10th Maths Quarterly Exam 2024 Question Paper & Solutions
QUARTERLY EXAMINATION - 2024
Subject: MATHEMATICS | Marks: 100 | Time: 3.00 Hours
Part I: Choose the best answer (14 x 1 = 14)
1. If there are 1024 relations from a Set A = {1,2,3,4,5} to a set B, then the number of elements in B is
Solution:
Given, n(A) = 5.
Number of relations from A to B is \(2^{n(A) \times n(B)}\).
We are given that the number of relations is 1024.
So, \(2^{n(A) \times n(B)} = 1024\).
We know that \(1024 = 2^{10}\).
Therefore, \(2^{5 \times n(B)} = 2^{10}\).
Equating the powers, \(5 \times n(B) = 10\).
\(n(B) = \frac{10}{5} = 2\).
The number of elements in B is 2.
Answer: b) 2
2. If the ordered pairs (a+2,4) and (5, 2a+b) are equal then (a,b) is
Solution:
Given that the ordered pairs are equal: (a+2, 4) = (5, 2a+b).
Equating the corresponding elements:
a + 2 = 5 => a = 5 - 2 => a = 3.
4 = 2a + b.
Substitute a = 3 in the second equation:
4 = 2(3) + b => 4 = 6 + b => b = 4 - 6 => b = -2.
So, (a,b) is (3, -2).
Answer: d) (3,-2)
3. If \(f(x)=2x^2\) and \(g(x)=\frac{1}{3x}\), then fog is
Solution:
\(f(x) = 2x^2\), \(g(x) = \frac{1}{3x}\).
\(fog(x) = f(g(x))\).
Substitute g(x) into f(x):
\(f(g(x)) = f(\frac{1}{3x}) = 2\left(\frac{1}{3x}\right)^2 = 2\left(\frac{1}{9x^2}\right) = \frac{2}{9x^2}\).
Answer: c) \(\frac{2}{9x^2}\)
4. Using Euclid's division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are,
Solution:
Let 'n' be any positive integer. By Euclid's division lemma, n can be of the form 3q, 3q+1, or 3q+2.
Case 1: n = 3q. \(n^3 = (3q)^3 = 27q^3 = 9(3q^3)\). Remainder is 0.
Case 2: n = 3q+1. \(n^3 = (3q+1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3+3q^2+q) + 1\). Remainder is 1.
Case 3: n = 3q+2. \(n^3 = (3q+2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3+6q^2+4q) + 8\). Remainder is 8.
The possible remainders are 0, 1, 8.
Answer: a) 0,1,8
5. The sum of exponents of the prime factors in the prime factorization of 1729 is
Solution:
Prime factorization of 1729:
1729 is divisible by 7: \(1729 = 7 \times 247\).
247 is divisible by 13: \(247 = 13 \times 19\).
So, \(1729 = 7^1 \times 13^1 \times 19^1\).
The exponents of the prime factors are 1, 1, and 1.
Sum of exponents = 1 + 1 + 1 = 3.
Answer: c) 3
6. The next term of the sequence \(\frac{1}{16}, \frac{1}{8}, \frac{1}{12}, \frac{1}{18}, ...\) is
Solution:
Let's check the ratio between consecutive terms.
\(r_1 = \frac{1/8}{1/16} = 2\).
\(r_2 = \frac{1/12}{1/8} = \frac{8}{12} = \frac{2}{3}\).
\(r_3 = \frac{1/18}{1/12} = \frac{12}{18} = \frac{2}{3}\).
From the second term onwards, the sequence is a Geometric Progression (GP) with a common ratio \(r = \frac{2}{3}\).
The next term is \(\frac{1}{18} \times r = \frac{1}{18} \times \frac{2}{3} = \frac{2}{54} = \frac{1}{27}\).
Answer: b) \(\frac{1}{27}\)
7. If (x-6) is the HCF of \(x^2-2x-24\) and \(x^2-kx-6\), then K is
Solution:
If (x-6) is the HCF, it must be a factor of both polynomials.
For P(x) = \(x^2-kx-6\), P(6) must be 0.
\(P(6) = (6)^2 - k(6) - 6 = 0\).
\(36 - 6k - 6 = 0\).
\(30 - 6k = 0\).
\(30 = 6k\).
\(k = \frac{30}{6} = 5\).
Answer: b) 5
8. Which of the following should be added to make \(x^4+64\) a perfect square
Solution:
We have \(x^4+64 = (x^2)^2 + 8^2\).
This is in the form of \(a^2+b^2\). To make it a perfect square \((a+b)^2 = a^2+2ab+b^2\), we need to add 2ab.
Here, a = \(x^2\) and b = 8.
\(2ab = 2(x^2)(8) = 16x^2\).
Adding \(16x^2\) gives \(x^4 + 16x^2 + 64 = (x^2+8)^2\), which is a perfect square.
Answer: b) \(16x^2\)
9. The solution of \((2x-1)^2 = 9\) is Equal to (Note: The original paper has a typo of \((2x-1)^2=0\), we solve for the likely intended question \((2x-1)^2=9\) or the printed value.)
Solution based on the printed paper \((2x-1)^2 = 0\):
\((2x-1)^2 = 0\)
Taking square root on both sides: \(2x - 1 = 0\).
\(2x = 1\).
\(x = \frac{1}{2}\).
This solution is not among options a, b, or c.
Answer: d) None of these
Solution for likely intended question \((2x-1)^2 = 9\):
\((2x-1)^2 = 9\)
Taking square root on both sides: \(2x - 1 = \pm\sqrt{9}\).
\(2x - 1 = \pm 3\).
Case 1: \(2x - 1 = 3 \Rightarrow 2x = 4 \Rightarrow x = 2\).
Case 2: \(2x - 1 = -3 \Rightarrow 2x = -2 \Rightarrow x = -1\).
The solutions are -1 and 2.
10. If in \(\triangle ABC\), DE || BC, AB = 3.6cm, AC = 2.4 cm, and AD = 2.1 cm then the length of AE is
Solution:
Given DE || BC in \(\triangle ABC\). By Basic Proportionality Theorem (Thales' Theorem), the sides are proportional.
\(\frac{AD}{AB} = \frac{AE}{AC}\).
Given: AB = 3.6 cm, AC = 2.4 cm, AD = 2.1 cm.
\(\frac{2.1}{3.6} = \frac{AE}{2.4}\).
\(AE = \frac{2.1 \times 2.4}{3.6} = \frac{2.1 \times 24}{36} = \frac{2.1 \times 2}{3} = 0.7 \times 2 = 1.4\).
The length of AE is 1.4 cm.
Answer: a) 1.4 cm
11. The Point of intersection of 3x-y=4 and x+y=8 is
Solution:
We have two linear equations:
1) \(3x - y = 4\)
2) \(x + y = 8\)
Add equation (1) and (2):
\((3x - y) + (x + y) = 4 + 8\)
\(4x = 12 \Rightarrow x = 3\).
Substitute x=3 into equation (2):
\(3 + y = 8 \Rightarrow y = 5\).
The point of intersection is (3,5).
Answer: c) (3,5)
12. The straight line given by the equation x=11 is
Solution:
The equation x = c (where c is a constant) represents a vertical line. All points on this line have an x-coordinate of 11. A vertical line is always parallel to the Y-axis.
Answer: b) Parallel to Y-axis
13. The Slope of the line which is perpendicular to a line joining the points (0,0) and (-8,8) is
Solution:
First, find the slope (\(m_1\)) of the line joining (0,0) and (-8,8).
\(m_1 = \frac{y_2-y_1}{x_2-x_1} = \frac{8-0}{-8-0} = \frac{8}{-8} = -1\).
The slope of a line perpendicular to this line (\(m_2\)) satisfies the condition \(m_1 \times m_2 = -1\).
\(-1 \times m_2 = -1 \Rightarrow m_2 = 1\).
Answer: b) 1
14. \(Tan\theta Cosec^2\theta - Tan\theta\) is Equal to
Solution:
Factor out \(Tan\theta\):
\(Tan\theta(Cosec^2\theta - 1)\).
Using the trigonometric identity \(1 + Cot^2\theta = Cosec^2\theta\), we get \(Cosec^2\theta - 1 = Cot^2\theta\).
Substitute this back: \(Tan\theta \times Cot^2\theta\).
Since \(Cot\theta = \frac{1}{Tan\theta}\), we have \(Tan\theta \times \frac{1}{Tan^2\theta} = \frac{1}{Tan\theta} = Cot\theta\).
Answer: d) Cot \(\theta\)
Part II: Answer any 10 from the following (Q.No: 28 is compulsory) (10 x 2 = 20)
15. If AxB = {(3,2), (3,4), (5,2), (5,4)} then find A and B.
Solution:
Set A is the collection of all first elements in the ordered pairs of AxB. A = {3, 5}.
Set B is the collection of all second elements in the ordered pairs of AxB. B = {2, 4}.
16. If \(f(x) = x^2 - 5x + 6\) then evaluate f(2).
Solution:
Given \(f(x) = x^2 - 5x + 6\).
To find f(2), substitute x = 2 in the expression.
\(f(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0\).
17. Find k if fof(k) = 5 Where f(k) = 2k-1.
Solution:
Given f(k) = 2k-1.
fof(k) = f(f(k)) = f(2k-1).
Substitute (2k-1) into f(k):
f(2k-1) = 2(2k-1) - 1 = 4k - 2 - 1 = 4k - 3.
Given fof(k) = 5.
So, 4k - 3 = 5.
4k = 8.
k = 2.
18. If \(800 = 2^a \times 5^b\), then find a and b. (Note: Corrected from OCR)
Solution:
We need to find the prime factorization of 800.
\(800 = 8 \times 100 = 2^3 \times 10^2 = 2^3 \times (2 \times 5)^2 = 2^3 \times 2^2 \times 5^2 = 2^{3+2} \times 5^2 = 2^5 \times 5^2\).
Comparing this with \(2^a \times 5^b\), we get a = 5 and b = 2.
19. Find the sum of 6+13+20+......+97.
Solution:
The given series is an Arithmetic Progression (AP).
First term (a) = 6.
Common difference (d) = 13 - 6 = 7.
Last term (l) = 97.
First, find the number of terms (n):
\(l = a + (n-1)d \Rightarrow 97 = 6 + (n-1)7\).
\(91 = (n-1)7 \Rightarrow n-1 = 13 \Rightarrow n = 14\).
Sum of the series \(S_n = \frac{n}{2}(a+l)\).
\(S_{14} = \frac{14}{2}(6+97) = 7(103) = 721\).
20. Find x so that x+6, x+12, and x+15 are Consecutive Terms of a geometric progression.
Solution:
If three terms are in GP, the square of the middle term is equal to the product of the other two terms.
\((x+12)^2 = (x+6)(x+15)\).
\(x^2 + 24x + 144 = x^2 + 15x + 6x + 90\).
\(x^2 + 24x + 144 = x^2 + 21x + 90\).
\(24x - 21x = 90 - 144\).
\(3x = -54 \Rightarrow x = -18\).
21. Find the Lcm of \(8x^4y^2\), \(48x^2y^4\).
Solution:
LCM of the coefficients (8, 48) is 48.
LCM of the variables with the highest powers:
For x: \(LCM(x^4, x^2) = x^4\).
For y: \(LCM(y^2, y^4) = y^4\).
The LCM is \(48x^4y^4\).
22. Simplify: \(\frac{y}{x-y} - \frac{x}{y-x}\)
Solution:
\(\frac{y}{x-y} - \frac{x}{y-x} = \frac{y}{x-y} - \frac{x}{-(x-y)}\)
\( = \frac{y}{x-y} + \frac{x}{x-y}\)
\( = \frac{y+x}{x-y}\).
23. Determine the quadratic Equation, Whose Sum and Product of roots are -9, 20.
Solution:
The general form of a quadratic equation is \(x^2 - (sum \ of \ roots)x + (product \ of \ roots) = 0\).
Sum of roots = -9.
Product of roots = 20.
The equation is \(x^2 - (-9)x + 20 = 0\).
\(x^2 + 9x + 20 = 0\).
24. If \(\triangle ABC\) is Similar to \(\triangle DEF\) Such that BC = 3cm, EF = 4 cm, and area of \(\triangle ABC\) = 54 Cm\(^2\) find the area of \(\triangle DEF\).
Solution:
For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
\(\frac{Area(\triangle ABC)}{Area(\triangle DEF)} = \left(\frac{BC}{EF}\right)^2\).
\(\frac{54}{Area(\triangle DEF)} = \left(\frac{3}{4}\right)^2 = \frac{9}{16}\).
\(Area(\triangle DEF) = \frac{54 \times 16}{9} = 6 \times 16 = 96 \ cm^2\).
25. Find the slope of a line joining (-6,1) and (-3,2).
Solution:
Slope \(m = \frac{y_2-y_1}{x_2-x_1}\).
Let \((x_1, y_1) = (-6,1)\) and \((x_2, y_2) = (-3,2)\).
\(m = \frac{2-1}{-3-(-6)} = \frac{1}{-3+6} = \frac{1}{3}\).
26. Find the equation of a line whose inclination is 30° and making an intercept - 3 on the Y - axis.
Solution:
Inclination \(\theta = 30^\circ\).
Slope \(m = \tan(\theta) = \tan(30^\circ) = \frac{1}{\sqrt{3}}\).
Y-intercept (c) = -3.
The equation of the line is y = mx + c.
\(y = \frac{1}{\sqrt{3}}x - 3\).
Multiplying by \(\sqrt{3}\): \(\sqrt{3}y = x - 3\sqrt{3}\).
Standard form: \(x - \sqrt{3}y - 3\sqrt{3} = 0\).
27. Prove that \(\sqrt{\frac{1+\cos\theta}{1-\cos\theta}} = Cosec\theta + Cot\theta\).
Solution:
LHS = \(\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}\).
Multiply numerator and denominator inside the square root by \( (1+\cos\theta) \):
\( = \sqrt{\frac{(1+\cos\theta)(1+\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}} = \sqrt{\frac{(1+\cos\theta)^2}{1-\cos^2\theta}}\).
Using \(sin^2\theta + cos^2\theta = 1 \Rightarrow 1-\cos^2\theta = sin^2\theta\):
\( = \sqrt{\frac{(1+\cos\theta)^2}{sin^2\theta}} = \frac{1+\cos\theta}{sin\theta}\).
\( = \frac{1}{sin\theta} + \frac{\cos\theta}{sin\theta} = Cosec\theta + Cot\theta = \) RHS. Hence Proved.
28. (Compulsory) Show that the straight lines 5x+23y+14=0 and 23x-5y+9=0 are perpendicular.
Solution:
For two lines \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\) to be perpendicular, the condition is \(a_1a_2 + b_1b_2 = 0\). Or, their slopes \(m_1m_2 = -1\).
Slope of the first line (\(m_1\)) = \(-\frac{\text{coeff of x}}{\text{coeff of y}} = -\frac{5}{23}\).
Slope of the second line (\(m_2\)) = \(-\frac{23}{-5} = \frac{23}{5}\).
Product of slopes = \(m_1 \times m_2 = \left(-\frac{5}{23}\right) \times \left(\frac{23}{5}\right) = -1\).
Since the product of their slopes is -1, the lines are perpendicular.
Part III: Answer any 10 from the following (Q.No: 42 is compulsory) (10 x 5 = 50)
29. If A = {5,6}, B = {4,5,6}, C = {5,6,7}, Shows that AxA = (BxB) \(\cap\) (CxC).
Solution:
Given A = {5,6}, B = {4,5,6}, C = {5,6,7}.
LHS: AxA
AxA = {5,6} x {5,6} = {(5,5), (5,6), (6,5), (6,6)}.
RHS: (BxB) \(\cap\) (CxC)
BxB = {4,5,6} x {4,5,6} = {(4,4),(4,5),(4,6), (5,4),(5,5),(5,6), (6,4),(6,5),(6,6)}.
CxC = {5,6,7} x {5,6,7} = {(5,5),(5,6),(5,7), (6,5),(6,6),(6,7), (7,5),(7,6),(7,7)}.
(BxB) \(\cap\) (CxC) is the set of common elements in BxB and CxC.
(BxB) \(\cap\) (CxC) = {(5,5), (5,6), (6,5), (6,6)}.
Since LHS = RHS, it is proved that AxA = (BxB) \(\cap\) (CxC).
30. Let A = {1,2,3,4} and B = {2,5,8,11,14} be two sets. Let f: A→B be a function given by f(x) = 3x-1. Represent this function as (1) set of ordered pairs, (2) a table, (3) an arrow diagram, (4) a graphical form.
Solution:
f(x) = 3x-1. Domain A = {1,2,3,4}.
f(1) = 3(1)-1 = 2
f(2) = 3(2)-1 = 5
f(3) = 3(3)-1 = 8
f(4) = 3(4)-1 = 11
(1) Set of ordered pairs:
f = {(1,2), (2,5), (3,8), (4,11)}
(2) Table form:
| x | f(x) |
|---|---|
| 1 | 2 |
| 2 | 5 |
| 3 | 8 |
| 4 | 11 |
Draw two ovals. In the first (A), write 1,2,3,4. In the second (B), write 2,5,8,11,14. Draw arrows from 1 to 2, 2 to 5, 3 to 8, and 4 to 11.
(4) Graphical form:
Plot the points (1,2), (2,5), (3,8), and (4,11) on a Cartesian coordinate system.
31. A function f is defined by f(x) = 3-2x. Find x such that \(f(x^2) = [f(x)]^2\).
Solution:
Given f(x) = 3-2x.
LHS: \(f(x^2) = 3 - 2x^2\).
RHS: \([f(x)]^2 = (3-2x)^2 = 3^2 - 2(3)(2x) + (2x)^2 = 9 - 12x + 4x^2\).
Equating LHS and RHS:
\(3 - 2x^2 = 9 - 12x + 4x^2\).
Rearranging the terms to form a quadratic equation:
\(4x^2 + 2x^2 - 12x + 9 - 3 = 0\).
\(6x^2 - 12x + 6 = 0\).
Divide by 6: \(x^2 - 2x + 1 = 0\).
This is a perfect square: \((x-1)^2 = 0\).
Therefore, x - 1 = 0, which gives x = 1.
32. If the highest common factor of 210 and 55 is Expressible in the form 55x-325. Then find x.
Solution:
First, find the HCF of 210 and 55 using Euclid's Division Algorithm.
210 = 3 × 55 + 45
55 = 1 × 45 + 10
45 = 4 × 10 + 5
10 = 2 × 5 + 0
The HCF is 5.
Given that the HCF is expressible as 55x - 325.
So, 55x - 325 = 5.
55x = 5 + 325 = 330.
x = 330 / 55 = 6.
33. The ratio of 6th and 8th term of an A.P is 7:9. Find the ratio of 9th term to 13th term.
Solution:
Let the AP have first term 'a' and common difference 'd'.
Given \(\frac{a_6}{a_8} = \frac{7}{9}\).
\(\frac{a+5d}{a+7d} = \frac{7}{9}\).
\(9(a+5d) = 7(a+7d)\).
\(9a + 45d = 7a + 49d\).
\(2a = 4d \Rightarrow a = 2d\).
We need to find the ratio \(\frac{a_9}{a_{13}}\).
\(\frac{a_9}{a_{13}} = \frac{a+8d}{a+12d}\).
Substitute a = 2d:
\(\frac{2d+8d}{2d+12d} = \frac{10d}{14d} = \frac{10}{14} = \frac{5}{7}\).
The ratio is 5:7.
34. Rekha has 15 squares colour papers of sizes 10cm, 11cm, 12cm,........ 24cm how much area can be decorated with these colour papers?
Solution:
The total area is the sum of the areas of all square papers.
Total Area = \(10^2 + 11^2 + 12^2 + ... + 24^2\).
We can write this as \((1^2 + 2^2 + ... + 24^2) - (1^2 + 2^2 + ... + 9^2)\).
Using the formula for the sum of squares of first n natural numbers, \(\sum n^2 = \frac{n(n+1)(2n+1)}{6}\).
Sum up to 24: \(\frac{24(24+1)(2(24)+1)}{6} = \frac{24(25)(49)}{6} = 4 \times 25 \times 49 = 4900\).
Sum up to 9: \(\frac{9(9+1)(2(9)+1)}{6} = \frac{9(10)(19)}{6} = 3 \times 5 \times 19 = 285\).
Total Area = 4900 - 285 = 4615 cm\(^2\).
35. If \(x = \frac{a^2+3a-4}{3a^2-3}\) and \(y = \frac{a^2+2a-8}{2a^2-2a-4}\) then find the value of \(x^2y^{-2}\)
Solution:
First, simplify x and y by factoring the polynomials.
\(x = \frac{(a+4)(a-1)}{3(a^2-1)} = \frac{(a+4)(a-1)}{3(a-1)(a+1)} = \frac{a+4}{3(a+1)}\).
\(y = \frac{(a+4)(a-2)}{2(a^2-a-2)} = \frac{(a+4)(a-2)}{2(a-2)(a+1)} = \frac{a+4}{2(a+1)}\).
We need to find \(x^2y^{-2} = \frac{x^2}{y^2} = \left(\frac{x}{y}\right)^2\).
\(\frac{x}{y} = \frac{\frac{a+4}{3(a+1)}}{\frac{a+4}{2(a+1)}} = \frac{a+4}{3(a+1)} \times \frac{2(a+1)}{a+4} = \frac{2}{3}\).
Therefore, \(\left(\frac{x}{y}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\).
36. Find the square root of \(64x^4-16x^3+17x^2-2x+1\)
Solution:
Using the long division method for finding the square root of a polynomial:
\[
\begin{array}{r|l}
\multicolumn{2}{r}{8x^2 - x + 1} \\
\cline{2-2}
8x^2 & 64x^4 - 16x^3 + 17x^2 - 2x + 1 \\
\multicolumn{2}{r}{-64x^4} \\
\cline{2-2}
16x^2-x & -16x^3 + 17x^2 \\
\multicolumn{2}{r}{-(-16x^3 + x^2)} \\
\cline{2-2}
16x^2-2x+1 & 16x^2 - 2x + 1 \\
\multicolumn{2}{r}{-(16x^2 - 2x + 1)} \\
\cline{2-2}
\multicolumn{2}{r}{0} \\
\end{array}
\]
The square root is \(|8x^2 - x + 1|\).
37. If \(\alpha, \beta\) are the roots of \(2x^2-7x+5=0\). Find the value of 1) \(\frac{1}{\alpha} + \frac{1}{\beta}\) 2) \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\)
Solution:
For the quadratic equation \(2x^2-7x+5=0\):
Sum of roots, \(\alpha + \beta = -\frac{b}{a} = -\frac{-7}{2} = \frac{7}{2}\).
Product of roots, \(\alpha\beta = \frac{c}{a} = \frac{5}{2}\).
1) \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta+\alpha}{\alpha\beta} = \frac{7/2}{5/2} = \frac{7}{5}\).
2) \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha\beta}\).
First find \(\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (\frac{7}{2})^2 - 2(\frac{5}{2}) = \frac{49}{4} - 5 = \frac{49-20}{4} = \frac{29}{4}\).
So, \(\frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{29/4}{5/2} = \frac{29}{4} \times \frac{2}{5} = \frac{29}{10}\).
38. State and prove basic Proportionality theorem.
Solution:
Statement (Thales' Theorem): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Proof:
Given: In \(\triangle ABC\), a line DE is parallel to BC, intersecting AB at D and AC at E.
To Prove: \(\frac{AD}{DB} = \frac{AE}{EC}\).
Construction: Join BE and CD. Draw DM \(\perp\) AC and EN \(\perp\) AB.
Proof:
Area of \(\triangle ADE = \frac{1}{2} \times base \times height = \frac{1}{2} \times AD \times EN\).
Area of \(\triangle BDE = \frac{1}{2} \times DB \times EN\).
\(\frac{Area(\triangle ADE)}{Area(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB}\) --- (1)
Similarly,
Area of \(\triangle ADE = \frac{1}{2} \times AE \times DM\).
Area of \(\triangle CDE = \frac{1}{2} \times EC \times DM\).
\(\frac{Area(\triangle ADE)}{Area(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}\) --- (2)
\(\triangle BDE\) and \(\triangle CDE\) are on the same base DE and between the same parallel lines DE and BC.
So, Area(\(\triangle BDE\)) = Area(\(\triangle CDE\)).
From (1) and (2), we get \(\frac{AD}{DB} = \frac{AE}{EC}\). Hence Proved.
39. Find the area of the Quadrilateral formed by the points (8,6), (5,11), (-5,12) and (-4,3).
Solution:
Let the vertices be A(8,6), B(5,11), C(-5,12), and D(-4,3).
Area of quadrilateral = \(\frac{1}{2} |(x_1y_2+x_2y_3+x_3y_4+x_4y_1) - (y_1x_2+y_2x_3+y_3x_4+y_4x_1)|\).
= \(\frac{1}{2} |(8 \times 11 + 5 \times 12 + (-5) \times 3 + (-4) \times 6) - (6 \times 5 + 11 \times (-5) + 12 \times (-4) + 3 \times 8)|\).
= \(\frac{1}{2} |(88 + 60 - 15 - 24) - (30 - 55 - 48 + 24)|\).
= \(\frac{1}{2} |(148 - 39) - (54 - 103)|\).
= \(\frac{1}{2} |(109) - (-49)| = \frac{1}{2} |109 + 49| = \frac{1}{2} |158|\).
= 79 sq. units.
40. A(-3,0), B(10,-2), C(12,3) are the vertices of \(\triangle ABC\). Find the equation of the Altitude through A and B.
Solution:
Altitude through A:
This altitude is perpendicular to BC. First, find the slope of BC.
Slope of BC (\(m_{BC}\)) = \(\frac{3 - (-2)}{12 - 10} = \frac{5}{2}\).
Slope of altitude from A (\(m_A\)) = \(-\frac{1}{m_{BC}} = -\frac{2}{5}\).
Equation of altitude from A, passing through (-3,0):
\(y - 0 = -\frac{2}{5}(x - (-3)) \Rightarrow 5y = -2(x+3) \Rightarrow 2x + 5y + 6 = 0\).
Altitude through B:
This altitude is perpendicular to AC. First, find the slope of AC.
Slope of AC (\(m_{AC}\)) = \(\frac{3 - 0}{12 - (-3)} = \frac{3}{15} = \frac{1}{5}\).
Slope of altitude from B (\(m_B\)) = \(-\frac{1}{m_{AC}} = -5\).
Equation of altitude from B, passing through (10,-2):
\(y - (-2) = -5(x - 10) \Rightarrow y + 2 = -5x + 50 \Rightarrow 5x + y - 48 = 0\).
41. Prove that \(\frac{SinA}{1+CosA} + \frac{SinA}{1-CosA} = 2 Cosec A\).
Solution:
LHS = \(\frac{SinA}{1+CosA} + \frac{SinA}{1-CosA}\).
Take Sin A common: \(SinA \left[ \frac{1}{1+CosA} + \frac{1}{1-CosA} \right]\).
Take LCM inside the bracket: \(SinA \left[ \frac{(1-CosA) + (1+CosA)}{(1+CosA)(1-CosA)} \right]\).
\( = SinA \left[ \frac{2}{1-Cos^2A} \right]\).
Using \(Sin^2A + Cos^2A = 1 \Rightarrow 1-Cos^2A = Sin^2A\):
\( = SinA \left[ \frac{2}{Sin^2A} \right] = \frac{2}{SinA} = 2 Cosec A = \) RHS. Hence Proved.
42. (Compulsory) Solve: x+y+z=5; 2x-y+z=9; x-2y+3z = 16.
Solution:
Let the equations be:
(1) x + y + z = 5
(2) 2x - y + z = 9
(3) x - 2y + 3z = 16
Add (1) and (2) to eliminate y:
(x+y+z) + (2x-y+z) = 5+9 \(\Rightarrow\) 3x + 2z = 14 --- (4)
Multiply (1) by 2 and add to (3) to eliminate y:
2(x+y+z) = 10 \(\Rightarrow\) 2x + 2y + 2z = 10.
(2x+2y+2z) + (x-2y+3z) = 10+16 \(\Rightarrow\) 3x + 5z = 26 --- (5)
Now solve (4) and (5):
Subtract (4) from (5):
(3x + 5z) - (3x + 2z) = 26 - 14 \(\Rightarrow\) 3z = 12 \(\Rightarrow\) z = 4.
Substitute z=4 into (4):
3x + 2(4) = 14 \(\Rightarrow\) 3x + 8 = 14 \(\Rightarrow\) 3x = 6 \(\Rightarrow\) x = 2.
Substitute x=2 and z=4 into (1):
2 + y + 4 = 5 \(\Rightarrow\) y + 6 = 5 \(\Rightarrow\) y = -1.
The solution is x=2, y=-1, z=4.
Part IV: Answer any one from given two questions (EACH) (2 x 8 = 16)
43. (a) Construct a Triangle similar to a given triangle PQR with its sides equal to \(\frac{3}{5}\) of the corresponding sides of the triangle PQR [Scale factor \(< 1\)]. (or)
(b) Construct a Triangle similar to a given Triangle PQR with its sides equal to \(\frac{6}{5}\) of the corresponding sides of the Triangle PQR [scale factor \(> 1\)].
Solution for 43(a):
Steps of Construction:
- Construct a triangle PQR with any given measurements.
- Draw a ray QX starting from Q, making an acute angle with QR on the side opposite to vertex P.
- Locate 5 points (the greater of 3 and 5 in the ratio) Q₁, Q₂, Q₃, Q₄, Q₅ on the ray QX such that QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₄Q₅.
- Join Q₅ with R.
- Draw a line through Q₃ (the smaller number in the ratio) parallel to Q₅R. This line intersects QR at a point R'.
- Draw a line through R' parallel to PR. This line intersects PQ at a point P'.
- \(\triangle\)P'QR' is the required triangle, similar to \(\triangle\)PQR, with sides that are \(\frac{3}{5}\) of the corresponding sides of \(\triangle\)PQR.
Solution for 43(b):
Steps of Construction:
- Construct a triangle PQR with any given measurements.
- Draw a ray QX starting from Q, making an acute angle with QR on the side opposite to vertex P.
- Locate 6 points (the greater of 6 and 5) Q₁, ..., Q₆ on QX such that all segments are equal.
- Join Q₅ (the smaller number in the ratio) with R.
- Extend the line segment QR beyond R.
- Draw a line through Q₆ parallel to Q₅R. This line intersects the extended line segment QR at a point R'.
- Extend the line segment QP beyond P.
- Draw a line through R' parallel to PR. This line intersects the extended line segment QP at a point P'.
- \(\triangle\)P'QR' is the required triangle, similar to \(\triangle\)PQR, with sides that are \(\frac{6}{5}\) of the corresponding sides of \(\triangle\)PQR.
44. (a) A bus is travelling at a uniform speed of 50 km/hr. Draw a distance-time graph and find:
- How far will it go in 90 minutes?
- Time required to cover the distance of 300 km.
(b) Draw the graph of xy = 24, x,y>0. Using the graph find
- y when x=3
- x when y=6
Solution for 44(a):
The relationship is Distance = Speed × Time, so \(d = 50t\). This is a linear relationship.
Table of values:
| Time (t) in hours | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Distance (d) in km | 50 | 100 | 150 | 200 |
From the graph:
- Distance in 90 minutes: 90 minutes = 1.5 hours. Locate 1.5 on the X-axis. Move up to the line and then across to the Y-axis. The value will be 75. So, the bus travels 75 km.
- Time for 300 km: Locate 300 on the Y-axis. Move across to the line and then down to the X-axis. The value will be 6. So, the time required is 6 hours.
Solution for 44(b):
The equation is xy = 24, or \(y = \frac{24}{x}\).
Table of values:
| x | 1 | 2 | 3 | 4 | 6 | 8 | 12 |
|---|---|---|---|---|---|---|---|
| y | 24 | 12 | 8 | 6 | 4 | 3 | 2 |
From the graph:
- y when x=3: Find x=3 on the X-axis. Move vertically up to the curve. From that point, move horizontally to the Y-axis. The y-value is 8.
- x when y=6: Find y=6 on the Y-axis. Move horizontally to the curve. From that point, move vertically down to the X-axis. The x-value is 4.