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CBSE X
Delhi
MATHS PAPER 2013
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. The common difference of the A.P. $\frac{1}{p}$, $\frac{1\xe2\u02c6\u2019p}{p}$, $\frac{1\xe2\u02c6\u20192p}{p}$..................is:
Subtract the fractions by finding their LCM.
1) Find the difference between the first and the second term of the A.P. by finding the LCM of the fractions.
2) The difference between two terms gives the common difference of the A.P.
The common difference of the A.P $\frac{1}{p}$, $\frac{1\xe2\u02c6\u2019p}{p}$, $\frac{1\xe2\u02c6\u20192p}{p}$................... is the difference between the second term and first term i.e.
$\frac{1\xe2\u02c6\u2019p}{p}$ - $\frac{1}{p}$ = $\frac{1\xe2\u02c6\u2019p\xe2\u02c6\u20191}{p}$= $\frac{\xe2\u02c6\u2019p}{p}$ = -1.
âˆ´The common difference is - 1.
2. In Fig. 1, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA âŠ¥ PB, then the length of each tangent is:
If each angle of a quadrilateral is a right angle and the adjacent sides are equal, then the quadrilateral is a square.
1) PA and PB are two tangents to circle with centre C drawn from an external point P and they are equal.
2) CA and CB are the radii at the point of contact of the tangents and are equal.
3) Both the radii are perpendicular to the tangents at the point of contacts.
4) Given that the angle between the tangents is a right angle.
5) Angle between the tangents at the centre of the circle is a right angle according to the angle sum property.
6) If all the four angles of a quadrilateral are right angles and the adjacent sides are equal, then the quadrilateral is a square.
7) Each side of a square measures 4 cm which is equal to the radius of the circle.
8) Length of each of the tangents is 4 cm.
Given that, AP âŠ¥PB, CA âŠ¥ AP , CB âŠ¥ BP.
⇒âˆ ACB = 90Â°.
And AC = CB ( radius of the circle)
âˆ´APBC is a square .
Each side of the square is equal to 4 cm.
Therefore, length of each tangent is 4 cm.
3. In Fig.2, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches the sides BC, AB, AD and CD at points P, Q, R and S respectively, If AB = 29 cm, AD = 23 cm, âˆ B = 90Â° and DS = 5 cm, then the radius of the circle (in cm.) is:
If all the angles of a quadrilateral are right angles, it is a square.
1) Find the lengths of AR and QB by subtracting the lengths of DR and AQ from AD and AB respectively.
2) Length of QB and QP are equal as they are tangents to the circle from an the external point B.
3) Angles OQB and OPB are right angles as they are the angles between the radius and the tangents.
4) OQBP is a square.
5) Length of all the sides of the square are equal. So, the radius of the circle is equal to BQ.
AB, BC, CD and AD are tangents to the circle with centre O at Q, P, S and R respectively.
Given that, AB = 29 cm, AD = 23, DS = 5 cm and âˆ B = 90Â°.
The lengths of the tangents drawn from an external point to a circle are equal.
DS = DR = 5 cm
âˆ´ AR = AD - DR = 23âˆ’5 = 18 cm
AQ = AR = 18 cm
âˆ´ QB = AB - AQ = 29 âˆ’ 18 = 11 cm
QB = BP = 11 cm.
âˆ PBQ = 90Â° [Given]
âˆ OPB = 90Â° [Angle between the tangent and the radius at the point of contact is a right angle.]
âˆ OQB = 90Â° [Angle between the tangent and the radius at the point of contact is a right angle.]
âˆ POQ = 90Â° [Angle sum property of a quadrilateral.]
So, OQBP is a square.
âˆ´QB = BP = r = 11 cm
âˆ´ the radius of the circle is 11 cm.
4. The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30Â°. The distance of the car from the base of the tower (in m.) is:
Apply appropriate trigonometric ratio to the angle of depression of the car.
1) Draw the figure representing the data given in the question.
2) Mark the alternate angle of the angle of depression of the car.
3) Find the cotangent of the alternate angle of the angle of depression.
4) Substitute the given values and simplify to find the horizontal distance between the foot of the tower and the car.
Assume AB as the tower of height 75 m.
âˆ ABD = âˆ ACB = 30Â°
Assume C as the position of the car from the foot of the tower.
In the right angled triangle ABC,
Cot 30Â° = $\frac{\mathrm{AC}}{\mathrm{AB}}$
⇒AC = AB Cot 30Â°
⇒AC = 75 ×$\sqrt{3}$
âˆ´ Horizontal distance between the car and the foot of the tower is 75$\sqrt{3}m$.
5. The probability of getting an even number, when a die is thrown once, is :
Consider 6 in the list of favourable outcomes.
1) List the possible outcomes when a die is thrown once.
2) Write the number possible outcomes.
3) List the number of events with even numbers in the possible outcomes.
4) Write the number of favourable outcomes.
5) Ratio of number of favourable outcomes and the total number of possible outcomes is the required probability.
When a die is thrown once, the sample space S is given by { 1, 2, 3, 4, 5, 6}.
Number of possible outcomes = 6
Let E be the event of getting an even number. E = {2, 4, 6}
Number of favourable outcomes = 3
âˆ´ Probability of getting an even number = P (E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{possible}\mathrm{out}\mathrm{comes}}$= $\frac{3}{6}$ = $\frac{1}{2}$
6. A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears a prime-number less than 23, is :
1 is neither a prime nor a composite number.
1) List the numbers on the discs in the box.
2) Note the number of possible outcomes.
3) List the prime numbers between 1 and 90.
4) Assume the favourable outcomes as E and note the number of favourable outcomes.
5) Ratio of the number of favourable outcomes to the number of possible outcomes is the required probability.
Given that the box contains 90 discs.
As one disc is drawn at random from the box, numbered from 1 to 90 i.e 1,2,3,.......90.
Number of possible outcomes = 90
Let E be the event of getting prime number less than 23 are 2, 3, 5, 7, 11, 13, 17, and 19.
Number of favourable outcomes = 8
âˆ´ P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{possible}\mathrm{out}\mathrm{comes}}$= $\frac{8}{90}$= $\frac{4}{45}$.
7. In Fig. 3, Find the area of triangle ABC (in sq. units) is:
Consider the absolute value of the horizontal distance between the points B and C.
1) Draw a perpendicular AM from the vertex A to the base BC of the triangle.
2) Consider the absolute value of the horizontal distance between the points B and C as the length of the base.
3) Length of AM as measured on y-axis is considered as the altitude of the triangle corresponding to the base BC.
4) Find the area of the triangle ABC using the formula involving the base and the corresponding altitude.
Construction: Draw AM âŠ¥ BC.
Given : BC = 5 unit and AM = 3 unit.
In Î”ABC, BC is the base and AM is the height.
âˆ´ Area of triangle ABC = $\frac{1}{2}$× base × height
= $\frac{1}{2}$×BC ×AM
= $\frac{1}{2}$× 5 × 3
= 7.5 sq .units
8. If the difference between the circumference and the radius of a circle is 37 cm, then using Ï€ = $\frac{22}{7}$, the circumference (in cm) of the circle is:
(A) 154
(B) 44
(C) 14
(D) 7
Simplify the difference between the fractions by finding the LCM.
1) Assume the radius of the circle as 'r'.
2) Equate the difference between the circumference and the radius to the given value.
3) Substitute the value of Ï€ and simplify to find the radius of the circle.
4) Find the circumference of the circle using the radius obtained.
Given that the difference between circumference and radius of the circle is 37 cm.
Assume the radius of the circle as 'r'.
⇒2Ï€r - r = 37
⇒r(2Ï€ - 1) = 37
⇒ r(2 × $\frac{22}{7}$ - 1) = 37
⇒ r × $\frac{37}{7}$ = 37
⇒r = 7 cm
âˆ´ Circumference of the circle = 2Ï€r = 2 × $\frac{22}{7}$ × 7 = 44 cm
9. Solve the following quadratic equation for x:
4$\sqrt{3}{x}^{2}+5x\xe2\u02c6\u20192\sqrt{3}=0$
Use the factors of both the constant term and the coefficient of ${x}^{2}$ to spit the middle term of the equation.
1) Find the product of the constant term and the coefficient of ${x}^{2}.$
2) Use the factor pair of the product which gives the middle term on subtraction.
3) Find the factors of the equation by taking common in the first two terms and the last two terms.
4) Find the values of x be equating the each factor to zero.
Given quadratic equation is
4$\sqrt{3}{x}^{2}$+ 5x - 2$\sqrt{3}$= 0
⇒4$\sqrt{3}{x}^{2}$+ 8x - 3x - 2$\sqrt{3}$= 0
⇒4x($\sqrt{3}x$ + 2) - $\sqrt{3}$( $\sqrt{3}x$ + 2) = 0
⇒(4x - $\sqrt{3})$($\sqrt{3}x$ + 2) = 0
âˆ´ x = $\frac{\sqrt{3}}{4}$ or x = - $\frac{2}{\sqrt{3}}$
10. How many threeâˆ’digit natural numbers are divisible by 7?
Write the data using the inequalities and simplify.
1) Numbers which are divisible by 7 are in the form of 7n.
2) Write the condition using the inequalities.
3) Divide all the terms of the inequality by 7.
4) Write the fractions in the form of mixed fractions.
5) Find the difference between the integral part of the mixed numbers to find the number of natural numbers which are divisible by 7.
Let three digit numbers from 100 to 999.
All the three-digit natural numbers that are divisible by 7 will be of the form 7n.
âˆ´ 100 â‰¤ 7n â‰¤ 999
$\frac{100}{7}$â‰¤n â‰¤$\frac{999}{7}$
14$\frac{2}{7}$â‰¤n â‰¤ 142$\frac{5}{7}$
As n is an integer, number of 3-digit natural numbers that are divisible by 7 are 142 âˆ’ 14 i.e. 128.
11. In Fig. 4, a circle inscribed in triangle ABC touches its sides AB, BC and AC at points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF.
Length of CF i.e. r can be found by subtracting the sum of p and q from the sum of p, q and r.
1) Tangents drawn from the vertices of the triangle are equal.
2) Assume the lengths of the equal tangents from three sides as p, q and r.
3) Each side of the triangle is the sum of two different tangents.
4) Find the sum of the sides and equate it to the perimeter of the triangle.
5) Substitute the values of p, q and r and simplify to find the sum of p, q and r.
6) Find the length of p, q and r by subtracting the sum of two of them from the sum of all the three.
7) Lengths of p, q and r gives the required lengths.
Lengths of sides of the triangle are given as: AB = 12 cm, BC = 8 cm and AC = 10 cm.
Assume AD = AF = p cm, BD = BE = q cm and CE = CF = r cm
(Tangents drawn from an external point to the circle are equal )
⇒2(p + q + r) = AB + BC + AC = 30 cm
⇒(p + q + r) = 15 cm
⇒AB = AD + DB = p + q = 12 cm
âˆ´ r = CF = 15 - 12 = 3 cm.
⇒ AC = AF + FC = p + r = 10 cm
âˆ´q = BE = 15 - 10 = 5 cm.
⇒ BC = BE + EC = q + r = 8 cm
âˆ´ p = AD = 15 âˆ’ 8 = 7 cm.
12. Prove that the parallelogram circumscribing a circle is a rhombus.
Rearrange the lengths of tangents to form the sides of the rhombus.
1) Draw the figure showing a parallelogram circumscribing a circle.
2) Tangents to a circle from an external point are equal. So, write the tangents from each vertex of the parallelogram and equate the corresponding tangents.
3) Add the tangents such that they form the sides of the parallelogram.
4) Simplify the equation using the property of the parallelogram and prove that it is a rhombus.
Given : Draw ABCD a parallelogram circumscribing a circle with centre O.
Required to prove : ABCD is a rhombus.
Length of the tangents drawn to a circle from an exterior point are equal.
âˆ´ AP = AS, BP = BQ, CR = CQ and DR = DS.
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
âˆ´ AB + CD = AD + BC or 2AB = 2BC (Since, AB = DC and AD = BC)
âˆ´ AB = BC = DC = AD.
âˆ´ ABCD is a rhombus.
Hence, it is proved.
13. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability that the drawn card is neither a king nor a queen.
Consider the cards that are neither a queen nor a king of both the colours in the pack of playing cards.
1) Find the number of cards in the pack of playing cards that are either a king or a queen.
2) Total number of cards is 52. So, the number of cards that are neither a king nor a queen is 44.
3) Note the total number of possible outcomes as 52 and the number of favourable outcomes as 44.
4) Ratio of the number of favourable outcomes and the total number of outcomes gives the required probability.
Let E be the event of that the drawn card is neither a king nor a queen.
Total number of possible outcomes = 52.
Total number of cards that are king or queen in the pack of playing cards = 4 + 4 = 8.
âˆ´ Number of cards that are neither a king nor a queen = 52 âˆ’ 8 = 44.
Total number of favourable out comes = 44.
âˆ´ Required probability = P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$ = $\frac{44}{52}$ = $\frac{11}{13}$
âˆ´ The probability that the drawn card is neither a king nor a queen is $\frac{11}{13}$.
14. Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions 14 cm × 7 cm. Find the area of the remaining card board. [ Use Ï€ = $\frac{22}{7}$]
Circles of the maximum area has the diameter equal to the width of the rectangle and half of the length.
1) The circle of the maximum area has the diameter equal to the width of the rectangle.
2) Find the radius of the circle from the diameter.
3) Find the area of two circles.
4) Subtract the area of two circles from the area of the rectangle to get the area of the remaining card board.
Dimension of the rectangular card board = 14 cm × 7 cm
Two circular pieces of equal radii and maximum area touching each other are cut from the rectangular card board, therefore, the diameter of each circular piece is $\frac{14}{2}$ = 7 cm.
Radius of each circular piece = $\frac{d}{2}$=$\frac{7}{2}$ cm
âˆ´ Sum of area of two circular pieces = 2×Ï€ ×${\left(\frac{7}{2}\right)}^{2}$= 2×$\frac{22}{7}$×$\frac{49}{4}$ = 77 ${\mathrm{cm}}^{2}$.
Area of the remaining card board = Area of the card board âˆ’ Area of two circular pieces.
= (14 × 7) âˆ’ (77)$$= 98âˆ’77 = 21 ${\mathrm{cm}}^{2}$.
âˆ´The area of the remaining card board is 21 ${\mathrm{cm}}^{2}$.
15. For what value of k, are the roots of the quadratic equation kx (x âˆ’ 2) + 6 = 0 equal?
Consider the coefficients of the quadratic equation along with their signs.
1) Roots of the quadratic equation are equal, so the discriminant of the equation is zero.
2) Write the coefficients of the quadratic equation and find the discriminant.
3) Equate the value of the discriminant to zero and find simplify to find the value of k.
Given quadratic equation is kx(x âˆ’ 2) + 6 = 0.
The equation can be simplified as k${x}^{2}$âˆ’2kx + 6 = 0.
If the roots of a quadratic equation are equal, itâ€™s discriminant is equal to zero. i.e. D = 0.
⇒ ${b}^{2}$âˆ’4ac = 0, where a = k, b = âˆ’2k and c = 6.
⇒ 4${k}^{2}$âˆ’24k = 0
⇒ 4k(kâˆ’ 6 ) = 0
⇒ k = 0 or k = 6
But k cannot be 0, so the value of k is 6.
16. Find the number of terms of the A.P. 18, 15$\frac{1}{2}$, 13, â€¦...., - 49$\frac{1}{2}$ and find the sum of all its terms.
Consider the last term of the A.P. along with its sign.
1) Find the common difference of the A.P.
2) Equate the n th term to the last term of the A.P. and substitute the values.
3) Simplify to find the value of n which gives the number of terms of the A.P.
4) Find the sum of all the terms of the A.P. using the given values and the value of n.
The terms AP is given by 18, 15$\frac{1}{2}$, 13, ..........., - 49$\frac{1}{2}$.
First term a = 18, common difference, d = 15$\frac{1}{2}$ - 18 = -2$\frac{1}{2}$ and the last term of the A.P. = - 49$\frac{1}{2}$
Assume that A.P. has n terms.
âˆ´${a}_{n}$= a + (n âˆ’ 1) d
⇒- 49$\frac{1}{2}$ = 18 - ( n -1) × 2$\frac{1}{2}$
⇒- $\frac{99}{2}$ = 18 - $\frac{5}{2}$( n - 1)
⇒- 99 = 36 - 5( n -1)
⇒5( n -1) = 135
⇒5n = 140
âˆ´ n = $\frac{140}{5}$= 28
âˆ´The given A.P. has 28 terms.
Sum of all the terms (${S}_{n}$):
${S}_{n}$= $\frac{n}{2}$[ 2a + ( n -1) d ]
${S}_{28}$= $\frac{28}{2}$$[2\times 18+(28-1)\times \xe2\u02c6\u2019(\frac{5}{2})]$
${S}_{28}$= 14 $[36\xe2\u02c6\u2019\frac{135}{2}]$
${S}_{28}$= - 441
âˆ´The sum of all the terms of the A.P. is = - 441.
17. Construct a triangle with sides 5 cm, 4 cm and 6 cm. Then construct another triangle whose sides are $\frac{2}{3}$ times the corresponding sides of first triangle.
Locate three points on AX such that they are equidistant.
1) Construct a triangle ABC using the measurements of three sides. 2) Draw a ray AX making an acute angle with AB.
3) Mark 3 points on AX and join the third point to B.
4) Draw a line ${A}_{2}$B' parallel to ${A}_{3}$B from ${A}_{2}$.
5) Draw a line B'C' parallel to BC.
6) AB'C' is the required triangle.
Step1:
Draw a line segment AB = 4 cm. Consider the point A as centre, draw an arc of 5 cm radius.
Similarly, consider the point B as its centre, draw an arc of 6 cm radius.
Both the arcs will intersect each other at C and the required triangle ABC is formed.
Step 2:
Draw a ray AX making an acute angle with the line segment AB on the opposite side of the vertex C.
Step 3:
Locate 3 points ${A}_{1}$, ${A}_{2}$, ${A}_{3}$on line AX such that A${A}_{1}$ = ${A}_{1}{A}_{2}$= ${A}_{2}{A}_{3}$.
Step 4:
Join B and ${A}_{3}$i.e. B${A}_{3}$and draw a line through ${A}_{2}$parallel to B${A}_{3}$ to intersect AB at point B'.
Step 5:
Draw a line segment through B' parallel to BC to intersect AC at C'.
âˆ´Î”AB'C' is the required triangle.
18. The horizontal distance between two poles is 15 m. The angle of depression of the top of first pole as seen from the top of second pole is 30Â°. If the height of the second pole is 24 m, find the height of the first pole. [ $\sqrt{3}$= 1.732 ]
Find the tangent of the alternate angle of the angle of the depression.
1) Draw the figure representing the data given in the question.
2) Draw a horizontal line at the top of the first pole.
3) Mark the alternate angle of the angle of depression of the top of the first pole from the top of the second pole.
4) Assume the height of the first pole as 'h'.
5) Obtain the measures of the sides of the right angled triangle involving the alternate angle of the angle of depression.
6) Find the tangent of the alternate angle of the angle of depression. Substitute the values and simplify to find the height of the first pole.
Assume AB and CD as two poles and CD = 24 m.
Given that the angle of depression of the top of the pole AB from the top of the pole CD is 30Â° and the distance between the two poles is 15 m i.e. BD = 15 m
âˆ´ âˆ CAL = 30Â° and BD = 15 m.
Required to find the height of the pole AB.
Let the height of the pole AB be 'h' metres.
AL = BD = 15 m and AB = LD = h
âˆ´CL = CD - LD = 24 - h
In the right angled triangle ACL,
tan âˆ CAL = $\frac{\mathrm{Opposite}\mathrm{side}}{\mathrm{Adjacent}\mathrm{side}}$= $\frac{\mathrm{CL}}{\mathrm{AL}}$
tan 30Â° = $\frac{24\xe2\u02c6\u2019h}{15}$
$\frac{1}{\sqrt{3}}$= $\frac{24\xe2\u02c6\u2019h}{15}$
24 - h = $\frac{15}{\sqrt{3}}$
24 - h = 5$\sqrt{3}$
h = 24 - (5 × 1.732)
h = 15.34
âˆ´ Height of the pole (AB) = h = 15.34 m.
19. Prove that the points (7, 10), (âˆ’2, 5) and (3, âˆ’4) are the vertices of an isosceles right triangle.
Angle opposite to the longest side i.e. hypotenuse is the right angle in a right angled triangle.
1) Assume the three given points as A, B and C.
2) Find the distances between each pair of points using the distance formula.
3) Note that the lengths of two sides are equal and the triangle formed by the points is an isosceles triangle.
4) Note that the sum of the squares of two sides is equal to the square of the other side.
5) The sides obey the Pythagoras property so, they form a right angled triangle.
6) The angle opposite to the longest side i.e. the hypotenuse is the right angle.
7) So, the triangle formed by the given points is an isosceles right angled triangle.
The given points are A (7, 10), B (âˆ’ 2, 5) and C (3, âˆ’ 4).
By Using distance formula, we have
AB = $\sqrt{{(7+2)}^{2}+{(10\xe2\u02c6\u20195)}^{2}}$= $\sqrt{81+25}$= $\sqrt{106}$
BC = $\sqrt{{(\xe2\u02c6\u20192\xe2\u02c6\u20193)}^{2}+{(5+4)}^{2}}$= $\sqrt{25+81}$= $\sqrt{106}$
CA = $\sqrt{{(7\xe2\u02c6\u20193)}^{2}+{(10+4)}^{2}}$= $\sqrt{16+196}$= $\sqrt{212}$
AB = BC ⇒ Î”ABC is an isosceles triangle.
${\mathrm{AB}}^{2}$+ ${\mathrm{BC}}^{2}$= 106 + 106 = 212 = ${\mathrm{AC}}^{2}$
âˆ´ Î”ABC is a right angled triangle right angled at âˆ B.
Therefore, Î”ABC is an isosceles right angled triangle.
âˆ´ The points (7, 10), (âˆ’ 2, 5) and (3, âˆ’ 4) are the vertices of an isosceles right triangle.
20. Find the ratio in which the y-axis divides the line segment joining the points (âˆ’4, âˆ’ 6) and (10, 12). Also find the coordinates of the point of division.
Equate the y-coordinates of the point to find the value of k.
1) Assume the ratio in which the point on y-axis divides the line segment joining the given points as k:1.
2) Find the coordinates of the point dividing using the section formula.
3) Every point on the y-axis is in the form of (0, y).
4) Equate the x-coordinates of the points and simplify to find the value of k.
5) Find the ratio in which the point on x-axis divides using the value of k.
6) Substitute the value of k in the fraction obtained by using the section formula.
7) Simplify to find the coordinates of the point on y-axis.
Let a point on y-axis divide the line segment joining the points (âˆ’4, âˆ’6) and (10, 12) in the ratio k : 1.
The point of the intersection be (0, y).
So, by putting the values of coordinates and the ratio in the section formula, we get
$(\frac{10k+(\xe2\u02c6\u20194)}{k+1},\frac{12k+(\xe2\u02c6\u20196)}{k+1})$ = (0, y)
âˆ´$\frac{10k\xe2\u02c6\u20194}{k+1}$= 0 ⇒10k - 4 = 0
⇒ k = $\frac{4}{10}$= $\frac{2}{5}$
âˆ´ y = $\frac{12k\xe2\u02c6\u20196}{k+1}$= $\frac{12\times \frac{2}{5}\xe2\u02c6\u20196}{\frac{2}{5}+1}$= $\frac{\frac{24\xe2\u02c6\u201930}{5}}{\frac{2+5}{5}}$ = - $\frac{6}{7}$
âˆ´ y-axis divides the line segment joining the given points in the ratio of 2 : 5 and the point of intersection is$\left(0,\xe2\u02c6\u2019\frac{6}{7}\right)$.
21. In Fig.5, AB and CD are two diameters of a circle with centre O, which are perpendicular to each other. OB is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.[ use Ï€ = $\frac{22}{7}$]
Radius of the bigger circle is the diameter of the smaller circle and the altitude of the right angled triangle.
1) Radius of the bigger circle is the diameter of the smaller circle.
2) Find the radius of the smaller circle.
3) Radius of the bigger circle is also the altitude of the right angled triangle corresponding to the side BC.
4) Subtract the sum of the areas of the smaller circle and the area of the right angled triangle from the area of the bigger circle to find the area of the shaded region.
Given that AB and CD are the diameters of a circle with centre O.
âˆ´ OA = OB = OC = OD = 7 cm (Radius of the circle)
Area of the shaded region
= Area of the circle with diameter OB + (Area of the semi-circle ACDA âˆ’ Area of Î”ACD)
= Ï€${r}^{2}$ + ( $\frac{1}{2}$Ï€${r}^{2}$ -$\frac{1}{2}$ x CD x OA)
= $\frac{22}{7}$×${\left(\frac{7}{2}\right)}^{2}$+$(\frac{1}{2}\times \frac{22}{7}\times 49\xe2\u02c6\u2019\frac{1}{2}\times 14\times 7)$
= ($\frac{77}{2}$+ 77 - 49) ${\mathrm{cm}}^{2}$
= 66.5 ${\mathrm{cm}}^{2}$
âˆ´The area of the shaded region is 66.5${\mathrm{cm}}^{2}$.
22. A vessel is in the form of hemispherical bowl surmounted by a hollow cylinder of same diameter. The diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the total surface area of the vessel.[Ï€ = $\frac{22}{7}$]
1) Find the sum of the inner and the outer surface areas of the vessel.
2) Ignore the thickness of the vessel.
1) Find the radius of the bowl and the cylinder from the diameter.
2) Find the height of the cylindrical part of the vessel.
3) Find two times of the sum of the curved surface area of the cylinder and the surface area of the hemisphere.
4) Simplify to find the total surface area of the vessel.
Assume the radius and the height of the cylinder as r cm and h cm respectively.
The diameter of the hemispherical bowl = 14 cm
âˆ´ Radius of the hemispherical bowl = Radius of the cylinder = r =$\frac{14}{2}$ cm = 7 cm
Total height of the vessel = 13 cm
âˆ´ Height of the cylinder = Total height of the vessel âˆ’ Radius of the hemispherical bowl
= 13âˆ’ 7 = 6 cm
Total surface area of the vessel = 2 (Curved surface area of the cylinder + Surface area of the hemisphere)
= 2(2Ï€rh + 2Ï€${r}^{2}$)
= 4Ï€r(h + r)
= 4 × $\frac{22}{7}$ × 7 × (6 + 7)
= 1144 ${\mathrm{cm}}^{2}$.
âˆ´ Total surface area of the vessel is 1144 ${\mathrm{cm}}^{2}$.
23. A wooden toy was made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the volume of wood in the toy. [Ï€ = $\frac{22}{7}$]
Radius of the hemisphere and the cylinder are equal.
1) Find the radius of the hemisphere from the diameter.
2) Subtract two time of the volume of the hemisphere from the volume of the cylinder to find the volume of the wood in the toy.
3) Substitute the values in the formula and simplify to find the total volume.
Given that the height of the cylinder, h = 10 cm
And radius of the cylinder = Radius of each hemisphere = r = 3.5 cm
Volume of wood in the toy = Volume of the cylinder âˆ’ 2 × Volume of each hemisphere.
= (Ï€${r}^{2}$h) âˆ’ (2 × $\frac{2}{3}$ × Ï€${r}^{3}$)
= [$\frac{22}{7}$×${\left(3.5\right)}^{2}$×10] âˆ’[$\frac{4}{3}$×$\frac{22}{7}$×${\left(3.5\right)}^{3}$]
= 385 âˆ’ $\frac{539}{3}$
= $\frac{616}{3}$
= 205.33 ${\mathrm{cm}}^{3}$(Approximately)
âˆ´ The volume of the wood in the toy is approximately 205.33 ${\mathrm{cm}}^{3}$.
24. In a circle of radius 21 cm, an arc subtends an angle of 60Â° at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc. [ Use Ï€ = $\frac{22}{7}$]
Units of the area of the sector are ${\mathrm{cm}}^{2}$.
1) Note the angle subtended by the arc at the centre of the circle and the radius of the circle.
2) Write the formula to find the length of the arc of the circle.
3) Substitute the values and simplify to find the result.
4) Write the formula to find the area of the sector formed by the arc.
5) Substitute the values and simplify to find the result.
Given that the radius of the circle = 21 cm.
Measure of the angle subtended by the arc at the centre = 60Â°.
(i) Length of the arc
l = $\frac{\mathrm{\xce\xb8}}{360\xc2\xb0}$ × 2Ï€r
where r = 21 cm and Î¸ = 60Â°
= $\frac{60\xc2\xb0}{360\xc2\xb0}$ × 2 × $\frac{22}{7}$ × 21
= 22 cm.
(ii) Area of the sector formed by the arc
A = $\frac{\mathrm{\xce\xb8}}{360\xc2\xb0}$×Ï€${r}^{2}$
r = 21 cm and Î¸ = 60Â°
= $\frac{60\xc2\xb0}{360\xc2\xb0}$ × $\frac{22}{7}$ × ${\left(21\right)}^{2}$
= 231${\mathrm{cm}}^{2}$.
25. Solve the following for x:
$\frac{1}{2a+b+2x}=\frac{1}{2a}$ + $\frac{1}{b}$ + $\frac{1}{2x}$
Factorise the equation by rearranging the terms of the equation.
1) Rearrange the fractions of the equation such that the terms with variable 'x'
2) Find the LCM of the fractions and simplify.
3) Find the factors of the quadratic equation obtained by taking common.
4) Equate each factor to zero and simplify to find the values of x.
The given equation is
$\frac{1}{2a+b+2x}=\frac{1}{2a}$ + $\frac{1}{b}$ + $\frac{1}{2x}$
⇒$\frac{1}{2a+b+2x}\xe2\u02c6\u2019\frac{1}{2x}=\frac{1}{2a}$ + $\frac{1}{b}$$$
⇒$\frac{2x\xe2\u02c6\u20192a\xe2\u02c6\u2019b\xe2\u02c6\u20192x}{2x(2a+b+2x)}$ = $\frac{b+2a}{2\mathrm{ab}}$
⇒$\frac{\xe2\u02c6\u2019(2a+b)}{2x(2a+b+2x)}$ = $\frac{2a+b}{2\mathrm{ab}}$
⇒$\frac{\xe2\u02c6\u20191}{x(2a+b+2x)}$ = $\frac{1}{\mathrm{ab}}$
⇒2${x}^{2}$+ 2ax + bx + ab = 0
⇒2x(x + a) + b(x +a) = 0
⇒(x +a)(2x +b) = 0
⇒x + a = 0 or 2x + b = 0
⇒ x = âˆ’a or x = âˆ’$\frac{b}{2}$.
26. Sum of the areas of two squares is 400 ${\mathrm{cm}}^{2}$. If the difference of their perimeters is 16 cm, find the sides of the two squares.
1) Substitute the value of either x or y obtained from equation (1) in equation (2).
2) Factorise the quadratic equation using the factors of the constant.
3) Do not consider the negative values as the lengths.
1) Assume the lengths of sides of the square as x and y where x>y.
2) Find the sum of the areas of the squares and obtain equation (1).
3) Find the difference between the perimeters of the squares and obtain equation (2).
4) Find the value of x in terms of y from equation (2)
5) Substitute the value of x in equation (1) and simplify to get a quadratic equation.
6) Solve the quadratic equation by splitting the middle term and find the value of y.
7) Find the value of x using equation (2). x and y are the required lengths of sides of the squares.
Assume that the sides of the two squares be x cm and y cm where x > y.
The areas of the squares are ${x}^{2}$ and ${y}^{2}$and their perimeters are 4x and 4y.
Given that, ${x}^{2}$+ ${y}^{2}$= 400 â€¦ (1)
and 4x âˆ’ 4y = 16
⇒ 4(x âˆ’ y) = 16 ⇒ x - y = 4 â€¦ (2)
⇒ x = y + 4
Substituting the value of x in eqn (1), we get ${(y+4)}^{2}$+ ${y}^{2}$= 400
⇒ ${y}^{2}$+16 + 8y +${y}^{2}$ = 400
⇒2${y}^{2}$+ 8y + 16 = 400
⇒ ${y}^{2}$+ 4y - 192 = 0
⇒ ${y}^{2}$ + 16y âˆ’12y - 192 = 0
⇒ y(y + 16) âˆ’12(y + 16) = 0
⇒ (y + 16) (y âˆ’12) = 0
⇒ y = âˆ’16 or y = 12
Since the value of y cannot be negative so the value of y = 12.
Value of x = y + 4 = 12 + 4 = 16.
âˆ´The sides of the two squares are 16 cm and 12 cm.
27. If the sum of first 7 terms of an A.P. is 49 and that of first 17 terms is 289, find the sum of its first n terms.
Use the formula to find the sum of n terms with n and d terms.
1) Find the sum of the first 7 terms of the A.P. and obtain equation (1)
2) Find the sum of the first 17 terms of the A.P. and obtain equation (2)
3) Solve equation (1) and equation (2) and find the values of a and d.
4) Substitute the values of a and d in the formula to find the sum of n terms of an A.P. and simplify.
Given that, the sum of first 7 terms of an A.P. is 49 i.e.${S}_{7}$ = 49 and the sum of first 17terms of an AP is 289 i.e${S}_{17}$ = 289.
${S}_{n}$=$\frac{n}{2}$[2a + (n - 1)d ]
⇒${S}_{7}$=$\frac{7}{2}$[ 2a + 6d ]
⇒49 = $\frac{7}{2}$[2a + 6d]
⇒2a + 6d = 14
⇒ a + 3d = 7 ------------ (1)
Similarly,
${S}_{17}$= $\frac{17}{2}$[2a +( 17 - 1) d ]
289 = $\frac{17}{2}$[2a + 16d ]
⇒ a + 8d = 17 ------------ (2)
Solving equations (1) and equation (2), we get 5d = 10.
âˆ´ d = 2
âˆ´ If d = 2 then, a = 7 âˆ’ 3d = 1.
${S}_{n}$= $\frac{n}{2}$[2a + (n - 1)d]
${S}_{n}$= $\frac{n}{2}$[ 2 + ( n - 1)2 ]
${S}_{n}$= $\frac{n}{2}$[ 2n]
âˆ´ ${S}_{n}$= ${n}^{2}$.
âˆ´The sum of n terms of the A.P. is ${n}^{2}$.
28. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
The shortest distance between two points is the perpendicular distance.
1) Draw a figure representing the data given in the question.
2) Mark any point B on the tangent 'l' and draw a line segment joining the centre O and the point B.
3) Let the line segment intersect the circle at C.
4) OB is written as the sum of two parts OC and CB.
5) OB is greater than OC. So, OB is greater than OA since OA and OC are radii of the circle.
6) OA is shorter than OB.
7) OA is shorter than any other segment joining the centre of the circle and a point on the tangent.
8) This proves that OA, the radius at the point of contact of the tangent is perpendicular to the tangent.
Given:In a circle C (0, r) and a tangent l touches the circle at point A.
To prove: OA âŠ¥ l
Construction: Consider a point B on l, other than A, on the tangent l . Join OB. Let OB intersect the circle in C.
Proof: Among all line segments joining the point O to a point on l, the perpendicular is shortest to l.
OA = OC (Radius of the circle)
Now, OB = OC + BC.
âˆ´ OB > OC
⇒ OB > OA
⇒ OA < OB
As B is any point on the tangent l, OA is shorter than any other line segment joining O to any point on l.
There fore, OA âŠ¥ l.
Hence, the tangent at any point of a circle is perpendicular to the radius through the point of contact.
29. In fig. 6, l and m are two parallel tangents to a circle with centre O, touching the circle at A and B respectively. Another tangent at C intersects the line l at D and m at E. Prove that âˆ DOE = 90Â°
Prove that the corresponding sides of the triangles are equal to prove that they are congruent.
Sum of the angles of a straight line is 180Â°.
1) Join O and C.
2) Compare the triangles OAD and OCD and prove that they are congruent.
3) Angles DOA and COD are equal according to CPCT.
4) Similarly prove that the triangles BOE and COE are congruent.
5) Angles BOE and EOC are equal according to CPCT.
6) Equate the sum of the angles on the diameter AOB and equate to 180Â°.
7) Simplify using the equality of angles and prove that the sum of âˆ COD and âˆ COE is a right angle.
8) Sum of âˆ COD and âˆ COE isâˆ DOE.
Given: l and m are two tangents to the circle with centre O parallel to each other and touching the circle at A and B respectively. DE is another tangent at the point C, which intersects l at D and m at E.
Required to prove: âˆ DOE = 90Â°
Construction: Join OC.
Proof:
Consider Î”ODA and Î”ODC,
OA = OC (Radii of the same circle)
AD = DC (Length of tangents drawn from an external point to a circle are equal)
DO = OD (Common side)
Î”ODA â‰… Î”ODC (SSS congruence property)
âˆ´ âˆ DOA = âˆ COD â€¦........ (i) (C.P.C.T)
Similarly, we can prove that Î”OEB â‰… Î”OEC
âˆ EOB = âˆ COE â€¦ (ii)
AOB is a diameter of the circle which is a straight line and the angle on a straight line is a straight angle.
âˆ´ âˆ DOA + âˆ COD + âˆ COE + âˆ EOB = 180Âº.
From (i) and (ii), we get,
2âˆ COD + 2 âˆ COE = 180Âº
⇒ âˆ COD + âˆ COE = 90Âº
⇒ âˆ DOE = 90Â°
Hence, it is proved.
30. The angle of elevation of the top of a building from the foot of the 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 60 m high, find the height of the building.
Apply appropriate trigonometric ratio to the angles of the elevation.
1) Draw the figure representing the data given in the question.
2) Complete the right angled triangles involving each of the angles of elevation.
3) Find the cotangent of the angle of elevation of the tower and simplify to find the distance between the building and the tower.
4) Find the tangent of the angle of elevation of the building and simplify to find the height of the building.
Assume AB as the building and CD as the tower.
Suppose the height of the building AB as 'h' m.
Given that, âˆ ACB = 30Â°, âˆ CBD = 60Â° and CD = 60 m.
Î”BCD is a right angled triangle then,
cot 60Â° = $\frac{\mathrm{BC}}{\mathrm{CD}}$ ⇒ BC = CD cot 60Â°
⇒ BC = 60 ×$\frac{1}{\sqrt{3}}$= 20$\sqrt{3}$m
Î”ACB is a right angled triangle then,
tan 30Â° = $\frac{\mathrm{AB}}{\mathrm{BC}}$ ⇒ AB = BC tan 30Â°
⇒ h = 20$\sqrt{3}$ ×$\frac{1}{\sqrt{3}}$= 20m
âˆ´ The height of the building is 20 m.
31. A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person from the group is selected at random. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is (i) extremely patient (ii) extremely kind or honest. Which of the above values you prefer more?
Find the total number of persons who are extremely honest or kind.
1) Note the total number of persons in the group which is considered as the total number of possible outcomes.
2) Assume the number of persons who are extremely patient as ${E}_{1}.$
3) Note the number of persons favourable to ${E}_{1}$.
4) Assume the number of persons who are extremely extremely kind or honest as ${E}_{2}$.
5) Find the number of persons who are extremely kind or honest.
6) Note the number of persons favourable to ${E}_{2}$.
7) Find the ratio of the number of outcomes favourable to ${E}_{1}$and the total number of possible outcomes to get the probability of the person selected is extremely patient.
8) Find ratio of the number of outcomes favourable to ${E}_{2}$and the total number of possible outcomes to get the probability of the person selected is extremely kind or honest.
Given that the group consists of 12 persons.
âˆ´Total number of possible outcomes = 12
Assume ${E}_{1}$as the event of selecting persons who are extremely patient.
âˆ´ Number of favourable outcomes to ${E}_{1}$is 3.
Assume ${E}_{2}$as the event of selecting persons who are extremely kind or honest.
Number of persons who are extremely honest is 6.
Number of persons who are extremely kind = 12 âˆ’ (6 + 3) = 3
âˆ´ Number of favourable outcomes to${E}_{2}$ = 6 + 3 = 9.
(i) Probability of selecting a person who is extremely patient is P(${E}_{1}$).
P(${E}_{1}$) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{E}_{1}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$ = $\frac{3}{12}$ = $\frac{1}{4}$.
(ii) Probability of selecting a person who is extremely kind or honest is
P (${E}_{2}$) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{E}_{2}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$ = $\frac{9}{12}$ = $\frac{3}{4}$.
32. The three vertices of a parallelogram ABCD are A(3, âˆ’4), B(âˆ’1, âˆ’3) and C(âˆ’6, 2). Find the coordinates of vertex D and find the area of ABCD.
Find the area of one of the triangle formed by the diagonals of the parallelogram and multiply it by 2.
1) Assume the coordinates of the fourth vertex as (x, y).
2) Equate the midpoints of the diagonals of the parallelogram as the diagonals of a parallelogram bisect each other.
3) Simplify to find the coordinates of the fourth vertex.
4) Area of a quadrilateral is equal to twice the area of the triangles formed by the diagonal of the parallelogram.
5) Find the area of one of the triangles using the area of the triangle formula.
6) Multiply the area of the triangle by 2 to obtain the area of the parallelogram.
Given that three vertices of the parallelogram ABCD are A (3, âˆ’4), B (âˆ’1, âˆ’3) and C (âˆ’6, 2).
Assume that the coordinates of the vertex D be (x, y).
In a parallelogram, the diagonals bisect each other.
âˆ´Mid point of AC = Mid point of BD
⇒$(\frac{3\xe2\u02c6\u20196}{2},\frac{\xe2\u02c6\u20194+2}{2})$= $(\frac{\xe2\u02c6\u20191+x}{2},\frac{\xe2\u02c6\u20193+y}{2})$
⇒$(\frac{\xe2\u02c6\u20193}{2},\frac{\xe2\u02c6\u20192}{2})$= $(\frac{\xe2\u02c6\u20191+x}{2},\frac{\xe2\u02c6\u20193+y}{2})$
⇒$\frac{\xe2\u02c6\u20193}{2}$= $\frac{\xe2\u02c6\u20191+x}{2}$, $\frac{\xe2\u02c6\u20192}{2}$= $\frac{\xe2\u02c6\u20193+y}{2}$
⇒ x = âˆ’2, y = 1.
So that, the coordinates of the fourth vertex D is (âˆ’2, 1).
Now, area of parallelogram ABCD
= area of triangle ABC + area of triangle ACD
= 2 × area of triangle ABC [Diagonal divides the parallelogram into two triangles of equal area]
The area of a triangle whose vertices are (${x}_{1}$, ${y}_{1}$), (${x}_{2}$, ${y}_{2}$) and (${x}_{3}$, ${y}_{3}$) is $\frac{1}{2}$[${x}_{1}({y}_{2}\xe2\u02c6\u2019{y}_{3})+{x}_{2}({y}_{3}\xe2\u02c6\u2019{y}_{1})+{x}_{3}({y}_{1}\xe2\u02c6\u2019{y}_{2})$]
Area of triangle ABC
= $\frac{1}{2}$|[3(-3 - 2) + (-1)(2 - (-4)) + (-6)( -4 -(-3) )]|
= $\frac{1}{2}$| [-15 -6 + 6]| = $\frac{\xe2\u02c6\pounds \xe2\u02c6\u201915\xe2\u02c6\pounds}{2}$square units.
âˆ´Area of the triangle ABC = $\frac{15}{2}$ square units (Area of the triangle cannot be negative)
âˆ´ Area of the parallelogram ABCD = 2 × $\frac{15}{2}$ = 15 square units.
33. Water is flowing through a cylindrical pipe, of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m/s. Determine the rise in level of water in the tank in half an hour.
1) Convert the diameter of the cylindrical pipe into metres.
2) Find the speed of the water in metres per minute.
1) Find the radius of the circular end of the pipe in metres.
2) Find the area of the cross section of the pipe in terms of Ï€.
3) Find the speed of the water in metres per minute.
4) Multiply the speed of the water by the area of the cross section of the pipe to get the volume of the water flown in one minute.
5) Find the volume of the water flown through the pipe in 30 minutes.
6) Assume the level of the water raised in the tank as 'h'.
7) Equate the volume of the water flown through the pie in 30 minutes and the volume of the water filled in the tank.
8) Simplify to find the rise in level of water in the tank by the water filled in half an hour.
Given that, internal diameter of circular end of pipe = 2 cm
âˆ´ Radius (${r}_{1}$) of circular end of pipe = $\frac{2}{100\times 2}$ m = 0.01 m
Area of cross-section of the circular end of pipe = Ï€${{r}_{1}}^{2}$= Ï€${\left(0.01\right)}^{2}$= 0.0001Ï€ ${m}^{2}$
Speed of water = 0.4 m/s = 0.4 ×60 = 24 metre / min.
Volume of water that flows through the pipe in 1 minute = 24 × 0.0001 Ï€${m}^{3}$= 0.0024Ï€ ${m}^{3}$
Volume of water that flows through the pipe in 30 minutes = 30 × 0.0024 Ï€${m}^{3}$ = 0.072Ï€ ${m}^{3}$
Radius (${r}_{2}$) of base of cylindrical tank = 40 cm = 0.4 m
Let the rise in level of the water in the cylindrical tank filled in 30 minutes be h m.
Volume of water filled in tank in 30 minutes = Volume of water flowed in 30 minutes from the pipe.
âˆ´ Ï€ ×${\left({r}_{2}\right)}^{2}$×h = 0.072Ï€
⇒ ${\left(0.4\right)}^{2}$×h = 0.072
⇒ 0.16 × h = 0.072
⇒h = $\frac{0.072}{0.16}$
⇒ h = 0.45 m = 45 cm
âˆ´ The rise in level of water in the tank in half an hour is 45 cm.
34. A bucket open at the top, and made up of a metal sheet is in the form of a frustum of a cone. The depth of the bucket is 24 cm and the diameters of its upper and lower circular ends are 30 cm and 10 cm respectively. Find the cost of metal sheet used in it at the rate of Rs 10 per 100${\mathrm{cm}}^{2}$. [Use Ï€ = 3.14]
Consider the area of the base of the bucket only as it is open at the top.
1) Draw a figure representing the data given in the question.
2) Find the slant height of the bucket in the shape of frustum of a cone using the Pythagoras theorem.
3) Area of metal sheet required to make the bucket is equal to the sum of the curved surface area of the frustum of the cone and the area of the base of the bucket.
4) Substitute the values in the formula and simplify to get the final answer.
5) Multiply the area of the sheet required by the cost of the sheet to find the total cost of the sheet.
Given that diameter of upper end of bucket = 30 cm
âˆ´ Radius (${r}_{1}$) of upper end of bucket = 15 cm
Diameter of lower end of bucket = 10 cm
âˆ´ Radius (${r}_{2}$) of lower end of bucket = 5 cm
Height (h) of bucket = 24 cm
Slant height (l) of frustum = $\sqrt{{({r}_{1}\xe2\u02c6\u2019{r}_{2})}^{2}+{h}^{2}}$
= $\sqrt{{(15\xe2\u02c6\u20195)}^{2}+{24}^{2}}$ = $\sqrt{100+576}$
= $\sqrt{676}$= 26 cm
Area of metal sheet required to make the bucket = Ï€(${r}_{1}$+ ${r}_{2}{)}^{2}$l
+ Ï€${{r}_{2}}^{2}$
= Ï€ (15 + 5) 26 + Ï€(${5}^{2}$)
= 520 Ï€ + 25 Ï€ = 545 Ï€ ${\mathrm{cm}}^{2}$
Cost of 100 ${\mathrm{cm}}^{2}$metal sheet = Rs 10
Cost of 545Ï€ ${\mathrm{cm}}^{2}$metal sheet = Rs $\frac{545\times 3.14\times 10}{100}$ = Rs.171.13
âˆ´Total cost of metal sheet used to make the bucket is Rs 171.13
Section A | Section B | Section C | Section D |
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