10th Maths - Quarterly Exam 2024 - English Medium Original Question Paper | Covai District
QUARTERLY EXAMINATION - 2024
MATHEMATICS - SOLUTIONS
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, Set A = {1, 2, 3, 4, 5}. Number of elements in A, n(A) = 5.
Let the number of elements in Set B be n(B) = m.
The number of relations from A to B is given by the formula \(2^{n(A) \times n(B)}\).
Given, the number of relations is 1024.
So, \(2^{5 \times m} = 1024\).
We know that \(1024 = 2^{10}\).
Therefore, \(2^{5m} = 2^{10}\).
Equating the powers, we get: \(5m = 10 \implies m = \frac{10}{5} = 2\).
The number of elements in B is 2.
Given, Set A = {1, 2, 3, 4, 5}. Number of elements in A, n(A) = 5.
Let the number of elements in Set B be n(B) = m.
The number of relations from A to B is given by the formula \(2^{n(A) \times n(B)}\).
Given, the number of relations is 1024.
So, \(2^{5 \times m} = 1024\).
We know that \(1024 = 2^{10}\).
Therefore, \(2^{5m} = 2^{10}\).
Equating the powers, we get: \(5m = 10 \implies m = \frac{10}{5} = 2\).
The number of elements in B is 2.
b) 2
2. If the ordered pairs (a+2,4) and (5, 2a+b) are equal then (a,b) is
Solution:
If two ordered pairs are equal, their corresponding elements are equal.
(a+2, 4) = (5, 2a+b)
Equating the first elements: \(a + 2 = 5 \implies a = 5 - 2 \implies a = 3\).
Equating the second elements: \(4 = 2a + b\).
Substitute the value of a = 3 into the second equation:
\(4 = 2(3) + b \implies 4 = 6 + b \implies b = 4 - 6 \implies b = -2\).
So, (a, b) = (3, -2).
If two ordered pairs are equal, their corresponding elements are equal.
(a+2, 4) = (5, 2a+b)
Equating the first elements: \(a + 2 = 5 \implies a = 5 - 2 \implies a = 3\).
Equating the second elements: \(4 = 2a + b\).
Substitute the value of a = 3 into the second equation:
\(4 = 2(3) + b \implies 4 = 6 + b \implies b = 4 - 6 \implies b = -2\).
So, (a, b) = (3, -2).
d) (3,-2)
3. If \(f(x)=2x^2\) and \(g(x)=\frac{1}{3x}\), then fog is
Solution:
Given \(f(x) = 2x^2\) and \(g(x) = \frac{1}{3x}\).
We need to find fog, which is f(g(x)).
\(fog(x) = f(g(x)) = f(\frac{1}{3x})\).
Now, substitute \(\frac{1}{3x}\) in place of x in f(x):
\(f(\frac{1}{3x}) = 2 \left( \frac{1}{3x} \right)^2 = 2 \left( \frac{1}{9x^2} \right) = \frac{2}{9x^2}\).
Given \(f(x) = 2x^2\) and \(g(x) = \frac{1}{3x}\).
We need to find fog, which is f(g(x)).
\(fog(x) = f(g(x)) = f(\frac{1}{3x})\).
Now, substitute \(\frac{1}{3x}\) in place of x in f(x):
\(f(\frac{1}{3x}) = 2 \left( \frac{1}{3x} \right)^2 = 2 \left( \frac{1}{9x^2} \right) = \frac{2}{9x^2}\).
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 written in the form \(3q, 3q+1,\) or \(3q+2\).
Case 1: \(n = 3q\).
\(n^3 = (3q)^3 = 27q^3 = 9(3q^3)\). This is divisible by 9, so the remainder is 0.
Case 2: \(n = 3q+1\).
\(n^3 = (3q+1)^3 = (3q)^3 + 3(3q)^2(1) + 3(3q)(1)^2 + 1^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1\). The remainder is 1.
Case 3: \(n = 3q+2\).
\(n^3 = (3q+2)^3 = (3q)^3 + 3(3q)^2(2) + 3(3q)(2)^2 + 2^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8\). The remainder is 8.
The possible remainders are 0, 1, and 8.
Let 'n' be any positive integer. By Euclid's division lemma, n can be written in the form \(3q, 3q+1,\) or \(3q+2\).
Case 1: \(n = 3q\).
\(n^3 = (3q)^3 = 27q^3 = 9(3q^3)\). This is divisible by 9, so the remainder is 0.
Case 2: \(n = 3q+1\).
\(n^3 = (3q+1)^3 = (3q)^3 + 3(3q)^2(1) + 3(3q)(1)^2 + 1^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1\). The remainder is 1.
Case 3: \(n = 3q+2\).
\(n^3 = (3q+2)^3 = (3q)^3 + 3(3q)^2(2) + 3(3q)(2)^2 + 2^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8\). The remainder is 8.
The possible remainders are 0, 1, and 8.
a) 0,1,8
5. The sum of exponents of the prime factors in the prime factorization of 1729 is
Solution:
We need to find the prime factorization of 1729.
1729 is divisible by 7: \(1729 = 7 \times 247\).
247 is divisible by 13: \(247 = 13 \times 19\).
So, the prime factorization is \(1729 = 7^1 \times 13^1 \times 19^1\).
The exponents of the prime factors are 1, 1, and 1.
The sum of the exponents = 1 + 1 + 1 = 3.
We need to find the prime factorization of 1729.
1729 is divisible by 7: \(1729 = 7 \times 247\).
247 is divisible by 13: \(247 = 13 \times 19\).
So, the prime factorization is \(1729 = 7^1 \times 13^1 \times 19^1\).
The exponents of the prime factors are 1, 1, and 1.
The sum of the exponents = 1 + 1 + 1 = 3.
c) 3
6. The next term of the sequence \(\frac{3}{16}, \frac{1}{8}, \frac{1}{12}, \frac{1}{18}, ...\) is
Solution:
Let's find the ratio between consecutive terms to see if it's a Geometric Progression (GP).
\(r = \frac{T_2}{T_1} = \frac{1/8}{3/16} = \frac{1}{8} \times \frac{16}{3} = \frac{2}{3}\).
\(r = \frac{T_3}{T_2} = \frac{1/12}{1/8} = \frac{1}{12} \times \frac{8}{1} = \frac{8}{12} = \frac{2}{3}\).
The common ratio is \(r = \frac{2}{3}\). The sequence is a GP.
The next term is the 5th term, \(T_5 = T_4 \times r\).
\(T_5 = \frac{1}{18} \times \frac{2}{3} = \frac{2}{54} = \frac{1}{27}\).
Let's find the ratio between consecutive terms to see if it's a Geometric Progression (GP).
\(r = \frac{T_2}{T_1} = \frac{1/8}{3/16} = \frac{1}{8} \times \frac{16}{3} = \frac{2}{3}\).
\(r = \frac{T_3}{T_2} = \frac{1/12}{1/8} = \frac{1}{12} \times \frac{8}{1} = \frac{8}{12} = \frac{2}{3}\).
The common ratio is \(r = \frac{2}{3}\). The sequence is a GP.
The next term is the 5th term, \(T_5 = T_4 \times r\).
\(T_5 = \frac{1}{18} \times \frac{2}{3} = \frac{2}{54} = \frac{1}{27}\).
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.
This means that when x = 6, the value of both polynomials should be 0.
Let's check the first polynomial: \(P(x) = x^2 - 2x - 24\).
\(P(6) = 6^2 - 2(6) - 24 = 36 - 12 - 24 = 0\). This confirms (x-6) is a factor.
Now for the second polynomial: \(Q(x) = x^2 - kx - 6\).
Since (x-6) is a factor, \(Q(6) = 0\).
\(Q(6) = 6^2 - k(6) - 6 = 0\).
\(36 - 6k - 6 = 0 \implies 30 - 6k = 0 \implies 30 = 6k \implies k = 5\).
If (x-6) is the HCF, it must be a factor of both polynomials.
This means that when x = 6, the value of both polynomials should be 0.
Let's check the first polynomial: \(P(x) = x^2 - 2x - 24\).
\(P(6) = 6^2 - 2(6) - 24 = 36 - 12 - 24 = 0\). This confirms (x-6) is a factor.
Now for the second polynomial: \(Q(x) = x^2 - kx - 6\).
Since (x-6) is a factor, \(Q(6) = 0\).
\(Q(6) = 6^2 - k(6) - 6 = 0\).
\(36 - 6k - 6 = 0 \implies 30 - 6k = 0 \implies 30 = 6k \implies k = 5\).
b) 5
8. Which of the following should be added to make \(x^4+64\) a perfect square?
Solution:
We want to make \(x^4+64\) a perfect square. This can be written as \((x^2)^2 + 8^2\).
This is in the form \(a^2 + b^2\), where \(a=x^2\) and \(b=8\).
To make it a perfect square \((a+b)^2\) or \((a-b)^2\), we need a \(\pm 2ab\) term.
\((a+b)^2 = a^2 + 2ab + b^2\).
\(2ab = 2 \times x^2 \times 8 = 16x^2\).
So, we need to add \(16x^2\) to get \((x^2+8)^2\).
\(x^4+64 + 16x^2 = x^4 + 16x^2 + 64 = (x^2+8)^2\).
We want to make \(x^4+64\) a perfect square. This can be written as \((x^2)^2 + 8^2\).
This is in the form \(a^2 + b^2\), where \(a=x^2\) and \(b=8\).
To make it a perfect square \((a+b)^2\) or \((a-b)^2\), we need a \(\pm 2ab\) term.
\((a+b)^2 = a^2 + 2ab + b^2\).
\(2ab = 2 \times x^2 \times 8 = 16x^2\).
So, we need to add \(16x^2\) to get \((x^2+8)^2\).
\(x^4+64 + 16x^2 = x^4 + 16x^2 + 64 = (x^2+8)^2\).
b) \(16x^2\)
9. The solution of \((2x-1)^2 = 0\) is Equal to
Solution:
Given \((2x-1)^2 = 0\).
Taking the square root on both sides:
\(2x - 1 = 0\).
\(2x = 1\).
\(x = \frac{1}{2}\).
The solution is \(x = 1/2\). This is not among options a, b, or c.
Given \((2x-1)^2 = 0\).
Taking the square root on both sides:
\(2x - 1 = 0\).
\(2x = 1\).
\(x = \frac{1}{2}\).
The solution is \(x = 1/2\). This is not among options a, b, or c.
d) None of these
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:
In \(\triangle ABC\), we are given that DE || BC.
By the Basic Proportionality Theorem (Thales' Theorem), if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides proportionally.
Therefore, \(\frac{AD}{AB} = \frac{AE}{AC}\).
Given: AB = 3.6 cm, AC = 2.4 cm, AD = 2.1 cm.
\(\frac{2.1}{3.6} = \frac{AE}{2.4}\).
\(AE = \frac{2.1 \times 2.4}{3.6} = \frac{5.04}{3.6}\).
\(AE = \frac{50.4}{36} = 1.4\) cm.
In \(\triangle ABC\), we are given that DE || BC.
By the Basic Proportionality Theorem (Thales' Theorem), if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides proportionally.
Therefore, \(\frac{AD}{AB} = \frac{AE}{AC}\).
Given: AB = 3.6 cm, AC = 2.4 cm, AD = 2.1 cm.
\(\frac{2.1}{3.6} = \frac{AE}{2.4}\).
\(AE = \frac{2.1 \times 2.4}{3.6} = \frac{5.04}{3.6}\).
\(AE = \frac{50.4}{36} = 1.4\) cm.
a) 1.4 cm
11. The Point of intersection of 3x-y = 4 and x+y=8 is
Solution:
We need to solve the system of linear equations:
1) \(3x - y = 4\)
2) \(x + y = 8\)
Add equation (1) and (2) to eliminate y:
\((3x - y) + (x + y) = 4 + 8\).
\(4x = 12 \implies x = 3\).
Substitute x = 3 into equation (2):
\(3 + y = 8 \implies y = 8 - 3 \implies y = 5\).
The point of intersection is (3, 5).
We need to solve the system of linear equations:
1) \(3x - y = 4\)
2) \(x + y = 8\)
Add equation (1) and (2) to eliminate y:
\((3x - y) + (x + y) = 4 + 8\).
\(4x = 12 \implies x = 3\).
Substitute x = 3 into equation (2):
\(3 + y = 8 \implies y = 8 - 3 \implies y = 5\).
The point of intersection is (3, 5).
c) (3,5)
12. The straight line given by the equation x=11 is
Solution:
The equation \(x = 11\) represents a line where the x-coordinate of every point is 11. This is a vertical line. A vertical line is always parallel to the Y-axis.
The equation \(x = 11\) represents a line where the x-coordinate of every point is 11. This is a vertical line. A vertical line is always parallel to the Y-axis.
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 of the line joining (0,0) and (-8,8). Let this be \(m_1\).
\(m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 0}{-8 - 0} = \frac{8}{-8} = -1\).
Let the slope of the perpendicular line be \(m_2\).
The condition for two lines to be perpendicular is \(m_1 \times m_2 = -1\).
\(-1 \times m_2 = -1 \implies m_2 = \frac{-1}{-1} = 1\).
First, find the slope of the line joining (0,0) and (-8,8). Let this be \(m_1\).
\(m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 0}{-8 - 0} = \frac{8}{-8} = -1\).
Let the slope of the perpendicular line be \(m_2\).
The condition for two lines to be perpendicular is \(m_1 \times m_2 = -1\).
\(-1 \times m_2 = -1 \implies m_2 = \frac{-1}{-1} = 1\).
b) 1
14. \(Tan \theta Cosec^2 \theta - Tan \theta\) is Equal to
Solution:
Given expression: \(Tan \theta Cosec^2 \theta - Tan \theta\).
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 into the expression:
\(= Tan \theta \times Cot^2 \theta\).
Since \(Cot \theta = \frac{1}{Tan \theta}\), we have:
\(= Tan \theta \times \frac{1}{Tan^2 \theta} = \frac{1}{Tan \theta} = Cot \theta\).
Given expression: \(Tan \theta Cosec^2 \theta - Tan \theta\).
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 into the expression:
\(= Tan \theta \times Cot^2 \theta\).
Since \(Cot \theta = \frac{1}{Tan \theta}\), we have:
\(= Tan \theta \times \frac{1}{Tan^2 \theta} = \frac{1}{Tan \theta} = Cot \theta\).
d) Cot \(\theta\)
Part II
Answer any 10 from the following and 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:
The Cartesian product AxB is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B.
Set A is the set of all first elements in the ordered pairs of AxB. A = {3, 5}
Set B is the set of all second elements in the ordered pairs of AxB. B = {2, 4}
The Cartesian product AxB is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B.
Set A is the set of all first elements in the ordered pairs of AxB. A = {3, 5}
Set B is the set 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 function:
\(f(2) = (2)^2 - 5(2) + 6\)
\(f(2) = 4 - 10 + 6\)
\(f(2) = 10 - 10 = 0\).
Given \(f(x) = x^2 - 5x + 6\).
To find f(2), substitute x = 2 in the function:
\(f(2) = (2)^2 - 5(2) + 6\)
\(f(2) = 4 - 10 + 6\)
\(f(2) = 10 - 10 = 0\).
17. Find k if fof(k) = 5 Where f(k) = 2k-1.
Solution:
Given \(f(k) = 2k - 1\).
fof(k) means f(f(k)).
f(f(k)) = f(2k - 1).
To find f(2k-1), replace k with (2k-1) in the function f(k):
f(2k-1) = 2(2k - 1) - 1
f(f(k)) = 4k - 2 - 1 = 4k - 3.
Given fof(k) = 5.
So, \(4k - 3 = 5 \implies 4k = 8 \implies k = 2\).
Given \(f(k) = 2k - 1\).
fof(k) means f(f(k)).
f(f(k)) = f(2k - 1).
To find f(2k-1), replace k with (2k-1) in the function f(k):
f(2k-1) = 2(2k - 1) - 1
f(f(k)) = 4k - 2 - 1 = 4k - 3.
Given fof(k) = 5.
So, \(4k - 3 = 5 \implies 4k = 8 \implies k = 2\).
18. If \(800 = a^b \times b^a\), then find a and b.
Solution:
First, find the prime factorization of 800.
\(800 = 8 \times 100 = 2^3 \times 10^2 = 2^3 \times (2 \times 5)^2 = 2^3 \times 2^2 \times 5^2 = 2^5 \times 5^2\).
We are given \(800 = a^b \times b^a\).
Comparing \(2^5 \times 5^2\) with \(a^b \times b^a\), we can identify:
a = 2 and b = 5 (or a = 5 and b = 2).
First, find the prime factorization of 800.
\(800 = 8 \times 100 = 2^3 \times 10^2 = 2^3 \times (2 \times 5)^2 = 2^3 \times 2^2 \times 5^2 = 2^5 \times 5^2\).
We are given \(800 = a^b \times b^a\).
Comparing \(2^5 \times 5^2\) with \(a^b \times b^a\), we can identify:
a = 2 and b = 5 (or 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, we find the number of terms (n).
\(T_n = a + (n-1)d\)
\(97 = 6 + (n-1)7 \implies 91 = (n-1)7 \implies n-1 = 13 \implies n = 14\).
Now, we find the sum using the formula \(S_n = \frac{n}{2}(a+l)\).
\(S_{14} = \frac{14}{2}(6+97) = 7(103) = 721\).
The given series is an Arithmetic Progression (AP).
First term, a = 6.
Common difference, d = 13 - 6 = 7.
Last term, l = 97.
First, we find the number of terms (n).
\(T_n = a + (n-1)d\)
\(97 = 6 + (n-1)7 \implies 91 = (n-1)7 \implies n-1 = 13 \implies n = 14\).
Now, we find the sum using the formula \(S_n = \frac{n}{2}(a+l)\).
\(S_{14} = \frac{14}{2}(6+97) = 7(103) = 721\).
20. Find x so that x+6, x+12, and x+15 are Consecutive Terms of a geometric progression.
Solution:
If three terms a, b, c are in GP, then \(b^2 = ac\).
Here, a = x+6, b = x+12, c = x+15.
\((x+12)^2 = (x+6)(x+15)\)
\(x^2 + 24x + 144 = x^2 + 15x + 6x + 90\)
\(x^2 + 24x + 144 = x^2 + 21x + 90\)
\(24x - 21x = 90 - 144\)
\(3x = -54 \implies x = -18\).
If three terms a, b, c are in GP, then \(b^2 = ac\).
Here, a = x+6, b = x+12, c = x+15.
\((x+12)^2 = (x+6)(x+15)\)
\(x^2 + 24x + 144 = x^2 + 15x + 6x + 90\)
\(x^2 + 24x + 144 = x^2 + 21x + 90\)
\(24x - 21x = 90 - 144\)
\(3x = -54 \implies x = -18\).
21. Find the Lcm of \(8x^4y^2\), \(48x^2y^4\).
Solution:
First, find the prime factorization of the coefficients:
\(8 = 2^3\)
\(48 = 16 \times 3 = 2^4 \times 3\)
The LCM of the coefficients is the highest power of each prime factor: \(2^4 \times 3 = 16 \times 3 = 48\).
For the variables, take the highest power of each variable:
Highest power of x is \(x^4\).
Highest power of y is \(y^4\).
Combining them, the LCM is \(48x^4y^4\).
First, find the prime factorization of the coefficients:
\(8 = 2^3\)
\(48 = 16 \times 3 = 2^4 \times 3\)
The LCM of the coefficients is the highest power of each prime factor: \(2^4 \times 3 = 16 \times 3 = 48\).
For the variables, take the highest power of each variable:
Highest power of x is \(x^4\).
Highest power of y is \(y^4\).
Combining them, the LCM is \(48x^4y^4\).
22. Simplify: \(\frac{x}{x-y} - \frac{y}{y-x}\)
Solution:
Notice that \(y-x = -(x-y)\).
\(\frac{x}{x-y} - \frac{y}{-(x-y)}\)
\(= \frac{x}{x-y} + \frac{y}{x-y}\)
Since the denominators are the same, we can add the numerators:
\(= \frac{x+y}{x-y}\).
Notice that \(y-x = -(x-y)\).
\(\frac{x}{x-y} - \frac{y}{-(x-y)}\)
\(= \frac{x}{x-y} + \frac{y}{x-y}\)
Since the denominators are the same, we can add the numerators:
\(= \frac{x+y}{x-y}\).
23. Determine the quadratic Equation, Whose Sum and Product of roots are -9, 20.
Solution:
Sum of roots (\(\alpha + \beta\)) = -9.
Product of roots (\(\alpha \beta\)) = 20.
The general form of a quadratic equation is:
\(x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0\)
\(x^2 - (-9)x + 20 = 0\)
\(x^2 + 9x + 20 = 0\).
Sum of roots (\(\alpha + \beta\)) = -9.
Product of roots (\(\alpha \beta\)) = 20.
The general form of a quadratic equation is:
\(x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0\)
\(x^2 - (-9)x + 20 = 0\)
\(x^2 + 9x + 20 = 0\).
24. If \(\triangle ABC\) is similar to \(\triangle DEF\) such that BC = 3cm, EF = 4 cm, and area of \(\triangle ABC\) = 54 Cm², find the area of \(\triangle DEF\).
Solution:
If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
\(\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{BC}{EF}\right)^2\)
\(\frac{54}{\text{Area}(\triangle DEF)} = \left(\frac{3}{4}\right)^2 = \frac{9}{16}\)
\(\text{Area}(\triangle DEF) = \frac{54 \times 16}{9} = 6 \times 16 = 96\) cm².
If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
\(\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{BC}{EF}\right)^2\)
\(\frac{54}{\text{Area}(\triangle DEF)} = \left(\frac{3}{4}\right)^2 = \frac{9}{16}\)
\(\text{Area}(\triangle DEF) = \frac{54 \times 16}{9} = 6 \times 16 = 96\) cm².
25. Find the slope of a line joining (-6,1) and (-3,2).
Solution:
The formula for the slope (m) of a line joining points \((x_1, y_1)\) and \((x_2, y_2)\) is:
\(m = \frac{y_2 - y_1}{x_2 - x_1}\)
\(m = \frac{2 - 1}{-3 - (-6)} = \frac{1}{-3 + 6} = \frac{1}{3}\).
The formula for the slope (m) of a line joining points \((x_1, y_1)\) and \((x_2, y_2)\) is:
\(m = \frac{y_2 - y_1}{x_2 - x_1}\)
\(m = \frac{2 - 1}{-3 - (-6)} = \frac{1}{-3 + 6} = \frac{1}{3}\).
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 in slope-intercept form is \(y = mx + c\).
\(y = \frac{1}{\sqrt{3}}x - 3\).
Multiplying by \(\sqrt{3}\), we get \(\sqrt{3}y = x - 3\sqrt{3}\).
The equation is \(x - \sqrt{3}y - 3\sqrt{3} = 0\).
Inclination, \(\theta = 30^\circ\).
Slope, \(m = \tan \theta = \tan 30^\circ = \frac{1}{\sqrt{3}}\).
Y-intercept, c = -3.
The equation of the line in slope-intercept form is \(y = mx + c\).
\(y = \frac{1}{\sqrt{3}}x - 3\).
Multiplying by \(\sqrt{3}\), we get \(\sqrt{3}y = x - 3\sqrt{3}\).
The equation is \(x - \sqrt{3}y - 3\sqrt{3} = 0\).
27. Prove that \(\sqrt{\frac{1+\cos\theta}{1-\cos\theta}} = \csc\theta + \cot\theta\).
Solution:
LHS = \(\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}\)
Multiply numerator and denominator inside the square root by the conjugate of the denominator, which is \(1+\cos\theta\).
LHS = \(\sqrt{\frac{1+\cos\theta}{1-\cos\theta} \times \frac{1+\cos\theta}{1+\cos\theta}}\)
= \(\sqrt{\frac{(1+\cos\theta)^2}{1-\cos^2\theta}}\)
Using the identity \(\sin^2\theta + \cos^2\theta = 1 \implies 1 - \cos^2\theta = \sin^2\theta\).
= \(\sqrt{\frac{(1+\cos\theta)^2}{\sin^2\theta}}\)
= \(\frac{1+\cos\theta}{\sin\theta}\)
= \(\frac{1}{\sin\theta} + \frac{\cos\theta}{\sin\theta}\)
= \(\csc\theta + \cot\theta\) = RHS. (Hence proved)
LHS = \(\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}\)
Multiply numerator and denominator inside the square root by the conjugate of the denominator, which is \(1+\cos\theta\).
LHS = \(\sqrt{\frac{1+\cos\theta}{1-\cos\theta} \times \frac{1+\cos\theta}{1+\cos\theta}}\)
= \(\sqrt{\frac{(1+\cos\theta)^2}{1-\cos^2\theta}}\)
Using the identity \(\sin^2\theta + \cos^2\theta = 1 \implies 1 - \cos^2\theta = \sin^2\theta\).
= \(\sqrt{\frac{(1+\cos\theta)^2}{\sin^2\theta}}\)
= \(\frac{1+\cos\theta}{\sin\theta}\)
= \(\frac{1}{\sin\theta} + \frac{\cos\theta}{\sin\theta}\)
= \(\csc\theta + \cot\theta\) = RHS. (Hence proved)
28. Show that the straight lines 5x+23y+14=0 and 23x-5y+9=0 are perpendicular. (Compulsory)
Solution:
For the first line, \(5x+23y+14=0\), the slope \(m_1\) is given by \(-(\text{coefficient of x})/(\text{coefficient of y})\).
\(m_1 = -\frac{5}{23}\).
For the second line, \(23x-5y+9=0\), the slope \(m_2\) is:
\(m_2 = -\frac{23}{-5} = \frac{23}{5}\).
Two lines are perpendicular if the product of their slopes is -1 (i.e., \(m_1 \times m_2 = -1\)).
\(m_1 \times m_2 = \left(-\frac{5}{23}\right) \times \left(\frac{23}{5}\right) = -1\).
Since the product of their slopes is -1, the lines are perpendicular.
For the first line, \(5x+23y+14=0\), the slope \(m_1\) is given by \(-(\text{coefficient of x})/(\text{coefficient of y})\).
\(m_1 = -\frac{5}{23}\).
For the second line, \(23x-5y+9=0\), the slope \(m_2\) is:
\(m_2 = -\frac{23}{-5} = \frac{23}{5}\).
Two lines are perpendicular if the product of their slopes is -1 (i.e., \(m_1 \times m_2 = -1\)).
\(m_1 \times m_2 = \left(-\frac{5}{23}\right) \times \left(\frac{23}{5}\right) = -1\).
Since the product of their slopes is -1, the lines are perpendicular.
Part III
Answer any 10 from the following and 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)∩(CxC).
Solution:
LHS: AxA
A = {5,6}
AxA = {5,6} x {5,6} = {(5,5), (5,6), (6,5), (6,6)}
RHS: (BxB)∩(CxC)
First, find BxB:
B = {4,5,6}
BxB = {(4,4), (4,5), (4,6), (5,4), (5,5), (5,6), (6,4), (6,5), (6,6)}
Next, find CxC:
C = {5,6,7}
CxC = {(5,5), (5,6), (5,7), (6,5), (6,6), (6,7), (7,5), (7,6), (7,7)}
Now, find the intersection (common elements) of BxB and CxC:
(BxB)∩(CxC) = {(5,5), (5,6), (6,5), (6,6)}
Comparing LHS and RHS, we see that AxA = (BxB)∩(CxC). Hence proved.
LHS: AxA
A = {5,6}
AxA = {5,6} x {5,6} = {(5,5), (5,6), (6,5), (6,6)}
RHS: (BxB)∩(CxC)
First, find BxB:
B = {4,5,6}
BxB = {(4,4), (4,5), (4,6), (5,4), (5,5), (5,6), (6,4), (6,5), (6,6)}
Next, find CxC:
C = {5,6,7}
CxC = {(5,5), (5,6), (5,7), (6,5), (6,6), (6,7), (7,5), (7,6), (7,7)}
Now, find the intersection (common elements) of BxB and CxC:
(BxB)∩(CxC) = {(5,5), (5,6), (6,5), (6,6)}
Comparing LHS and RHS, we see that AxA = (BxB)∩(CxC). Hence proved.
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) Table form, (3) arrow diagram, (4) Graphical form.
Solution:
Given A = {1,2,3,4}, B = {2,5,8,11,14} and f(x) = 3x-1.
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:
(3) Arrow diagram:
Plot the points (1,2), (2,5), (3,8), and (4,11) on a Cartesian coordinate plane.
Given A = {1,2,3,4}, B = {2,5,8,11,14} and f(x) = 3x-1.
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 |
A B
1 ------> 2
2 ------> 5
3 ------> 8
4 ------> 11
14
(Draw two ovals, one for set A and one for set B. List the elements. Draw arrows from each element in A to its corresponding image in B).
(4) Graphical form:Plot the points (1,2), (2,5), (3,8), and (4,11) on a Cartesian coordinate plane.
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: \(f(x^2)\). Replace x with \(x^2\) in the function definition. \(f(x^2) = 3 - 2(x^2) = 3 - 2x^2\).
RHS: \([f(x)]^2\). Square the entire function. \([f(x)]^2 = (3 - 2x)^2 = 3^2 - 2(3)(2x) + (2x)^2 = 9 - 12x + 4x^2\).
Given \(f(x^2) = [f(x)]^2\). \(3 - 2x^2 = 9 - 12x + 4x^2\).
Rearrange to form a quadratic equation: \(4x^2 + 2x^2 - 12x + 9 - 3 = 0\).
\(6x^2 - 12x + 6 = 0\).
Divide the entire equation by 6: \(x^2 - 2x + 1 = 0\).
This is the expansion of \((x-1)^2\).
\((x-1)^2 = 0\).
\(x - 1 = 0 \implies x = 1\).
Given f(x) = 3 - 2x.
LHS: \(f(x^2)\). Replace x with \(x^2\) in the function definition. \(f(x^2) = 3 - 2(x^2) = 3 - 2x^2\).
RHS: \([f(x)]^2\). Square the entire function. \([f(x)]^2 = (3 - 2x)^2 = 3^2 - 2(3)(2x) + (2x)^2 = 9 - 12x + 4x^2\).
Given \(f(x^2) = [f(x)]^2\). \(3 - 2x^2 = 9 - 12x + 4x^2\).
Rearrange to form a quadratic equation: \(4x^2 + 2x^2 - 12x + 9 - 3 = 0\).
\(6x^2 - 12x + 6 = 0\).
Divide the entire equation by 6: \(x^2 - 2x + 1 = 0\).
This is the expansion of \((x-1)^2\).
\((x-1)^2 = 0\).
\(x - 1 = 0 \implies x = 1\).
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.
Step 1: \(210 = 55 \times 3 + 45\)
Step 2: \(55 = 45 \times 1 + 10\)
Step 3: \(45 = 10 \times 4 + 5\)
Step 4: \(10 = 5 \times 2 + 0\)
The last non-zero remainder is the HCF. So, HCF(210, 55) = 5.
It is given that the HCF is expressible as 55x - 325.
Therefore, \(55x - 325 = 5\).
\(55x = 5 + 325\).
\(55x = 330\).
\(x = \frac{330}{55} = 6\).
Thus, x = 6.
First, find the HCF of 210 and 55 using Euclid's division algorithm.
Step 1: \(210 = 55 \times 3 + 45\)
Step 2: \(55 = 45 \times 1 + 10\)
Step 3: \(45 = 10 \times 4 + 5\)
Step 4: \(10 = 5 \times 2 + 0\)
The last non-zero remainder is the HCF. So, HCF(210, 55) = 5.
It is given that the HCF is expressible as 55x - 325.
Therefore, \(55x - 325 = 5\).
\(55x = 5 + 325\).
\(55x = 330\).
\(x = \frac{330}{55} = 6\).
Thus, x = 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 A.P. be a, a+d, a+2d, ...
The nth term is \(T_n = a + (n-1)d\).
Given, \(\frac{T_6}{T_8} = \frac{7}{9}\).
\(\frac{a + (6-1)d}{a + (8-1)d} = \frac{a+5d}{a+7d} = \frac{7}{9}\).
Cross-multiply: \(9(a+5d) = 7(a+7d)\).
\(9a + 45d = 7a + 49d\).
\(9a - 7a = 49d - 45d\).
\(2a = 4d \implies a = 2d\).
We need to find the ratio \(\frac{T_9}{T_{13}}\).
\(\frac{T_9}{T_{13}} = \frac{a+8d}{a+12d}\).
Substitute a = 2d into this expression:
\(\frac{2d+8d}{2d+12d} = \frac{10d}{14d} = \frac{10}{14} = \frac{5}{7}\).
The ratio of the 9th term to the 13th term is 5:7.
Let the A.P. be a, a+d, a+2d, ...
The nth term is \(T_n = a + (n-1)d\).
Given, \(\frac{T_6}{T_8} = \frac{7}{9}\).
\(\frac{a + (6-1)d}{a + (8-1)d} = \frac{a+5d}{a+7d} = \frac{7}{9}\).
Cross-multiply: \(9(a+5d) = 7(a+7d)\).
\(9a + 45d = 7a + 49d\).
\(9a - 7a = 49d - 45d\).
\(2a = 4d \implies a = 2d\).
We need to find the ratio \(\frac{T_9}{T_{13}}\).
\(\frac{T_9}{T_{13}} = \frac{a+8d}{a+12d}\).
Substitute a = 2d into this expression:
\(\frac{2d+8d}{2d+12d} = \frac{10d}{14d} = \frac{10}{14} = \frac{5}{7}\).
The ratio of the 9th term to the 13th term is 5:7.
34. Rekha has 15 square colour papers of sizes 10cm, 11cm, 12cm,..., 24cm. How much area can be decorated with these colour papers?
Solution:
The sides of the square papers are 10, 11, 12, ..., 24 cm.
The area of a square is side². Total area = \(10^2 + 11^2 + 12^2 + ... + 24^2\).
We can find this sum using the formula for the sum of the first n squares: \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\).
Total area = \( (1^2 + 2^2 + ... + 24^2) - (1^2 + 2^2 + ... + 9^2) \).
Sum of first 24 squares: \(S_{24} = \frac{24(24+1)(2 \times 24+1)}{6} = \frac{24 \times 25 \times 49}{6} = 4 \times 25 \times 49 = 100 \times 49 = 4900\).
Sum of first 9 squares: \(S_9 = \frac{9(9+1)(2 \times 9+1)}{6} = \frac{9 \times 10 \times 19}{6} = \frac{1710}{6} = 285\).
Total area = \(S_{24} - S_9 = 4900 - 285 = 4615\) cm².
The sides of the square papers are 10, 11, 12, ..., 24 cm.
The area of a square is side². Total area = \(10^2 + 11^2 + 12^2 + ... + 24^2\).
We can find this sum using the formula for the sum of the first n squares: \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\).
Total area = \( (1^2 + 2^2 + ... + 24^2) - (1^2 + 2^2 + ... + 9^2) \).
Sum of first 24 squares: \(S_{24} = \frac{24(24+1)(2 \times 24+1)}{6} = \frac{24 \times 25 \times 49}{6} = 4 \times 25 \times 49 = 100 \times 49 = 4900\).
Sum of first 9 squares: \(S_9 = \frac{9(9+1)(2 \times 9+1)}{6} = \frac{9 \times 10 \times 19}{6} = \frac{1710}{6} = 285\).
Total area = \(S_{24} - S_9 = 4900 - 285 = 4615\) cm².
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 expressions.
For x: \(x = \frac{a^2+3a-4}{3a^2-3} = \frac{(a+4)(a-1)}{3(a^2-1)} = \frac{(a+4)(a-1)}{3(a-1)(a+1)} = \frac{a+4}{3(a+1)}\).
For y: \(y = \frac{a^2+2a-8}{2a^2-2a-4} = \frac{(a+4)(a-2)}{2(a^2-a-2)} = \frac{(a+4)(a-2)}{2(a-2)(a+1)} = \frac{a+4}{2(a+1)}\).
We need to find \(x^2y^{-2}\), which is \(\frac{x^2}{y^2} = \left(\frac{x}{y}\right)^2\).
First, find \(\frac{x}{y}\):
\(\frac{x}{y} = \frac{\frac{a+4}{3(a+1)}}{\frac{a+4}{2(a+1)}} = \frac{a+4}{3(a+1)} \times \frac{2(a+1)}{a+4} = \frac{2}{3}\).
Now, find \(\left(\frac{x}{y}\right)^2\):
\(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\).
First, simplify x and y by factoring the expressions.
For x: \(x = \frac{a^2+3a-4}{3a^2-3} = \frac{(a+4)(a-1)}{3(a^2-1)} = \frac{(a+4)(a-1)}{3(a-1)(a+1)} = \frac{a+4}{3(a+1)}\).
For y: \(y = \frac{a^2+2a-8}{2a^2-2a-4} = \frac{(a+4)(a-2)}{2(a^2-a-2)} = \frac{(a+4)(a-2)}{2(a-2)(a+1)} = \frac{a+4}{2(a+1)}\).
We need to find \(x^2y^{-2}\), which is \(\frac{x^2}{y^2} = \left(\frac{x}{y}\right)^2\).
First, find \(\frac{x}{y}\):
\(\frac{x}{y} = \frac{\frac{a+4}{3(a+1)}}{\frac{a+4}{2(a+1)}} = \frac{a+4}{3(a+1)} \times \frac{2(a+1)}{a+4} = \frac{2}{3}\).
Now, find \(\left(\frac{x}{y}\right)^2\):
\(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\).
36. Find the square root of \(64x^4 - 16x^3 + 17x^2 - 2x + 1\).
Solution:
We use the long division method for finding the square root of a polynomial.
We use the long division method for finding the square root of a polynomial.
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
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) \(\frac{1}{\alpha} + \frac{1}{\beta}\) and 2) \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\).
Solution:
For the quadratic equation \(2x^2-7x+5=0\), a=2, b=-7, c=5.
Sum of roots, \(\alpha + \beta = -\frac{b}{a} = -\frac{-7}{2} = \frac{7}{2}\).
Product of roots, \(\alpha \beta = \frac{c}{a} = \frac{5}{2}\).
1) \(\frac{1}{\alpha} + \frac{1}{\beta}\)
\(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \beta} = \frac{\alpha + \beta}{\alpha \beta}\).
\( = \frac{7/2}{5/2} = \frac{7}{5}\).
2) \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\)
\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta}\).
First, find \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\).
\( = \left(\frac{7}{2}\right)^2 - 2\left(\frac{5}{2}\right) = \frac{49}{4} - 5 = \frac{49-20}{4} = \frac{29}{4}\).
Now substitute this into the expression:
\(\frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{29/4}{5/2} = \frac{29}{4} \times \frac{2}{5} = \frac{29}{10}\).
For the quadratic equation \(2x^2-7x+5=0\), a=2, b=-7, c=5.
Sum of roots, \(\alpha + \beta = -\frac{b}{a} = -\frac{-7}{2} = \frac{7}{2}\).
Product of roots, \(\alpha \beta = \frac{c}{a} = \frac{5}{2}\).
1) \(\frac{1}{\alpha} + \frac{1}{\beta}\)
\(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \beta} = \frac{\alpha + \beta}{\alpha \beta}\).
\( = \frac{7/2}{5/2} = \frac{7}{5}\).
2) \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\)
\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta}\).
First, find \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\).
\( = \left(\frac{7}{2}\right)^2 - 2\left(\frac{5}{2}\right) = \frac{49}{4} - 5 = \frac{49-20}{4} = \frac{29}{4}\).
Now substitute this into the expression:
\(\frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{29/4}{5/2} = \frac{29}{4} \times \frac{2}{5} = \frac{29}{10}\).
38. State and prove basic Proportionality theorem.
Solution:
Statement (Basic Proportionality Theorem or Thales' 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 \(\triangle ABC\), a line DE is parallel to BC and intersects 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 a triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\).
Area(\(\triangle ADE\)) = \(\frac{1}{2} \times AD \times EN\). (i)
Area(\(\triangle BDE\)) = \(\frac{1}{2} \times DB \times EN\). (ii)
Divide (i) by (ii): \(\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB}\). (iii)
Similarly,
Area(\(\triangle ADE\)) = \(\frac{1}{2} \times AE \times DM\). (iv)
Area(\(\triangle CDE\)) = \(\frac{1}{2} \times EC \times DM\). (v)
Divide (iv) by (v): \(\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}\). (vi)
Now, \(\triangle BDE\) and \(\triangle CDE\) are on the same base DE and between the same parallel lines DE and BC. Therefore, Area(\(\triangle BDE\)) = Area(\(\triangle CDE\)).
From (iii) and (vi), using this equality:
\(\frac{AD}{DB} = \frac{AE}{EC}\).
Hence, the theorem is proved.
Statement (Basic Proportionality Theorem or Thales' 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 \(\triangle ABC\), a line DE is parallel to BC and intersects 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 a triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\).
Area(\(\triangle ADE\)) = \(\frac{1}{2} \times AD \times EN\). (i)
Area(\(\triangle BDE\)) = \(\frac{1}{2} \times DB \times EN\). (ii)
Divide (i) by (ii): \(\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB}\). (iii)
Similarly,
Area(\(\triangle ADE\)) = \(\frac{1}{2} \times AE \times DM\). (iv)
Area(\(\triangle CDE\)) = \(\frac{1}{2} \times EC \times DM\). (v)
Divide (iv) by (v): \(\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}\). (vi)
Now, \(\triangle BDE\) and \(\triangle CDE\) are on the same base DE and between the same parallel lines DE and BC. Therefore, Area(\(\triangle BDE\)) = Area(\(\triangle CDE\)).
From (iii) and (vi), using this equality:
\(\frac{AD}{DB} = \frac{AE}{EC}\).
Hence, the theorem is 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).
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 = \(\frac{1}{2} \left| \begin{matrix} 8 & 5 & -5 & -4 & 8 \\ 6 & 11 & 12 & 3 & 6 \end{matrix} \right|\)
= \(\frac{1}{2} |((8)(11) + (5)(12) + (-5)(3) + (-4)(6)) - ((5)(6) + (-5)(11) + (-4)(12) + (8)(3))|\)
= \(\frac{1}{2} |(88 + 60 - 15 - 24) - (30 - 55 - 48 + 24)|\)
= \(\frac{1}{2} |(148 - 39) - (54 - 103)|\)
= \(\frac{1}{2} |(109) - (-49)|\)
= \(\frac{1}{2} |109 + 49|\)
= \(\frac{1}{2} |158| = 79\) sq. units.
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 = \(\frac{1}{2} \left| \begin{matrix} 8 & 5 & -5 & -4 & 8 \\ 6 & 11 & 12 & 3 & 6 \end{matrix} \right|\)
= \(\frac{1}{2} |((8)(11) + (5)(12) + (-5)(3) + (-4)(6)) - ((5)(6) + (-5)(11) + (-4)(12) + (8)(3))|\)
= \(\frac{1}{2} |(88 + 60 - 15 - 24) - (30 - 55 - 48 + 24)|\)
= \(\frac{1}{2} |(148 - 39) - (54 - 103)|\)
= \(\frac{1}{2} |(109) - (-49)|\)
= \(\frac{1}{2} |109 + 49|\)
= \(\frac{1}{2} |158| = 79\) 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 (to side BC):
First, find the slope of BC. \(m_{BC} = \frac{3 - (-2)}{12 - 10} = \frac{5}{2}\).
The altitude from A is perpendicular to BC. So, its slope \(m_{AD}\) is \(-\frac{1}{m_{BC}} = -\frac{2}{5}\).
The altitude passes through A(-3,0). Using the point-slope form \(y - y_1 = m(x - x_1)\):
\(y - 0 = -\frac{2}{5}(x - (-3))\)
\(5y = -2(x+3) \implies 5y = -2x - 6 \implies 2x + 5y + 6 = 0\).
Altitude through B (to side AC):
First, find the slope of AC. \(m_{AC} = \frac{3 - 0}{12 - (-3)} = \frac{3}{15} = \frac{1}{5}\).
The altitude from B is perpendicular to AC. So, its slope \(m_{BE}\) is \(-\frac{1}{m_{AC}} = -5\).
The altitude passes through B(10,-2). Using the point-slope form:
\(y - (-2) = -5(x - 10)\)
\(y + 2 = -5x + 50 \implies 5x + y + 2 - 50 = 0 \implies 5x + y - 48 = 0\).
Altitude through A (to side BC):
First, find the slope of BC. \(m_{BC} = \frac{3 - (-2)}{12 - 10} = \frac{5}{2}\).
The altitude from A is perpendicular to BC. So, its slope \(m_{AD}\) is \(-\frac{1}{m_{BC}} = -\frac{2}{5}\).
The altitude passes through A(-3,0). Using the point-slope form \(y - y_1 = m(x - x_1)\):
\(y - 0 = -\frac{2}{5}(x - (-3))\)
\(5y = -2(x+3) \implies 5y = -2x - 6 \implies 2x + 5y + 6 = 0\).
Altitude through B (to side AC):
First, find the slope of AC. \(m_{AC} = \frac{3 - 0}{12 - (-3)} = \frac{3}{15} = \frac{1}{5}\).
The altitude from B is perpendicular to AC. So, its slope \(m_{BE}\) is \(-\frac{1}{m_{AC}} = -5\).
The altitude passes through B(10,-2). Using the point-slope form:
\(y - (-2) = -5(x - 10)\)
\(y + 2 = -5x + 50 \implies 5x + y + 2 - 50 = 0 \implies 5x + y - 48 = 0\).
41. Prove that \(\frac{\sin A}{1+\cos A} + \frac{\sin A}{1-\cos A} = 2 \csc A\).
Solution:
LHS = \(\frac{\sin A}{1+\cos A} + \frac{\sin A}{1-\cos A}\)
Factor out \(\sin A\):
= \(\sin A \left( \frac{1}{1+\cos A} + \frac{1}{1-\cos A} \right)\)
Find a common denominator:
= \(\sin A \left( \frac{(1-\cos A) + (1+\cos A)}{(1+\cos A)(1-\cos A)} \right)\)
= \(\sin A \left( \frac{2}{1-\cos^2 A} \right)\)
Using the identity \(\sin^2 A + \cos^2 A = 1 \implies 1 - \cos^2 A = \sin^2 A\).
= \(\sin A \left( \frac{2}{\sin^2 A} \right)\)
= \(\frac{2}{\sin A}\)
Since \(\csc A = \frac{1}{\sin A}\), we have:
= \(2 \csc A\) = RHS. (Hence proved)
LHS = \(\frac{\sin A}{1+\cos A} + \frac{\sin A}{1-\cos A}\)
Factor out \(\sin A\):
= \(\sin A \left( \frac{1}{1+\cos A} + \frac{1}{1-\cos A} \right)\)
Find a common denominator:
= \(\sin A \left( \frac{(1-\cos A) + (1+\cos A)}{(1+\cos A)(1-\cos A)} \right)\)
= \(\sin A \left( \frac{2}{1-\cos^2 A} \right)\)
Using the identity \(\sin^2 A + \cos^2 A = 1 \implies 1 - \cos^2 A = \sin^2 A\).
= \(\sin A \left( \frac{2}{\sin^2 A} \right)\)
= \(\frac{2}{\sin A}\)
Since \(\csc A = \frac{1}{\sin A}\), we have:
= \(2 \csc A\) = RHS. (Hence proved)
42. Solve: x+y+z=5; 2x-y+z=9; x-2y+3z = 16. (Compulsory)
Solution:
(1) \(x+y+z=5\)
(2) \(2x-y+z=9\)
(3) \(x-2y+3z=16\)
Add (1) and (2) to eliminate y: \((x+y+z) + (2x-y+z) = 5+9\)
(4) \(3x+2z=14\)
Multiply (1) by 2 and add to (3) to eliminate y: \(2(x+y+z) = 2(5) \implies 2x+2y+2z=10\)
\((2x+2y+2z) + (x-2y+3z) = 10+16\)
(5) \(3x+5z=26\)
Now solve the system of equations (4) and (5):
(4) \(3x+2z=14\)
(5) \(3x+5z=26\)
Subtract (4) from (5):
\((3x+5z) - (3x+2z) = 26 - 14\)
\(3z = 12 \implies z=4\).
Substitute z=4 into (4):
\(3x+2(4)=14 \implies 3x+8=14 \implies 3x=6 \implies x=2\).
Substitute x=2 and z=4 into (1):
\(2+y+4=5 \implies y+6=5 \implies y=-1\).
The solution is x=2, y=-1, z=4.
(1) \(x+y+z=5\)
(2) \(2x-y+z=9\)
(3) \(x-2y+3z=16\)
Add (1) and (2) to eliminate y: \((x+y+z) + (2x-y+z) = 5+9\)
(4) \(3x+2z=14\)
Multiply (1) by 2 and add to (3) to eliminate y: \(2(x+y+z) = 2(5) \implies 2x+2y+2z=10\)
\((2x+2y+2z) + (x-2y+3z) = 10+16\)
(5) \(3x+5z=26\)
Now solve the system of equations (4) and (5):
(4) \(3x+2z=14\)
(5) \(3x+5z=26\)
Subtract (4) from (5):
\((3x+5z) - (3x+2z) = 26 - 14\)
\(3z = 12 \implies z=4\).
Substitute z=4 into (4):
\(3x+2(4)=14 \implies 3x+8=14 \implies 3x=6 \implies x=2\).
Substitute x=2 and z=4 into (1):
\(2+y+4=5 \implies y+6=5 \implies y=-1\).
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 \(\frac{3}{5} < 1\)] (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 \(\frac{6}{5} > 1\)]
Solution for 43(a):
Steps of Construction:
Solution for 43(b):
Steps of Construction:
Steps of Construction:
- Construct the given triangle PQR with any measurements.
- Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
- Since the scale factor is \(\frac{3}{5}\), the greater number is 5. Locate 5 points \(Q_1, Q_2, Q_3, Q_4, Q_5\) on QX such that \(QQ_1 = Q_1Q_2 = Q_2Q_3 = Q_3Q_4 = Q_4Q_5\).
- Join \(Q_5\) to R.
- The numerator of the scale factor is 3. From point \(Q_3\), draw a line parallel to \(Q_5R\) to intersect QR at R'.
- From R', draw a line parallel to PR to intersect PQ at P'.
- The triangle P'QR' is the required similar triangle with sides equal to \(\frac{3}{5}\) of the corresponding sides of \(\triangle PQR\).
Solution for 43(b):
Steps of Construction:
- Construct the given triangle PQR with any measurements.
- Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
- Since the scale factor is \(\frac{6}{5}\), the greater number is 6. Locate 6 points \(Q_1, Q_2, Q_3, Q_4, Q_5, Q_6\) on QX such that the segments are equal.
- The denominator of the scale factor is 5. Join \(Q_5\) to R.
- Extend the line segment QR.
- From point \(Q_6\), draw a line parallel to \(Q_5R\) to intersect the extended line QR at R'.
- Extend the line segment QP. From R', draw a line parallel to PR to intersect the extended line QP at P'.
- The triangle P'QR' is the required similar triangle with sides equal to \(\frac{6}{5}\) of the corresponding sides of \(\triangle PQR\).
44. (a) A bus is travelling at a uniform speed of 50 km/h. Draw 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. Given Speed = 50 km/h, so \(d = 50t\). This is a direct variation.
Table of values:
Graph: Plot the points (1, 50), (2, 100), (3, 150), (4, 200) on a graph paper. Join the points to get a straight line passing through the origin. (Scale: X-axis 1 cm = 1 hour, Y-axis 1 cm = 50 km).
From the graph:
Solution for 44(b):
The relationship is \(xy=24\), which is an inverse variation (\(y = \frac{24}{x}\)).
Table of values (for x, y > 0):
Graph: Plot these points on a graph paper and join them with a smooth curve. This curve is a rectangular hyperbola in the first quadrant. (Scale: X-axis 1 cm = 2 units, Y-axis 1 cm = 2 units).
From the graph:
The relationship is Distance = Speed × Time. Given Speed = 50 km/h, so \(d = 50t\). This is a direct variation.
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. On the X-axis, locate t = 1.5. Draw a vertical line to meet the graph. From that point, draw a horizontal line to meet the Y-axis. The value on the Y-axis is 75.
Answer: The bus will travel 75 km in 90 minutes. - Time for 300 km: On the Y-axis, locate d = 300. Draw a horizontal line to meet the graph. From that point, draw a vertical line to meet the X-axis. The value on the X-axis is 6.
Answer: The time required to cover 300 km is 6 hours.
Solution for 44(b):
The relationship is \(xy=24\), which is an inverse variation (\(y = \frac{24}{x}\)).
Table of values (for x, y > 0):
| x | 1 | 2 | 3 | 4 | 6 | 8 | 12 |
|---|---|---|---|---|---|---|---|
| y | 24 | 12 | 8 | 6 | 4 | 3 | 2 |
From the graph:
- Find y when x=3: On the X-axis, locate x=3. Draw a vertical line up to the curve. From that point on the curve, draw a horizontal line to the Y-axis. The value on the Y-axis is 8.
Answer: When x=3, y=8. - Find x when y=6: On the Y-axis, locate y=6. Draw a horizontal line to the curve. From that point on the curve, draw a vertical line down to the X-axis. The value on the X-axis is 4.
Answer: When y=6, x=4.