Virudhunagar District
Common Quarterly Examination - September 2025
Standard 10 Mathematics - Complete Solutions
Part - I (Marks: 14)
Choose the best answer: (14×1=14)
1) If \(n(A \times B) = 6\) and \(A = \{1, 3\}\) then \(n(B)\) is
Answer:
c) 3
Explanation: Given \(n(A \times B) = 6\) and \(A = \{1, 3\}\). The number of elements in set A is \(n(A) = 2\). We know that \(n(A \times B) = n(A) \times n(B)\). So, \(6 = 2 \times n(B)\), which gives \(n(B) = \frac{6}{2} = 3\).2) If \(f(x) = x^m\) and \(g(x) = x^n\) does fog = ?
Answer:
c) \(x^{mn}\)
Explanation: \(fog(x) = f(g(x)) = f(x^n) = (x^n)^m = x^{mn}\).3) Using Euclid's division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are
Answer:
a) 0, 1, 8
Explanation: Any positive integer can be written in the form \(3q, 3q+1,\) or \(3q+2\). - \((3q)^3 = 27q^3 = 9(3q^3)\). Remainder is 0. - \((3q+1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9(\dots) + 1\). Remainder is 1. - \((3q+2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9(\dots) + 8\). Remainder is 8. The possible remainders are 0, 1, and 8.4) The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
Answer:
d) 2520
Explanation: We need to find the LCM of numbers from 1 to 10. LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) = \(2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520\).5) The value of \((1^3+2^3+3^3+...+15^3) - (1+2+3+...+15)\) is
Answer:
c) 14280
Explanation: We use the formulas \( \sum n^3 = \left(\frac{n(n+1)}{2}\right)^2 \) and \( \sum n = \frac{n(n+1)}{2} \). For n=15, \( \sum_{k=1}^{15} k^3 = \left(\frac{15 \times 16}{2}\right)^2 = 120^2 = 14400 \). And \( \sum_{k=1}^{15} k = \frac{15 \times 16}{2} = 120 \). The value is \(14400 - 120 = 14280\).6) Which of the following should be added to make \(x^4+64\) a perfect square?
Answer:
b) \(16x^2\)
Explanation: \(x^4+64 = (x^2)^2 + (8)^2\). To make it a perfect square of the form \((a+b)^2 = a^2+2ab+b^2\), we need to add the term \(2ab\). Here \(a=x^2\) and \(b=8\). So, \(2ab = 2(x^2)(8) = 16x^2\).7) Graph of a linear equation is a ____
Answer:
a) Straight line
Explanation: A linear equation of the form \(ax+by+c=0\) always represents a straight line on a graph.8) What is the value of x in \(3\sqrt{x} = 9\)?
Answer:
b) 9
Explanation: \(3\sqrt{x} = 9 \implies \sqrt{x} = \frac{9}{3} = 3\). Squaring both sides, \((\sqrt{x})^2 = 3^2 \implies x = 9\).9) If \(\triangle ABC\) is an isosceles triangle with \(\angle C = 90^\circ\) and AC = 5 cm, then AB is
Answer:
d) \(5\sqrt{2}\) cm
Explanation: In an isosceles right-angled triangle, if \(\angle C = 90^\circ\), then \(AC = BC = 5\) cm. By Pythagoras theorem, \(AB^2 = AC^2 + BC^2 = 5^2 + 5^2 = 50\). Therefore, \(AB = \sqrt{50} = 5\sqrt{2}\) cm.10) If in \(\triangle ABC\), DE || BC. AB = 3.6 cm, AC = 2.4 cm and AD = 2.1 cm then the length of AE is
Answer:
a) 1.4 cm
Explanation: By Basic Proportionality Theorem (Thales Theorem), if DE || BC, then \(\frac{AD}{AB} = \frac{AE}{AC}\). \(\frac{2.1}{3.6} = \frac{AE}{2.4} \implies AE = \frac{2.1 \times 2.4}{3.6} = 1.4\) cm.11) The slope of the line which is perpendicular to a line joining the points (0,0) and (-8, 8) is
Answer:
b) 1
Explanation: The slope of the line joining (0,0) and (-8,8) is \(m_1 = \frac{8-0}{-8-0} = -1\). The slope of the perpendicular line is \(m_2 = -\frac{1}{m_1} = -\frac{1}{-1} = 1\).12) The point of intersection of \(x-y = 4\) and \(x+y = 8\) is
Answer:
c) (6, 2)
Explanation: Adding the two equations: \((x-y) + (x+y) = 4+8 \implies 2x = 12 \implies x = 6\). Substituting \(x=6\) into the second equation: \(6+y=8 \implies y=2\). The intersection point is (6, 2).13) \(\tan\theta \text{cosec}^2\theta - \tan\theta\) is equal to
Answer:
d) \(\cot\theta\)
Explanation: \(\tan\theta(\text{cosec}^2\theta - 1)\). We know \(1+\cot^2\theta = \text{cosec}^2\theta\), so \(\text{cosec}^2\theta - 1 = \cot^2\theta\). The expression becomes \(\tan\theta \times \cot^2\theta = \tan\theta \times \frac{1}{\tan^2\theta} = \frac{1}{\tan\theta} = \cot\theta\).14) When will the values of \(\sin\theta\) and \(\cos\theta\) be equal?
Answer:
d) \(\theta = 45^\circ\)
Explanation: For \(\theta = 45^\circ\), \(\sin 45^\circ = \frac{1}{\sqrt{2}}\) and \(\cos 45^\circ = \frac{1}{\sqrt{2}}\).Part - II (Marks: 20)
Answer any 10 questions. [Q.No. 28 is compulsory] (10×2=20)
15) If B×A = {(-2, 3), (-2, 4), (0, 3), (0, 4), (3, 3), (3, 4)} find A and B.
Solution:
Set A is the set of all second elements: A = {3, 4}.
Set B is the set of all first elements: B = {-2, 0, 3}.
Set B is the set of all first elements: B = {-2, 0, 3}.
16) A relation f : X→Y is defined by f(x) = x²-2 where, X = {-2, -1, 0, 3} and Y = R. i) List the elements of f. ii) Is f a function?
Solution:
f(-2) = (-2)² - 2 = 2
f(-1) = (-1)² - 2 = -1
f(0) = (0)² - 2 = -2
f(3) = (3)² - 2 = 7
i) The elements of f are {(-2, 2), (-1, -1), (0, -2), (3, 7)}.
ii) Yes, f is a function because every element in the domain X has a unique image in the codomain Y.
f(-1) = (-1)² - 2 = -1
f(0) = (0)² - 2 = -2
f(3) = (3)² - 2 = 7
i) The elements of f are {(-2, 2), (-1, -1), (0, -2), (3, 7)}.
ii) Yes, f is a function because every element in the domain X has a unique image in the codomain Y.
17) Let f be a function from R to R defined by f(x) = 3x-5, find the values of a and b given that (a, 4) and (1, b) belong to f.
Solution:
Given f(x) = 3x - 5.
For (a, 4): f(a) = 4 ⇒ 3a - 5 = 4 ⇒ 3a = 9 ⇒ a = 3.
For (1, b): f(1) = b ⇒ b = 3(1) - 5 ⇒ b = -2.
Thus, a = 3 and b = -2.
For (a, 4): f(a) = 4 ⇒ 3a - 5 = 4 ⇒ 3a = 9 ⇒ a = 3.
For (1, b): f(1) = b ⇒ b = 3(1) - 5 ⇒ b = -2.
Thus, a = 3 and b = -2.
18) Find the 4 digit pin number 'pqrs' of an ATM card such that p²xq¹xr⁴xs³ = 3,15,000.
Solution:
Prime factorize 3,15,000:
3,15,000 = 315 × 1000 = (3² × 5 × 7) × (10³) = (3² × 5 × 7) × (2 × 5)³ = 3² × 5 × 7 × 2³ × 5³ = 2³ × 3² × 5⁴ × 7¹.
Comparing p²q¹r⁴s³ with 3² × 7¹ × 5⁴ × 2³:
p² = 3² ⇒ p=3
q¹ = 7¹ ⇒ q=7
r⁴ = 5⁴ ⇒ r=5
s³ = 2³ ⇒ s=2
The pin 'pqrs' is 3752.
3,15,000 = 315 × 1000 = (3² × 5 × 7) × (10³) = (3² × 5 × 7) × (2 × 5)³ = 3² × 5 × 7 × 2³ × 5³ = 2³ × 3² × 5⁴ × 7¹.
Comparing p²q¹r⁴s³ with 3² × 7¹ × 5⁴ × 2³:
p² = 3² ⇒ p=3
q¹ = 7¹ ⇒ q=7
r⁴ = 5⁴ ⇒ r=5
s³ = 2³ ⇒ s=2
The pin 'pqrs' is 3752.
19) Find the 8th term of the G.P. 9, 3, 1, ...
Solution:
Here, first term a = 9, common ratio r = 3/9 = 1/3.
The 8th term is \(t_8 = ar^{8-1} = 9 \times (\frac{1}{3})^7 = 3^2 \times \frac{1}{3^7} = \frac{1}{3^5} = \frac{1}{243}\).
The 8th term is \(t_8 = ar^{8-1} = 9 \times (\frac{1}{3})^7 = 3^2 \times \frac{1}{3^7} = \frac{1}{3^5} = \frac{1}{243}\).
20) Solve: 2x-3y = 6, x+y = 1.
Solution:
(1) 2x - 3y = 6
(2) x + y = 1 ⇒ x = 1 - y
Substitute (2) into (1): 2(1 - y) - 3y = 6 ⇒ 2 - 2y - 3y = 6 ⇒ -5y = 4 ⇒ y = -4/5.
Substitute y back into (2): x + (-4/5) = 1 ⇒ x = 1 + 4/5 ⇒ x = 9/5.
Solution is x = 9/5, y = -4/5.
(2) x + y = 1 ⇒ x = 1 - y
Substitute (2) into (1): 2(1 - y) - 3y = 6 ⇒ 2 - 2y - 3y = 6 ⇒ -5y = 4 ⇒ y = -4/5.
Substitute y back into (2): x + (-4/5) = 1 ⇒ x = 1 + 4/5 ⇒ x = 9/5.
Solution is x = 9/5, y = -4/5.
21) Reduce the rational expression to its lowest form: \(\frac{x^2-1}{x^2+x}\).
Solution:
\(\frac{x^2-1}{x^2+x} = \frac{(x-1)(x+1)}{x(x+1)} = \frac{x-1}{x}\).
22) Determine the nature of roots for the quadratic equation: 2x²-x-10 = 0.
Solution:
The discriminant is \(\Delta = b^2 - 4ac\). Here a=2, b=-1, c=-10.
\(\Delta = (-1)^2 - 4(2)(-10) = 1 + 80 = 81\).
Since \(\Delta = 81 > 0\) and is a perfect square, the roots are real, unequal, and rational.
\(\Delta = (-1)^2 - 4(2)(-10) = 1 + 80 = 81\).
Since \(\Delta = 81 > 0\) and is a perfect square, the roots are real, unequal, and rational.
23) Prove the following identity: \(\sqrt{\frac{1+\sin\theta}{1-\sin\theta}} = \sec\theta + \tan\theta\).
Solution:
LHS = \(\sqrt{\frac{1+\sin\theta}{1-\sin\theta} \times \frac{1+\sin\theta}{1+\sin\theta}} = \sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}} = \sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}} = \frac{1+\sin\theta}{\cos\theta} = \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} = \sec\theta + \tan\theta\) = RHS. Hence proved.
24) In figure ∠A = ∠CED. Prove that ΔCAB ~ ΔCED. Also find the value of x.
Solution:
Proof: In ΔCAB and ΔCED, ∠C is common (∠BCA = ∠DCE) and ∠A = ∠CED (Given). By AA similarity, ΔCAB ~ ΔCED.
Finding x: Since triangles are similar, the ratio of corresponding sides is equal. \(\frac{CB}{CD} = \frac{AB}{ED}\). From the figure, CB = CE+EB = 10+2=12, CD=8, AB=9, ED=x. \(\frac{12}{8} = \frac{9}{x} \implies 12x = 72 \implies x = 6\) cm.
Proof: In ΔCAB and ΔCED, ∠C is common (∠BCA = ∠DCE) and ∠A = ∠CED (Given). By AA similarity, ΔCAB ~ ΔCED.
Finding x: Since triangles are similar, the ratio of corresponding sides is equal. \(\frac{CB}{CD} = \frac{AB}{ED}\). From the figure, CB = CE+EB = 10+2=12, CD=8, AB=9, ED=x. \(\frac{12}{8} = \frac{9}{x} \implies 12x = 72 \implies x = 6\) cm.
25) In the figure AD is the bisector of ∠BAC, if AB = 6 cm, BD = 4 cm and DC = 3 cm. Find AC.
Solution:
By the Angle Bisector Theorem, the internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle.
Therefore, we have:
$$ \frac{AB}{AC} = \frac{BD}{DC} $$Given values are AB = 6 cm, BD = 4 cm, and DC = 3 cm. Substituting these into the formula:
$$ \frac{6}{AC} = \frac{4}{3} $$Now, we cross-multiply to solve for AC:
$$ 4 \times AC = 6 \times 3 $$ $$ 4 \times AC = 18 $$ $$ AC = \frac{18}{4} $$ $$ AC = 4.5 \text{ cm} $$Therefore, the length of AC is 4.5 cm.
26) Find the slope of a line joining the points (5, √5) with the origin.
Solution:
The points are (0, 0) and (5, √5).
Slope \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\sqrt{5} - 0}{5 - 0} = \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}}\).
27) Find the equation of a line passing through the point (3, -4) and having slope -5/7.
Solution:
Using point-slope form \(y - y_1 = m(x - x_1)\).
\(y - (-4) = -\frac{5}{7}(x - 3)\).
\(7(y+4) = -5(x-3) \implies 7y+28 = -5x+15 \implies 5x+7y+13 = 0\).
28) Show that the straight lines 2x+3y-8 = 0 and 4x+6y+18 = 0 are parallel.
Solution:
Slope of the first line \(m_1 = -\frac{\text{coeff of x}}{\text{coeff of y}} = -\frac{2}{3}\).
Slope of the second line \(m_2 = -\frac{4}{6} = -\frac{2}{3}\).
Since \(m_1 = m_2\), the lines are parallel.
Part - III (Marks: 50)
Answer any 10 questions. [Q.No. 42 is compulsory] (10×5=50)
29) A function f is defined by f(x) = 2x-3. Find i) \(\frac{f(0)+f(1)}{2}\), ii) x such that f(x)=0, iii) x such that f(x)=x, iv) x such that f(x)=f(1-x).
Solution:
i) f(0)= -3, f(1)= -1. \(\frac{-3-1}{2} = -2\).
ii) 2x-3=0 ⇒ x=3/2.
iii) 2x-3=x ⇒ x=3.
iv) 2x-3 = 2(1-x)-3 ⇒ 2x-3 = 2-2x-3 ⇒ 4x=2 ⇒ x=1/2.
ii) 2x-3=0 ⇒ x=3/2.
iii) 2x-3=x ⇒ x=3.
iv) 2x-3 = 2(1-x)-3 ⇒ 2x-3 = 2-2x-3 ⇒ 4x=2 ⇒ x=1/2.
30) Let f: A→B be a function defined by f(x) = x/2 - 1, where A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9}. Represent f by i) set of ordered pairs, ii) a table, iii) an arrow diagram, iv) a graph.
Solution:
f(2)=0, f(4)=1, f(6)=2, f(10)=4, f(12)=5.
i) Ordered pairs: {(2,0), (4,1), (6,2), (10,4), (12,5)}.
ii) Table:
iii) Arrow Diagram:
iv) Graph: Plot the points (2,0), (4,1), (6,2), (10,4), (12,5) on a graph paper.
i) Ordered pairs: {(2,0), (4,1), (6,2), (10,4), (12,5)}.
ii) Table:
iii) Arrow Diagram:
iv) Graph: Plot the points (2,0), (4,1), (6,2), (10,4), (12,5) on a graph paper.
31) If f(x) = 3x-2, g(x) = 2x+k and if fog = gof, then find the value of k.
Solution:
fog(x) = f(g(x)) = 3(2x+k) - 2 = 6x+3k-2.
gof(x) = g(f(x)) = 2(3x-2) + k = 6x-4+k.
6x+3k-2 = 6x-4+k ⇒ 2k = -2 ⇒ k = -1.
gof(x) = g(f(x)) = 2(3x-2) + k = 6x-4+k.
6x+3k-2 = 6x-4+k ⇒ 2k = -2 ⇒ k = -1.
32) How many terms of the series 1+5+9+... must be taken so that their sum is 190?
Solution:
This is an AP with a=1, d=4, Sₙ=190.
\(190 = \frac{n}{2}[2(1) + (n-1)4] \implies 380 = n(2+4n-4) = 4n^2-2n\).
\(4n^2-2n-380 = 0 \implies 2n^2-n-190=0\).
\((2n+19)(n-10)=0\). Since n must be positive, n=10.
33) Find the sum to n terms of the series 3+33+333+...+ to n terms.
Solution:
Sₙ = 3(1+11+111+...) = \(\frac{3}{9}(9+99+999+...)\)
= \(\frac{1}{3}[(10-1)+(100-1)+(1000-1)+...]\)
= \(\frac{1}{3}[(10+10^2+...)-(1+1+...)]\)
= \(\frac{1}{3}\left[\frac{10(10^n-1)}{9} - n\right]\).
34) Rekha has 15 square colour papers of sizes 10 cm, 11 cm, ..., 24 cm. How much area can be decorated with these colour papers?
Solution:
Total area = \(10^2+11^2+...+24^2 = (1^2+...+24^2) - (1^2+...+9^2)\).
Using \(\sum n^2 = \frac{n(n+1)(2n+1)}{6}\),
Area = \(\frac{24(25)(49)}{6} - \frac{9(10)(19)}{6} = 4900 - 285 = 4615\) cm².
35) Solve: 6x+2y-5z = 13, 3x+3y-2z = 13, 7x+5y-3z = 26.
Solution:
(1) 6x + 2y - 5z = 13
(2) 3x + 3y - 2z = 13
(3) 7x + 5y - 3z = 26
2×(2) - (1) ⇒ (6x+6y-4z) - (6x+2y-5z) = 26-13 ⇒ 4y+z = 13 (4)
7×(2) - 3×(3) ⇒ (21x+21y-14z) - (21x+15y-9z) = 91-78 ⇒ 6y-5z = 13 (5)
From (4), z = 13-4y. Sub into (5): 6y - 5(13-4y) = 13 ⇒ 26y = 78 ⇒ y=3.
Then z = 13-4(3) = 1. Sub y,z into (2): 3x+3(3)-2(1)=13 ⇒ 3x=6 ⇒ x=2.
Solution: x=2, y=3, z=1.
(2) 3x + 3y - 2z = 13
(3) 7x + 5y - 3z = 26
2×(2) - (1) ⇒ (6x+6y-4z) - (6x+2y-5z) = 26-13 ⇒ 4y+z = 13 (4)
7×(2) - 3×(3) ⇒ (21x+21y-14z) - (21x+15y-9z) = 91-78 ⇒ 6y-5z = 13 (5)
From (4), z = 13-4y. Sub into (5): 6y - 5(13-4y) = 13 ⇒ 26y = 78 ⇒ y=3.
Then z = 13-4(3) = 1. Sub y,z into (2): 3x+3(3)-2(1)=13 ⇒ 3x=6 ⇒ x=2.
Solution: x=2, y=3, z=1.
36) Find the square root of 64x⁴-16x³+17x²-2x+1.
Solution:
We can find the square root using the long division method as follows:
8x² -x +1
_________________________
8x² | 64x⁴ -16x³ +17x² -2x +1
- (64x⁴)
_________________________
16x²-x | -16x³ +17x²
- (-16x³ + x²)
_________________________
16x²-2x+1 | 16x² -2x +1
- ( 16x² -2x +1)
_________________________
0
Therefore, the square root of the given polynomial is \(|8x^2 - x + 1|\).
Answer: \(|8x^2 - x + 1|\).37) If one root of the equation 2y²-ay+64 = 0 is twice the other then find the values of a.
Solution:
Let the roots be α and 2α. Sum of roots: α+2α = 3α = -(-a)/2 = a/2. Product of roots: α(2α) = 2α² = 64/2 = 32.
From product, α² = 16 ⇒ α = ±4.
If α=4, 3(4) = a/2 ⇒ a=24.
If α=-4, 3(-4) = a/2 ⇒ a=-24.
Values of a are ±24.
38) State and prove Angle Bisector Theorem.
Solution:
Statement: The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle.
Proof:
Proof:
39) Find the area of the quadrilateral whose vertices are at (-3, -8), (6, -6), (4, 2) and (-8, 2).
Solution:
Order the points anti-clockwise: A(-3, -8), D(-8, 2), C(4, 2), B(6, -6).
Area = \(\frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|\)
= \(\frac{1}{2} |((-6) + (-16) + (-24) + (-48)) - (64 + 8 + 12 + 18)|\)
= \(\frac{1}{2} |(-94) - (102)| = \frac{1}{2} |-196| = 98\) sq. units.
40) If the points A(2, 2), B(-2, -3), C(1, -3) and D(x, y) form a parallelogram then find the value of x and y.
Solution:
Midpoint of diagonal AC = Midpoint of diagonal BD.
Midpoint of AC = \((\frac{2+1}{2}, \frac{2-3}{2}) = (\frac{3}{2}, -\frac{1}{2})\).
Midpoint of BD = \((\frac{-2+x}{2}, \frac{-3+y}{2})\).
Equating coordinates: \(\frac{-2+x}{2}=\frac{3}{2} \Rightarrow x=5\). And \(\frac{-3+y}{2}=-\frac{1}{2} \Rightarrow y=2\).
41) Find the equation of a straight line passing through the point P(-5, 2) and parallel to the line joining the points Q(3, -2) and R(-5, 4).
Solution:
Slope of QR, \(m = \frac{4-(-2)}{-5-3} = \frac{6}{-8} = -\frac{3}{4}\).
The parallel line has the same slope. Using point-slope form with P(-5,2):
\(y-2 = -\frac{3}{4}(x - (-5)) \implies 4y-8 = -3x-15 \implies 3x+4y+7=0\).
42) If sinθ+cosθ = √3 then prove tan 3θ = \(\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\).
Solution:
Note: The condition sinθ+cosθ = √3 is impossible, as the maximum value is √2. The question likely intends to just prove the identity.
The identity \( \tan(3\theta) = \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta} \) is a standard trigonometric formula.
Proof: \( \tan(3\theta) = \tan(2\theta+\theta) = \frac{\tan2\theta+\tan\theta}{1-\tan2\theta\tan\theta} \). Substitute \( \tan2\theta = \frac{2\tan\theta}{1-\tan^2\theta} \) and simplify to get the required expression.
Part - IV (Marks: 16)
Answer the following: (2×8=16)
43) Draw a triangle ABC of base BC = 8 cm, ∠A = 60° and the bisector of ∠A meets BC at D such that BD = 6 cm. (OR) Construct a triangle similar to a given triangle PQR with its sides equal to 7/3 of the corresponding sides of the triangle PQR (scale factor 7/3 > 1).
Solution for Triangle Construction:
Solution for Similar Triangle Construction:
Steps of Construction:
Solution for Similar Triangle Construction:
Steps of Construction:
Since the scale factor is \(\frac{7}{3}\), which is greater than 1, the new triangle will be larger than the original triangle PQR.
- Draw a triangle PQR with any suitable measurements.
- Draw a ray QX making an acute angle with QR, on the side opposite to vertex P.
- Locate 7 points \(Q_1, Q_2, Q_3, Q_4, Q_5, Q_6, Q_7\) on the ray QX such that the distances between them are equal (\(QQ_1 = Q_1Q_2 = \dots = Q_6Q_7\)).
- Join \(Q_3\) (the 3rd point, as 3 is the denominator) to R.
- Draw a line through \(Q_7\) parallel to \(Q_3R\). This line will intersect the extended line segment QR at a point R'.
- Draw a line through R' parallel to PR. This line will intersect the extended line segment QP at a point P'.
- \(\triangle P'QR'\) is the required similar triangle, with each side being \(\frac{7}{3}\) times the corresponding side of \(\triangle PQR\).
44) Draw the graph of xy = 24, x, y > 0. Using the graph find, (i) y when x = 3 and (ii) x when y = 6. (OR) A bus is travelling at a uniform speed 50 km/hr. Draw the time-distance graph and hence find i) the constant of variation ii) how far will it travel in 90 minutes? iii) the time required to cover a distance of 300 km from the graph.
Solution for xy=24 Graph:
Solution for Time-Distance Graph:
From the graph: (i) When x=3, y=8. (ii) When y=6, x=4.
Solution for Time-Distance Graph:
i) Constant of variation (speed) k = y/x = 50 km/hr.
ii) In 90 min (1.5 hrs), distance = 1.5 * 50 = 75 km.
iii) Time to cover 300 km = Distance/Speed = 300/50 = 6 hours.