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BRINDHAVAN HR SEC SCHOOL, SUKKIRANPATTI
QUARTERLY EXAM -SEP 2024 ANSWER KEY
10th Standard Maths
PART - A
Choose the correct answer (13 x 1 = 13)
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
2)If the order pairs (a, -1) and (5, b) belongs to {(x, y) | y = 2x + 3}, then a and b are
3)The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
4)If \(t_n\) is the \(n^{th}\) term of A.P, then \(t_{2n} - t_n\) is
5)The sequence \(\sqrt{11}, \sqrt{55}, 5\sqrt{11}, 25\sqrt{11}, \dots\) represents
6)If (x - 6) is the HCF of \(x^2 - 2x - 24\) and \(x^2 - kx - 6\) then the value of k is
7)Which of the following should be added to make \(x^4 + 64\) a perfect square
8)A Quadratic equation whose one zero is 5 and sum of the zeroes is 0 is given by
9)The perimeters of two similar triangles ∆ABC and ∆PQR are 36 cm and 24 cm respectively. If PQ = 10 cm, then the length of AB is
10)In a ∆ABC, AD is the bisector ∠BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm. The length of the side AC is
11)The straight line given by the equation x = 11 is
12)If (5, 7), (3, p) and (6, 6) are collinear, then the value of p is
13)(2, 1) is the point of intersection of two lines.
14)If \(5x = \sec\theta\) and \(\frac{5}{x} = \tan\theta\), then \(x^2 - \frac{1}{x^2}\) is equal to
PART - B
Answer any 10 questions. Question no 28 is compulsory (10 x 2 = 20)
15)A Relation R is given by the set \(\{(x,y) / y = x + 3, x \in \{0, 1, 2, 3, 4, 5\}\}\). Determine its domain and range.
Answer: Given Set = \(\{(x,y) / y = x + 3, x \in \{0, 1, 2, 3, 4, 5\}\}\)
When x = 0, y = 0 + 3 = 3
When x = 1, y = 1 + 3 = 4
When x = 2, y = 2 + 3 = 5
When x = 3, y = 3 + 3 = 6
When x = 4, y = 4 + 3 = 7
When x = 5, y = 5 + 3 = 8
Relation R = \(\{(0, 3), (1,4), (2,5), (3,6), (4,7), (5,8)\}\)
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {3, 4, 5, 6, 7, 8}
16)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.
Answer: \(f(x) = 3x - 5\) can be written as \(f = \{(x, 3x-5) | x \in R\}\).
(a, 4) means the image of a is 4. That is, f(a) = 4
\(3a - 5 = 4 \Rightarrow 3a = 9 \Rightarrow a = 3\)
(1, b) means the image of 1 is b. That is, f(1) = b
\(3(1) - 5 = b \Rightarrow b = -2\)
17)If \(f(x) = x^2 - 1\), \(g(x) = x - 2\) find a, if \(g \circ f(a) = 1\).
Answer: \(f(x) = x^2 - 1\), \(g(x) = x - 2\)
\(f(a) = a^2 - 1\)
Given, \((g \circ f)(a) = 1\)
\(g[f(a)] = 1\)
\(g[a^2 - 1] = 1\)
\((a^2 - 1) - 2 = 1\)
\(a^2 - 3 = 1\)
\(a^2 = 1 + 3 = 4\)
\(a = \pm 2\)
18)Solve \(5x \equiv 4 \pmod{6}\)
Answer: \(5x \equiv 4 \pmod{6}\)
\(5x - 4 = 6k\) for some integer k.
\(x = \frac{6k+4}{5}\)
When we put k = 1, 6, 11, 16...
then \(6k + 4\) is divisible by 5.
If k=1, \(x = \frac{6(1)+4}{5} = \frac{10}{5} = 2\)
If k=6, \(x = \frac{6(6)+4}{5} = \frac{40}{5} = 8\)
If k=11, \(x = \frac{6(11)+4}{5} = \frac{70}{5} = 14\)
If k=16, \(x = \frac{6(16)+4}{5} = \frac{100}{5} = 20\)
Therefore, the solution are 2, 8, 14, 20,...
19)In a G.P. 729, 243, 81,... find \(t_7\)
Answer: \(n^{th}\) term of G.P = \(ar^{n-1}\)
Here a = 729
\(r = \frac{t_2}{t_1} = \frac{243}{729} = \frac{1}{3}\)
\(t_7 = ar^{7-1} = ar^6 = 729 \left(\frac{1}{3}\right)^6\)
\(= 729 \times \frac{1}{729} = 1\)
\(t_7 = 1\)
20)If \(1 + 2 + 3 + ... + k = 325\), then find \(1^3 + 2^3 + 3^3 + ... + k^3\).
Answer: Sum of first k natural numbers = \(\frac{k(k+1)}{2} = 325\)
Sum of cube of first k natural numbers = \(\left(\frac{k(k+1)}{2}\right)^2\)
\(= (325)^2 = 1,05,625\)
\(1^3 + 2^3 + 3^3 + \dots + k^3 = 1,05,625\)
21)Determine the nature of the roots for the following quadratic equations \(15x^2 + 11x + 2 = 0\)
Answer: \(15x^2 + 11x + 2 = 0\) comparing with \(ax^2 + bx + c = 0\).
Here a = 15, b = 11, c = 2.
Discriminant \(\Delta = b^2 - 4ac\)
\(= 11^2 - 4(15)(2)\)
\(= 121 - 120\)
\(= 1 > 0\)
∴ The roots are real and unequal.
22)Find the excluded values, if any of the expression \(\frac{t}{t^2-5t+6}\)
Answer: \(\frac{t}{t^2-5t+6}\) is undefined when \(t^2-5t+6=0\) i.e.
\((t-3)(t-2)=0 \Rightarrow t=3, 2\).
∴ The excluded values are 3, 2.
23)Simplify \(\frac{x^3}{x-y} + \frac{y^3}{y-x}\)
Answer: \(\frac{x^3}{x-y} + \frac{y^3}{y-x} = \frac{x^3}{x-y} - \frac{y^3}{x-y}\)
\(= \frac{x^3 - y^3}{x-y}\)
\(= \frac{(x-y)(x^2+xy+y^2)}{x-y}\)
\(= x^2 + xy + y^2\)
24)If ∆ABC is similar to ∆DEF such that BC = 3 cm, EF = 4 cm and area of ∆ABC = 54 cm². Find the area of ∆DEF.
Answer: Since the ratio of area of two similar triangles is equal to the ratio of the squares of any two corresponding sides, we have
\(\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \frac{BC^2}{EF^2}\)
\(\frac{54}{\text{Area}(\triangle DEF)} = \frac{3^2}{4^2} = \frac{9}{16}\)
\(\text{Area}(\triangle DEF) = \frac{16 \times 54}{9} = 16 \times 6 = 96 \text{ cm}^2\)
25)Find the slope of a line joining the given points (- 6, 1) and (-3, 2)
Answer: (-6, 1) and (-3, 2)
The slope \(m = \frac{y_2-y_1}{x_2-x_1} = \frac{2-1}{-3-(-6)} = \frac{1}{-3+6} = \frac{1}{3}\)
26)Show that the straight lines x - 2y + 3 = 0 and 6x + 3y + 8 = 0 are perpendicular.
Answer: Slope of the straight line x - 2y + 3 = 0 is
\(m_1 = -\frac{\text{coeff of x}}{\text{coeff of y}} = -\frac{1}{-2} = \frac{1}{2}\)
Slope of the straight line 6x + 3y + 8 = 0 is
\(m_2 = -\frac{6}{3} = -2\)
Now, \(m_1 \times m_2 = \frac{1}{2} \times (-2) = -1\)
Hence, the two straight lines are perpendicular.
27)Prove the following identity. \(\sqrt{\frac{1+\sin\theta}{1-\sin\theta}} = \sec\theta + \tan\theta\)
Answer:
LHS = \(\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}\)
Multiplying the Numerator and denominator by \(\sqrt{1+\sin\theta}\)
\(= \sqrt{\frac{1+\sin\theta}{1-\sin\theta} \times \frac{1+\sin\theta}{1+\sin\theta}} = \sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}\)
\(= \sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}}\)
\(= \frac{1+\sin\theta}{\cos\theta} = \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\)
\(= \sec\theta + \tan\theta = \text{RHS}\)
28)Find the equation of straight line whose slope -4 and passing through the point (1, 2)
Answer: point = (1, 2); m = -4
Using point-slope form: \(y - y_1 = m(x - x_1)\)
\(y - 2 = -4(x - 1)\)
\(y - 2 = -4x + 4\)
\(4x + y - 2 - 4 = 0\)
\(4x + y - 6 = 0\)
PART - C
Answer any 10 questions. Question no 42 is compulsory (10 x 5 = 50)
29) Let A = {x ∈ W | x < 2}, B = {x ∈ N | 1 < x ≤ 4} and C = {3,5}. Verify that A x (B U C) = (A x B) U (A x C)
Answer: Given A = {x ∈ W | x < 2} ⇒ A = {0,1}
B = {x ∈ N | 1 < x ≤ 4} ⇒ B = {2,3,4}
C = {3,5}
To verify: A x (B U C) = (A x B) U (A x C)
LHS:
B U C = {2,3,4} U {3,5} = {2,3,4,5}
A x (B U 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 = {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) U (A x C) = {(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)} ... (2)
From (1) and (2), it is clear that LHS = RHS.
Hence verified.
30)Let f: A→B be a function defined by \(f(x) = \frac{x}{2} - 1\), 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.
Answer: A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9}, \(f(x) = \frac{x}{2} - 1\)
\(f(2) = \frac{2}{2} - 1 = 1 - 1 = 0\)
\(f(4) = \frac{4}{2} - 1 = 2 - 1 = 1\)
\(f(6) = \frac{6}{2} - 1 = 3 - 1 = 2\)
\(f(10) = \frac{10}{2} - 1 = 5 - 1 = 4\)
\(f(12) = \frac{12}{2} - 1 = 6 - 1 = 5\)
(i) Set of ordered pairs:
f = {(2,0), (4,1), (6,2), (10,4), (12,5)}
(ii) Table:
| x | 2 | 4 | 6 | 10 | 12 |
|---|---|---|---|---|---|
| f(x) | 0 | 1 | 2 | 4 | 5 |
(iii) Arrow diagram:
(iv) Graph:
31) If f(x) = x², g(x) = 3x and h(x) = x - 2, Prove that (f∘g)∘h = f∘(g∘h).
Answer:
LHS: (f∘g)∘h
f∘g = f[g(x)] = f(3x) = (3x)² = 9x²
(f∘g)∘h = (f∘g)[h(x)] = (f∘g)[x - 2] = 9(x - 2)²
RHS: f∘(g∘h)
g∘h = g[h(x)] = g[x - 2] = 3(x - 2)
f∘(g∘h) = f[g(h(x))] = f[3(x - 2)] = [3(x - 2)]² = 9(x - 2)²
Since LHS = RHS, (f∘g)∘h = f∘(g∘h) is proved.
32) The 13th term of an A.P is 3 and the sum of the first 13 terms is 234. Find the common difference and the sum of first 21 terms.
Answer:
Given the 13th term = 3, so \(t_{13} = a + 12d = 3\) ..... (1)
Sum of first 13 terms = 234, so \(S_{13} = \frac{13}{2}[2a + 12d] = 234\)
\(13(a + 6d) = 234 \implies a + 6d = 18\) ..... (2)
Solving (1) and (2):
(1) - (2) \(\implies (a+12d) - (a+6d) = 3 - 18\)
\(6d = -15 \implies d = -\frac{15}{6} = -\frac{5}{2}\)
Substitute d in (2): \(a + 6(-\frac{5}{2}) = 18 \implies a - 15 = 18 \implies a = 33\)
Common difference \(d = -\frac{5}{2}\)
Sum of first 21 terms, \(S_{21} = \frac{21}{2}[2a + (21-1)d]\)
\(S_{21} = \frac{21}{2}[2(33) + 20(-\frac{5}{2})] = \frac{21}{2}[66 - 50] = \frac{21}{2}[16] = 21 \times 8 = 168\).
33) Find the sum to n terms of the series 3 + 33 + 333 + ... to n terms.
Answer:
Let \(S_n = 3 + 33 + 333 + \dots\) to n terms.
Take 3 as a common factor:
\(S_n = 3(1 + 11 + 111 + \dots)\)
Multiply and divide by 9:
\(S_n = \frac{3}{9}(9 + 99 + 999 + \dots)\)
\(S_n = \frac{1}{3}[(10-1) + (10^2-1) + (10^3-1) + \dots \text{to n terms}]\)
Separate the terms into two series:
\(S_n = \frac{1}{3}[(10 + 10^2 + 10^3 + \dots \text{to n terms}) - (1 + 1 + 1 + \dots \text{to n terms})]\)
The first series is a Geometric Progression (G.P.) with first term \(a=10\) and common ratio \(r=10\). The sum is \(S_{GP} = \frac{a(r^n - 1)}{r-1}\).
\(S_n = \frac{1}{3} \left[ \frac{10(10^n - 1)}{10-1} - n \right]\)
\(S_n = \frac{1}{3} \left[ \frac{10}{9}(10^n - 1) - n \right]\)
Therefore, the sum is:
\[ S_n = \frac{10}{27}(10^n - 1) - \frac{n}{3} \]
34) There are 12 pieces of five, ten and twenty rupee currencies whose total value is Rs. 105. When first 2 sorts are interchanged in their numbers its value will be increased by Rs. 20. Find the number of currencies in each sort.
Answer: Let the number of five, ten and twenty rupee currencies be x, y and z respectively.
Given:
Total number of currencies: \(x + y + z = 12\) ... (1)
Total value: \(5x + 10y + 20z = 105\) ... (2)
Value after interchanging x and y: \(10x + 5y + 20z = 105 + 20 = 125\) ... (3)
Solving the system of linear equations:
From the calculations:
Subtracting (3) from (2): \(-5x + 5y = -20\) ... (4)
Solving (1) and (2) by eliminating z, we get: \(15x + 10y = 135\) ... (6)
Solving equations (4) and (6), we get \(y = 3\).
Substituting \(y=3\) in (4): \(-5x + 15 = -20 \implies -5x = -35 \implies x = 7\).
Substituting \(x=7, y=3\) in (1): \(7 + 3 + z = 12 \implies z = 2\).
- The number of five rupee currencies is 7.
- The number of ten rupee currencies is 3.
- The number of twenty rupee currencies is 2.
35) The roots of the equation \(x^2 + 6x - 4 = 0\) are α, β. Find the quadratic equation whose roots are \(\alpha^2\) and \(\beta^2\).
Answer: If the roots are given, the quadratic equation is \(X^2 - (\text{sum of the roots})X + (\text{product of the roots}) = 0\).
For the given equation, \(x^2 + 6x - 4 = 0\):
Sum of roots: \(\alpha + \beta = -\frac{b}{a} = -6\)
Product of roots: \(\alpha\beta = \frac{c}{a} = -4\)
For the new equation with roots \(\alpha^2\) and \(\beta^2\):
New Sum of Roots:
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
\(= (-6)^2 - 2(-4) = 36 + 8 = 44\)
New Product of Roots:
\(\alpha^2\beta^2 = (\alpha\beta)^2 = (-4)^2 = 16\)
The required equation is:
\[ x^2 - (44)x + 16 = 0 \]
\[ x^2 - 44x + 16 = 0 \]
36) Find the square root of the following polynomials by division method: \(37x^2 - 28x^3 + 4x^4 + 42x + 9\).
Answer: First, arrange the polynomial in descending order of powers: \(4x^4 - 28x^3 + 37x^2 + 42x + 9\).
Using the long division method:
The square root of the polynomial is the quotient obtained from the division.
Therefore, \(\sqrt{4x^4 - 28x^3 + 37x^2 + 42x + 9} = |2x^2 - 7x - 3|\).
37) Basic Proportionality Theorem (BPT) or State and prove Thales theorem?
Answer:
Statement: A straight line drawn parallel to a side of a triangle intersecting the other two sides, divides the sides in the same ratio.
Proof:
In ∆ABC, D is a point on AB and E is a point on AC.
To prove: \(\frac{AD}{DB} = \frac{AE}{EC}\)
Construction: Draw a line DE || BC
| No. | Statement | Reason |
|---|---|---|
| 1. | ∠ABC = ∠ADE = ∠1 | Corresponding angles are equal because DE || BC |
| 2. | ∠ACB = ∠AED = ∠2 | Corresponding angles are equal because DE || BC |
| 3. | ∠DAE = ∠BAC = ∠3 | Both triangles have a common angle |
| ∆ABC ~ ∆ADE | By AAA similarity | |
| \(\frac{AB}{AD} = \frac{AC}{AE}\) | Corresponding sides are proportional | |
| \(\frac{AD+DB}{AD} = \frac{AE+EC}{AE}\) | Split AB and AC using the points D and E. | |
| \(1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}\) | On simplification | |
| \(\frac{DB}{AD} = \frac{EC}{AE}\) | Cancelling 1 on both sides | |
| \(\frac{AD}{DB} = \frac{AE}{EC}\) | Taking reciprocals. Hence proved. |
38) Find the area of the quadrilateral formed by the points (8, 6), (5, 11), (-5, 12) and (-4, 3).
Answer: Let the vertices be A(8, 6), B(5, 11), C(-5, 12) and D(-4, 3).
First, we plot the vertices on a coordinate plane to ensure they are taken in the correct order (counter-clockwise) for the area calculation.
Area of the quadrilateral ABCD
\(= \frac{1}{2} \{(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (x_2y_1 + x_3y_2 + x_4y_3 + x_1y_4)\}\)
\(= \frac{1}{2} \{ (8 \cdot 11 + 5 \cdot 12 + (-5) \cdot 3 + (-4) \cdot 6) - (5 \cdot 6 + (-5) \cdot 11 + (-4) \cdot 12 + 8 \cdot 3) \}\)
\(= \frac{1}{2} \{ (88 + 60 - 15 - 24) - (30 - 55 - 48 + 24) \}\)
\(= \frac{1}{2} \{ (109) - (-49) \}\)
\(= \frac{1}{2} \{ 109 + 49 \} = \frac{1}{2} \{ 158 \} = 79 \text{ sq.units}\)
39) Find the equation of a straight line Passing through (1, -4) and has intercepts which are in the ratio 2:5.
Answer:
Let the intercepts be \(a\) (x-intercept) and \(b\) (y-intercept).
The equation of the line in intercept form is \(\frac{x}{a} + \frac{y}{b} = 1\).
Given that the intercepts are in the ratio 2:5, we have \(\frac{a}{b} = \frac{2}{5}\), which implies \(a = \frac{2b}{5}\).
Substitute this into the intercept form:
\[ \frac{x}{(2b/5)} + \frac{y}{b} = 1 \implies \frac{5x}{2b} + \frac{y}{b} = 1 \]
Multiply by \(2b\) to clear the denominators:
\[ 5x + 2y = 2b \]
The line passes through the point (1, -4). Substitute \(x=1\) and \(y=-4\):
\[ 5(1) + 2(-4) = 2b \]
\[ 5 - 8 = 2b \implies -3 = 2b \implies b = -\frac{3}{2} \]
Now, substitute the value of \(b\) back into the equation \(5x + 2y = 2b\):
\[ 5x + 2y = 2\left(-\frac{3}{2}\right) \]
\[ 5x + 2y = -3 \]
The required equation is \(5x + 2y + 3 = 0\).
40) Find the equation of the perpendicular bisector of the line joining the points A(-4, 2) and B(6, -4).
Answer:
The perpendicular bisector passes through the midpoint of AB and is perpendicular to AB.
Step 1: Find the midpoint of 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 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 is the negative reciprocal of \(m_{AB}\).
Slope \(m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-3/5} = \frac{5}{3}\).
Step 4: Find the equation of the line.
Using the point-slope form \(y - y_1 = m(x - x_1)\) with the midpoint M(1, -1) and slope \(m = 5/3\):
\[ y - (-1) = \frac{5}{3}(x - 1) \]
\[ y + 1 = \frac{5}{3}(x - 1) \]
\[ 3(y + 1) = 5(x - 1) \]
\[ 3y + 3 = 5x - 5 \]
The required equation is \(5x - 3y - 8 = 0\).
41) If \(\cot\theta + \tan\theta = x\) and \(\sec\theta - \cos\theta = y\), then prove that \((x^2y)^{2/3} - (xy^2)^{2/3} = 1\).
Answer:
Step 1: Simplify x and y.
\[ x = \cot\theta + \tan\theta = \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} = \frac{\cos^2\theta + \sin^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} \]
\[ y = \sec\theta - \cos\theta = \frac{1}{\cos\theta} - \cos\theta = \frac{1 - \cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta} \]
Step 2: Calculate the term \(x^2y\).
\[ x^2y = \left(\frac{1}{\sin\theta\cos\theta}\right)^2 \left(\frac{\sin^2\theta}{\cos\theta}\right) = \frac{1}{\sin^2\theta\cos^2\theta} \cdot \frac{\sin^2\theta}{\cos\theta} = \frac{1}{\cos^3\theta} \]
Now, find \((x^2y)^{2/3}\):
\[ (x^2y)^{2/3} = \left(\frac{1}{\cos^3\theta}\right)^{2/3} = \frac{1}{(\cos^3\theta)^{2/3}} = \frac{1}{\cos^2\theta} = \sec^2\theta \]
Step 3: Calculate the term \(xy^2\).
\[ xy^2 = \left(\frac{1}{\sin\theta\cos\theta}\right) \left(\frac{\sin^2\theta}{\cos\theta}\right)^2 = \frac{1}{\sin\theta\cos\theta} \cdot \frac{\sin^4\theta}{\cos^2\theta} = \frac{\sin^3\theta}{\cos^3\theta} = \tan^3\theta \]
Now, find \((xy^2)^{2/3}\):
\[ (xy^2)^{2/3} = (\tan^3\theta)^{2/3} = \tan^2\theta \]
Step 4: Prove the identity.
\[ (x^2y)^{2/3} - (xy^2)^{2/3} = \sec^2\theta - \tan^2\theta \]
Using the Pythagorean identity \(\sec^2\theta - \tan^2\theta = 1\),
\[ \sec^2\theta - \tan^2\theta = 1 \]
Hence, proved.
42) Swathi has 15 ice cubes different sizes 9cm, 10 cm, 11cm......... 23 cm. How much volume of ice cubes can be used to prepare some fruit juice with these ice cubes?
Answer:
The total volume is the sum of the volumes of all cubes. The volume of a cube with side 'a' is \(a^3\). We need to find the sum:
\[ S = 9^3 + 10^3 + 11^3 + \dots + 23^3 \]
We can write this sum as the difference of two sums of cubes starting from 1:
\[ S = (1^3 + 2^3 + \dots + 23^3) - (1^3 + 2^3 + \dots + 8^3) \]
The formula for the sum of the cubes of the first n natural numbers is \(\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2\).
Part 1: Sum up to 23.
\[ (1^3 + \dots + 23^3) = \left[ \frac{23(23+1)}{2} \right]^2 = \left[ \frac{23 \times 24}{2} \right]^2 = (23 \times 12)^2 = (276)^2 = 76176 \]
Part 2: Sum up to 8.
\[ (1^3 + \dots + 8^3) = \left[ \frac{8(8+1)}{2} \right]^2 = \left[ \frac{8 \times 9}{2} \right]^2 = (4 \times 9)^2 = (36)^2 = 1296 \]
Final Calculation:
\[ S = 76176 - 1296 = 74880 \]
The total volume of the ice cubes is 74,880 cm³.
PART - D
Answer all the questions (2 x 8 = 16)
43) a) Construct a triangle similar to a given triangle ABC with its sides equal to \(\frac{6}{5}\) of the corresponding sides of the triangle ABC (scale factor \(\frac{6}{5} > 1\)).
Answer:
Steps of construction:
- Constructed a ∆ABC with any measurement.
- Drawn a ray BX making an acute angle with BC on the side opposite to the vertex A.
- Located 6 points B₁, B₂, B₃, B₄, B₅, B₆ on BX such that BB₁ = B₁B₂ = ... = B₅B₆.
- Joined B₅ to C and drawn a line through B₆ parallel to B₅C intersecting the extended line segment BC at C'.
- Drawn a line through C' parallel to CA intersecting the extended BA at A'.
- Then A'BC' is the required triangle each of whose sides is six fifths of the corresponding sides of ∆ABC.
(OR)
b) Construct a ∆PQR such that QR = 6.5 cm, ∠P = 60° and the altitude from P to QR is of length 4.5 cm.
Answer:
Construction:
- Draw a line segment QR = 5 cm.
- At Q, draw QE such that ∠RQE = 30°.
- At Q, draw QF such that ∠EQF = 90°.
- Draw the perpendicular bisector XY to QR, which intersects QF at O and QR at G.
- With O as centre and OQ as radius, draw a circle.
- The major arc QKR contains the angle 30°.
- From G, mark a point H on XY such that GH = 4.2 cm.
- Draw a line AB through H parallel to QR. This line intersects the circle at P and P'.
- Join QP and RP. Then ∆PQR is the required triangle.
44) a) A bus is travelling at a uniform speed of 50 km/hr. Draw the distance-time graph and hence find (i) the constant of variation (ii) how far will it travel in \(1\frac{1}{2}\) hr (iii) the time required to cover a distance of 300 km from the graph.
Answer:
Let x be the time taken in minutes and y be the distance travelled in km.
| Time taken x (in minutes) | 60 | 120 | 180 | 240 |
|---|---|---|---|---|
| Distance y (in km) | 50 | 100 | 150 | 200 |
(i) Observe that as time increases, the distance travelled also increases. Therefore, the variation is a direct variation. It is of the form y = kx.
Constant of variation:
\(k = \frac{y}{x} = \frac{50}{60} = \frac{100}{120} = \frac{150}{180} = \frac{200}{240} = \frac{5}{6}\)
Hence, the relation may be given as \(y = \frac{5}{6}x\).
(ii) From the graph, if x = 90 minutes (\(1\frac{1}{2}\) hours), then \(y = \frac{5}{6} \times 90 = 75\) km. The distance travelled is 75 km.
(iii) From the graph, if y = 300 km, then \(300 = \frac{5}{6}x \implies x = \frac{300 \times 6}{5} = 360\) minutes (or) 6 hours. The time taken to cover 300 km is 6 hours.
(OR)
b) The following table shows the data about the number of pipes and the time taken to fill the same tank.
| No. of pipes (x) | 2 | 3 | 6 | 9 |
|---|---|---|---|---|
| Time Taken (in min) (y) | 45 | 30 | 15 | 10 |
Draw the graph for the above data and hence (i) find the time taken to fill the tank when five pipes are used (ii) Find the number of pipes when the time is 9 minutes.
Answer:
1. Variation: As the number of pipes (x) increases, the time taken (y) decreases. This is an indirect variation.
2. Equation: The equation is of the form \(xy = k\).
\(k = 2 \times 45 = 90\)
\(k = 3 \times 30 = 90\)
So, the equation is \(xy = 90\).
3. Points: (2, 45), (3, 30), (6, 15), (9, 10)
4. Solution from Graph:
(i) When 5 pipes are used (x=5), from the graph, the time taken (y) is 18 minutes.
Verification: \(5 \times y = 90 \implies y = \frac{90}{5} = 18\)
(ii) When the time is 9 minutes (y=9), from the graph, the number of pipes (x) is 10.
Verification: \(x \times 9 = 90 \implies x = \frac{90}{9} = 10\)