10th Maths Quarterly Exam Question Paper with Solutions
TIRUVANNAMALAI DISTRICT - 2024 | X - Std
PART - A (14 x 1 = 14)
I. Choose the correct answer :-
1. A = {a, b, c}, B = {2, 3}, C = {p, q, r, s} then n[(A U C) X C] is
Answer: b) 20
Note: There appears to be a typo in the question. It likely meant to ask for \(n[(A \cup B) \times C]\). Let's solve based on that assumption.
Given: \(A = \{a, b, c\} \implies n(A) = 3\)
\(B = \{2, 3\} \implies n(B) = 2\)
\(C = \{p, q, r, s\} \implies n(C) = 4\)
\(A \cup B = \{a, b, c, 2, 3\}\). So, \(n(A \cup B) = 5\).
\(n[(A \cup B) \times C] = n(A \cup B) \times n(C) = 5 \times 4 = 20\).
(If we strictly follow the question `n[(A U C) x C]`: `A U C = {a,b,c,p,q,r,s}`, so `n(A U C) = 7`. Then `n[(A U C) x C] = 7 x 4 = 28`, which is not in the options.)
2. If the ordered pairs (a + 2, 4) and (5, 2a + b) are equal than (a, b) is
Answer: d) (3, -2)
Given \((a + 2, 4) = (5, 2a + b)\).
Equating the corresponding elements:
\(a + 2 = 5 \implies a = 3\)
\(2a + b = 4\). Substitute \(a = 3\).
\(2(3) + b = 4 \implies 6 + b = 4 \implies b = -2\)
So, (a, b) is (3, -2).
3. \(f(x) = (x + 1)^2 - (x - 1)^3\) represents a function which is
Answer: b) Cubic
\(f(x) = (x^2 + 2x + 1) - (x^3 - 3x^2 + 3x - 1)\)
\(f(x) = x^2 + 2x + 1 - x^3 + 3x^2 - 3x + 1\)
\(f(x) = -x^3 + 4x^2 - x + 2\)
The highest power of x is 3, so it is a cubic function.
4. If the HCF of 65 and 117 is expressible in the form of 65m - 117, then the value of m is
Answer: b) 2
First, find the HCF of 65 and 117 using Euclid's algorithm:
\(117 = 1 \times 65 + 52\)
\(65 = 1 \times 52 + 13\)
\(52 = 4 \times 13 + 0\). The HCF is 13.
Now, set the given expression equal to the HCF:
\(65m - 117 = 13\)
\(65m = 130\)
\(m = 130 / 65 = 2\)
5. If 6 times of 6th term of an A.P. is equal to 7 times 7th term, then 13th term of the A.P. is
Answer: a) 0
Given: \(6 \times t_6 = 7 \times t_7\)
\(6(a + 5d) = 7(a + 6d)\)
\(6a + 30d = 7a + 42d\)
\(0 = 7a - 6a + 42d - 30d\)
\(0 = a + 12d\)
The 13th term is \(t_{13} = a + 12d\). Therefore, \(t_{13} = 0\).
6. If \(A = 2^{65}\) and \(B = 2^{64} + 2^{63} + 2^{62} + \dots + 2^0\), which of the following is true?
Answer: d) A is larger than 1 by B
B is a geometric series with first term \(a=2^0=1\), common ratio \(r=2\), and number of terms \(n=65\).
Sum of a G.P. is \(S_n = a(r^n - 1) / (r-1)\)
\(B = 1(2^{65} - 1) / (2-1) = 2^{65} - 1\)
Since \(A = 2^{65}\), we have \(B = A - 1\), or \(A = B + 1\).
This means A is larger than B by 1.
7. The solution of \((2x – 1)^2 = 9\) is equal to
Answer: c) -1, 2
\((2x - 1)^2 = 9\)
Taking the square root on both sides: \(2x - 1 = \pm 3\)
Case 1: \(2x - 1 = 3 \implies 2x = 4 \implies x = 2\)
Case 2: \(2x - 1 = -3 \implies 2x = -2 \implies x = -1\)
The solutions are -1 and 2.
8. The square root of \( \frac{256x^8y^4z^{10}}{25x^6y^6z^6} \) is
Answer: d) \( \frac{16}{5} \frac{xz^2}{y} \)
\(\sqrt{\frac{256x^8y^4z^{10}}{25x^6y^6z^6}} = \sqrt{\frac{256}{25} \cdot x^{8-6} \cdot y^{4-6} \cdot z^{10-6}}\)
\(= \sqrt{\frac{16^2}{5^2} \cdot x^2 \cdot y^{-2} \cdot z^4} = \left| \frac{16}{5} \cdot x \cdot y^{-1} \cdot z^2 \right|\)
\(= \frac{16}{5} \left| \frac{xz^2}{y} \right|\). Assuming positive variables, the answer is \( \frac{16xz^2}{5y} \).
9. \( \frac{3y-3}{y} \div \frac{7y-7}{3y^2} \) is
Answer: a) \( \frac{9y}{7} \)
\(\frac{3y-3}{y} \div \frac{7y-7}{3y^2} = \frac{3(y-1)}{y} \times \frac{3y^2}{7(y-1)}\)
Cancel out the common term \((y-1)\):
\(= \frac{3}{y} \times \frac{3y^2}{7} = \frac{9y^2}{7y} = \frac{9y}{7}\)
10. If in ΔABC, DE || BC, AB = 3.6cm, AC = 2.4cm and AD = 2.1 cm then the length of AE is
Answer: a) 1.4 cm
By Thales Theorem (Basic Proportionality Theorem), if DE || BC, then \(\frac{AD}{AB} = \frac{AE}{AC}\).
\(\frac{2.1}{3.6} = \frac{AE}{2.4}\)
\(AE = \frac{2.1 \times 2.4}{3.6} = \frac{2.1 \times 24}{36} = \frac{2.1 \times 2}{3} = 0.7 \times 2 = 1.4\) cm.
11. The straight line given by the equation x = 11 is
Answer: b) Parallel to Y axis
The equation \(x=11\) represents a vertical line where the x-coordinate of every point is 11. A vertical line is parallel to the Y-axis.
12. If slope of the line PQ is \( \frac{1}{\sqrt{3}} \) then the slope of perpendicular bisector of PQ is
Answer: b) \( -\sqrt{3} \)
The perpendicular bisector is perpendicular to the line PQ. If the slope of a line is \(m_1\), the slope of a line perpendicular to it is \(m_2 = -1/m_1\).
Given \(m_1 = \frac{1}{\sqrt{3}}\).
\(m_2 = -1 / (\frac{1}{\sqrt{3}}) = -\sqrt{3}\).
13. (2, 1) is the point of intersection of two lines.
Answer: b) x + y = 3; 3x + y = 7
We substitute the point (2, 1) into each pair of equations.
For option b):
Line 1: \(x + y = 2 + 1 = 3\). (Satisfied)
Line 2: \(3x + y = 3(2) + 1 = 6 + 1 = 7\). (Satisfied)
Since the point satisfies both equations, it is their point of intersection.
14. If \(x = a \tan\theta\) and \(y = b \sec\theta\) then
Answer: b) \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \)
From the given equations, we have:
\(\tan\theta = \frac{x}{a}\) and \(\sec\theta = \frac{y}{b}\).
Using the trigonometric identity \(\sec^2\theta - \tan^2\theta = 1\), we substitute the expressions:
\(\left(\frac{y}{b}\right)^2 - \left(\frac{x}{a}\right)^2 = 1\)
\(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\)
PART - B (10 x 2 = 20)
II. Answer any ten questions. Q.No. 28 is compulsory.
15. Let A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A X B and B X A.
Given A = {1, 2, 3}.
B = {prime numbers less than 10} = {2, 3, 5, 7}.
A x B:
\(\{(1,2), (1,3), (1,5), (1,7), (2,2), (2,3), (2,5), (2,7), (3,2), (3,3), (3,5), (3,7)\}\)
B x A:
\(\{(2,1), (2,2), (2,3), (3,1), (3,2), (3,3), (5,1), (5,2), (5,3), (7,1), (7,2), (7,3)\}\)
16. Given \(f(x) = 2x - x^2\), find (i) f(1) and (ii) f(x+1).
Given \(f(x) = 2x - x^2\).
(i) f(1):
\(f(1) = 2(1) - (1)^2 = 2 - 1 = 1\)
(ii) f(x+1):
\(f(x+1) = 2(x+1) - (x+1)^2\)
\(= (2x+2) - (x^2 + 2x + 1)\)
\(= 2x + 2 - x^2 - 2x - 1\)
\(= -x^2 + 1\)
17. The function 't' which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by \(t(C) = F\) where \(F = \frac{9}{5}C + 32\). Find, (i) t(0) (ii) t(-10).
Given \(F = \frac{9}{5}C + 32\).
(i) t(0):
\(F = \frac{9}{5}(0) + 32 = 32\). So, t(0) = 32°F.
(ii) t(-10):
\(F = \frac{9}{5}(-10) + 32 = 9(-2) + 32 = -18 + 32 = 14\). So, t(-10) = 14°F.
18. Compute x, such that \(10^4 \equiv x \pmod{19}\).
\(10^2 = 100\). When 100 is divided by 19, \(100 = 5 \times 19 + 5\). So, \(10^2 \equiv 5 \pmod{19}\).
Now, \(10^4 = (10^2)^2 \equiv 5^2 \pmod{19}\).
\(10^4 \equiv 25 \pmod{19}\).
When 25 is divided by 19, \(25 = 1 \times 19 + 6\). So, \(25 \equiv 6 \pmod{19}\).
Therefore, \(x = 6\).
19. Find x, y and z given that the numbers x, 10, y, 24, z are in A.P.
Let the common difference be 'd'.
The terms are \(t_1=x, t_2=10, t_3=y, t_4=24, t_5=z\).
We know \(t_4 - t_2 = (a+3d) - (a+d) = 2d\).
\(24 - 10 = 2d \implies 14 = 2d \implies d = 7\).
\(x = t_2 - d = 10 - 7 = 3\).
\(y = t_2 + d = 10 + 7 = 17\).
\(z = t_4 + d = 24 + 7 = 31\).
So, x = 3, y = 17, z = 31.
20. Find the sum \(3+1+\frac{1}{3}+...+\infty\).
This is an infinite Geometric Progression (G.P.).
First term \(a = 3\).
Common ratio \(r = 1/3\).
Since \(|r| = |1/3| < 1\), the sum to infinity exists.
Sum \(S_\infty = \frac{a}{1-r} = \frac{3}{1 - 1/3} = \frac{3}{2/3} = 3 \times \frac{3}{2} = \frac{9}{2}\).
21. Find the excluded values of the expression \( \frac{x+10}{8x} \).
The expression is undefined when its denominator is zero.
Set the denominator to zero: \(8x = 0\).
\(x = 0\).
The excluded value is 0.
22. Determine the nature of the roots of quadratic equation \(9a^2b^2x^2 - 24abcdx + 16c^2d^2 = 0\), where \(a \ne 0, b \ne 0\).
We use the discriminant \(\Delta = B^2 - 4AC\).
Here, \(A = 9a^2b^2\), \(B = -24abcd\), \(C = 16c^2d^2\).
\(\Delta = (-24abcd)^2 - 4(9a^2b^2)(16c^2d^2)\)
\(= 576a^2b^2c^2d^2 - 576a^2b^2c^2d^2 = 0\).
Since \(\Delta = 0\), the roots are real and equal.
23. Simplify \( \frac{x^3}{x-y} + \frac{y^3}{y-x} \).
\(\frac{x^3}{x-y} + \frac{y^3}{y-x} = \frac{x^3}{x-y} - \frac{y^3}{x-y}\)
\(= \frac{x^3 - y^3}{x-y}\)
Using the identity \(a^3 - b^3 = (a-b)(a^2+ab+b^2)\), we get:
\(= \frac{(x-y)(x^2+xy+y^2)}{x-y} = x^2+xy+y^2\).
24. AD is the bisector of ∠A. If BD = 4cm, DC = 3cm and AB = 6cm find AC.
By the Angle Bisector Theorem, \(\frac{AB}{AC} = \frac{BD}{DC}\).
\(\frac{6}{AC} = \frac{4}{3}\)
\(4 \times AC = 6 \times 3 = 18\)
\(AC = \frac{18}{4} = 4.5\) cm.
25. Show that the points P(-1.5, 3), Q(6, -2), R(-3, 4) are collinear.
We can prove collinearity by showing that the slopes of any two segments are equal.
Slope of PQ = \(\frac{y_2-y_1}{x_2-x_1} = \frac{-2-3}{6-(-1.5)} = \frac{-5}{7.5} = \frac{-50}{75} = -\frac{2}{3}\).
Slope of QR = \(\frac{y_3-y_2}{x_3-x_2} = \frac{4-(-2)}{-3-6} = \frac{6}{-9} = -\frac{2}{3}\).
Since the slope of PQ is equal to the slope of QR, and Q is a common point, the points P, Q, and R are collinear.
26. Find the equation of a line whose intercepts on the x and y axes are 4 and -6 respectively.
The intercept form of a line is \(\frac{x}{a} + \frac{y}{b} = 1\), where 'a' is the x-intercept and 'b' is the y-intercept.
Given \(a = 4\) and \(b = -6\).
\(\frac{x}{4} + \frac{y}{-6} = 1\)
To eliminate fractions, multiply by the LCM of 4 and 6, which is 12:
\(12(\frac{x}{4}) - 12(\frac{y}{6}) = 12(1)\)
\(3x - 2y = 12\)
The equation is \(3x - 2y - 12 = 0\).
27. Prove the identity: \( \frac{\cos\theta}{1+\sin\theta} = \sec\theta - \tan\theta \).
LHS = \( \frac{\cos\theta}{1+\sin\theta} \)
Multiply numerator and denominator by the conjugate \((1-\sin\theta)\):
\(= \frac{\cos\theta(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)} = \frac{\cos\theta(1-\sin\theta)}{1-\sin^2\theta}\)
Since \(1-\sin^2\theta = \cos^2\theta\):
\(= \frac{\cos\theta(1-\sin\theta)}{\cos^2\theta} = \frac{1-\sin\theta}{\cos\theta}\)
\(= \frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta} = \sec\theta - \tan\theta \) = RHS.
Hence, proved.
28. (Compulsory) Show that the straight lines x - 2y + 3 = 0 and 6x + 3y + 8 = 0 are perpendicular.
For the first line, \(x - 2y + 3 = 0\), the slope \(m_1 = -\frac{\text{coefficient of x}}{\text{coefficient of y}} = -\frac{1}{-2} = \frac{1}{2}\).
For the second line, \(6x + 3y + 8 = 0\), the slope \(m_2 = -\frac{6}{3} = -2\).
The condition for perpendicular lines is \(m_1 \times m_2 = -1\).
Product of slopes = \((\frac{1}{2}) \times (-2) = -1\).
Since the product of their slopes is -1, the lines are perpendicular.
PART - C (10 x 5 = 50)
III. Answer any TEN questions. (Q.No. 42 is compulsory)
29. Let A = {x ∈ W : x < 2}, B = {x ∈ N : 1 < x ≤ 4}, C = {3,5} . Verify that A X (B ∪ C) = (A X B) ∪ (A X C).
A = {Whole numbers less than 2} = {0, 1}
B = {Natural numbers from 2 to 4} = {2, 3, 4}
C = {3, 5}
LHS: A X (B ∪ C)
B ∪ C = {2, 3, 4} ∪ {3, 5} = {2, 3, 4, 5}
A X (B ∪ C) = {0, 1} X {2, 3, 4, 5}
= {(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)} --- (1)
RHS: (A X B) ∪ (A X C)
A X B = {0, 1} X {2, 3, 4} = {(0,2), (0,3), (0,4), (1,2), (1,3), (1,4)}
A X C = {0, 1} X {3, 5} = {(0,3), (0,5), (1,3), (1,5)}
(A X B) ∪ (A X C) = {(0,2), (0,3), (0,4), (1,2), (1,3), (1,4), (0,5), (1,5)} --- (2)
From (1) and (2), LHS = RHS. Hence verified.
30. If the function f : R -> R is defined by \( f(x) = \begin{cases} 2x+7 & \text{if } x < -2 \\ x^2-2 & \text{if } -2 \le x < 3 \\ 3x-2 & \text{if } x \ge 3 \end{cases} \) , then find the value of (i) f(4) (ii) f(-2) (iii) f(4) + 2f(1) (iv) \( \frac{f(1)-3f(4)}{f(-3)} \)
(i) f(4): Since \(4 \ge 3\), use \(f(x) = 3x - 2\).
\(f(4) = 3(4) - 2 = 12 - 2 = 10\).
(ii) f(-2): Since \(-2 \le -2 < 3\), use \(f(x) = x^2 - 2\).
\(f(-2) = (-2)^2 - 2 = 4 - 2 = 2\).
(iii) f(4) + 2f(1): We need f(1). Since \(-2 \le 1 < 3\), use \(f(x) = x^2 - 2\).
\(f(1) = (1)^2 - 2 = 1 - 2 = -1\).
\(f(4) + 2f(1) = 10 + 2(-1) = 10 - 2 = 8\).
(iv) \( \frac{f(1)-3f(4)}{f(-3)} \): We need f(-3). Since \(-3 < -2\), use \(f(x) = 2x + 7\).
\(f(-3) = 2(-3) + 7 = -6 + 7 = 1\).
The expression is \( \frac{-1 - 3(10)}{1} = \frac{-1 - 30}{1} = -31\).
PART - C (10 x 5 = 50)
III. Answer any TEN questions. (Q.No. 42 is compulsory)
31. If f(x) = x - 4, g(x) = x², h(x) = 3x – 5 then prove that (fog)oh = fo(goh).
Given, f(x) = x - 4, g(x) = x², h(x) = 3x – 5.
This is to prove the associative property of function composition.
LHS = (fog)oh
First, find fog(x):
\(fog(x) = f(g(x)) = f(x²) = x² - 4\)
Now, find (fog)oh(x):
\((fog)oh(x) = (fog)(h(x)) = (fog)(3x-5)\)
Substitute (3x-5) into the expression for fog(x):
\(= (3x-5)² - 4 = (9x² - 30x + 25) - 4\)
\(= 9x² - 30x + 21\) --- (1)
RHS = fo(goh)
First, find goh(x):
\(goh(x) = g(h(x)) = g(3x-5) = (3x-5)²\)
Now, find fo(goh)(x):
\(fo(goh)(x) = f(goh(x)) = f((3x-5)²)\)
Substitute (3x-5)² into the expression for f(x):
\(= (3x-5)² - 4 = (9x² - 30x + 25) - 4\)
\(= 9x² - 30x + 21\) --- (2)
From (1) and (2), LHS = RHS.
Hence, (fog)oh = fo(goh) is proved.
32. A mother divides Rs. 207 into three parts such that the amount are in A.P. and gives it to her three children. The product of the two least amounts that the children had Rs. 4623. Find the amount received by each child.
Let the three amounts in A.P. be \(a-d\), \(a\), and \(a+d\).
Sum of the amounts:
\((a-d) + a + (a+d) = 207\)
\(3a = 207 \implies a = \frac{207}{3} = 69\)
The two least amounts are \(a-d\) and \(a\).
Product of the two least amounts:
\((a-d) \times a = 4623\)
Substitute \(a=69\):
\((69-d) \times 69 = 4623\)
\(69-d = \frac{4623}{69}\)
\(69-d = 67 \implies d = 69 - 67 = 2\)
Now, find the three amounts:
- First child's amount = \(a-d = 69 - 2 = 67\)
- Second child's amount = \(a = 69\)
- Third child's amount = \(a+d = 69 + 2 = 71\)
The amounts received by the three children are Rs. 67, Rs. 69, and Rs. 71.
33. Find the G.P. in which the 2nd term is √6 and 6th term is 9√6.
The general form of a G.P. is \(a, ar, ar^2, \dots\), where \(t_n = ar^{n-1}\).
Given, 2nd term \(t_2 = ar = \sqrt{6}\) --- (1)
Given, 6th term \(t_6 = ar^5 = 9\sqrt{6}\) --- (2)
Divide equation (2) by equation (1):
\(\frac{ar^5}{ar} = \frac{9\sqrt{6}}{\sqrt{6}}\)
\(r^4 = 9\)
\(r^2 = \sqrt{9} = 3 \implies r = \pm\sqrt{3}\)
Case 1: If \(r = \sqrt{3}\)
Substitute \(r\) in equation (1):
\(a(\sqrt{3}) = \sqrt{6} \implies a = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}\)
The G.P. is: \(\sqrt{2}, \sqrt{2}(\sqrt{3}), \sqrt{2}(\sqrt{3})^2, \dots\)
\(\implies \sqrt{2}, \sqrt{6}, 3\sqrt{2}, \dots\)
Case 2: If \(r = -\sqrt{3}\)
Substitute \(r\) in equation (1):
\(a(-\sqrt{3}) = \sqrt{6} \implies a = -\frac{\sqrt{6}}{\sqrt{3}} = -\sqrt{2}\)
The G.P. is: \(-\sqrt{2}, -\sqrt{2}(-\sqrt{3}), -\sqrt{2}(-\sqrt{3})^2, \dots\)
\(\implies -\sqrt{2}, \sqrt{6}, -3\sqrt{2}, \dots\)
34. If \(9x^4 + 12x^3 + 28x^2 + ax + b\) is a perfect square, find the value of a and b.
We use the long division method to find the square root.
3x² + 2x + 4
____________________
3x² | 9x⁴ + 12x³ + 28x² + ax + b
-(9x⁴)
____________________
6x²+2x | 12x³ + 28x²
-(12x³ + 4x²)
____________________
6x²+4x+4 | 24x² + ax + b
-(24x² + 16x + 16)
____________________
(a-16)x + (b-16)
Since the given polynomial is a perfect square, the remainder must be zero.
\((a-16)x + (b-16) = 0\)
Equating the coefficients to zero:
\(a-16 = 0 \implies a = 16\)
\(b-16 = 0 \implies b = 16\)
Therefore, the values are a = 16 and b = 16.
35. The roots of the equation \(2x^2 - 7x + 5 = 0\) are α and β. Without solving for the roots find (i) \( \frac{1}{\alpha} + \frac{1}{\beta} \) (ii) \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \).
For the quadratic equation \(2x^2 - 7x + 5 = 0\), we have a=2, b=-7, c=5.
Sum of roots: \(\alpha + \beta = -\frac{b}{a} = -\frac{-7}{2} = \frac{7}{2}\)
Product of roots: \(\alpha\beta = \frac{c}{a} = \frac{5}{2}\)
(i) \( \frac{1}{\alpha} + \frac{1}{\beta} \)
\(= \frac{\beta + \alpha}{\alpha\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{7/2}{5/2} = \frac{7}{5}\)
(ii) \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \)
\(= \frac{\alpha^2 + \beta^2}{\alpha\beta}\)
First, find \(\alpha^2 + \beta^2\):
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
\(= \left(\frac{7}{2}\right)^2 - 2\left(\frac{5}{2}\right) = \frac{49}{4} - 5 = \frac{49-20}{4} = \frac{29}{4}\)
Now, substitute this back:
\(\frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{29/4}{5/2} = \frac{29}{4} \times \frac{2}{5} = \frac{29}{10}\)
36. Solve the following system of linear equations in three variables: \(3x - 2y + z = 2\), \(2x + 3y - z = 5\), \(x + y + z = 6\).
(1) \(3x - 2y + z = 2\)
(2) \(2x + 3y - z = 5\)
(3) \(x + y + z = 6\)
Add (1) and (2) to eliminate z:
\((3x+2x) + (-2y+3y) + (z-z) = 2+5\)
\(5x + y = 7\) --- (4)
Add (2) and (3) to eliminate z:
\((2x+x) + (3y+y) + (-z+z) = 5+6\)
\(3x + 4y = 11\) --- (5)
Now we solve the system of two equations (4) and (5).
Multiply equation (4) by 4:
\(4(5x + y) = 4(7) \implies 20x + 4y = 28\) --- (6)
Subtract equation (5) from (6):
\((20x+4y) - (3x+4y) = 28-11\)
\(17x = 17 \implies x = 1\)
Substitute \(x=1\) into equation (4):
\(5(1) + y = 7 \implies y = 2\)
Substitute \(x=1\) and \(y=2\) into equation (3):
\(1 + 2 + z = 6 \implies 3 + z = 6 \implies z = 3\)
The solution is x = 1, y = 2, z = 3.
37. State and prove Thales theorem.
Statement (Thales Theorem or Basic Proportionality Theorem):
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
Given: In ΔABC, 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 ⊥ AC and EN ⊥ AB.
Proof:
Area of a triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\)
In ΔADE, Area(ΔADE) = \(\frac{1}{2} \times AD \times EN\)
In ΔBDE, Area(ΔBDE) = \(\frac{1}{2} \times DB \times EN\)
Dividing the areas: \(\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB}\) --- (1)
Similarly, considering base on AC:
Area(ΔADE) = \(\frac{1}{2} \times AE \times DM\)
Area(ΔCDE) = \(\frac{1}{2} \times EC \times DM\)
Dividing the areas: \(\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}\) --- (2)
Now, ΔBDE and ΔCDE are on the same base DE and between the same parallel lines DE and BC. Therefore, their areas are equal.
Area(ΔBDE) = Area(ΔCDE)
From (1) and (2), using this equality, we get:
\(\frac{AD}{DB} = \frac{AE}{EC}\)
Hence, the theorem is proved.
38. Find the area of the quadrilateral formed by the points (8, 6), (5, 11), (-5, 12), and (-4, 3).
Let the vertices be A(8, 6), B(5, 11), C(-5, 12), and D(-4, 3).
The area of a quadrilateral with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4)\) is given by:
Area = \(\frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (x_2y_1 + x_3y_2 + x_4y_3 + x_1y_4)|\)
Area = \(\frac{1}{2} \left| \begin{matrix} 8 & 5 & -5 & -4 & 8 \\ 6 & 11 & 12 & 3 & 6 \end{matrix} \right|\)
Calculate the downward products:
\(x_1y_2 = 8 \times 11 = 88\)
\(x_2y_3 = 5 \times 12 = 60\)
\(x_3y_4 = (-5) \times 3 = -15\)
\(x_4y_1 = (-4) \times 6 = -24\)
Sum 1 = \(88 + 60 - 15 - 24 = 109\)
Calculate the upward products:
\(x_2y_1 = 5 \times 6 = 30\)
\(x_3y_2 = (-5) \times 11 = -55\)
\(x_4y_3 = (-4) \times 12 = -48\)
\(x_1y_4 = 8 \times 3 = 24\)
Sum 2 = \(30 - 55 - 48 + 24 = -49\)
Area = \(\frac{1}{2} | \text{Sum 1} - \text{Sum 2} | = \frac{1}{2} |109 - (-49)| = \frac{1}{2} |109 + 49| = \frac{1}{2} (158) = 79\)
The area of the quadrilateral is 79 square units.
39. Find the equation of the perpendicular bisector of the line joining the points A(-4, 2) & B(6, -4).
Step 1: Find the midpoint of the line segment AB.
Midpoint M = \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) = \left(\frac{-4+6}{2}, \frac{2-4}{2}\right) = \left(\frac{2}{2}, \frac{-2}{2}\right) = (1, -1)\)
Step 2: Find the slope of the line segment AB.
Slope \(m_{AB} = \frac{y_2-y_1}{x_2-x_1} = \frac{-4-2}{6-(-4)} = \frac{-6}{10} = -\frac{3}{5}\)
Step 3: Find the slope of the perpendicular bisector.
The slope of the perpendicular line, \(m_{\perp}\), is the negative reciprocal of \(m_{AB}\).
\(m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-3/5} = \frac{5}{3}\)
Step 4: Find the equation of the perpendicular bisector.
The line passes through the midpoint M(1, -1) and has a slope of 5/3. Using the point-slope form \(y - y_1 = m(x - x_1)\):
\(y - (-1) = \frac{5}{3}(x - 1)\)
\(y + 1 = \frac{5}{3}(x - 1)\)
Multiply by 3 to clear the fraction:
\(3(y+1) = 5(x-1)\)
\(3y + 3 = 5x - 5\)
\(5x - 3y - 8 = 0\)
The equation of the perpendicular bisector is \(5x - 3y - 8 = 0\).
40. Two buildings of different height are located at opposite sides of each other. If a heavy rod is attached joining the terrace of the buildings from (6, 10) to (14, 12), find the equation of the rod joining the buildings?
We need to find the equation of the line passing through the points A(6, 10) and B(14, 12).
Using the two-point form of a line equation: \(\frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1}\)
\(\frac{y-10}{12-10} = \frac{x-6}{14-6}\)
\(\frac{y-10}{2} = \frac{x-6}{8}\)
Cross-multiply:
\(8(y-10) = 2(x-6)\)
Divide by 2:
\(4(y-10) = x-6\)
\(4y - 40 = x - 6\)
Rearrange into the standard form \(Ax+By+C=0\):
\(x - 4y - 6 + 40 = 0\)
\(x - 4y + 34 = 0\)
The equation of the rod is \(x - 4y + 34 = 0\).
41. If \( \frac{\cos\theta}{1+\sin\theta} = \frac{1}{a} \) then prove that \( \frac{a^2-1}{a^2+1} = \sin\theta \). (Note: Solved based on a likely correction of the provided text for the proof to hold)
Given, \( \frac{\cos\theta}{1+\sin\theta} = \frac{1}{a} \). This implies \( a = \frac{1+\sin\theta}{\cos\theta} \).
We need to prove that \( \frac{a^2-1}{a^2+1} = \sin\theta \).
First, let's find expressions for \(a^2-1\) and \(a^2+1\).
\(a^2 = \left(\frac{1+\sin\theta}{\cos\theta}\right)^2 = \frac{(1+\sin\theta)^2}{\cos^2\theta} = \frac{1 + 2\sin\theta + \sin^2\theta}{\cos^2\theta}\)
Numerator: \(a^2 - 1\)
\(a^2 - 1 = \frac{1 + 2\sin\theta + \sin^2\theta}{\cos^2\theta} - 1 = \frac{1 + 2\sin\theta + \sin^2\theta - \cos^2\theta}{\cos^2\theta}\)
Using the identity \(\cos^2\theta = 1 - \sin^2\theta\):
\(= \frac{1 + 2\sin\theta + \sin^2\theta - (1-\sin^2\theta)}{\cos^2\theta} = \frac{2\sin\theta + 2\sin^2\theta}{\cos^2\theta} = \frac{2\sin\theta(1+\sin\theta)}{\cos^2\theta}\)
Denominator: \(a^2 + 1\)
\(a^2 + 1 = \frac{1 + 2\sin\theta + \sin^2\theta}{\cos^2\theta} + 1 = \frac{1 + 2\sin\theta + \sin^2\theta + \cos^2\theta}{\cos^2\theta}\)
Using the identity \(\sin^2\theta + \cos^2\theta = 1\):
\(= \frac{1 + 2\sin\theta + 1}{\cos^2\theta} = \frac{2 + 2\sin\theta}{\cos^2\theta} = \frac{2(1+\sin\theta)}{\cos^2\theta}\)
Now, find the ratio:
\(\frac{a^2-1}{a^2+1} = \frac{\frac{2\sin\theta(1+\sin\theta)}{\cos^2\theta}}{\frac{2(1+\sin\theta)}{\cos^2\theta}}\)
Cancel out common terms \( \frac{2(1+\sin\theta)}{\cos^2\theta} \) from numerator and denominator:
\(= \sin\theta\)
Hence, proved.
42. (Compulsory) Rekha has 15 square colour papers of sizes 10cm, 11cm, 12cm, ......, 24 cm. How much area can be decorated with these colour papers?
The sides of the square papers are 10 cm, 11 cm, ..., 24 cm.
The area of each paper will be the square of its side.
Total area = \(10^2 + 11^2 + 12^2 + \dots + 24^2\)
We can write this as the sum of squares from 1 to 24 minus the sum of squares from 1 to 9.
Total Area = \((1^2 + 2^2 + \dots + 24^2) - (1^2 + 2^2 + \dots + 9^2)\)
Using the formula for the sum of the first n squares, \(\sum n^2 = \frac{n(n+1)(2n+1)}{6}\):
Sum up to 24: \(S_{24} = \frac{24(24+1)(2 \times 24+1)}{6} = \frac{24 \times 25 \times 49}{6} = 4 \times 25 \times 49 = 100 \times 49 = 4900\)
Sum up to 9: \(S_{9} = \frac{9(9+1)(2 \times 9+1)}{6} = \frac{9 \times 10 \times 19}{6} = 3 \times 5 \times 19 = 285\)
Total Area = \(S_{24} - S_{9} = 4900 - 285 = 4615\) cm².
Therefore, 4615 cm² of area can be decorated.
(Solutions for other questions in Part C and D will be provided upon request as they involve detailed proofs, constructions, and graph plotting which are best explained with diagrams.)
PART - D (2 x 8 = 16)
IV. Answer all the questions :-
43. a) Construct a triangle similar to a given triangle PQR with its sides equal to 4/5 of the corresponding sides of the triangle PQR (scale factor 4/5 < 1). (OR) b) Construct a PQR which the base PQ = 4.5cm, ∠R = 35° and the median RG from R to PQ is 6cm.
Solution: These questions require geometric constructions. The steps would be provided in a geometry class.
For 43(a):
- Draw the triangle PQR.
- Draw a ray PX making an acute angle with PQ on the side opposite to vertex R.
- Locate 5 points (the greater of 4 and 5) P₁, P₂, P₃, P₄, P₅ on PX so that PP₁ = P₁P₂ = ... = P₄P₅.
- Join P₅Q and draw a line through P₄ parallel to P₅Q to intersect PQ at Q'.
- Draw a line through Q' parallel to QR to intersect PR at R'.
- Then, ΔPQ'R' is the required similar triangle.
For 43(b):
- Draw the base PQ = 4.5 cm.
- At P, draw a line PX such that ∠QPX = 35°.
- Draw a line PY perpendicular to PX at P.
- Draw the perpendicular bisector of PQ, which intersects PY at O and PQ at G.
- With O as center and OP as radius, draw a circle. All points on the major arc of this circle will subtend 35° at PQ.
- With G (midpoint of PQ) as center and radius 6 cm (length of median), draw an arc to intersect the major arc of the circle at R.
- Join PR and QR. ΔPQR is the required triangle.
44. a) A bus is travelling at a uniform speed of 50km/hr. Draw the distance - time 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 300km from the graph. (OR) b) Draw the graph of xy = 24, x, y > 0. Using the graph find (i) y when x = 3 and (ii) x when y = 6.
Solution: These questions require plotting graphs.
For 44(a):
- The relationship is Distance = Speed × Time, so \(d = 50t\). This is a direct variation.
- (i) Constant of variation (k): k = 50 km/hr.
- (ii) Distance in 90 minutes (1.5 hours): \(d = 50 \times 1.5 = 75\) km.
- (iii) Time for 300 km: \(300 = 50t \implies t = 6\) hours.
- The graph is a straight line passing through the origin (0,0) and other points like (1, 50), (2, 100), etc. The answers are found by locating the corresponding points on the plotted line.
For 44(b):
- The equation is \(xy = 24\) or \(y = 24/x\). This is an inverse variation.
- Create a table of values: (2, 12), (3, 8), (4, 6), (6, 4), (8, 3), (12, 2).
- Plot these points on a graph to get a rectangular hyperbola curve.
- (i) Find y when x=3: From the graph, locate x=3 on the x-axis, move up to the curve, and then across to the y-axis. You will find y = 8.
- (ii) Find x when y=6: From the graph, locate y=6 on the y-axis, move across to the curve, and then down to the x-axis. You will find x = 4.