(\text{arc } SQ)$ + $m(\text{arc } PR)$ $]$
2. If $\sec \theta = \frac{25}{7}$, find the value of $\tan \theta$.
$1 + \tan^2\theta = \sec^2\theta$
$1 + \tan^2\theta = $ $(\frac{25}{7})^2$
$\tan^2\theta = \frac{625}{49} - $ 1
$\tan^2\theta = \frac{625 - 49}{49}$
$\tan^2\theta = \frac{576}{49}$
$\tan \theta = $ $\frac{24}{7}$
$1 + \tan^2\theta = $ $(\frac{25}{7})^2$
$\tan^2\theta = \frac{625}{49} - $ 1
$\tan^2\theta = \frac{625 - 49}{49}$
$\tan^2\theta = \frac{576}{49}$
$\tan \theta = $ $\frac{24}{7}$
3. Find the volume of a cone if radius is 7 cm and height is 6 cm.
Volume of cone = $\frac{1}{3} \times \pi \times r^2 \times h$
$= \frac{1}{3} \times \frac{22}{7} \times $ $7^2$ $\times 6$
$= \frac{1}{3} \times \frac{22}{7} \times $ $49$ $\times 6$
Volume of cone = 308 cm$^3$
$= \frac{1}{3} \times \frac{22}{7} \times $ $7^2$ $\times 6$
$= \frac{1}{3} \times \frac{22}{7} \times $ $49$ $\times 6$
Volume of cone = 308 cm$^3$
Q.2 (B) Solve the following (Any 4). (8 Marks)
1. Construct a tangent to a circle with centre P and radius 3.2 cm at any point M on it.
Steps of construction:1. Draw a circle with centre P and radius 3.2 cm.
2. Take any point M on the circle.
3. Draw ray PM.
4. Draw a line perpendicular to ray PM passing through point M. This line is the required tangent.
2. In $\triangle PQR$, seg RS bisects $\angle PRQ$. If $PR=15$, $RQ=20$, $PS=12$, then find SQ.
By the Angle Bisector Theorem:
$\frac{PR}{RQ} = \frac{PS}{SQ}$
$\frac{15}{20} = \frac{12}{SQ}$
$SQ = \frac{20 \times 12}{15}$
$SQ = \frac{240}{15}$
$SQ = 16$ units
$\frac{PR}{RQ} = \frac{PS}{SQ}$
$\frac{15}{20} = \frac{12}{SQ}$
$SQ = \frac{20 \times 12}{15}$
$SQ = \frac{240}{15}$
$SQ = 16$ units
3. Find the surface area of a sphere of diameter 14 cm.
Diameter = 14 cm $\therefore$ Radius ($r$) = 7 cm.
Surface Area = $4\pi r^2$
$= 4 \times \frac{22}{7} \times 7 \times 7$
$= 4 \times 22 \times 7$
$= 88 \times 7$
Surface Area = 616 cm$^2$
Surface Area = $4\pi r^2$
$= 4 \times \frac{22}{7} \times 7 \times 7$
$= 4 \times 22 \times 7$
$= 88 \times 7$
Surface Area = 616 cm$^2$
4. In $\triangle PQR, \angle PQR=90^{\circ}$, seg $QN \perp$ seg PR, $PN=9, NR=16$, find QN.
By Theorem of Geometric Mean:
$QN^2 = PN \times NR$
$QN^2 = 9 \times 16$
$QN^2 = 144$
Taking square root,
$QN = 12$ units
$QN^2 = PN \times NR$
$QN^2 = 9 \times 16$
$QN^2 = 144$
Taking square root,
$QN = 12$ units
5. Find the slope of the line passing through the points A(3,3) and B(5,7).
Slope $m = \frac{y_2 - y_1}{x_2 - x_1}$
$m = \frac{7 - 3}{5 - 3}$
$m = \frac{4}{2}$
Slope = 2
$m = \frac{7 - 3}{5 - 3}$
$m = \frac{4}{2}$
Slope = 2
Q.3 (A) Complete the activities (Any 1). (3 Marks)
1. Line AB || Line CD || Line EF. If $AC=12, CE=9, BD=8$, find DF.
Property of three parallel lines and their transversal:
$\frac{AC}{CE} = \frac{BD}{DF}$
$\frac{12}{9} = \frac{8}{DF}$
$DF = \frac{8 \times 9}{12}$
$DF = $ 6
$\frac{AC}{CE} = \frac{BD}{DF}$
$\frac{12}{9} = \frac{8}{DF}$
$DF = \frac{8 \times 9}{12}$
$DF = $ 6
2. Cylinder: Radius 7 cm, Height 21 cm. Find Volume and Circumference of base.
Volume of cylinder = $\pi r^2 h$
$= \frac{22}{7} \times 7 \times 7 \times $ 21
$= 154 \times 21$
Volume = 3234 cm$^3$
Circumference of base = $2\pi r$
$= 2 \times \frac{22}{7} \times $ 7
Circumference = 44 cm
$= \frac{22}{7} \times 7 \times 7 \times $ 21
$= 154 \times 21$
Volume = 3234 cm$^3$
Circumference of base = $2\pi r$
$= 2 \times \frac{22}{7} \times $ 7
Circumference = 44 cm
Q.3 (B) Solve the following (Any 2). (6 Marks)
1. Prove that, 'Opposite angles of a cyclic quadrilateral are supplementary'.
Given: $\square ABCD$ is cyclic.
To Prove: $\angle B + \angle D = 180^{\circ}$ and $\angle A + \angle C = 180^{\circ}$.
Proof:
$\angle ADC$ is an inscribed angle intercepting arc ABC.
$\therefore \angle ADC = \frac{1}{2} m(\text{arc } ABC)$ ... (I)
Similarly, $\angle ABC$ is an inscribed angle intercepting arc ADC.
$\therefore \angle ABC = \frac{1}{2} m(\text{arc } ADC)$ ... (II)
Adding (I) and (II):
$\angle ADC + \angle ABC = \frac{1}{2} [m(\text{arc } ABC) + m(\text{arc } ADC)]$
$\angle D + \angle B = \frac{1}{2} [360^{\circ}]$ (Since arc ABC + arc ADC constitutes the complete circle)
$\mathbf{\angle D + \angle B = 180^{\circ}}$
Similarly, we can prove $\angle A + \angle C = 180^{\circ}$.
To Prove: $\angle B + \angle D = 180^{\circ}$ and $\angle A + \angle C = 180^{\circ}$.
Proof:
$\angle ADC$ is an inscribed angle intercepting arc ABC.
$\therefore \angle ADC = \frac{1}{2} m(\text{arc } ABC)$ ... (I)
Similarly, $\angle ABC$ is an inscribed angle intercepting arc ADC.
$\therefore \angle ABC = \frac{1}{2} m(\text{arc } ADC)$ ... (II)
Adding (I) and (II):
$\angle ADC + \angle ABC = \frac{1}{2} [m(\text{arc } ABC) + m(\text{arc } ADC)]$
$\angle D + \angle B = \frac{1}{2} [360^{\circ}]$ (Since arc ABC + arc ADC constitutes the complete circle)
$\mathbf{\angle D + \angle B = 180^{\circ}}$
Similarly, we can prove $\angle A + \angle C = 180^{\circ}$.
2. Draw a circle with centre P and radius 3.5 cm. Take point Q at a distance 8 cm from the centre. Construct tangents to the circle from point Q.
Construction Steps:1. Draw circle with centre P, $r = 3.5$ cm.
2. Draw segment PQ = 8 cm.
3. Draw perpendicular bisector of seg PQ to find midpoint M.
4. Draw a circle (or arcs) with centre M and radius MP to cut the original circle at points A and B.
5. Draw line QA and line QB.
QA and QB are the required tangents.
3. Show that points $P(-2,3)$, $Q(1,2)$, $R(4,1)$ are collinear.
Slope of line = $\frac{y_2 - y_1}{x_2 - x_1}$
Slope of PQ = $\frac{2 - 3}{1 - (-2)} = \frac{-1}{3}$
Slope of QR = $\frac{1 - 2}{4 - 1} = \frac{-1}{3}$
Since Slope of PQ = Slope of QR and point Q is common,
Points P, Q, and R are collinear.
Slope of PQ = $\frac{2 - 3}{1 - (-2)} = \frac{-1}{3}$
Slope of QR = $\frac{1 - 2}{4 - 1} = \frac{-1}{3}$
Since Slope of PQ = Slope of QR and point Q is common,
Points P, Q, and R are collinear.
4. If $\triangle PQR \sim \triangle LMN$, $9 \times A(\triangle PQR) = 16 \times A(\triangle LMN)$ and $QR=20$, then find MN.
Given: $9 \times A(\triangle PQR) = 16 \times A(\triangle LMN)$
$\therefore \frac{A(\triangle PQR)}{A(\triangle LMN)} = \frac{16}{9}$
Since triangles are similar, ratio of areas = ratio of squares of corresponding sides.
$\frac{A(\triangle PQR)}{A(\triangle LMN)} = \frac{QR^2}{MN^2}$
$\frac{16}{9} = (\frac{20}{MN})^2$
Taking square root of both sides:
$\frac{4}{3} = \frac{20}{MN}$
$MN = \frac{20 \times 3}{4}$
$MN = 15$ units
$\therefore \frac{A(\triangle PQR)}{A(\triangle LMN)} = \frac{16}{9}$
Since triangles are similar, ratio of areas = ratio of squares of corresponding sides.
$\frac{A(\triangle PQR)}{A(\triangle LMN)} = \frac{QR^2}{MN^2}$
$\frac{16}{9} = (\frac{20}{MN})^2$
Taking square root of both sides:
$\frac{4}{3} = \frac{20}{MN}$
$MN = \frac{20 \times 3}{4}$
$MN = 15$ units
Q.4 Solve the following (Any 2). (8 Marks)
1. $\triangle ABC$ is an equilateral triangle. Point D is on seg BC such that $BD = \frac{1}{5} BC$. Then prove that $\frac{AD^2}{AB^2} = \frac{21}{25}$.
Proof:
Draw $AM \perp BC$. Since $\triangle ABC$ is equilateral, AM is the median.
$\therefore BM = \frac{1}{2} BC$.
Given $BD = \frac{1}{5} BC$.
$DM = BM - BD = \frac{1}{2}BC - \frac{1}{5}BC = \frac{5-2}{10}BC = \frac{3}{10}BC$.
In right angled $\triangle AMC$, height $AM = \frac{\sqrt{3}}{2} AB$ (Altitude of equilateral triangle).
Now, in right angled $\triangle AMD$, by Pythagoras theorem:
$AD^2 = AM^2 + DM^2$
Since $AB = BC$ (Equilateral), substitute BC with AB:
$AD^2 = (\frac{\sqrt{3}}{2} AB)^2 + (\frac{3}{10} AB)^2$
$AD^2 = \frac{3}{4} AB^2 + \frac{9}{100} AB^2$
$AD^2 = AB^2 (\frac{75}{100} + \frac{9}{100})$
$AD^2 = AB^2 (\frac{84}{100})$
$AD^2 = AB^2 (\frac{21}{25})$
$\therefore \mathbf{\frac{AD^2}{AB^2} = \frac{21}{25}}$
Draw $AM \perp BC$. Since $\triangle ABC$ is equilateral, AM is the median.
$\therefore BM = \frac{1}{2} BC$.
Given $BD = \frac{1}{5} BC$.
$DM = BM - BD = \frac{1}{2}BC - \frac{1}{5}BC = \frac{5-2}{10}BC = \frac{3}{10}BC$.
In right angled $\triangle AMC$, height $AM = \frac{\sqrt{3}}{2} AB$ (Altitude of equilateral triangle).
Now, in right angled $\triangle AMD$, by Pythagoras theorem:
$AD^2 = AM^2 + DM^2$
Since $AB = BC$ (Equilateral), substitute BC with AB:
$AD^2 = (\frac{\sqrt{3}}{2} AB)^2 + (\frac{3}{10} AB)^2$
$AD^2 = \frac{3}{4} AB^2 + \frac{9}{100} AB^2$
$AD^2 = AB^2 (\frac{75}{100} + \frac{9}{100})$
$AD^2 = AB^2 (\frac{84}{100})$
$AD^2 = AB^2 (\frac{21}{25})$
$\therefore \mathbf{\frac{AD^2}{AB^2} = \frac{21}{25}}$
2. $\triangle LMN \sim \triangle LQP$. In $\triangle LMN$, $LM=3.6$ cm, $\angle L=50^{\circ}$, $LN=4.2$ cm and $\frac{LM}{LQ} = \frac{4}{7}$. Construct $\triangle LQP$.
Analysis:
Since vertex L is common, we treat this as constructing similar triangles with a common vertex.
Ratio $\frac{LM}{LQ} = \frac{4}{7}$ implies sides of $\triangle LQP$ are larger than $\triangle LMN$.
Steps:
1. Draw $\triangle LMN$ with $LM=3.6$, $\angle L=50^{\circ}$, $LN=4.2$.
2. Draw a ray from L at an acute angle to LM.
3. Mark 7 equal points on the ray ($A_1$ to $A_7$).
4. Join $A_4$ to M (Since LM corresponds to 4 parts).
5. Draw a line parallel to $A_4M$ from $A_7$ intersecting line LM extended at Q.
6. From Q, draw a line parallel to MN intersecting line LN extended at P.
$\triangle LQP$ is the required triangle.
Since vertex L is common, we treat this as constructing similar triangles with a common vertex.
Ratio $\frac{LM}{LQ} = \frac{4}{7}$ implies sides of $\triangle LQP$ are larger than $\triangle LMN$.
Steps:
1. Draw $\triangle LMN$ with $LM=3.6$, $\angle L=50^{\circ}$, $LN=4.2$.
2. Draw a ray from L at an acute angle to LM.
3. Mark 7 equal points on the ray ($A_1$ to $A_7$).
4. Join $A_4$ to M (Since LM corresponds to 4 parts).
5. Draw a line parallel to $A_4M$ from $A_7$ intersecting line LM extended at Q.
6. From Q, draw a line parallel to MN intersecting line LN extended at P.
$\triangle LQP$ is the required triangle.
3. To find the width of the river, a man observes the top of a tree on the opposite bank making an angle of elevation of $60^{\circ}$. When he moves 24 meter backward... angle becomes $30^{\circ}$. Find height of tree and width of river. ($\sqrt{3}=1.73$)
Let height of tree = $h$ m. Let initial width of river = $x$ m.
Case 1: Angle $60^{\circ}$.
$\tan 60^{\circ} = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3}$ ... (I)
Case 2: Angle $30^{\circ}$ from distance $(x + 24)$.
$\tan 30^{\circ} = \frac{h}{x+24} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x+24}$
$h\sqrt{3} = x + 24$
Substitute $h$ from (I):
$(x\sqrt{3})\sqrt{3} = x + 24$
$3x = x + 24$
$2x = 24 \Rightarrow \mathbf{x = 12 \text{ m}}$ (Width of river)
Now, $h = 12\sqrt{3} = 12 \times 1.73 = 20.76$ m.
Height of tree = 20.76 m.
Case 1: Angle $60^{\circ}$.
$\tan 60^{\circ} = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3}$ ... (I)
Case 2: Angle $30^{\circ}$ from distance $(x + 24)$.
$\tan 30^{\circ} = \frac{h}{x+24} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x+24}$
$h\sqrt{3} = x + 24$
Substitute $h$ from (I):
$(x\sqrt{3})\sqrt{3} = x + 24$
$3x = x + 24$
$2x = 24 \Rightarrow \mathbf{x = 12 \text{ m}}$ (Width of river)
Now, $h = 12\sqrt{3} = 12 \times 1.73 = 20.76$ m.
Height of tree = 20.76 m.
Q.5 Solve the following (Any 1). (3 Marks)
1. The diameter of a circular garden is 13 m. The distance between two gates is 13 m. An electric pole is to be erected on the circumference such that the difference between distance of the pole from each gate is 7 m. Can such a pole be erected? If yes, find distances.
Let gates be A and B. $AB = 13$ m. Diameter = 13 m.
Since chord AB = Diameter, the gates are at opposite ends of the diameter.
Let Pole be at P. $\angle APB = 90^{\circ}$ (Angle in a semicircle).
Let $PA = x$ and $PB = y$.
Given difference is 7m: $|x - y| = 7$.
In right $\triangle APB$: $x^2 + y^2 = 13^2 = 169$.
We know: $(x-y)^2 = x^2 + y^2 - 2xy$
$7^2 = 169 - 2xy \Rightarrow 49 = 169 - 2xy \Rightarrow 2xy = 120$.
Now, $(x+y)^2 = x^2 + y^2 + 2xy = 169 + 120 = 289$.
$\therefore x + y = 17$.
Solving $x+y=17$ and $x-y=7$:
$2x = 24 \Rightarrow x = 12$.
$y = 5$.
Yes, the pole can be erected at distances 12 m and 5 m from the gates.
Since chord AB = Diameter, the gates are at opposite ends of the diameter.
Let Pole be at P. $\angle APB = 90^{\circ}$ (Angle in a semicircle).
Let $PA = x$ and $PB = y$.
Given difference is 7m: $|x - y| = 7$.
In right $\triangle APB$: $x^2 + y^2 = 13^2 = 169$.
We know: $(x-y)^2 = x^2 + y^2 - 2xy$
$7^2 = 169 - 2xy \Rightarrow 49 = 169 - 2xy \Rightarrow 2xy = 120$.
Now, $(x+y)^2 = x^2 + y^2 + 2xy = 169 + 120 = 289$.
$\therefore x + y = 17$.
Solving $x+y=17$ and $x-y=7$:
$2x = 24 \Rightarrow x = 12$.
$y = 5$.
Yes, the pole can be erected at distances 12 m and 5 m from the gates.
2. The line $x - 6y + 11 = 0$ bisects the segment joining points (8, -1) and (0, k), find k.
Let $A=(8, -1)$ and $B=(0, k)$.
Since the line bisects segment AB, the midpoint of AB lies on the line.
Midpoint $M = (\frac{8+0}{2}, \frac{-1+k}{2}) = (4, \frac{k-1}{2})$.
Substitute M in equation $x - 6y + 11 = 0$:
$4 - 6(\frac{k-1}{2}) + 11 = 0$
$4 - 3(k-1) + 11 = 0$
$4 - 3k + 3 + 11 = 0$
$18 - 3k = 0$
$3k = 18$
$k = 6$
Since the line bisects segment AB, the midpoint of AB lies on the line.
Midpoint $M = (\frac{8+0}{2}, \frac{-1+k}{2}) = (4, \frac{k-1}{2})$.
Substitute M in equation $x - 6y + 11 = 0$:
$4 - 6(\frac{k-1}{2}) + 11 = 0$
$4 - 3(k-1) + 11 = 0$
$4 - 3k + 3 + 11 = 0$
$18 - 3k = 0$
$3k = 18$
$k = 6$
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