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CBSE X
All India
MATHS PAPER 2009
Time allowed: 180 minutes; Maximum Marks: 90
General Instructions: | |
1) | All questions are compulsory. |
2) | The question paper consists of thirty questions divided into 4 sections A, B, C and D. Section A comprises of ten questions of 01 mark each, Section B comprises of five questions of 02 marks each, Section C comprises ten questions of 03 marks each and Section D comprises of five questions of 06 marks each. |
3) | All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. |
4) | There is no overall choice. However, internal choice has been provided in one question of 02 marks each, three questions of 03 marks each and two questions of 06 marks each. You have to attempt only one of the alternatives in all such questions. |
5) | In question on construction, drawing should be near and exactly as per the given measurements. |
6) | Use of calculators is not permitted. |
Section A | Section B | Section C | Section D |
1. Write whether the rational number $\frac{51}{1500}$ will have a terminating decimal expansion or a non-terminating repeating decimal expansion.
1) Write the denominator of the rational number in the exponential form using prime factors with exact exponents.
2) Verify the exponents of the exponential form of the denominator whether they are non-negative integers.
1) Write the denominator of the rational number in exponential form using prime factors.
2) If the denominator is of the form ${2}^{n}{5}^{m}$, where n and m are non-negative integers, then the rational number has a decimal expansion which terminates.
3) If the denominator is not of the form ${2}^{n}{5}^{m}$, where n and m are non-negative integers, then the rational number has a decimal expansion which is non-terminating repeating (recurring).
Given rational number is $\frac{51}{1500}$. It can also be written as
$\frac{51}{1500}$ = $\frac{(3\times 17)}{(2\times 2\times 3\times 5\times 5\times 5)}$= $\frac{17}{(2\times 2\times 5\times 5\times 5)}$= $\frac{17}{({2}^{2}\times {5}^{3})}$
The number is of type $\frac{17}{{2}^{x}{5}^{y}}$, in which x = 2 and y = 3.
Since x and y are non-negative integers, the rational number $\frac{51}{1500}$ or $\frac{17}{{2}^{2}\times {5}^{3}}$ will have a terminating decimal expansion.
2. Write the polynomial, the product and sum of whose zeroes are $\frac{\xe2\u02c6\u20199}{2}$ and$\frac{\xe2\u02c6\u20193}{2}$ respectively.
1) Write the general form of the quadratic polynomial with correct signs.
2) Substitute the values of sum and product of the roots according to the order.
3) Avoid calculation mistakes.
1) Write the general form of the quadratic polynomial.
2) Substitute the sum and product of the zeroes in the general form of the quadratic polynomial.
3) Simplify the polynomial for fractions.
4) Assume the value of 'k' to eliminate the denominator of the fractions.
Assume α and Î² as the zeroes of the polynomial.
The product of the zeroes = αÎ² = $\frac{\xe2\u02c6\u20199}{2}$
The sum of the zeroes = α + Î² = $\frac{\xe2\u02c6\u20193}{2}$
The required polynomial can be written as
p(x) = k[${x}^{2}$âˆ’(α + Î²)x + αÎ²]
p(x) = k[${x}^{2}$âˆ’(âˆ’$\frac{3}{2}$)x + (âˆ’$\frac{9}{2}$)]
p(x) = k[${x}^{2}$ + $\frac{3}{2}$ x âˆ’ $\frac{9}{2}$]
If k = 2, p(x) = 2${x}^{2}$ + 3x âˆ’ 9.
So, one of the polynomials, where the product of zeroes is $\frac{\xe2\u02c6\u20199}{2}$ and sum of zeroes is $\frac{\xe2\u02c6\u20193}{2}$, is 2${x}^{2}$ + 3x âˆ’ 9.
3. Write whether the following pair of linear equations is consistent or not:
x + y = 14
x âˆ’ y = 4
1) Write both the given equations in the form of ax + by + c = 0.
2) Write the coefficients of the equations in correct order.
3) Consider the ratios of coefficients of x terms, y terms and constants in the same order.1) Write the given equations in the form of ax + by + c = 0.
2) Comapre the given equations to ${a}_{1}$x +${b}_{1}$y + ${c}_{1}$= 0 and ${a}_{2}$x + ${b}_{2}$y + ${c}_{2}$= 0, and note the values of coefficients.
3) Find the ratios of coefficients of corresponding terms.4) Compare the ratios for their equality and decide whether the equations are consistent, inconsistent or dependent.
The equation x + y = 14 can also be written as x + y âˆ’14 = 0
Similarly, x âˆ’ y = 4 can be written as x âˆ’ y âˆ’ 4 = 0
x + y âˆ’14 = 0 when compared to ${a}_{1}$x + ${b}_{1}$y + ${c}_{1}$= 0, ${a}_{1}$= 1, ${b}_{1}$= 1 and ${c}_{1}$= âˆ’14.
x âˆ’ y âˆ’ 4 = 0 when compared to ${a}_{2}$x + ${b}_{2}$y + ${c}_{2}$= 0, ${a}_{2}$= 1, ${b}_{2}$= âˆ’1 and ${c}_{2}$= âˆ’ 4.
$\frac{{a}_{1}}{{a}_{2}}$= $\frac{1}{1}$
$\frac{{b}_{1}}{{b}_{2}}$= $\frac{1}{\xe2\u02c6\u20191}$
We can notice from above that$$$$$\frac{{a}_{1}}{{a}_{2}}\xe2\u2030\frac{{b}_{1}}{{b}_{2}}$.
Therefore, the pair of linear equations given in the question is consistent.
4. Write the nature of roots of quadratic equation 4${x}^{2}$+ 4$\sqrt{3}$x + 3 = 0.
1) Write the coefficients of the equation with their signs.
2) Write the discriminant of the equation correctly.
3) Avoid calculation mistakes.
1) Compare the given quadratic equation with a${x}^{2}$+ bx + c = 0 and write the values of a, b and c.
2) Write the expression to find the discriminant of the equation.
3) Substitute the values of a, b and c in the expression to find the discriminant.
4) Simplify and check the value.
5) If D = 0, the equation has two equal real roots. If D < 0, the equation has no real roots. If D > 0, the equation has two distinct real roots.
The given equation is a quadratic equation.
When it is compared to a${x}^{2}$+ bx + c = 0, a = 4, b = 4$\sqrt{3}$âˆšand c = 3.
The discriminant or D is given by ${b}^{2}$-4ac.
Substituting a = 4, b = 4$\sqrt{3}$ and c = 3, we get
${b}^{2}$-4ac = $({4\sqrt{3})}^{2}$ - 4 × 4 × 3 = 48 - 48 = 0.
Value of the discriminant is equal to zero.
Therefore, there are two equal and real roots for the given quadratic equation.
5. For what value of k, are the numbers x, 2x + k and 3x + 6 three consecutive terms of an A.P.
1) Write the correct formula to find the arithmetic mean of an A.P.
2) Substitute the values of the terms in the formula correctly.
3) Avoid calculation mistakes.
1) Write the formula for arithmetic mean of three terms of an A.P.
2) Substitute the values of the terms in the formula.
3) Simplify to find the value of k.
Assume that a, b and c are the three consecutive terms of the A.P. where a = x, b = 2x + k and c = 3x + 6.
For an A.P. , 2b = a + c
Substituting a = x, b = 2x + k and c = 3x + 6 in the above equation, we get
2 × (2x + k) = x + 3x + 6
⇒4x + 2k = 4x + 6
⇒2k = 6
⇒k = 3
Thus, x, 2x + k and 3x + 6 are three consecutive terms of an A.P. if k = 3.
6. In a Î”ABC, DE || BC. If DE = $\frac{2}{3}$ BC and area of Î”ABC = 81${\mathrm{cm}}^{2}$, find the area of Î”ADE.
1) Compare the corresponding angles of the triangles.
2) Choose the correct criterion of similarity.
3) Choose the corresponding sides correctly to write the ratio of their squares.
1) Prove that Î”ADE is similar to Î”ABC by proving that their corresponding angles are equal.
2) Equate the ratio of the areas of similar triangle to the ratio of squares of corresponding sides.
3) Find the ratio of ${\mathrm{DE}}^{2}$ and ${\mathrm{BC}}^{2}$.
4) Find area of â–³ADE using the above ratio and area of â–³ABC.
âˆ DAE = âˆ BAC [Since it is common angle ]
âˆ ADE = âˆ ABC [Since they are corresponding angles]
Hence according to the AA similarity criterion, Î”ADE ~ Î”ABC
The ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
âˆ´ $\frac{\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{ADE}}{\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{ABC}}$ = $\frac{{\mathrm{AD}}^{2}}{{\mathrm{AB}}^{2}}$= $\frac{{\mathrm{AE}}^{2}}{{\mathrm{AC}}^{2}}$= $\frac{{\mathrm{DE}}^{2}}{{\mathrm{BC}}^{2}}$
Since, DE = $\frac{2}{3}$ BC,
⇒ $\frac{\mathrm{DE}}{\mathrm{BC}}$= $\frac{2}{3}$
⇒ $\frac{{\mathrm{DE}}^{2}}{{\mathrm{BC}}^{2}}$= $\frac{4}{9}$
âˆ´ Area of â–³ADE = $\frac{4}{9}$ × Area of â–³ABC = $\frac{4}{9}$ × 81 = 36 ${\mathrm{cm}}^{2}$.
Hence, area of â–³ADE is 36 ${\mathrm{cm}}^{2}$.
7. If sec A = $\frac{15}{7}$ and A + B = 90Â°, find the value of cosec B.
1) Apply the correct trigonometric ratio of complementary angle.
1) A and B are complementary angles. Write A as complementary angle of B.
2) Apply secant for both the complementary angles.
3) Find secant of complementary angle of B.
Since A + B = 90Â°, A = 90Â° - B
⇒sec A = sec (90Â° - B)
⇒sec A = cosec B [Since sec (90Â° - Î¸) = cosec Î¸]
Given that sec A = $\frac{15}{7}$
âˆ´ cosec B = $\frac{15}{7}$
Hence, value of cosec B is $\frac{15}{7}$.
8. If the mid-point of the line segment joining the points P (6, b âˆ’ 2) and Q (âˆ’2, 4) is (2, âˆ’3), find the value of b.
1) Write the correct formula to find the mid-point of a line segment.
2) Equate the y-coordinates of the mid-points.
1) Find the mid-point of the line segment joining the points (6, b - 2) and (-2, 4).
2) Equate the y-coordinate of the mid-points to find the value of b.
Coordinates of the midpoint of the line segment joining (${x}_{1},{x}_{2}$) and (${y}_{1},{y}_{2}$) = $[\frac{({x}_{1}+{x}_{2})}{2},\frac{({y}_{1}+{y}_{2})}{2}]$
Using mid-point formula, $\frac{(b\xe2\u02c6\u20192)+4}{2}$= - 3
⇒ $\frac{(b+2)}{2}$ = - 3
⇒b + 2 = - 6
⇒b = - 8
Hence, the value of b is - 8.
9. The length of the minute hand of a wall clock is 7 cm. How much area does it sweep in 20 minutes?
1) Find the angle made by the minute hand in 20 min.
2) Use the correct formula to find the area of the sector.
1) Find the angle made by the minute hand in 20 minutes.
2) Consider the length of the minute hand as the radius of the sector.
3) Find the area of the sector using the formula.
In 60 minutes, the angle swept by the minute hand = 360Â°
âˆ´ In 20 minutes, the angle swept by the minute hand = $\frac{360\xc2\xb0}{60}$ × 20 = 120Â°
Radius of the sector formed by the minute hand = length of the minute hand.
Area swept by the minute hand in 20 minutes = $\frac{120\xc2\xb0}{360\xc2\xb0}$ × Ï€ × ${\left(7\right)}^{2}$
= $\frac{1}{3}$ × $\frac{22}{7}$ × ${\left(7\right)}^{2}$
= $\frac{154}{3}$ ${\mathrm{cm}}^{2}$
Hence, the area swept by the minute hand in 20 minutes is $\frac{154}{3}$ ${\mathrm{cm}}^{2}$.
10. What is the lower limit of the modal class of the following frequency distribution?
Choose the exact class with the highest frequency.
1) Choose the class with the highest frequency.
2) Choose the lower limit of the class interval of the model class.
A modal class is the class interval having maximum frequency.
In the above frequency distribution, 27 is the maximum frequency and it comes in class interval 40 - 50.
Hence, 40 - 50 is the model class.
âˆ´ 40 is the lower limit of the model class in the given frequency distribution.
11. Without drawing the graph, find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident:
9x âˆ’10y = 21, $\frac{3}{2}$x - $\frac{5}{3}$ y = $\frac{7}{2}$
1) Find the ratio of corresponding coefficients of the equations.
2) Relate the exact condition for finding the relation between the pair of lines.
1) Write the coefficients of the equations by comparing the standard form of the linear equations.
2) Find the ratios of the corresponding coefficients.
3) Find the relation between the pair of given lines by comparing the ratios of coefficients of the linear equations.
If ${a}_{1}$x +${b}_{1}$y + ${c}_{1}$= 0 and ${a}_{2}$x+ ${b}_{2}$y + ${c}_{2}$= 0 are two linear equations.
If $\frac{{a}_{1}}{{a}_{2}}$ â‰ $\frac{{b}_{1}}{{b}_{2}}$, then the above equations represent intersecting lines.
If$\frac{{a}_{1}}{{a}_{2}}$ = $\frac{{b}_{1}}{{b}_{2}}$, then the equations represent coincident lines.
If $\frac{{a}_{1}}{{a}_{2}}$ = $\frac{{b}_{1}}{{b}_{2}}$â‰ $\frac{{c}_{1}}{{c}_{2}}$, then the equations represent parallel lines.
The given equations are :-
9x - 10y = 21 or 9x -10y - 21= 0
$\frac{3}{2}$x -$\frac{5}{3}$ y = $\frac{7}{2}$ or $\frac{3}{2}$x -$\frac{5}{3}$ y - $\frac{7}{2}$ = 0
When these equations are compared, ${a}_{1}$= 9, ${b}_{1}$= - 10 and ${c}_{1}$= - 21
Similarly, ${a}_{2}$= $\frac{3}{2}$, ${b}_{2}$= $\frac{\xe2\u02c6\u20195}{3}$ and ${c}_{2}$= $\frac{\xe2\u02c6\u20197}{2}$
$\frac{{a}_{1}}{{a}_{2}}$ = $\frac{9}{\frac{3}{2}}$ = 6
$\frac{{b}_{1}}{{b}_{2}}$= $\frac{\xe2\u02c6\u201910}{\frac{\xe2\u02c6\u20195}{3}}$= 6
and $\frac{{c}_{1}}{{c}_{2}}$ = $\frac{\xe2\u02c6\u201921}{\frac{\xe2\u02c6\u20197}{2}}$= 6
Hence, $\frac{{a}_{1}}{{a}_{2}}$ = $\frac{{b}_{1}}{{b}_{2}}$= $\frac{{c}_{1}}{{c}_{2}}$
Therefore, the given pair of equations represents coincident lines.
12. The ${17}^{\mathrm{th}}$ term of an A.P. exceeds its ${10}^{\mathrm{th}}$term by 7. Find the common difference.
1) Write the correct formula to find the ${n}^{\mathrm{th}}$ term of the A.P.
2) Find the difference between the ${17}^{\mathrm{th}}$ term and the ${10}^{\mathrm{th}}$ term.
1) Find the ${17}^{\mathrm{th}}$ term and the ${10}^{\mathrm{th}}$ term of an A.P.
2) Find the difference between the ${17}^{\mathrm{th}}$ term and the ${10}^{\mathrm{th}}$ term.
3) Equate the difference to 7.
4) Simplify to find the common difference (d) between the terms of the A.P.
For an A.P. its nth term or ${a}_{n}$= a + (n âˆ’ 1)d where a is the first term and d is the common difference between the terms.
${17}^{\mathrm{th}}$ term of the A.P. or ${a}_{17}$= a + (17 - 1)d = a + 16d
${10}^{\mathrm{th}}$ term of the A.P. or ${a}_{10}$= a + (10 - 1)d = a + 9d
Given that the difference between the ${17}^{\mathrm{th}}$ term and the ${10}^{\mathrm{th}}$ term = 7
⇒(a + 16d) âˆ’ (a + 9d) = 7
⇒7d = 7
⇒d = 1
Therefore, the common difference (d) between the terms of the A.P. is 1.
13. Without using trigonometric tables, evaluate: $\frac{7}{2}$× $\frac{\mathrm{cos}70\xc2\xb0}{\mathrm{sin}20\xc2\xb0}$ + $\frac{3}{2}$× $\frac{\mathrm{cos}55\xc2\xb0\mathrm{cos}\mathrm{ec}35\xc2\xb0}{\mathrm{tan}5\xc2\xb0\mathrm{tan}25\xc2\xb0\mathrm{tan}45\xc2\xb0\mathrm{tan}85\xc2\xb0\mathrm{tan}65\xc2\xb0}$.
1) Use the correct trigonometric ratios of complimentary angles.
1) Use the trigonometric ratios of complementary angles such that all the trigonometric ratios are cancelled.
2) Apply the reciprocals of trigonometric ratios such that all the trigonometric ratios are cancelled.
3) Simplify to get the value of the expression.
According to trigonometric ratios of complementary angles, cos Î¸ = sin (90Â° - Î¸) and cot Î¸ = tan (90Â° - Î¸)
$\frac{7}{2}\times \frac{\mathrm{cos}70\xc2\xb0}{\mathrm{sin}20\xc2\xb0}$ + $\frac{3}{2}\times \frac{\mathrm{cos}55\xc2\xb0\mathrm{cos}\mathrm{ec}35\xc2\xb0}{\mathrm{tan}5\xc2\xb0\mathrm{tan}25\xc2\xb0\mathrm{tan}45\xc2\xb0\mathrm{tan}85\xc2\xb0\mathrm{tan}65\xc2\xb0}$
= $\frac{7}{2}\times \frac{\mathrm{cos}70\xc2\xb0}{\mathrm{sin}(90\xc2\xb0\xe2\u02c6\u201970\xc2\xb0)}+\frac{3}{2}\times \frac{\mathrm{cos}55\xc2\xb0\mathrm{cos}\mathrm{ec}(90\xc2\xb0\xe2\u02c6\u201955\xc2\xb0)}{\mathrm{tan}5\xc2\xb0\mathrm{tan}25\xc2\xb0\mathrm{tan}45\xc2\xb0\mathrm{tan}(90\xc2\xb0\xe2\u02c6\u20195\xc2\xb0)\mathrm{tan}(90\xe2\u02c6\u201925\xc2\xb0)}$
= $\frac{7}{2}$× $\frac{\mathrm{cos}70\xc2\xb0}{\mathrm{cos}70\xc2\xb0}$ + $\frac{3}{2}$× $\frac{\mathrm{cos}55\xc2\xb0\mathrm{cos}\mathrm{ec}(90\xc2\xb0\xe2\u02c6\u201955\xc2\xb0)}{\mathrm{tan}5\xc2\xb0\mathrm{tan}25\xc2\xb0\times 1\times \mathrm{cot}5\xc2\xb0\times \mathrm{cot}25\xc2\xb0}$
= $\frac{7}{2}$ + $\frac{3\times \mathrm{cos}55\xc2\xb0\mathrm{tan}5\xc2\xb0\mathrm{tan}25\xc2\xb0}{2\times \mathrm{tan}5\xc2\xb0\mathrm{tan}25\xc2\xb0\mathrm{cos}55\xc2\xb0}$ [cot Î¸ = $\frac{1}{\mathrm{tan}\mathrm{\xce\xb8}}$ and sec Î¸ = $\frac{1}{\mathrm{cos}\mathrm{\xce\xb8}}$]
= $\frac{7}{2}$ + $\frac{3}{2}$
= $\frac{10}{2}$
= 5.
Hence, the value of the given expression is 5.
14. Show that the points (âˆ’2, 5); (3, âˆ’4) and (7, 10) are the vertices of a right angled isosceles triangle.
OR
The centre of a circle is (2α âˆ’ 1, 7) and it passes through the point (âˆ’3, âˆ’1). If the diameter of the circle is 20 units, then find the values(s) of α.
Use Pythagoras theorem, to prove that the given points form a right angled triangle.
Find the diameter of the circle by multiplying the radius by 2.
1) Find the lengths of sides joining the given points using the distance formula.
2) Verify if the lengths of two sides are equal and the sides satisfy Pythagoras theorem.
3) Find the radius of the circle using the distance formula and find the diameter of the circle.
4) Equate the diameter of the circle to 20 units to find the value of α.
The coordinates of the given points are A(- 2, 5), B(3, -4) and C(7, 10).
Using the distance formula, we get
AB = $\sqrt{{[3\xe2\u02c6\u2019(\xe2\u02c6\u20192\left)\right]}^{2}+{(\xe2\u02c6\u20194\xe2\u02c6\u20195)}^{2}}$=$\sqrt{{5}^{2}+{(\xe2\u02c6\u20199)}^{2}}$ =$\sqrt{25+81}$ = $\sqrt{106}$
BC = $\sqrt{(7\xe2\u02c6\u20193{)}^{2}+[10\xe2\u02c6\u2019(\xe2\u02c6\u20194){]}^{2}}$= $\sqrt{{4}^{2}+{14}^{2}}$ =$\sqrt{16+196}$ = $\sqrt{212}$
CA = $\sqrt{(\xe2\u02c6\u20192\xe2\u02c6\u20197{)}^{2}+(5\xe2\u02c6\u201910{)}^{2}}$= $\sqrt{(\xe2\u02c6\u20199{)}^{2}+(\xe2\u02c6\u20195{)}^{2}}$= $\sqrt{81+25}$= $\sqrt{106}$
The triangle ABC is isosceles since AB = CA.
${\mathrm{AB}}^{2}+{\mathrm{CA}}^{2}$= 106 + 106 = 212 = ${\mathrm{BC}}^{2}$
âˆ´ According to Pythagoras theorem, the triangle is right-angled at A.
Hence, the points (âˆ’2, 5), (3, âˆ’4) and (7, 10) are the vertices of a right-angled triangle.
OR
The center of the circle is given as (2αâˆ’1, 7). The point through which the circle passes through is given as (âˆ’3, âˆ’1).
Using distance formula, the radius of the circle or OP
= $\sqrt{[\xe2\u02c6\u20193\xe2\u02c6\u2019(2\alpha \xe2\u02c6\u20191){]}^{2}+(\xe2\u02c6\u20191\xe2\u02c6\u20197{)}^{2}}$
= $\sqrt{(\xe2\u02c6\u20192\xe2\u02c6\u20192\alpha {)}^{2}+(\xe2\u02c6\u20198{)}^{2}}$
= $\sqrt{(2+2\alpha {)}^{2}+64}$
Since the diameter of circle is given as 20 units and diameter = 2 Ã— radius or 2 Ã— OP
2 Ã— OP = 20 units = 2$\sqrt{(2+2\alpha {)}^{2}+64}$
â‡’ $\sqrt{(2+2\alpha {)}^{2}+64}$= 10
â‡’ (2 + 2α${)}^{2}$+ 64 = 100
â‡’ (2 + 2α${)}^{2}$ = 36
â‡’ 2 + 2α = Â± 6
If 2 + 2α = 6 , then 2α = 4 â‡’
â‡’ α = 2 units
If 2 + 2α = - 6 , then 2α = - 8 â‡’
â‡’ α = - 4 units
âˆ´ α = 2 or - 4 units.
15. If C is a point lying on the line segment AB joining A (1, 1) and B (2, âˆ’3) such that 3AC = CB, then find the coordinates of C.
1) Find the ratio in which the point C divides AB using the given condition.
2) Assume the A(1, 1) as the first point and B(2, -3) as the second point.
1) Find the ratio in which the point C divides AB using the given condition.
2) Consider A as the first point and B as the second point.
3) Use the section formula to find the coordinates of point C.
Assume that the coordinates of C are x and y. Point C can be represented as C(x, y).
Given that 3AC = CB⇒AC:CB = 1:3
Coordinates of the point on a line joining the points (${x}_{1},{y}_{1})\mathrm{and}({x}_{2},{y}_{2})$in the ratio of m:n are
$(\frac{m{x}_{2}+n{x}_{1}}{m+n}$, $\frac{m{y}_{2}+n{y}_{1}}{m+n})$
Using the section formula,
C(x, y) = $[\frac{1\times 2+3\times 1}{1+3}$, $\frac{1\times (\xe2\u02c6\u20193)+3\times 1}{1+3}]$ = $[\frac{2+3}{4}$, $\frac{\xe2\u02c6\u20193+3}{4}]$ = $[\frac{5}{4},0]$
âˆ´ The coordinates of point C are ($\frac{5}{4}$, 0)
16. Show that the square of any positive odd integer is of the form 8m + 1, for some integer m.
OR
Prove that 7 + 3$\sqrt{2}$ is not a rational number.
1) Find the squares of the general form of the odd positive integers using the algebraic identity.
2) Write 7 + 3$\sqrt{2}$using integers as $\frac{p}{q}$.
1) Write the general form of the positive odd integers as 8n + r where 0â‰¤ r <8 and x is an integer.
2) Find the square of the each form and show that it is of the form 8m + 1.
OR
1) Assume that 7 + 3$\sqrt{2}$is a rational number and write it in the form of $\frac{p}{q}$.
2) Prove that it is not a rational number after simplification.
Assume a positive integer as â€˜aâ€™ and assume b = 8.
It can be represented as a = 8n + r where 0â‰¤ r <8 and x is an integer.
Hence 'a' will be (8n + 1) or (8n + 2) or (8n + 3) or (8n + 4) or (8n + 5) or (8n + 6) or (8n + 7).
Positive odd integer is of the form (8n + 1) or (8n + 3) or (8n + 5) or (8n + 7).
Consider the above 4 positive odd integer forms separately.
Case 1: If a = 8n + 1
${a}^{2}$=${(8n+1)}^{2}$= ${64n}^{2}$+ 16n + 1 = 8(8${n}^{2}$+ 2n) + 1 = 8m + 1 where m = (8${n}^{2}$ + 2n) is an integer.
Case 2: If a = 8n +3
${a}^{2}$= ${(8n+3)}^{2}$= ${(64n}^{2}+48n+9)$= 8(8n + 6 + 1) + 1 = 8m + 1 where m = (8${n}^{2}$+ 6n + 1) is an integer.
Case 3: If a = 8n + 5
${a}^{2}$ = (8n + 5${)}^{2}$= 64${n}^{2}$+ 80 n + 25 = 8(8${n}^{2}$ + 10n + 3) + 1 = 8m + 1 where m = (8${n}^{2}+10x+3$) is an integer.
Case 4: If a = 8n + 7
${a}^{2}$= (8n + 7${)}^{2}$= (64${n}^{2}$+ 112n + 49) = 8(8${n}^{2}$+ 14n + 6) + 1 = 8m + 1 where m = (8${n}^{2}$ + 14n + 6) is an integer.
From all these cases, it can be noticed that the square of a positive odd integer is of the form 8m + 1 where m is any integer.
OR
Assume that 7 + 3$\sqrt{2}$is a rational number.
It can be represented using integers p and q (q â‰ 0) as, 7 + 3$\sqrt{2}$= $\frac{p}{q}$
⇒$\sqrt{2}$âˆš= $\frac{1}{3}$ ($\frac{p}{q}$ - 7)
p and q are rational numbers. So, $\frac{1}{3}$($\frac{p}{q}$ - 7) should also be rational and $\sqrt{2}$âˆšshould also be a rational number. But we know that $\sqrt{2}$ is an irrational number and so our assumption is wrong.
So, our assumption is false and (7 + 3$\sqrt{2}$) is not a rational number.
17. If the polynomial 6${x}^{4}$+ 8${x}^{3}$âˆ’5${x}^{2}$+ ax + b is exactly divisible by the polynomial 2${x}^{2}$âˆ’5, the find the values of a and b.
1) Find the remainder by the long division method of polynomials.
2) Equate both the terms of the remainder to zero.
1) Divide the given polynomial by 2${x}^{2}$-5.
2) Equate the remainder to zero to find the values of 'a' and 'b'.
p(x) = 6${x}^{4}$+ 8${x}^{3}$âˆ’5${x}^{2}$+ ax + b
q(x) = 2${x}^{2}$ - 5
Dividing p(x) by q(x)
We get the remainder as (a + 20)x + (b + 25).
As p(x) is exactly divisible by q(x), remainder should be 0.
âˆ´(a + 20)x + (b + 25) = 0
⇒a + 20 = 0 and b + 25 = 0
⇒ a + 20 = 0 , a = - 20 and
⇒ b + 25 = 0, b = - 25
Therefore, the value of a = -20 and b = - 25.
18. If ${9}^{\mathrm{th}}$ term of an A.P. is zero, prove that its ${29}^{\mathrm{th}}$ term is double of its ${19}^{\mathrm{th}}$ term.
Find the relation between the${19}^{\mathrm{th}}$ term and the${29}^{\mathrm{th}}$ term of the progression.
1) Equate the ${9}^{\mathrm{th}}$ term of the progression to zero and find the value of 'a' in terms of d.
2) Find ${19}^{\mathrm{th}}$ term and the ${29}^{\mathrm{th}}$ term in terms of d.
3) Find the relation between ${19}^{\mathrm{th}}$ term and the ${29}^{\mathrm{th}}$ term.
Let a be the ${1}^{\mathrm{st}}$ term and d be the common difference of the A.P.
${n}^{\mathrm{th}}$ term of A.P. = ${a}_{n}$= a + (n - 1)d
âˆ´ ${9}^{\mathrm{th}}$ term of A.P. = ${a}_{9}$= a + (9 - 1)d = a + 8d
Given that, ${a}_{9}$= 0
âˆ´ ${a}_{9}$= a + 8d = 0
⇒a = - 8d
${19}^{\mathrm{th}}$ term of A.P. = ${a}_{19}$= a + (19 - 1)d = a + 18d = - 8d + 18d = 10d
${29}^{\mathrm{th}}$ term of A.P. = ${a}_{29}$= a + (29 - 1)d = a + 28d = - 8d + 28d = 20d
âˆ´ ${a}_{29}$= 20d = 2 × 10d = 2 ×${a}_{19}$
Hence, the ${29}^{\mathrm{th}}$ term of the A.P. is double the ${19}^{\mathrm{th}}$ term of the A.P.
19. Draw a circle of radius 3 cm. From a point P, 6 cm away from its centre, construct a pair of tangents to the circle. Measure the lengths of the tangents.
Mark the point at 6 cm from the center of the circle either to the left or right of the circle.
Join the point at 6 cm to points of intersection of circles exactly.
Draw a circle of given radius and mark a point O at 6 cm from the center.
Draw the perpendicular bisector to the line and mark the point at which it intersects the circle.
Draw another circle with the point of intersection as center.
Mark the points of intersections of circles and join with the point marked in step 1.
Draw a circle of radius 3 cm and mark the center point as O.
Mark a point P at a distance of 6 cm from O and join O and P.
Draw the perpendicular bisector to the line OP.
Mark the point at which it intersects the circle as Q.
With Q as center and QP as radius, draw another circle.
This will intersect the first circle at 2 points. Mark the points as M and N.
Join M with P to obtain line MP and N with P to obtain the line NP. MP and NP are the tangents to the circle with center O.
On measuring, the length of the tangents, MP and NP is 5.20 cm.
20. In the given figure, two triangles ABC and DBC lie on the same side of base BC. P is a point on BC such that PQ || BA and PR || BD. Prove that QR || AD.
1) Consider the exact corresponding sides of the triangles.
2) Apply converse of basic proportionality theorem to prove that the lines are parallel.
1) Consider â–³ABC and â–³QPC, and apply basic proportionality theorem.
2) Consider â–³BDC and â–³PQR, and apply basic proportionality theorem.
3) Use transitive property and converse of basic proportionality theorem to prove that line QR is parallel to AD.
Consider â–³ABC and â–³QPC
PQ || BA and hence according to the basic proportionality theorem, we get
$\frac{\mathrm{CP}}{\mathrm{PB}}$ = $\frac{\mathrm{CQ}}{\mathrm{QA}}$ --------------------- (1)
Consider â–³BDC and â–³PQR
PR || BD and hence according to the basic proportionality theorem, we get
$\frac{\mathrm{CP}}{\mathrm{PB}}$ = $\frac{\mathrm{CR}}{\mathrm{RD}}$ -----------------------(2)
Taking (1) and (2) together we get
$\frac{\mathrm{CQ}}{\mathrm{QA}}$ = $\frac{\mathrm{CR}}{\mathrm{RD}}$
âˆ´Considering â–³ADC and â–³QRC, from above, using converse of proportionality theorem, we get QR|| AD.
21. In the given figure, a triangle ABC is right angled at B. Side BC is trisected at points D and E. Prove that ${8\mathrm{AE}}^{2}=5{\mathrm{AD}}^{2}$.
OR
In the given figure, a circle is inscribed in a triangle ABC having side BC = 8 cm, AC = 10 cm and AB = 12 cm. Find AD, BE and CF.
1) Each point of trisection divides the side of the triangle in the ratio of 1:2 or 2:1.
2) Substitute the equations of the Pythagoras theorem of one triangle in the other.
1) Apply Pythagoras theorem in â–³ABD, â–³ABE and â–³ABC.
2) Substitute the value of BD according the trisection ratio as $\frac{\mathrm{BC}}{3}$.
3) Substitute the values of ${\mathrm{AC}}^{2}$and ${\mathrm{AD}}^{2}$ in ${3\mathrm{AC}}^{2}+5{\mathrm{AD}}^{2}$and simplify to get $8{\mathrm{AE}}^{2}$.
Given: BC is trisected at D and E.
âˆ´BD = DE = EC = $\frac{\mathrm{BC}}{3}$
⇒ BE = $\frac{2\mathrm{BC}}{3}$
Consider âˆ†ABD. Using Pythagoras Theorem, we get
${\mathrm{AB}}^{2}$+ ${\mathrm{BD}}^{2}$= ${\mathrm{AD}}^{2}$
${\mathrm{AB}}^{2}+{\left(\frac{\mathrm{BC}}{3}\right)}^{2}={\mathrm{AD}}^{2}$
${\mathrm{AB}}^{2}+\frac{\left({\mathrm{BC}}^{2}\right)}{9}={\mathrm{AD}}^{2}$------------------ (1)
Consider âˆ†ABE. Using Pythagoras theorem, we get
${\mathrm{AB}}^{2}+{\mathrm{BE}}^{2}={\mathrm{AE}}^{2}$
${\mathrm{AB}}^{2}+{\left(\frac{2\mathrm{BC}}{3}\right)}^{2}={\mathrm{AE}}^{2}$
${\mathrm{AB}}^{2}+\frac{{4\mathrm{BC}}^{2}}{9}={\mathrm{AE}}^{2}$ -----------------------(2)
Consider âˆ†ABC. Using Pythagoras theorem, we get
${\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}={\mathrm{AC}}^{2}$ -----------------------(3)
We have to prove that 3${\mathrm{AC}}^{2}$+ 5${\mathrm{AD}}^{2}$= 8${\mathrm{AE}}^{2}$
${3\mathrm{AC}}^{2}+5{\mathrm{AD}}^{2}=3({\mathrm{AB}}^{2}+{\mathrm{BC}}^{2})+5\{{\mathrm{AB}}^{2}+\left(\frac{{\mathrm{BC}}^{2}}{9}\right)\}$
=$3{\mathrm{AB}}^{2}+3{\mathrm{BC}}^{2}+5{\mathrm{AB}}^{2}+\frac{\left({5\mathrm{BC}}^{2}\right)}{9}$
= ${8\mathrm{AB}}^{2}+\frac{\left({32\mathrm{BC}}^{2}\right)}{9}$
= ${8[\mathrm{AB}}^{2}+\frac{\left({4\mathrm{BC}}^{2}\right)}{9}$]
From (2), we know that ${\mathrm{AB}}^{2}+\frac{{4\mathrm{BC}}^{2}}{9}={\mathrm{AE}}^{2}$.
Hence substituting in above, we get
$3{\mathrm{AC}}^{2}+5{\mathrm{AD}}^{2}=8{\mathrm{AE}}^{2}$
Hence proved.
OR
AD and AF are tangents drawn from point A to the circle. Similarly BD and BE are tangents from point B to the circle and CE and CF are tangents from point C
The tangents drawn to the circle from any point external to the circle are equal.
âˆ´ AD = AF
BD = BE and
CE = CF
Let us take AD = a, BE = b and CF = c
AB = AD + BD = a + b = 12 cm ------------- (1)
BC = BE + CE = b + c = 8 cm --------------- (2)
AC = AF + CF = a + c = 10 cm -------------- (3)
AB + BC + AC = 2(a + b + c) = 12 + 8 + 10 = 30 cm
⇒a + b + c = 15 cm --------------- (4)
Substituting (1) in (4), we get c = (15) âˆ’ (a + b) = 15 âˆ’ 12 = 3 cm
Substituting (2) in (4), we get a =15 âˆ’ (b + c) = 15 âˆ’ 8 = 7 cm
Substituting (3) in (4), we get b = 15 âˆ’ (a + c) = 15 âˆ’ 10 = 5 cm
Thus, we get, AD = a = 7 cm, BE = b = 5 cm and CF = c = 3 cm.
22. Prove that ${\mathrm{sec}}^{2}\mathrm{\xce\xb8}\xe2\u02c6\u2019\frac{{\mathrm{sin}}^{2}\mathrm{\xce\xb8}\xe2\u02c6\u2019{2\mathrm{sin}}^{4}\mathrm{\xce\xb8}}{{2\mathrm{cos}}^{4}\mathrm{\xce\xb8}\xe2\u02c6\u2019{\mathrm{cos}}^{2}\mathrm{\xce\xb8}}$ = 1.
1) Apply appropriate trigonometric identities.
2) Tangent is the ratio of sine and cosine.
1) Take ${\mathrm{sin}}^{2}\mathrm{\xce\xb8}$ as a common factor in the numerator and ${\mathrm{cos}}^{2}\mathrm{\xce\xb8}$ as a common factor in the denominator of the fraction of the second term of the equation.
2) Apply appropriate trigonometric identities and simplify.
${\mathrm{sec}}^{2}\mathrm{\xce\xb8}\xe2\u02c6\u2019\frac{{\mathrm{sin}}^{2}\mathrm{\xce\xb8}\xe2\u02c6\u2019{2\mathrm{sin}}^{4}\mathrm{\xce\xb8}}{2{\mathrm{cos}}^{4}\mathrm{\xce\xb8}\xe2\u02c6\u2019{\mathrm{cos}}^{2}\mathrm{\xce\xb8}}$
⇒${\mathrm{sec}}^{2}\mathrm{\xce\xb8}\xe2\u02c6\u2019\frac{{\mathrm{sin}}^{2}\mathrm{\xce\xb8}\xe2\u02c6\u20192{\mathrm{sin}}^{4}\mathrm{\xce\xb8}}{{\mathrm{cos}}^{2}\mathrm{\xce\xb8}(2{\mathrm{cos}}^{2}\mathrm{\xce\xb8}\xe2\u02c6\u20191)}$ [(${\mathrm{cos}}^{2}\mathrm{\xce\xb8}=1\xe2\u02c6\u2019{\mathrm{sin}}^{2}\mathrm{\xce\xb8}\left)\right]$
${\Rightarrow \mathrm{sec}}^{2}\mathrm{\xce\xb8}\xe2\u02c6\u2019\frac{{\mathrm{tan}}^{2}\mathrm{\xce\xb8}(1\xe2\u02c6\u20192{\mathrm{sin}}^{2}\mathrm{\xce\xb8})}{{\mathrm{cos}}^{2}\mathrm{\xce\xb8}(1\xe2\u02c6\u2019{\mathrm{sin}}^{2}\mathrm{\xce\xb8})\xe2\u02c6\u20191}$
(tan Î¸ = $\frac{\mathrm{sin}\mathrm{\xce\xb8}}{\mathrm{cos}\mathrm{\xce\xb8}}$)
⇒${\mathrm{sec}}^{2}\mathrm{\xce\xb8}\xe2\u02c6\u2019\frac{{\mathrm{tan}}^{2}\mathrm{\xce\xb8}(1\xe2\u02c6\u20192{\mathrm{sin}}^{2}\mathrm{\xce\xb8})}{(1\xe2\u02c6\u20192{\mathrm{sin}}^{2}\mathrm{\xce\xb8})}$
⇒${\mathrm{sec}}^{2}\mathrm{\xce\xb8}\xe2\u02c6\u2019{\mathrm{tan}}^{2}\mathrm{\xce\xb8}$
= 1 [${\mathrm{sec}}^{2}\mathrm{\xce\xb8}=1+{\mathrm{tan}}^{2}\mathrm{\xce\xb8}$ ]
Therefore, ${\mathrm{sec}}^{2}\mathrm{\xce\xb8}$ - $\frac{{\mathrm{sin}}^{2}\mathrm{\xce\xb8}\xe2\u02c6\u20192{\mathrm{sin}}^{4}\mathrm{\xce\xb8}}{2{\mathrm{cos}}^{4}\mathrm{\xce\xb8}\xe2\u02c6\u2019{\mathrm{cos}}^{2}\mathrm{\xce\xb8}}=1.$
23. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
1) Write the correct formula for the area of a triangle.
2) Substitute the corresponding coordinates in the formula.
1) Assume the three points as three vertices of a triangle.
2) Area of the triangle formed by three points is zero as the points are collinear.
3) Equate the area formed by three points to zero.
4) Simplify to find an equation which gives the relation between the coordinates of one of the points.Since the points (x, y), (1, 2) and (7, 0) are collinear, the area of the triangle formed with these three points as vertices will be zero.
Area of triangle with points (${x}_{1}$, ${y}_{1}$), (${x}_{2}$, ${y}_{2}$) , (${x}_{3}$, ${y}_{3}$) =
$\frac{1}{2}${${x}_{1}$(${y}_{2}$âˆ’${y}_{3}$) + ${x}_{2}$(${y}_{3}$âˆ’ ${y}_{1}$) + ${x}_{3}$(${y}_{1}$âˆ’${y}_{2}$ )}
Area of the triangle formed by the given points
= $\frac{1}{2}${x(2 âˆ’ 0) + 1(0 âˆ’ y) + 7(y âˆ’ 2)}
= $\frac{1}{2}${2x âˆ’ y + 7y âˆ’ 14}
= $\frac{1}{2}${2x + 6y âˆ’ 14}
= x + 3y âˆ’ 7
Thus, the area of the triangle is x + 3y âˆ’ 7. Since this area is zero, we get x + 3yâˆ’7 = 0.
âˆ´ x + 3yâˆ’7 = 0 is the equation which shows the relation between x and y, if the points have to be collinear.
24. In figure 4, the shape of the top of a table in a restaurant is that of a sector of a circle with centre O and âˆ BOD = 90Â°. If BO = OD = 60 cm, find
(i) the area of the top of the table.
(ii) the perimeter of the table top.
(Take Ï€ = 3.14)
OR
In figure 5, ABCD is a square of side 14 cm and APD and BPC are semicircles. Find the area of the shaded region. (Take Ï€ = $\frac{22}{7}$)
1) Use the reflex angle of 90Â° to find the area of the top of the table.
2) Perimeter of the table top includes both the radii and the length of the arc of the sector.
OR
1) Half of the length of the side of the square is equal to the radius of the semicircle.
2) Area of the shaded part of the square = Area of the square - Area of the unshaded part of the square.
1) Find the reflex angle of âˆ BOD which is the angle of the sector.
2) Use the formula to find the area of the sector to find the area of the table top.
OR
1) Find the area of the square.
2) Half of the side of the square is equal to the radius of the semicircle.
3) Find the area of the semicircle.
4) Subtract twice the area of the semicircle from the area of the square to find the area of the unshaded part of the circle.
(i) Area of table top
Area of a sector can be obtained by $\frac{\mathrm{\xce\xb8}}{360\xc2\xb0}\times \mathrm{\xcf\u20ac}{r}^{2}$where âˆ… is the angle of the sector and r is the radius.
Given that âˆ BOD = 90Â°.
âˆ´Reflex âˆ BOD = 360Â° - 90Â° = 270Â°
Area of the table top = $\frac{\mathrm{\xce\xb8}}{360\xc2\xb0}\times \mathrm{\xcf\u20ac}{r}^{2}$
where âˆ… = 270Â° and r = radius of the sector = 60 cm (BO = BD = 60 cm)
= $\frac{270\xc2\xb0}{360\xc2\xb0}$ × 3.14 × 60 × 60
= $\frac{3}{4}$ × 3.14 × 3600
= 8478 ${\mathrm{cm}}^{2}$
Hence, the area of the table top = 8478 ${\mathrm{cm}}^{2}$
(ii) Perimeter of the table top = length of the arc BD + OB + OD
Length of the arc BD = $\frac{270\xc2\xb0}{360\xc2\xb0}\times 2\mathrm{\xcf\u20acr}$
= $\frac{3}{4}$× 2 × 3.14 × 60
= 282.6 cm
âˆ´ Perimeter of the table top = 282.6 + 60 + 60 = 402.6 cm
OR
Area of the shaded region = Area of the square ABCD - (area of the semicircle APD + area of the semicircle BPC)
Area of the square ABCD
=14 ×14 =196 ${\mathrm{cm}}^{2}$(since the length of the sides of the square = 14 cm)
Area of the semicircle APD =$\frac{\left(\mathrm{\xcf\u20ac}{r}^{2}\right)}{2}$, where r = 7 cm. (AD = 14 cm, r = $\frac{\mathrm{AD}}{2}$)
Area = $\frac{(22\times 7\times 7)}{(7\times 2)}$ = 77 ${\mathrm{cm}}^{2}$
Area of semicircle BPC = 77 ${\mathrm{cm}}^{2}$since, r = 7 cm
âˆ´ Area of the shaded region = 196 -(77 + 77)
= 196 -154
= 42${\mathrm{cm}}^{2}$
Hence, area of the shaded region of the figure = 42 ${\mathrm{cm}}^{2}$.
25. A box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn from the bag at random. Find the probability that the number on the card, drawn from the box is an odd number, a perfect square number, a number divisible by 7.
1) Find the exact number of number of cards in the box.
2) Find the number of odd numbers, perfect squares and the numbers divisible by 7 between 14 and 99.
1) Find the number of cards in the box which are numbered from 14 to 99.
2) Find the number of odd numbers between 14 and 99 and find the probability of the card being an odd number using the probability formula.
3) Find the number of odd numbers between 14 and 99 and find the probability of the card being a perfect square using the probability formula.
4) Find the numbers which are divisible by 7 between 14 and 99 and find the probability of the card being a number divisible by 7 using the probability formula.
The probability of an event = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$
Number of cards in box = 99 âˆ’14 + 1 = 86
Number of cards having odd number = $\frac{86}{2}$ = 43
Hence, probability of getting an odd number = $\frac{43}{86}$ = $\frac{1}{2}$
Number of cards having perfect squares = 6 (Perfect squares between 14 and 99 are 16, 25, 36, 49, 64 and 81)
Hence, probability of getting perfect square = $\frac{6}{86}$ = $\frac{3}{43}$
Number of cards with numbers divisible by 7 = 13
(Numbers divisible by 7 between 14 and 99 are 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91 and 98)
Hence, probability of getting number divisible by 7 = $\frac{13}{86}$.
26. A trader bought a number of articles for Rs 900. Five articles were found damaged. He sold each of the remaining articles at Rs. 2 more than what he paid for it. He got a profit of Rs. 80 on the whole transaction. Find the number of articles he bought.
OR
Two years ago the manâ€™s age was three times the square of his sonâ€™s age. Three years hence his age will be four times his sonâ€™s age. Find their present ages.
1) If the total number of articles is x, then find the selling price of (x - 5) articles.
2) Equate the difference between the selling price of (x - 5) articles and Rs 900 to Rs 80.
OR
1) Subtract the number of years to find the age of son 2 years ago.
2) Add the number of years to find the age of the son 3 years later.
1) Assume the number of articles as 'x' and find the cost price of each article which is equal to Rs $\frac{900}{x}$.
2) Find the selling price of 1 articles which is equal to Rs ($\frac{900}{x}$ + 2).
3) Find the selling price of (x âˆ’ 5) articles.
4) Equate the difference between the selling price of (x âˆ’ 5) articles and Rs 900 to Rs 80.
5) Solve the quadratic equation obtained to find original number of articles.
OR
1) Assume the present age of the son as 'x'.
2) Find the age of the son and the father two years ago with the given condition.
3) Find the age present age of the father.
4) Find the age of the son and the the age of the father three years later.
5) Find the present age of the father.
6) Equate the present ages of the father to find the present age of the father by solving the quadratic equation obtained.
7) Obtain the present age of the father from the present age of the son.
Assume the total number of articles bought as â€˜xâ€™
Cost of x articles = Rs 900
âˆ´ Cost price of 1 article =Rs $\frac{900}{x}$
Number of articles sold = (x - 5)
Selling price of 1 article = Rs $\frac{900}{x}$ + 2
Profit after selling (x âˆ’ 5) articles = Rs 80
âˆ´ (x âˆ’ 5) × ($\frac{900}{x}$ + 2) âˆ’ 900 = 80
(x âˆ’ 5) × ($\frac{900}{x}$ + 2) = 980
(x âˆ’ 5) × ($\frac{900}{x}$ + 2) = 980
900 âˆ’ $\frac{4500}{x}$ + 2(x âˆ’ 5) = 980
450 âˆ’ $\frac{2250}{x}$ + x âˆ’ 5 = 490
450x âˆ’ 2250 + ${x}^{2}$ âˆ’ 5x = 490x
${x}^{2}$âˆ’45x âˆ’ 2250 = 0
${x}^{2}$âˆ’75x + 30x âˆ’ 2250 = 0
x(x âˆ’ 75) + 30(x âˆ’ 75) = 0
(x âˆ’ 75)(x + 30) = 0
âˆ´x = 75, âˆ’30
Since number of articles can not be negative, x = 75
Hence, the number of articles bought by the trader = 75
OR
Let the sonâ€™s present age be â€˜xâ€™
Sonâ€™s age 2 years ago = (x âˆ’ 2) years
Fatherâ€™s age 2 years ago = 3(x âˆ’ 2${)}^{2}$= 3(${x}^{2}$âˆ’4x + 4)
Fatherâ€™s present age = 3(${x}^{2}$âˆ’4x + 4) + 2
Fatherâ€™s age 3 years later from now = (x + 3)
Fatherâ€™s age 3 years later from now = 4(x + 3) = 4x +12
Fatherâ€™s present age = 4x + 12 âˆ’ 3 = 4x + 9
⇒4x + 9 = 3(${x}^{2}$âˆ’4x + 4) + 2
⇒4x = 3${x}^{2}$âˆ’12x + 12 + 2 âˆ’ 9
⇒3${x}^{2}$âˆ’16x + 5 = 0
⇒3${x}^{2}$âˆ’15x âˆ’ x + 5 = 0
⇒3x(x âˆ’ 5) âˆ’ 1(x âˆ’ 5) = 0
⇒(x âˆ’ 5)(3x âˆ’ 1) = 0
x = 5, $\frac{1}{3}$
Since age cannot be a fraction, x = 5
âˆ´ Sonâ€™s age = 5 years
Fatherâ€™s age = 4x + 9 = 4×5 + 9 = 29 years.
27. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Using the above theorem prove the following: The area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.
1) Consider the ratio of the exact corresponding sides of the similar triangles.
2) Diagonal of a square is $\sqrt{2}$ times the length of the side.
3) Each angle in every equilateral triangle is equal to 60Â°.
1) Draw perpendiculars AP and XQ in each of the similar triangles ABC and XYZ.
2) Find the ratio of the areas of similar triangles.
3) Prove that âˆ†APB ~ âˆ†XQY.
4) Find the ratio of corresponding sides AP and XQ as $\frac{\mathrm{AB}}{\mathrm{XQ}}$.
5) Using the ratio of the corresponding sides of â–³ABC and â–³XYZ to prove the theorem.
6) Assume the side of the square as 'a' units then the side of the equilateral triangle drawn on the side is 'a'.
7) Diagonal of the square isâˆš$\sqrt{2}$a, also the side of the equilateral triangle drawn on the diagonal is $\sqrt{2}$a.
8) Equate the ratio of the equilateral triangles to the ratio of the square of the corresponding sides to prove the theorem.
Consider two similar triangles âˆ†ABC and âˆ†XYZ.
To prove: $\frac{(\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{ABC})}{(\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{XYZ})}$=${\left(\frac{\mathrm{AB}}{\mathrm{XY}}\right)}^{2}$$$= ${\left(\frac{\mathrm{BC}}{\mathrm{YZ}}\right)}^{2}$=${\left(\frac{\mathrm{AC}}{\mathrm{XZ}}\right)}^{2}$
Draw APâŠ¥BC and XQâŠ¥YZ.
Area of âˆ†ABC = $\frac{1}{2}$× base × height = $\frac{1}{2}$ × BC × AP ------(1)
Area of âˆ†XYZ = $\frac{1}{2}$× base × height = $\frac{1}{2}$ × YZ × XQ ------(2)
âˆ´$\frac{(\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{ABC})}{(\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{XYZ})}$= $\frac{(\mathrm{BC}\times \mathrm{AP})}{(\mathrm{YZ}\times \mathrm{XQ})}$ -------------(3)
âˆ APB = âˆ XQY = 90Â°
âˆ´ âˆ ABP = âˆ XYQ since âˆ†ABC ~ âˆ†XYZ
âˆ´ âˆ†APB ~ âˆ†XQY based on AA similarity criterion
From above, we get
$\frac{\mathrm{AP}}{\mathrm{XQ}}$ = $\frac{\mathrm{AB}}{\mathrm{XY}}$ ------------ (4)
Since âˆ†ABC ~ âˆ†XYZ, $\frac{\mathrm{AB}}{\mathrm{XY}}$ = $\frac{\mathrm{BC}}{\mathrm{YZ}}$ = $\frac{\mathrm{AC}}{\mathrm{XZ}}$ -----------(5)
Using (4) and (5), we get
$\frac{\mathrm{AB}}{\mathrm{XY}}$ = $\frac{\mathrm{BC}}{\mathrm{YZ}}$ = $\frac{\mathrm{AC}}{\mathrm{XZ}}$ = $\frac{\mathrm{AP}}{\mathrm{XQ}}$ -----------------------------(6)
Substituting (6) in (3), we get
$\frac{(\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{ABC})}{(\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{XYZ})}$ = $\frac{(\mathrm{BC}\times \mathrm{AP})}{(\mathrm{YZ}\times \mathrm{XQ})}$ = $\frac{\mathrm{BC}}{\mathrm{YZ}}$ × $\frac{\mathrm{BC}}{\mathrm{YZ}}$= ${\left(\frac{\mathrm{BC}}{\mathrm{YZ}}\right)}^{2}$
From (6), we get
$\frac{(\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{ABC})}{(\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{XYZ})}$= ${\left(\frac{\mathrm{AB}}{\mathrm{XY}}\right)}^{2}$ = ${\left(\frac{\mathrm{BC}}{\mathrm{YZ}}\right)}^{2}$= ${\left(\frac{\mathrm{AC}}{\mathrm{XZ}}\right)}^{2}$
To prove: Area of âˆ†AEC = $\frac{1}{2}$ (Area of âˆ†BFC)
Let the length of each side of the square be â€˜aâ€™
âˆ´AC = EC = EA = a (since all the sides of an equilateral triangle are equal)
BC = $\sqrt{2}$ a (since BC is the diagonal)
âˆ´BC = CF = BF = $\sqrt{2}$ a
All the angles of an equilateral triangle are 60Â° and all the sides of an equilateral triangle are equal.
Hence, all equilateral triangles are similar.
âˆ´ âˆ†AEC ~ âˆ†BFC
âˆ´$\frac{(\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{AEC})}{(\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{BFC})}$= ${\left(\frac{\mathrm{AC}}{\mathrm{BC}}\right)}^{2}$= ${\left(\frac{\mathrm{EA}}{\mathrm{CF}}\right)}^{2}$= ${\left(\frac{\mathrm{EC}}{\mathrm{BF}}\right)}^{2}$= ${\left(\frac{a}{\sqrt{2}a}\right)}^{2}$= $\frac{{a}^{2}}{2{a}^{2}}$= $\frac{1}{2}$
Hence, area of âˆ†AEC= $\frac{1}{2}$ × Area of âˆ†BFC
Therefore, the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.
28. The angle of elevation of the top of a building from the foot of a tower is 30Â° and the angle of elevation of the top of the tower from the foot of the building is 60Â°. If the tower is 50 m high, find the height of the building.
1) Angle of elevation is measured at the foot of the tower and the building.
2) Apply the appropriate trigonometric ratio of the angle of elevations.
1) Draw a figure showing the angles of elevation to the top of the tower CD and the building AB from the foot of the building and the tower.
2) Apply tangent to the angle of elevation of the tower and find the length of BD.
3) Apply tangent to the angle of elevation of the building and find the length of AB.
4) Substitute the value of BD and find the value of AB which is equal to the height of the building.
Let the building be AB and the tower be CD.
Length of CD = 50 m
Consider âˆ†CBD
$\frac{\mathrm{CD}}{\mathrm{BD}}$= tan 60Â° = $\sqrt{3}$
$\frac{50m}{\mathrm{BD}}$= $\sqrt{3}$
âˆ´BD = $\frac{(50m)}{\sqrt{3}}$
Consider âˆ†ABD
$\frac{\mathrm{AB}}{\mathrm{BD}}$ = tan 30Â° = $\frac{1}{\sqrt{3}}$
AB = $\frac{\mathrm{BD}}{\sqrt{3}}$ = $\frac{(50m)}{\sqrt{3}}$ × $\frac{1}{\sqrt{3}}$ = $\frac{50}{3}$ m = 16 $\frac{2}{3}$ m
Thus, the height of the building is 16$\frac{2}{3}$ m.
29. A spherical copper shell, of external diameter 18 cm, is melted and recast into a solid cone of base radius 14 cm and height 4 $\frac{3}{7}$ cm. Find the inner diameter of the shell.
OR
A bucket is in the form of a frustum of a cone with a capacity of 12308.8 ${\mathrm{cm}}^{3}$. The radii of the top and bottom circular ends of the bucket are 20 cm and 12 cm respectively. Find the height of the bucket and also the area of metal sheet used in making it.
1) Find the volume of the shell in terms of Ï€ and 'r'.
2) Find the volume of the cone in terms of Ï€.
3) Avoid calculation mistakes.
OR
1) Find the slant height of the bucket using the radii and the vertical height.
2) The area of the metal sheet used in making the bucket is the sum of the curved surface area of the bucket and the area of the bottom circular end of the bucket.
1) Assume the inner radius of the shell as 'r' and find the volume of the shell in terms of 'r'.
2) Find the volume of the cone using the formula for the volume of the cone.
3) Equate the volume of the shell and the cone to find the value of 'r'.
OR
1) Find the height of a bucket in the form of a frustum of a cone using the volume and the radii of the top and bottom circular ends.
2) Find the slant height of the bucket using the vertical height and the radii.
3) Find the area of the metal sheet used in making the bucket by adding the CSA of the bucket and the area of the bottom circular end of the bucket.
Let â€˜râ€™ be the inner radius and â€˜Râ€™ be the external radius of the copper shell.
Given that the external diameter of the copper shell = 2R = 18 cm
âˆ´R = 9 cm
Volume of the spherical copper shell = $\frac{4}{3}$ × Ï€ × (${R}^{3}-{r}^{3})$cu units
= $\frac{4}{3}$ Ï€(${9}^{3}\xe2\u02c6\u2019{r}^{3})$
= $\frac{4}{3}$ Ï€(729 âˆ’ ${r}^{3}$) ${\mathrm{cm}}^{3}$
The spherical copper shell is made into a solid cone.
Hence, volume of solid cone = volume of spherical shell
Volume of cone = $\frac{1}{3}$ Ï€${(\mathrm{radius}\mathrm{of}\mathrm{cone})}^{2}$(height of cone) = $\frac{1}{3}$ Ï€${\left(14\right)}^{2}$(4$\frac{3}{7}$) ${\mathrm{cm}}^{3}$
= $\frac{1}{3}$ Ï€ × ${\left(14\right)}^{2}$× $\frac{31}{7}\text{}\text{}$
âˆ´ $\frac{4}{3}$ Ï€(729 âˆ’ ${r}^{3}$) = $\frac{1}{3}$ Ï€ × ${\left(14\right)}^{2}$× $\frac{31}{7}$
4(729 âˆ’ ${r}^{3}$) = ${\left(14\right)}^{2}$× $\frac{31}{7}$
729 âˆ’ ${r}^{3}$= 7 × 31
${r}^{3}$âˆ’512 = 0
${r}^{3}$= 512
Hence, r = cube root of 512 = 8 cm
Therefore, inner radius of the spherical copper shell = 8 cm and inner diameter = 16 cm.
OR
Let the top radius of the bucket be â€˜râ€™ and the bottom radius be â€˜Râ€™
r = 20 cm and R = 12 cm
Let the height of the bucket be â€˜hâ€™
Volume of the bucket = $\frac{1}{3}$ × Ï€ × (${r}^{2}$ + ${R}^{2}$ + rR) × h = 12308.8 ${\mathrm{cm}}^{3}$
$\frac{1}{3}$ × Ï€ × (${r}^{2}$ + ${R}^{2}$ + rR) × h = 12308.8
$\frac{1}{3}$ × 3.14 × (${20}^{2}$+ ${12}^{2}$+ 20 × 12) × h = 12308.8
784 h = $\frac{(12308.8\times 3)}{\left(3.14\right)}$
h = $\frac{11760}{784}$ = 15 cm
Let â€˜sâ€™ be the slant height of the frustum of the cone.
s = $\sqrt{\left[\right({h}^{2}+{\left(r\xe2\u02c6\u2019R\right)}^{2}]}$=$\sqrt{[{15}^{2}+{\left(20\xe2\u02c6\u201912\right)}^{2}]}$ = $\sqrt{\left(225+64\right)}$= $\sqrt{289}$= 17 cm
Area of the metal sheet used in making the bucket
= CSA of the bucket + Area of the bottom circular end of the bucket
= Ï€(r + R)s + Ï€${R}^{2}$$$sq units
= 3.14 × (20 + 12) × 17 + (3.14 ×${12}^{2}$) $$
= 3.14 × 32 × 17 + (3.14 ×144) $$
= 3.14(544 + 144) $$
= 3.14 × 688$$
= 2160.32 ${\mathrm{cm}}^{2}$
Therefore, area of the metal sheet used in making the bucket is 2160.32 ${\mathrm{cm}}^{2}$.
30. Find the mode, median and mean for the following data:
1) Model class is the class with highest frequency.
2) Median class is the class in which the middle value of the sum of the frequencies ($\frac{N}{2}$) falls.
3) The class marks for each interval can be calculated by using the formula: $\frac{\mathrm{upper}\mathrm{class}lim\mathrm{it}+\mathrm{lower}\mathrm{class}lim\mathrm{it}}{2}$.
1) Find the model class of the data which has the highest number of students.
2) Find the mode using the formula: l + $\frac{({f}_{1}\xe2\u02c6\u2019{f}_{0})}{(2{f}_{1}\xe2\u02c6\u2019{f}_{0}\xe2\u02c6\u2019{f}_{2})}$ Ã— h, where l = lower class limit of modal class,
${f}_{1}$= frequency of modal class, ${f}_{0}$= frequency of the class preceding the modal class, ${f}_{2}$= frequency of the class succeeding the modal class, h = class size.
3) Median class is the class in which the middle value of the sum of the frequencies ($\frac{N}{2}$) falls.
4) Find the median using the formula: l + $\frac{(\frac{n}{2}\xe2\u02c6\u2019\mathrm{cf})}{f}$Ã— h, where l = lower limit of median class, h = class size, f = frequency of median class, cf = cumulative frequency of the class preceding median class.
3) Find the mean of the data using the formula: Mean = a + $\left(\frac{\xe2\u02c6\u2018{f}_{i}{u}_{i}}{\xe2\u02c6\u2018{f}_{i}}\right)$Ã— h, where a = assumed mean of the data, ${f}_{i}$ = frequencies, ${u}_{i}$=$\frac{{d}_{i}}{h}$ and ${d}_{i}$= (${x}_{i}$- a) and ${x}_{i}$= $\frac{\left(\mathrm{upper}\mathrm{class}lim\mathrm{it}+\mathrm{lower}\mathrm{class}lim\mathrm{it}\right)}{2}$.
(a) Mode
The maximum frequency is 33 and the class having the maximum frequency is 45âˆ’55.
âˆ´ modal class = 45âˆ’55.
Lower class limit (l) of modal class = 45
Frequency (${f}_{1}$) of the modal class = 33
Frequency (${f}_{0}$) of the class preceding the modal class = 31
Frequency (${f}_{2}$) of the class succeeding the modal class = 17
Class size (h) = 10
Mode = l + $\frac{({f}_{1}\xe2\u02c6\u2019{f}_{0})}{(2{f}_{1}\xe2\u02c6\u2019{f}_{0}\xe2\u02c6\u2019{f}_{2})}\times h$
= 45 + $\frac{(33\xe2\u02c6\u201931)}{\left[2\right(33)\xe2\u02c6\u201931\xe2\u02c6\u201917]}\times 10$
= 45 + $\frac{2}{(66\xe2\u02c6\u201948)}\times 10$
= 45 + $\frac{2}{18}$× 10
= 45 + $\frac{10}{9}$
= 45 + 1.11
= 46.11(approximately)
Therefore, mode of the data is 46.11
(b) Median
Calculate the cumulate frequency of the data in another column:-
We obtain 'n' as 100
$\frac{n}{2}$ = 50
Cumulative frequency (cf) just greater than 50 = 71
71 falls under the class 45-55. Hence median class = 45-55.
Lower limit (l) of median class = 45
Class size (h) = 10
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 38
Median = l + $\frac{(\frac{n}{2}\xe2\u02c6\u2019cf)}{f}\times h$ = 45 + $\frac{(50\xe2\u02c6\u201938)}{33}\times 10$ = 45 + $\frac{40}{11}$
= 45 + 3.63
= 48.63 (approximately)
Therefore, median of the data is 48.63 (approximately)
(c) Mean
The class marks for each interval can be calculated by using the formula.
${x}_{i}$= $\frac{(\mathrm{Upper}\mathrm{class}lim\mathrm{it}+\mathrm{Lower}\mathrm{class}lim\mathrm{it})}{2}$
Considering 60 as assumed mean (a), ${d}_{i}$, ${u}_{i}$ and ${f}_{i}{u}_{i}$ can be calculated as follows:
Mean = a + $\left(\frac{\xe2\u02c6\u2018{f}_{i}{u}_{i}}{\xe2\u02c6\u2018{f}_{i}}\right)$ × h
= 60 + $\left(\frac{\xe2\u02c6\u2019103}{100}\right)$ × 10 = 60 - $\frac{103}{10}$ = 60 - 10.3 = 49.7
Therefore, mean of the data is 49.7
âˆ´ Mode = 46.11, median = 48.63 and mean = 49.7
Section A | Section B | Section C | Section D |
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