SESSION ENDING (SUPPLEMENTARY) EXAMINATION (2022-23)
CLASS-IX SUBJECT- MATHEMATICS
MARKING SCHEME / ANSWER KEY
SECTION A
(1 mark each)
(B) $\frac{2}{3}$
Explanation: $a^b + b^a = (-1)^2 + 2^{-1} = 1 + \frac{1}{2} = \frac{3}{2}$. Its inverse $(\frac{3}{2})^{-1} = \frac{2}{3}$.
(A) $\sqrt{3}$
Explanation: $\sqrt{3} \approx 1.732$, which lies between 1 and 2.
(A) 1/ $5\sqrt{5}$
Explanation: $4^{x} - 4^{x-1} = 24$. Let $4^{x-1} = t$, then $4t - t = 24 \implies 3t=24 \implies t=8$. So $4^{x-1}=8 \implies 2^{2x-2}=2^3 \implies 2x-2=3 \implies x=5/2$. Then $(2x)^{-x} = (5)^{-5/2} = 1 / (5^{5/2}) = 1 / (5^2 \cdot 5^{1/2}) = 1 / 25\sqrt{5}$. There is a misprint in the option. Option (a) is $1/5\sqrt{5}$, which is likely the intended simplified form if the calculation was slightly different. Let's re-check: $4^{x} - 4^{x-1} = 24 \implies 4^{x-1}(4-1)=24 \implies 4^{x-1}=8 \implies 2^{2(x-1)}=2^3 \implies 2x-2=3 \implies x=5/2$. $(2x)^{-x} = (5)^{-5/2} = 1/(5^{5/2}) = 1/(5^2 \cdot 5^{1/2}) = 1/25\sqrt{5}$. This is not in the options. If the question was $(2x)^{-x}$ and $x=2.5$, then $(5)^{-2.5} = 1/25\sqrt{5}$. However, option (a) is $1/5\sqrt{5}$. This indicates a possible error in the question or options. The closest correct working yields $1/25\sqrt{5}$.
(B) -1
Explanation: If (x+2) is a factor, then $f(-2)=0$. So $(-2)^2 - 3a(-2) - 2a = 4 + 6a - 2a = 4 + 4a = 0 \implies a = -1$.
(A) a+c+e = b+d
Explanation: If $(x+1)$ is a factor, then $f(-1)=0$. So $a(-1)^4 - b(-1)^3 + c(-1)^2 + d(-1) + e = a + b + c - d + e = 0 \implies a + c + e = d - b$. This does not match any option. If the polynomial was $ax^4 + bx^3 + cx^2 + dx + e$, then $f(-1)=0$ gives $a - b + c - d + e = 0 \implies a+c+e = b+d$. There is likely a typo in the question paper (the sign of the $bx^3$ term should be positive to get option A). Assuming the standard form, the intended answer is (A).
(C) 43
Explanation: We know $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$. So $9^2 = a^2+b^2+c^2 + 2(19) \implies 81 = a^2+b^2+c^2 + 38 \implies a^2+b^2+c^2 = 43$.
(A) 0
(C) (−3,0)
(C) 2x+y=-5
Explanation: $2(-2) + (-1) = -4 -1 = -5$.
(D) infinitely many
(C) a postulate
(C) $\sqrt{3}$ r
Explanation: The line joining the centers is perpendicular to the common chord and bisects it. The distance from the center to the chord is r/2. Half the chord length = $\sqrt{r^2 - (r/2)^2} = \sqrt{r^2 - r^2/4} = \sqrt{3r^2/4} = \frac{\sqrt{3}}{2}r$. So chord length = $2 \times \frac{\sqrt{3}}{2}r = \sqrt{3}r$.
(B) (20-x)°
Explanation: Complement = $90° - (70+x)° = (20-x)°$.
For Visually Impaired: Supplement = $180° - (90-x)° = (90+x)°$. So answer is (D).
(D) 4:9
Explanation: Ratio of volumes = $(r_1/r_2)^3 = 8/27 \implies r_1/r_2 = 2/3$. Ratio of surface areas = $(r_1/r_2)^2 = 4/9$.
(D) 120°
Explanation: In an equilateral triangle inscribed in a circle, the angle subtended by a side at the center is twice the angle at the circumference. $\angle BOC = 2 \times \angle BAC = 2 \times 60° = 120°$.
(C) $4+2\sqrt{2}$ cm
Explanation: For an isosceles right triangle, let equal sides be $a$. Area = $\frac{1}{2}a^2 = 2 \implies a^2 = 4 \implies a=2$ cm. Hypotenuse = $a\sqrt{2} = 2\sqrt{2}$ cm. Perimeter = $2+2+2\sqrt{2} = 4+2\sqrt{2}$ cm.
(B) 65°,115°,65°,115°
Explanation: Adjacent angles of a parallelogram are supplementary. So $(2x-5)+(3x+10)=180 \implies 5x+5=180 \implies 5x=175 \implies x=35$. Angles are $2(35)-5=65°$ and $3(35)+10=115°$.
(C) SSA
(D) Assertion (A) is false but Reason (R) is true.
Explanation: Euclidean geometry is valid for plane surfaces, not curved ones. So A is false. Reason R is Euclid's axiom, which is true.
(B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Explanation: In a histogram with equal class intervals, both area and height are proportional to frequency. So R is true. However, A is the fundamental property. R explains how the area is proportional (through the height), but for equal class intervals, they essentially mean the same thing. R is a specific detail of A, not a distinct explanation of why area is used. In a more precise sense, A is true. R is also true, but it's a condition that makes A hold, not the reason it is defined that way. The most appropriate choice is B, as R describes the method but doesn't explain the conceptual choice of area.
SECTION B
(2 marks each)
Represent $\sqrt{3}$ on the number line.
Step 1: On a number line, mark point O at 0 and point A at 1. Draw a perpendicular line AB of length 1 unit from A.
Step 2: Join O and B. By Pythagoras theorem, $OB = \sqrt{1^2 + 1^2} = \sqrt{2}$.
Step 3: Draw a perpendicular line BC of length 1 unit from B.
Step 4: Join O and C. By Pythagoras theorem, $OC = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2+1} = \sqrt{3}$.
Step 5: With O as center and radius OC, draw an arc to cut the number line at point P. Point P represents $\sqrt{3}$.
Find the least rationalizing factor of (i) $\frac{2}{\sqrt[3]{2}}$ (ii) $\frac{\sqrt{2}}{2\sqrt{5}}$
(i) $\frac{2}{\sqrt[3]{2}} = \frac{2}{2^{1/3}} = 2^{1 - 1/3} = 2^{2/3} = \sqrt[3]{2^2} = \sqrt[3]{4}$. The least rationalizing factor is $\sqrt[3]{2}$ because $\sqrt[3]{4} \times \sqrt[3]{2} = \sqrt[3]{8} = 2$.
(ii) $\frac{\sqrt{2}}{2\sqrt{5}}$. The denominator part is $\sqrt{5}$. The least rationalizing factor is $\sqrt{5}$ because $\sqrt{5} \times \sqrt{5} = 5$.
If (1 - 2k , k) is a solution of 10x - 9y = 12, then find the value of k.
Substitute $x = 1-2k$ and $y = k$ into the equation:
$10(1-2k) - 9(k) = 12$
$10 - 20k - 9k = 12$
$10 - 29k = 12$
$-29k = 2$
$k = -\frac{2}{29}$
Find the coordinates of the point:
(i) which lies on the y-axis at a distance of 17 units from the x-axis. $\implies$ (0, 17) or (0, -17) (Assuming both directions)
(ii) which lies on the y-axis at a distance of 9 units upwards from the x-axis? $\implies$ (0, 9)
OR
(i) Point of intersection of the coordinate axis is called origin and its coordinates are (0,0).
(ii) Point (-7,0) lies on the negative direction of x axis.
The volume of a cone is 18480 cm³. If the height of the cone is 40 cm. Find the radius of the cone.
Volume of cone, $V = \frac{1}{3}\pi r^2 h$
$\frac{1}{3} \times \frac{22}{7} \times r^2 \times 40 = 18480$
$r^2 = \frac{18480 \times 3 \times 7}{22 \times 40}$
$r^2 = \frac{18480 \times 21}{880}$ (Since $3 \times 7 =21$)
$r^2 = 21 \times 21$ (Since $18480/880 = 21$)
$r^2 = 441 \implies r = 21$ cm.
OR
Find the diameter of a sphere whose surface area is 154 cm².
Surface area of sphere, $4\pi r^2 = 154$
$4 \times \frac{22}{7} \times r^2 = 154$
$r^2 = \frac{154 \times 7}{4 \times 22} = \frac{154 \times 7}{88}$
$r^2 = \frac{7 \times 7}{4}$ (Since $154/22 = 7$ and then $7/4$? Let's do step by step: $154/22 = 7$, so numerator = $7 \times 7 = 49$, denominator = $4$. So $r^2 = 49/4$)
$r^2 = \frac{49}{4} \implies r = \frac{7}{2} = 3.5$ cm.
Diameter = $2r = 7$ cm.
SECTION C
(3 marks each)
Express $1.\overline{3} + 3.\overline{1}$ in the form of $\frac{p}{q}$
Let $x = 1.\overline{3} = 1.333...$. Multiply by 10: $10x = 13.333...$. Subtract: $10x - x = 13.333... - 1.333... \implies 9x = 12 \implies x = \frac{12}{9} = \frac{4}{3}$.
Let $y = 3.\overline{1} = 3.111...$. Multiply by 10: $10y = 31.111...$. Subtract: $10y - y = 31.111... - 3.111... \implies 9y = 28 \implies y = \frac{28}{9}$.
Sum = $x + y = \frac{4}{3} + \frac{28}{9} = \frac{12}{9} + \frac{28}{9} = \frac{40}{9}$.
So, $1.\overline{3} + 3.\overline{1} = \frac{40}{9}$.
FOR VISUALLY IMPAIRED:
If $x = 3 - 2\sqrt{2}$, find the value $x^{2} + \frac{1}{x^{2}}$.
$\frac{1}{x} = \frac{1}{3 - 2\sqrt{2}} = \frac{3 + 2\sqrt{2}}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})} = \frac{3 + 2\sqrt{2}}{9 - 8} = 3 + 2\sqrt{2}$.
$x + \frac{1}{x} = (3 - 2\sqrt{2}) + (3 + 2\sqrt{2}) = 6$.
We know $(x^2 + \frac{1}{x^2}) = (x + \frac{1}{x})^2 - 2 = 6^2 - 2 = 36 - 2 = 34$.
Factorise the polynomial $f(x) = x^{3} - 6x^{2} + 11x - 6$.
Check factors of the constant term (-6): $x=1$ gives $1 - 6 + 11 - 6 = 0$. So $(x-1)$ is a factor.
Perform polynomial division: $(x^{3} - 6x^{2} + 11x - 6) \div (x-1)$.
Quotient = $x^2 - 5x + 6$.
Factorise the quotient: $x^2 - 5x + 6 = x^2 - 2x - 3x + 6 = x(x-2) -3(x-2) = (x-2)(x-3)$.
Therefore, $f(x) = (x-1)(x-2)(x-3)$.
If a - b = 4 and ab = 21, find $a^{3} - b^{3}$
Use the identity: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.
First, find $a^2 + b^2$ using $(a-b)^2 = a^2 - 2ab + b^2$.
$a^2 + b^2 = (a-b)^2 + 2ab = (4)^2 + 2(21) = 16 + 42 = 58$.
Now, $a^2 + ab + b^2 = (a^2 + b^2) + ab = 58 + 21 = 79$.
Therefore, $a^3 - b^3 = (4) \times (79) = 316$.
If the point (2k−3, k+2) lies on the graph of the equation 2x + 3y + 15 = 0, find the value of k.
Substitute $x = 2k-3$ and $y = k+2$ into the equation:
$2(2k-3) + 3(k+2) + 15 = 0$
$4k - 6 + 3k + 6 + 15 = 0$
$7k + 15 = 0$
$7k = -15 \implies k = -\frac{15}{7}$
OR
Given: $F = \frac{9}{5}C + 32$
(i) C = 30°C. $F = \frac{9}{5} \times 30 + 32 = 9 \times 6 + 32 = 54 + 32 = 86°F$.
(ii) F = 95°F. $95 = \frac{9}{5}C + 32 \implies \frac{9}{5}C = 95 - 32 = 63 \implies C = 63 \times \frac{5}{9} = 7 \times 5 = 35°C$.
(iii) Let the numerically same temperature be $T$. Then $T = \frac{9}{5}T + 32 \implies T - \frac{9}{5}T = 32 \implies \frac{5T - 9T}{5} = 32 \implies \frac{-4T}{5} = 32 \implies -4T = 160 \implies T = -40$.
Yes, there is a temperature. It is $-40°$ (i.e., $-40°C = -40°F$).
(Geometry Question)
Given: $\angle XYZ = 64°$. XY is produced to point P. Ray YQ bisects $\angle ZYP$.
To find: $\angle XYQ$ and reflex $\angle QYP$.
Solution:
$\angle XYZ + \angle ZYP = 180°$ (Linear Pair)
$64° + \angle ZYP = 180° \implies \angle ZYP = 116°$.
YQ bisects $\angle ZYP$, so $\angle ZYQ = \angle QYP = \frac{116°}{2} = 58°$.
$\angle XYQ = \angle XYZ + \angle ZYQ = 64° + 58° = 122°$.
Reflex $\angle QYP = 360° - \angle QYP = 360° - 58° = 302°$.
For Visually Impaired:
Ray OR bisects $\angle POS$, so $\angle ROS = \frac{x}{2}$.
$\angle POS + \angle SOQ = 180°$ (Linear Pair). So $\angle SOQ = 180° - x$.
Ray OT bisects $\angle SOQ$, so $\angle SOT = \frac{180° - x}{2} = 90° - \frac{x}{2}$.
$\angle ROT = \angle ROS + \angle SOT = \frac{x}{2} + (90° - \frac{x}{2}) = 90°$.
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Construction: Join diagonal AC.
Proof:
In $\triangle ABC$, P and Q are midpoints of AB and BC. By Midpoint Theorem, $PQ \parallel AC$ and $PQ = \frac{1}{2}AC$.
In $\triangle ADC$, R and S are midpoints of CD and DA. By Midpoint Theorem, $SR \parallel AC$ and $SR = \frac{1}{2}AC$.
From the above, $PQ \parallel SR$ and $PQ = SR = \frac{1}{2}AC$. So quadrilateral PQRS is a parallelogram. (One pair of opposite sides is parallel and equal).
Similarly, join diagonal BD. In $\triangle BCD$, Q and R are midpoints of BC and CD. So $QR = \frac{1}{2}BD$.
In a rectangle, diagonals are equal, i.e., $AC = BD$.
Therefore, $PQ = QR = \frac{1}{2}AC = \frac{1}{2}BD$.
So, in parallelogram PQRS, we have $PQ = QR$. A parallelogram with a pair of adjacent sides equal is a rhombus.
Hence, PQRS is a rhombus.
SECTION D
(5 marks each)
(1) Prove that in an isosceles triangle, angles opposite to equal sides are equal.
Given: $\triangle ABC$ with $AB = AC$.
To Prove: $\angle B = \angle C$.
Construction: Draw the bisector AD of $\angle A$ meeting BC at D.
Proof:
Statement Reason In $\triangle ABD$ and $\triangle ACD$ 1. $AB = AC$ Given 2. $\angle BAD = \angle CAD$ By construction (AD is bisector) 3. $AD = AD$ Common side $\triangle ABD \cong \triangle ACD$ By SAS congruence rule $\angle B = \angle C$ Corresponding parts of congruent triangles (CPCT) (2) Prove that each angle of an equilateral triangle is 60°.
Given: $\triangle ABC$ is equilateral, so $AB = BC = AC$.
To Prove: $\angle A = \angle B = \angle C = 60°$.
Proof:
Since $AB = AC$, $\angle B = \angle C$ (Angles opposite equal sides).
Since $AB = BC$, $\angle A = \angle C$ (Angles opposite equal sides).
Therefore, $\angle A = \angle B = \angle C$.
$\angle A + \angle B + \angle C = 180°$ (Angle sum property of a triangle).
$3\angle A = 180° \implies \angle A = 60°$.
Hence, $\angle A = \angle B = \angle C = 60°$.
Prove that "The angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the remaining part of the circle."
Case 1: When the arc PQ is minor. (Center O lies inside angle PAQ)
Given: Arc PQ of a circle with center O. Point A is on the remaining part of the circle.
To Prove: $\angle POQ = 2\angle PAQ$.
Construction: Join AO and extend it to B.
Proof:
In $\triangle AOP$, $OA = OP$ (radii). So $\angle OAP = \angle OPA$.
$\angle POB = \angle OAP + \angle OPA$ (Exterior angle property).
So $\angle POB = 2\angle OAP$.
Similarly, in $\triangle AOQ$, $OA = OQ$ (radii). So $\angle OAQ = \angle OQA$.
$\angle QOB = \angle OAQ + \angle OQA$ (Exterior angle property).
So $\angle QOB = 2\angle OAQ$.
Adding, $\angle POB + \angle QOB = 2(\angle OAP + \angle OAQ)$.
Hence, $\angle POQ = 2\angle PAQ$.
Case 2: When arc PQ is a semicircle (Center O lies on chord PQ/Angle PAQ is in a semicircle)
Similarly, by drawing the diameter, we can prove that the angle at the center (which is 180°) is twice the angle at the circumference (90°).
Case 3: When arc PQ is major (Center O lies outside angle PAQ)
Similarly, by drawing the diameter, we can prove the result.
OR (Application Problem)
Let Ishita (I), Juhi (J), and Kajol (K) be on the circle of radius 20 m. IJ = JK = 24 m. We need to find IK.
Let the center be O. Drop perpendiculars OM and ON from O to chords IJ and JK respectively. Since IJ = JK, OM = ON.
In right $\triangle OMJ$, $MJ = 24/2 = 12$ m, radius OJ = 20 m.
$OM = \sqrt{OJ^2 - MJ^2} = \sqrt{20^2 - 12^2} = \sqrt{400 - 144} = \sqrt{256} = 16$ m.
Since OM = ON = 16 m, and $OJ = OK = 20$ m, we can find $NJ = 12$ m (same as MJ). So $\triangle OMJ \cong \triangle ONJ$.
$\angle MJO = \angle NJO$. In $\triangle OMJ$, $\sin(\angle MJO) = OM/OJ = 16/20 = 4/5$.
$\angle IJK = \angle MJO + \angle NJO = 2\angle MJO$. We need chord IK.
Alternatively, we can find $\angle IOJ$ and $\angle JOK$. $\cos \angle MOJ = OM/OJ = 16/20 = 4/5$. So $\angle MOJ = \cos^{-1}(4/5)$. Then $\angle IOJ = 2\angle MOJ$.
$\angle IOK = \angle IOJ + \angle JOK = 4\angle MOJ$.
We can use the cosine rule in $\triangle IOK$. A simpler approach: In $\triangle OIM$, $IM=12$, $OM=16$. $\tan(\angle IOM) = IM/OM = 12/16 = 3/4$. So $\angle IOM = \tan^{-1}(3/4)$. Then $\angle IOJ = 2\angle IOM$.
This seems complex. Let's use the property that the line joining the center to the midpoint of a chord is perpendicular.
Consider quadrilateral OMJN. It's a square? OM = ON = 16, MJ = NJ = 12. So it's not a square.
We can find the angle subtended by the chord at the center using the chord length formula: chord length = $2r\sin(\theta/2)$, where $\theta$ is the angle at the center.
$24 = 2 \times 20 \times \sin(\theta/2) \implies \sin(\theta/2) = 24/40 = 3/5 = 0.6$.
So $\theta/2 = \sin^{-1}(0.6) \approx 36.87°$. Thus $\angle IOJ = \angle JOK \approx 73.74°$.
Then $\angle IOK = 73.74° + 73.74° = 147.48°$.
Now, apply the chord length formula for IK: $IK = 2 \times 20 \times \sin(\angle IOK/2) = 40 \times \sin(147.48°/2) = 40 \times \sin(73.74°)$.
Since $\sin(73.74°) = \cos(90°-73.74°) = \cos(16.26°)$. Also, $\sin(\theta/2) = 0.6$, $\cos(\theta/2) = \sqrt{1-0.6^2}=0.8$. Then $\sin \theta = 2 \sin(\theta/2)\cos(\theta/2) = 2 \times 0.6 \times 0.8 = 0.96$. So $\sin 73.74° = 0.96$.
$IK = 40 \times 0.96 = 38.4$ m.
So, the distance between Ishita and Kajol is 38.4 m.
Construct a histogram and frequency polygon for the following data:
Note for construction: Since the class intervals are of varying widths, the histogram must be drawn using frequency density (frequency per unit width) to maintain the principle that the area of the rectangle is proportional to the frequency. The width of the first class (30-60) is 30, second (60-100) is 40, etc. To draw the histogram, we need to adjust the heights.
Let's take the smallest class width (e.g., 20 for 100-120) as the base width for adjustment. Adjusted frequency = (Frequency / Class Width) * Base Width.
Taking base width = 20.
30-60 (width 30): Adjusted Height = (5/30)*20 = 3.33
60-100 (width 40): Adjusted Height = (4/40)*20 = 2
100-120 (width 20): Adjusted Height = (3/20)*20 = 3
120-140 (width 20): Adjusted Height = (2/20)*20 = 2
140-180 (width 40): Adjusted Height = (4/40)*20 = 2
180-200 (width 20): Adjusted Height = (3/20)*20 = 3
200-240 (width 40): Adjusted Height = (4/40)*20 = 2
240-250 (width 10): Adjusted Height = (5/10)*20 = 10
The histogram is drawn with these adjusted heights.
For the frequency polygon, find the midpoints of each class and plot the frequency against the midpoint. For the polygon to close, we can add two imaginary classes with frequency 0 at the beginning and end (e.g., 0-30 and 250-280). The midpoints are: 45, 80, 110, 130, 160, 190, 220, 245.
The radius of the internal and external surfaces of a hollow spherical shell is 3 cm and 5 cm respectively. If it is melted and recast into a solid cone of height $2\frac{2}{3}$ cm, find the diameter of the cone.
Volume of hollow spherical shell = Volume of external sphere - Volume of internal sphere.
$V_{shell} = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (R^3 - r^3)$
$V_{shell} = \frac{4}{3}\pi (5^3 - 3^3) = \frac{4}{3}\pi (125 - 27) = \frac{4}{3}\pi \times 98$ cm³.
Height of cone, $h = 2\frac{2}{3} = \frac{8}{3}$ cm.
Volume of cone, $V_{cone} = \frac{1}{3}\pi r_c^2 h$, where $r_c$ is the radius of the cone.
On melting, volume remains the same. So $V_{cone} = V_{shell}$.
$\frac{1}{3}\pi r_c^2 \times \frac{8}{3} = \frac{4}{3}\pi \times 98$
Cancel $\frac{1}{3}\pi$ from both sides: $r_c^2 \times \frac{8}{3} = 4 \times 98$
$r_c^2 = \frac{4 \times 98 \times 3}{8} = \frac{98 \times 3}{2} = 49 \times 3 = 147$
$r_c = \sqrt{147} = \sqrt{49 \times 3} = 7\sqrt{3}$ cm.
Diameter of the cone = $2r_c = 14\sqrt{3}$ cm.
SECTION E
(Case Study Questions)
Beti Bachao, Beti Padhao
Teachers = $x + y = 10$ ...(1)
Girls = $x^2 + y^2 = 58$ ...(2)
(i) Identity for $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$
(ii) Find the value of $xy$.
From (1) and (2), we know $(x+y)^2 = x^2 + y^2 + 2xy$
$10^2 = 58 + 2xy$
$100 = 58 + 2xy$
$2xy = 42 \implies xy = 21$.
(iii) Find the number of boys.
Boys = $x^3 + y^3 = (x+y)(x^2 + y^2 - xy)$
$x^3 + y^3 = (10)(58 - 21) = 10 \times 37 = 370$.
So, number of boys = 370.
OR (Find the values of x and y)
We have $x+y=10$ and $xy=21$.
The quadratic equation whose roots are x and y is $t^2 - (x+y)t + xy = 0 \implies t^2 - 10t + 21 = 0$.
$(t-3)(t-7)=0$. So $t=3$ or $t=7$.
Therefore, the values of $x$ and $y$ are 3 and 7 (in any order).
Circular Park
Radius of park, $r = 20$ m. Three girls sit at equal distances on the boundary. So they form an equilateral triangle inscribed in the circle.
(1) Circumference of the park = $2\pi r = 2 \times 3.14 \times 20 = 40\pi$ m. (Exact: $40\pi$ m)
(2) Length of the string of each toy telephone = distance between two girls = side of the equilateral triangle.
Relationship between side of equilateral triangle ($a$) and circumradius ($R$): $R = \frac{a}{\sqrt{3}}$.
$a = R \times \sqrt{3} = 20\sqrt{3}$ m.
So, the length of each string = $\mathbf{20\sqrt{3}}$ m.
(3) Area of the equilateral triangle formed.
Area = $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times (20\sqrt{3})^2$
$= \frac{\sqrt{3}}{4} \times 400 \times 3 = \sqrt{3} \times 100 \times 3 = 300\sqrt{3}$ m².
OR (3) Height of the triangle.
Height of equilateral triangle = $\frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} \times 20\sqrt{3} = \frac{\sqrt{3} \times 20\sqrt{3}}{2} = \frac{20 \times 3}{2} = 10 \times 3 = 30$ m.
Triangular Wall
Sides of the triangular wall: $a = 122$ m, $b = 22$ m, $c = 120$ m.
(1) Perimeter of the wall = $a + b + c = 122 + 22 + 120 = 264$ m.
(2) Area of the triangular wall.
Semi-perimeter, $s = \frac{264}{2} = 132$ m.
Using Heron's formula: Area = $\sqrt{s(s-a)(s-b)(s-c)}$
Area = $\sqrt{132 \times (132-122) \times (132-22) \times (132-120)}$
Area = $\sqrt{132 \times 10 \times 110 \times 12}$
$= \sqrt{132 \times 10 \times 110 \times 12}$
Simplify: $132 = 11 \times 12$, $110 = 11 \times 10$.
So, inside sqrt: $(11 \times 12) \times 10 \times (11 \times 10) \times 12 = 11^2 \times 12^2 \times 10^2$.
Area = $\sqrt{(11 \times 12 \times 10)^2} = 11 \times 12 \times 10 = 1320$ m².
(3) Rent paid by the company for 3 months.
Earning = ₹ 50000 per 10 m² per year.
Rate per m² per year = $\frac{50000}{10} = ₹ 5000$ per year.
Rate per m² per month = $\frac{5000}{12}$.
Area of wall = $1320$ m².
Rent for 3 months = Area $\times$ Rate per m² per month $\times$ 3
$= 1320 \times \frac{5000}{12} \times 3 = 1320 \times 5000 \times \frac{3}{12}$
$= 1320 \times 5000 \times \frac{1}{4} = 1320 \times 1250 = ₹ 16,50,000$.
OR (3) Corresponding height if base is 122 m.
Area of triangle = $\frac{1}{2} \times base \times height$
$1320 = \frac{1}{2} \times 122 \times h$
$h = \frac{1320 \times 2}{122} = \frac{2640}{122} = \frac{1320}{61} \approx 21.64$ m
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