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CBSE X
All India
MATHS PAPER 2011
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 roots of the equation${x}^{2}\xe2\u02c6\u20193x\xe2\u02c6\u2019m(m+3)=0$, where m is a constant, are:
Substitute the coefficients of the equation in the quadratic formula along with the signs.
1) Find the factors of the constant term
2) Use the factors of the constant term and split the middle term of the equation.
3) Find the factors of the equation by taking common.
4) Equate each of the factors to find the roots of the equation.
Given quadratic equation: ${x}^{2}\xe2\u02c6\u20193x\xe2\u02c6\u2019m(m+3)=0$.
It can be written as
${x}^{2}\xe2\u02c6\u2019\left\{\right(m+3)\xe2\u02c6\u2019m\}x\xe2\u02c6\u2019m(m+3)=0$
${x}^{2}\xe2\u02c6\u2019(m+3)x+\mathrm{mx}\xe2\u02c6\u2019m(m+3)=0$
x{x - (m + 3)} + m{x - (m + 3)} = 0
(x + m){x - (m + 3)}=0
Therefore, x = - m and x = (m + 3) are the zeroes of the equation.
2. If the common differences of an A.P. is 3, then ${a}_{20}\xe2\u02c6\u2019{a}_{15}$is:
Substitute the value of d as 3 in the difference of the ${20}^{\mathrm{th}}$term and the ${15}^{\mathrm{th}}$term.
1) Find the ${20}^{\mathrm{th}}$term of the A.P. using the formula to find the ${n}^{\mathrm{th}}$term of the A.P.
2) Find the ${15}^{\mathrm{th}}$term of the A.P. using the formula to find the ${n}^{\mathrm{th}}$term of the A.P.
3) Find the difference between the terms found.
4) Substitute the value of d as 3 in the difference and find the required value.
Let a be the first term of the A.P. and d be the common difference.
${n}^{\mathrm{th}}$term of an A.P. = ${a}_{n}=a+(n\xe2\u02c6\u20191)d$
${a}_{20}\xe2\u02c6\u2019{a}_{15}$= [a + (20 âˆ’1)d] âˆ’ [a + (15 âˆ’1)d]
= 19d âˆ’ 14d
= 5d
= 5 x 3
= 15.
3. In figure 1, O is the centre of a circle, PQ is a chord and PT is the tangent at P. If âˆ POQ = 70Â°, then âˆ TPQ is equal to ________.
Measures of angles OPQ and OQP are equal according to isosceles triangle property.
1) Consider the triangle OPQ.
2) OP and OQ are radii of the circle. So, measures of angles OPQ and OQP are equal according to isosceles triangle property.
3) Find the measure of the base angles OPQ and OQP of the triangle OPQ.
4) âˆ OPT is a right angle as the radius at the point of contact of a tangent is perpendicular to it.
5) Find the measure of âˆ TPQ by subtracting the measure of âˆ OPQ from 90Â°.
Consider the triangle in the circle. i.e., Î”OPQ:
OP = OQ (as both are radii of the circle)
⇒ âˆ OQP = âˆ OPQ (Equal sides of a triangle have equal angles opposite to them)
âˆ POQ + âˆ OPQ + âˆ OQP = 180Â° (based on the angle sum property of triangles)
⇒ 70Â° + 2âˆ OPQ = 180Â°
⇒ 2âˆ OPQ = 180Â° âˆ’ 70Â° = 110Â°
⇒ âˆ OPQ = 55Â°
The tangent is perpendicular to the radius through the point of contact.
âˆ´ âˆ OPT = 90Â°
⇒ âˆ OPQ + âˆ TPQ = 90Â°
⇒ 55Â° + âˆ TPQ = 90Â°
⇒ âˆ TPQ = 90Â° âˆ’ 55Â° = 35Â°.
4. In Figure 2, AB and AC are tangents to the circle with centre O such that âˆ BAC = 40Â°. Then âˆ BOC is equal to ________.
âˆ BOC is an angle of the quadrilateral ABOC.
1) Measures of angles OBA and OCA is 90Â° each.
2) Consider the quadrilateral OBCA.
3) Find the fourth angle BOC of the quadrilateral using the angle sum property of the quadrilateral.
The tangent to a circle is perpendicular to the radius through the point of contact.
âˆ´ âˆ ABO = âˆ ACO = 90Â°
In quadrilateral ABOC, according to the angle sum property:
âˆ ABO + âˆ BOC + âˆ ACO + âˆ BAC = 360Â°
⇒ 90Â° + âˆ BOC + 90Â° + 40Â° = 360Â°
⇒ âˆ BOC + 220Â° = 360Â°
⇒ âˆ BOC = 360Â° âˆ’ 220Â° = 140Â°.
5. The perimeter (in cm) of a square circumscribing a circle of radius a cm, is _________.
Diameter of the circle is equal to the length of the side of the square circumscribing the circle.
1) Find the diameter of the circle from the radius.
2) Diameter of the circle is equal to the length of the side of the square circumscribing the circle.
3) Find the perimeter of the square using the length of the side.
Let the radius of the circle be â€˜aâ€™ cms
The diameter of the circle becomes two times the radius
2 × a cm = 2a cm.
Therefore, the side of the circumscribing square = Diameter of the circle = 2a cm.
âˆ´ Perimeter of the circumscribing square = 4 × 2a cm = 8a cm.
6. The radius (in cm) of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is _________.
Diameter of the base of the cone is equal to the length of the edge of the square.
1) Largest right circular cone that can be cut from a cube has the base diameter equal to the length of the edge of the cube.
2) Assume the radius of the base of the cone as 'r' cm. Then the diameter of the base is 2r cm.
3) Equate the diameter to the length of the edge of the cube and simplify to find the value of r.
In a cube, the largest right circular cone has a diameter that is equal to the side of a cube.
Similarly, its height is equal to the side of the cube.
So, if the radius of the cone is r cm,
âˆ´ 2r = 4.2 cm.
⇒r = $\frac{4.2}{2}$ cm = 2.1 cm.
7. A tower stands vertically on the ground. From a point on the ground which is 25 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 45Â°. Then the height (in meters) of the tower is _________ m.
Find the tangent of the angle of elevation of the tower.
1) Draw a figure to represent the data given in the question.
2) Find the tangent of the angle of elevation of the tower.
3) Substitute the values of the sides in the ratio and simplify to find the height of the tower.
In the figure above AB is the tower and C is the point on the ground 25 m away from the foot of the tower in such a way that âˆ ACB = 45Â°.
Therefore, in â–³ABC,
tan 45Â° = $\frac{\mathrm{AB}}{\mathrm{AC}}$
⇒1 = $\frac{\mathrm{AB}}{25}$
⇒AB = 25 m
Therefore, the height of the tower AB is 25 m.
8. If P($\frac{a}{2}$, 4) is the mid-point of the line-segment joining the points A (âˆ’6, 5) and B(âˆ’2, 3), then the value of a is
1) Write the correct formula to find the mid-point of a line segment.
2) Equate the x-coordinates of the mid-points.
1) Find the mid-point of the line segment joining the points (-6, 5) and (-2, 3).
2) Equate the x-coordinate of the mid-points to find the value of a.
P($\frac{a}{2}$, 4) is the mid-point of the line segment joining points A (âˆ’6, 5) and B (âˆ’2, 3).
âˆ´ Coordinates of the point P can be found by using the midpoint formula.
P = $(\frac{\xe2\u02c6\u20196+(\xe2\u02c6\u20192)}{2},\frac{5+3}{2})$= $(\frac{\xe2\u02c6\u20198}{2},\frac{8}{2})$ = (-4, 4)
Given that P$(\frac{a}{2},4)$ is the mid-point, therefore
(-4, 4) = $(\frac{a}{2},4)$
âˆ´ $\frac{a}{2}$ = - 4
⇒a = âˆ’8
So, the value of a is âˆ’8.
9. If A and B are the points (âˆ’6, 7) and (âˆ’1, âˆ’5) respectively, then the distance
2AB is equal to __________.
Substitute the corresponding coordinates of the points in the distance formula along with their signs.
1) Use the distance formula and find the distance between the points A and B.
2) Multiply the length of AB by 2 to find the distance 2AB.
Given that the coordinates of the points A and B are (âˆ’6, 7) and (âˆ’1, âˆ’5) respectively.
AB = $\sqrt{{(\xe2\u02c6\u20196\xe2\u02c6\u2019(\xe2\u02c6\u20191\left)\right)}^{2}+{(7\xe2\u02c6\u2019(\xe2\u02c6\u20195\left)\right)}^{2}}$
= $\sqrt{{(\xe2\u02c6\u20196+1)}^{2}+{(7+5)}^{2}}$
= $\sqrt{{(\xe2\u02c6\u20195)}^{2}+{\left(12\right)}^{2}}$
= $\sqrt{25+144}$
= $\sqrt{169}$
= 13
âˆ´ 2AB = 2 x 13 = 26.
Hence, the distance 2AB is 26 units.
10. A card is drawn from a well-shuffled deck of 52 playing cards. The probability that the card will not be an ace is ________.
Consider the ace cards of all the colours.
1) Note the number of ace cards in a deck of 52 playing cards.
2) Total number of cards in the deck is considered as the total number of outcomes.
3) Find the remaining number of cards other than ace cards.
4) Assume the event of drawing a non-ace card as E.
5) Ratio of the number of favourable outcomes and the total number of outcomes is the required probability.
Number of ace cards in a deck of 52 playing cards is 4.
âˆ´ Number of non-ace cards in the pack = 52 âˆ’ 4 = 48
Assume E as the event of drawing a non-ace card.
Total number of possible outcomes = 52
Number of favourable outcomes = 48
P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}$ = $\frac{48}{52}$= $\frac{12}{13}$.
11. Find the value of m so that the quadratic equation mx (x âˆ’ 7) + 49 = 0 has two equal roots.
Consider the coefficients of the given equation along with their signs.
1) Write the given equation in the standard form of the quadratic equation.
2) Compare the coefficients of the equation with the coefficients of the standard quadratic equation and note the values.
3) Find the discriminant of the quadratic equation and equate it to zero.
4) Substitute the values and simplify to find the value of m.
The quadratic equation is given as mx (x âˆ’ 7) + 49 = 0 and it can be written as:
mx (x âˆ’ 7) + 49 = 0
⇒ m${x}^{2}$âˆ’7mx + 49 = 0
The discriminant of a quadratic equation with equal roots is equal to zero.
(${b}^{2}$âˆ’4ac ) = 0
i.e., ${b}^{2}$âˆ’4ac = 0, where a = m, b =âˆ’7m and c = 49
i.e., ${(\xe2\u02c6\u20197m)}^{2}$âˆ’4 × m × 49 = 0
⇒ 49${m}^{2}$âˆ’4 × m × 49 = 0
⇒ 49 m (m âˆ’ 4) = 0
⇒ m = 0 or mâˆ’ 4 = 0
⇒ m = 0 or m = 4
âˆ´ m = 4 (m â‰ 0)
Hence, the value of m is 4.
12. Find how many two-digit numbers are divisible by 6.
Consider the two digit numbers which are multiples of 6 between 10 and 99.
1) List the two digit numbers which are divisible by 6.
2) Note the first term, last term of the A.P. and find the common difference of the A.P.
3) Equate the last term of the A.P. to the ${n}^{\mathrm{th}}$term of the A.P.
4) Substitute the values of a and d, and simplify to find the value of n.
5) Value of n gives the number of terms of the A.P.
Two-digit numbers that are divisible by 6 are 12, 18, 24, â€¦, 96.
The series forms an A.P. with first term (a) = 12 and the common difference (d) = 6
${n}^{\mathrm{th}}$term of an A.P. = ${a}_{n}$= a + (nâˆ’1)d
⇒12 + (nâˆ’1) × 6 = 96
⇒ 6 (nâˆ’1) = 96âˆ’12 = 84
⇒ nâˆ’1 = 14
⇒ n = 15
Hence, there are 15 two-digit numbers that are divisible by 6.
13. In Figure 3, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of the side AD.
AD has two parts AS and DS.
1) Consider the vertices of the quadrilateral ABCD as the external points and equate the tangents drawn from the points.
2) Consider the side AD as the sum of two parts AS and DS.
3) Replace the equivalent tangents of AS and DS.
4) Simplify and find the length of AD.
Let the points at which the circle touches the sides AB, BC, CD, and DA of the quadrilateral be P,Q,R, S respectively.
The lengths of tangents drawn from an external point to a circle are equal.
âˆ´ AS = AP, DS = DR, BP = BQ, CR = CQ -------- (1)
AD
= AS + DS
= AP + DR
= (6 âˆ’ PB) + (8 âˆ’ CR)
= 14 âˆ’ BQ âˆ’ CQ [(From (1)]
= 14 âˆ’ (BQ + CQ)
= 14 âˆ’ BC
= (14 âˆ’ 9)
= 5 cm
Hence, the length of AD = 5 cm.
14. Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB such that $\frac{\mathrm{AP}}{\mathrm{AB}}$= $\frac{3}{5}$.
Mark points on ray AX such that they are equidistant.
1) Find the ratio of AP and PB using the given data.
2) Draw a line segment AB of length 7 cm.
3) Draw a ray AX at A making an acute angle with AB.
4) Mark 5 points (sum of 3 and 4) ${A}_{1}$, ${A}_{2}$..... ${A}_{5}$ on AX such that they are equidistant.
5) Join ${A}_{5}$ to B.
6) Draw a line at ${A}_{3}$, parallel to ${A}_{5}B$.
7) The line intersects the line segment AB at P.
8) P is the required point which divides the line segment AB in the ratio of 3:2.
The given condition is that $\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{3}{5}$
âˆ´ $\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{AP}}{\mathrm{AB}\xe2\u02c6\u2019\mathrm{AP}}=\frac{3}{5\xe2\u02c6\u20193}=\frac{3}{2}$
So, the line segment AB is divided by the point P in the ratio of 3:2.
Our objective is to draw the line segment AB of length 7 cm and mark a point P such that $\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{3}{5}i.e.\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{3}{2}$. The following steps are to be followed:
Step 1: Draw line segment AB of length 7 cm and below the line segment AB from A draw a ray AX making an acute angle with line segment AB.
Step 2: Locate and mark 5 points (Sum of the terms of the ratio) ${A}_{1},{A}_{2},{A}_{3},{A}_{4},\mathrm{and}{A}_{5}$on AX such that ${\mathrm{AA}}_{1}={A}_{1}{A}_{2}={A}_{2}{A}_{3}$and so on.
Step 3: Join the points B and ${A}_{5}.$
Step 4: Draw a parallel line to ${\mathrm{BA}}_{5}$from point ${A}_{3}$to intersect AB at P.
The point P is the required point on the line segment AB which satisfies the given condition.
15. Find the perimeter of the shaded region in Figure 4, if ABCD is a square of side 14 cm and APB and CPD are semicircles. [Use Ï€ = $\frac{22}{7}$]
Perimeter of the semicircles includes two sides of the square.
1) Diameter of the semicircle is the side of the square.
2) Find the radius of the semicircle and the length of the arc of the semicircle.
3) Perimeter of the shaded region includes the lengths of arcs of two semicircles and the lengths of two sides of the squares.
4) Substitute the values and add to find the perimeter of the shaded region.
ABCD is a square of side 14 cm. âˆ´
âˆ´ AB = BC = CD = AD = 14 cm
There are two semicircles in the square of radius = $\frac{14}{2}$ = 7 cm
Perimeter of the shaded region = BC + AD + Perimeter of semi-circle APB + Perimeter of semi-circle CPD
Perimeter of semicircle APB = $\frac{1}{2}$ x 2Ï€ x radius = $\frac{1}{2}$ x 2Ï€ x 7 = $\frac{22}{7}$ x 7 = 22 cm.
Perimeter of semicircle CPD = $\frac{1}{2}$x 2Ï€ x radius = $\frac{1}{2}$x 2Ï€ x 7 = $\frac{22}{7}$ x 7 = 22 cm.
âˆ´ Hence, the perimeter of the shaded region = 22 + 22 + 14 + 14 = 72 cm.
16. Two cubes each of volume 27 ${\mathrm{cm}}^{3}$are joined end to end to form a solid. Find the surface area of the resulting cuboid.
OR
A cone of height 20 cm and radius of base 5 cm is made up of modeling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere.
Breadth and the height of the new cuboid formed are equal to the side of the cube.
OR
Volume of the sphere is equal to the volume of the cone.
1) Find the length of the side of the cube from the volume of the cube.
2) Length of the cuboid is twice the length of the side of the cube. Breadth and the height of the cuboid formed are equal to the length of the side of the cube.
3) Find the total surface area of the resulting cuboid using the corresponding formula.
OR
1) Find the volume of the cone using the given values of height and the radius of the base.
2) Assume the radius of the sphere formed as 'r'.
3) Equate the volume of the sphere to the volume of the cone.
4) Simplify and find the value of the radius of the sphere.
5) Find the diameter of the sphere from the radius.
Solution:
Assume the edge of each cube as 'a' cm.
Hence, volume of each cube = ${a}^{3}$ ${\mathrm{cm}}^{3}$
Since, the volume of each cube is 27 ${\mathrm{cm}}^{3}$,
${a}^{3}$ = 27= ${\left(3\right)}^{3}$
⇒ a = 3
So, the length of edge of the cube = 3 cm
If two cubes are joined end-to-end, we get a cuboid whose length, breadth and height are 6 cm (3 + 3), 3 cm and 3 cm respectively.
This can be depicted by the diagram given below:
Surface area of the cuboid = 2 (lb + bh + hl)
= 2 [(6 × 3) + (3 × 3) + (3 × 6)]
= 2 × 45
= 90 ${\mathrm{cm}}^{2}$
So, the surface area of the resulting cuboid is 90 ${\mathrm{cm}}^{2}$.
OR
Assume the radius of the sphere as r.
Radius of the cone = 5 cm and height of the cone = 20 cm
Since the clay is reshaped from cone to sphere, the volumes of both the shapes are equal.
$\frac{4}{3}$ x Ï€${r}^{3}$= $\frac{1}{3}$Ï€${\left(5\right)}^{2}$x (20) [Volume of the cone = $\frac{1}{3}$Ï€${r}^{2}$h]
⇒${r}^{3}$= 125
⇒${r}^{3}$= ${\left(5\right)}^{3}$
⇒r = 5 cm
Hence, the diameter of the sphere = 2r = 2 × 5 = 10 cm.
17. Find the value of y for which the distance between the points A(3, âˆ’1) and B(11, y) is 10 units.
Substitute the corresponding coordinates of the points in the distance formula along with the sign.
1) Find the distance between the points A and B using the distance formula.
2) Equate it to the given distance and simplify.
3) Solve the quadratic equation obtained and find the two values for the y-coordinate of B.
Coordinates of the given points are A (3, âˆ’1) and B (11, y).
Distance between the points A and B = AB = 10 units
âˆ´$\sqrt{{(11\xe2\u02c6\u20193)}^{2}+{[y\xe2\u02c6\u2019(\xe2\u02c6\u20191\left)\right]}^{2}}$ = 10
⇒$\sqrt{\left(64\right)+{(y+1)}^{2}}=10$
⇒ 64 + ${(y+1)}^{2}=100$
⇒${(y+1)}^{2}=100\xe2\u02c6\u201964$
⇒${(y+1)}^{2}$= 36
⇒(y + 1) = Â± 6
⇒(y + 1) = 6 or (y + 1) = - 6
⇒y = 6 âˆ’1 = 5 or âˆ’6 âˆ’1 = âˆ’7
Hence, the values of y = 5 or âˆ’7.
18. A ticket is drawn at random from a bag containing tickets numbered from 1 to 40. Find the probability that the selected ticket has a number which is a multiple of 5.
Consider the multiples of 5 between 1 and 40 only.
1) The bag has tickets numbered from 1 to 40. So, the number of possible outcomes is 40.
2) Assume the event that the ticket drawn has a number which is a multiple of 5 as E.
3) List the numbers which are multiples of 5 between 1 and 40.
4) Note the number of favourable outcomes.
5) Ratio of number of favourable outcomes and the total number of possible outcomes is the probability of E.
Assume the event of drawing a multiple of 5 as E.
Since there are 40 tickets, the total number of outcomes = 40
The outcomes of event E are 5, 10, 15, 20, 25, 30, 35 and 40.
So, the number of favourable outcomes of event E = 8
Probability that the selected ticket has a number that is a multiple of 5:
P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}$
= $\frac{8}{40}$
= $\frac{1}{5}$
Hence, the probability of drawing a ticket which is a multiple of 5 = $\frac{1}{5}$.
19. Find the roots of the following quadratic equation:${x}^{2}\xe2\u02c6\u20193\sqrt{5}x+10=0$
Substitute the coefficients of the equation in the quadratic formula along with the signs.
1) Compare the given quadratic equation with the standard form of the quadratic equation and write the values of the coefficients.
2) Find the value of the discriminant of the equation using the formula.
3) Use the quadratic formula to find the roots of the quadratic equation and find the roots of the given equation.
Compare the given quadratic equation with the standard form of quadratic equation i.e., a${x}^{2}$ + bx + c = 0.
a = 1, b = âˆ’3$\sqrt{5}$, c = 10
$\sqrt{D}$ =$\sqrt{{b}^{2}\xe2\u02c6\u20194\mathrm{ac}}$
= $\sqrt{{(\xe2\u02c6\u20193\sqrt{5})}^{2}\xe2\u02c6\u20194\times 1\times 10}$
= $\sqrt{45\xe2\u02c6\u201940}\text{}$
= $\sqrt{5}$
âˆ´ x = $\frac{\xe2\u02c6\u2019b\xc2\pm \sqrt{D}}{2a}$
= $\frac{3\sqrt{5}\xc2\pm \sqrt{5}}{2\times 1}$
=$\frac{4\sqrt{5}}{2}$ or $\frac{2\sqrt{5}}{2}$
= 2$\sqrt{5}$or $\sqrt{5}$
Hence, the roots of the given quadratic equation are 2$\sqrt{5}$or $\sqrt{5}$.
20. Find the A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.
Find the values of a and d of the A.P. using the given conditions.
1) Find the fourth term of the A.P. using the formula to find the ${n}^{\mathrm{th}}$term of the A.P. Assume it as equation 1.
2) Find the sixth term and the thirteenth term of the A.P.
3) Find the sum of the sixth term and the thirteenth term of the A.P. and equate it to the given value. Assume it as equation 2.
4) Solve equations 1 and 2, and find the values of a and d.
5) Use the values of a and d, and find the terms of the A.P.
Assume the first term of the A.P. as 'a' and the common difference as 'd'.
${4}^{\mathrm{th}}$term of the A.P. = ${t}_{4}$= 9.
The sum of the sixth and thirteenth term is 40 ⇒${t}_{6}+{t}_{13}$= 40.
If ${t}_{4}$= 9
⇒a + (4âˆ’1)d = 9 ${[t}_{n}=a+(n\xe2\u02c6\u20191)d]$
⇒a + 3d = 9 ---------- (1)
${t}_{6}$+ ${t}_{13}$= 40
⇒{a + (6 - 1)d} + {a + (13 - 1)d} = 40
⇒{a + 5d} + {a + 12d} = 40
⇒2a + 17d = 40 ---------- (2)
Solving the linear equations (1) and (2)
From (1):
a = 9 âˆ’ 3d
Substituting the value of a in (2):
2 (9 âˆ’ 3d) + 17d = 40
⇒ 18 + 11d = 40
⇒ 11d = 22
⇒ d = 2
âˆ´ a = 9 âˆ’ 3 × 2 = 3
Hence, the given A.P. = a, a + d, a + 2d â€¦, where a = 3 and d = 2.
So, the A.P. is 3, 5, 7, 9 â€¦
21. In Figure 5, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contact T, are of lengths 12 cm and 9 cm respectively. If the area of Î”PQR = 189 ${\mathrm{cm}}^{2}$, then find the lengths of sides PQ and PR.
Radius at the point of contact of the tangent of a circle is perpendicular to the tangent.
1) Join the points O and P, O and Q, O and R, O and B, and O and C.
2) OT is perpendicular on QR as the radius at the point of contact of a tangent of a circle is perpendicular to the tangent.
3) Similarly, OS and OU are perpendiculars on PQ and PR respectively.
4) Area of triangle ABC is the sum of the areas of triangles OPQ, OQR and OPR.
5) Assume the lengths PS and PU as x.
6) Find the lengths of the sides of the triangle in terms of x.
7) Equate the sum of the areas of the triangles OPQ, OQR and OPR to the area of triangle PQR.
8) Simplify to find the value of x.
9) Find the lengths of sides PQ and PR using the value of x.
Let PQ and PR touch the circle at points S and U respectively. Join O with points P, Q, R, S and U.
The radius of the circle is 6 cm and hence we have, OS = OT = OU = 6 cm, QT = 12 cm and TR = 9 cm
âˆ´ QR = QT + TR = 12 cm + 9 cm = 21 cm
Since, QT and QS are tangents from the same point Q, QT = QS = 12 cm
Similarly, TR = RU = 9 cm
Let us assume that PS = PU = x cm
Then, PQ = PS + SQ = (12 + x) cm and PR = PU + RU = (9 + x) cm
Hence,
area of (Î”OQR) + area of (Î”OPR) + area of (Î”OPQ) = area of (Î”PQR)
⇒$\frac{1}{2}$× QR × OT ] + [$\frac{1}{2}$ × PR × OU] + [$\frac{1}{2}$× PQ × OS] = 189
⇒$\frac{1}{2}$× 21 × 6] + [$\frac{1}{2}$ × (9 + x) × 6] + [$\frac{1}{2}$ × (12 + x) × 6] = 189
⇒$\frac{1}{2}$× 6 (21 + 9 + x + 12 + x) = 189
⇒ 3 (42 + 2x) = 189
⇒ 42 + 2x = 63
⇒ 2x = 21
⇒ x = 10.5
Hence, PQ = (12 + 10.5) = 22.5 cm and PR = (9 + 10.5) = 19.5 cm
22. Draw a pair of tangents to a circle of radius 3 cm, which are inclined to each other at an angle of 60Â°.
OR
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are $\frac{3}{5}$ times the corresponding sides of the given triangle.
Angle between the tangents and the angle between the radii drawn at the points of contact of the tangents are supplementary.
OR
Locate five points on AX such that they are equidistant.
1) Draw a circle of radius 3 cm with centre O.
2) Draw a radius OA and perpendicular to the radius at point A.
3) Angle between the tangents and the angle between the radii drawn at the points of contact of the tangents are supplementary. So, the angle between the radii is 120Â°.
4) Draw another radius OB at angle of 120Â° with OA and perpendicular the radius at point B.
5) Let both the tangents intersect the point P. PA and PB are the required tangents.
OR
1) Construct a right angled triangle ABC using the measurements of two sides.
2) Draw a ray AX making an acute angle with AB.
3) Mark 5 points on AX and join the fifth point to B.
4) Draw a line segment ${A}_{3}$B' parallel to ${A}_{5}$B from ${A}_{3}$.
5) Draw a line segment B'C' parallel to BC.
6) AB'C' is the required triangle.
The Procedure to construct the tangent is given below
(i) Draw a circle with a radius of 3 cm and centre as O.
(ii) Consider a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A.
(iii) Draw OB the radius such that it makes an angle of 120Â° [(180Â° âˆ’ 60Â°)] with OA.
(iv) From B, draw a perpendicular to OB. Let both the perpendiculars intersect at point P.
Hence, PA and PB are the required tangents at an angle of 60Â°.
OR
To draw the right angled triangle -
(i) Lengths of the sides of the right angled triangle are 4 cm and 3 cm. They are perpendicular to each other. Draw a line segment AB = 4 cm. Draw a ray SA making an angle of 90Â° with it.
(ii) Taking A the center, draw an arc of radius 3 cm which intersects SA at point C. Join BC. ABC is the required right triangle.
(iii) From C, draw a ray AX which makes an acute angle with AB.
(iv) Locate and mark 5 points (since 5 is greater in the numbers 5 and 3) i.e., ${A}_{1}$, ${A}_{2}$, ${A}_{3}$, ${A}_{4}$and ${A}_{5}$on line segment AX such that${\mathrm{AA}}_{1}$ = ${A}_{1}{A}_{2}$= ${A}_{2}{A}_{3}$= ${A}_{3}{A}_{4}$= ${A}_{4}{A}_{5}$.
(v) Join ${A}_{5}$and B to form ${A}_{5}$B. Draw a line through ${A}_{3}$parallel to ${A}_{5}$B which intersects the line segment AB at B'.
(vi) Draw a line parallel to BC through B' and this line will intersect the line segment AC at C'. AB'C' is the required triangle.
23. A chord of a circle of radius 14 cm subtends an angle of 120Â° at the centre. Find the area of the corresponding minor segment of the circle. Use Ï€ = $\frac{22}{7}\mathrm{and}\sqrt{3}=1.73$
Prove that the triangles OAM and OBM are are congruent by RHS congruence criterion.
1) Draw a figure representing the data given in the figure.
2) Let OA and OB be the radii whose corresponding arc makes an angle of 120Â° at the centre.
3) Join AB and draw OMâŠ¥AB.
4) Find the area of the sector OAB.
5) Prove that the triangles OAM and OBM are congruent by RHS congruence criterion.
6) Prove that âˆ AOM and âˆ BOM are equal by CPCT.
7) Find the lengths of OM and AB using the trigonometric ratios of angles AOM and BOM.
8) Find the area of the triangle using the lengths of AM and BM.
9) Subtract the area of the triangle AOB from the area of the sector AOB to find the area of the minor segment.
Draw a circle with O as the centre and AB the chord that subtends an angle of 120Â° at the centre.
OM âŠ¥ AB.
Given that the radius of the circle, r = 14 cm
Area of the minor segment = Area of sector OAB âˆ’ Area of Î”AOB
Area of sector OAB = $\frac{\mathrm{\xce\xb8}}{360\xc2\xb0}$× Ï€${r}^{2}$
= $\frac{120\xc2\xb0}{360\xc2\xb0}\times \frac{22}{7}\times {\left(14\right)}^{2}$
= $\frac{1}{3}$ × $\frac{22}{7}$ × ${\left(14\right)}^{2}$
= $\frac{1}{3}$ × $\frac{22}{7}$ × 196
= $\frac{616}{6}{\mathrm{cm}}^{3}$
Since OM âŠ¥ AB,
AM = MB = $\frac{1}{2}$AB
⇒ AB = 2AM
Consider Î”OAM and Î”OBM:
OA = OB [Radii of the circle]
OM = OM [Common side]
âˆ OMA = âˆ OMB [Right angles]
âˆ´ Î”OAM â‰… Î”OBM [RHS congruence criterion]
⇒ âˆ AOM = âˆ BOM [C.P.C.T]
âˆ´ âˆ AOM = âˆ BOM = $\frac{120\xc2\xb0}{2}=60\xc2\xb0.$
Consider Î”AOD:
sin 60Â° = $\frac{\mathrm{AM}}{\mathrm{OA}}$, cos 60Â° = $\frac{\mathrm{OM}}{\mathrm{OA}}$
⇒$\frac{\sqrt{3}}{2}$= $\frac{\mathrm{AM}}{14}$, $\frac{1}{2}$= $\frac{\mathrm{OM}}{14}$
⇒AM = 14$\sqrt{3}$m, OM = 7 cm
âˆ´ AB = 2AM = 14$\sqrt{3}$cm
Area of (â–³AOB) = $\frac{1}{2}$× AB × OM = $\frac{1}{2}$ × 14$\sqrt{3}$× 7 = 49$\sqrt{3}{\mathrm{cm}}^{2}$
Area of minor segment = $\frac{616}{3}$âˆ’49$\sqrt{3}$
= 205.33 âˆ’49 × 1.73
= 205.33 âˆ’84.77
= 120.56 ${\mathrm{cm}}^{2}$
Hence, the area of the minor segment = 120.56 ${\mathrm{cm}}^{2}$.
24. An open metal bucket is in the shape of a frustum of a cone of height 21 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the cost of milk which can completely fill the bucket at Rs. 30 per litre. [Use Ï€ = $\frac{22}{7}$]
Convert the volume of the bucket into litres.
1) Find the volume of the bucket using the formula to find the volume of the frustum of a cone.
2) Convert the volume of the bucket into litres from cubic centimetres by dividing with 1000.
3) Multiply the volume of the bucket in litres by the cost of the milk.
Height of frustum of cone, h = 21 cm
Radius of lower end, r = 10 cm
Radius of upper end, R = 20 cm
Volume of frustum of cone = $\frac{1}{3}$Ï€h(${R}^{2}$ + ${r}^{2}$ + Rr)
Volume of milk required to fill the bucket
= $\frac{1}{3}$ × $\frac{22}{7}$ × 21 × [${\left(20\right)}^{2}+{\left(10\right)}^{2}+20\times 10]{\mathrm{cm}}^{3}$
= 22 × (400 + 100 + 200)
= 22 × 700
= 15400 ${\mathrm{cm}}^{3}$
= $\frac{15400}{1000}$ litres
= 15.4 litres
Cost of milk = Rs 30 per litre
Hence, the cost of milk for filling the bucket is Rs (30 × 15.4) i.e. Rs 462.
25. Point P(x, 4) lies on the line segment joining the points A(âˆ’5, 8) and B(4, âˆ’10). Find the ratio in which point P divides the line segment AB. Also find the value of x.
Assume the ratio in which the point P divides the line segment AB as k:1.
1) Assume the ratio in which the point P divides the line segment AB as k:1.
2) Find the coordinates of the point P using the section formula.
3) Equate the y-coordinate of the point P obtained with the given value of the coordinate.
4) Simplify and find the value of k.
5) Equate the x-coordinates of the point P and substitute the value of k.
6) Simplify and find the value of x.
Assume that the line segment AB is divided by point P(x, 4) in the ratio of k:1.
Coordinates of A = (âˆ’5, 8)
Coordinates of B = (4, âˆ’10)
According to the section formula,
(x, 4) = $[\frac{k\times 4+1(\xe2\u02c6\u20195)}{k+1},\frac{k\times (\xe2\u02c6\u201910)+1\times 8}{k+1}]$
(x, 4) = $[\frac{4k\xe2\u02c6\u20195}{k+1},\frac{\xe2\u02c6\u201910k+8}{k+1}]\text{}$
⇒$\frac{4k\xe2\u02c6\u20195}{k+1}=x$ ------------- (1)
and $\frac{\xe2\u02c6\u201910k+8}{k+1}=4$ ------------- (2)
Solving the linear equations:
Considering equation (2):
âˆ’ 10k + 8 = 4(k + 1)
⇒ âˆ’ 10k + 8 = 4k + 4
⇒ 14k = 4
⇒k = $\frac{2}{7}$
Hence, point P divides AB in the ratio of $\frac{2}{7}$ i.e., 2:7.
Considering equation (1): $\frac{4\times \frac{2}{7}\xe2\u02c6\u20195}{\frac{2}{7}+1}=x$
⇒$\frac{\frac{\xe2\u02c6\u201927}{7}}{\frac{9}{7}}=x$
⇒x = - 3.
26. Find the area of the quadrilateral ABCD, whose vertices are A(âˆ’3, âˆ’1), B (âˆ’2, âˆ’4), C(4, âˆ’ 1) and D (3, 4).
OR
Find the area of triangle formed by joining the mid-points of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5).
Substitute the values of the coordinates in the formula to find the area of the triangle in order.
OR
Mid points of the sides of the triangle are the vertices of the triangle whose area is required to be found.
1) Draw a quadrilateral ABCD and diagonal BC.
2) Find the area of the triangle ABD using the formula to find the area of the triangle.
3) Find the area of the triangle CBD using the formula to find the area of the triangle.
4) Sum of the areas of the triangles ABD and CBD is the area of the quadrilateral ABCD.
OR
1) Draw a triangle ABC and mark the midpoints of sides P, Q and R.
2) Find the coordinates of the points P, Q and R using the midpoint formula.
3) Find the area of the triangle PQR using the formula to find the area of the triangle.
To find the area of quadrilateral ABCD whose vertices are A (âˆ’3, âˆ’1), B (âˆ’2, âˆ’4), C (4, âˆ’1) and D (3, 4):
Draw a sketch to represent the data given in the question.
BD is the diagonal.
The area of a triangle whose vertices are (${x}_{1}$, ${y}_{1}$), (${x}_{2}$ , ${y}_{2}$) and (${x}_{3}$, ${y}_{3}$) is
$\frac{1}{2}\left[{x}_{1}\right({y}_{2}\xe2\u02c6\u2019{y}_{3})+{x}_{2}({y}_{3}\xe2\u02c6\u2019{y}_{1})+{x}_{3}({y}_{1}\xe2\u02c6\u2019{y}_{2}\left)\right]$
âˆ´ Area of (Î”ABD)
= $\frac{1}{2}[\xe2\u02c6\u20193(\xe2\u02c6\u20194\xe2\u02c6\u20194)+(\xe2\u02c6\u20192\left)\right(4+1)+3(\xe2\u02c6\u20191+4\left)\right]\text{}$
= $\frac{1}{2}[\xe2\u02c6\u20193(\xe2\u02c6\u20198)+(\xe2\u02c6\u20192\left)\right(5)+3(3\left)\right]$
= $\frac{1}{2}[24\xe2\u02c6\u201910+9]$
= $\frac{23}{2}$
= 11.5 sq units
Area of (Î”CDB)
= $\frac{1}{2}$[4(4 + 4) + 3(âˆ’4 + 1) + (âˆ’2)(âˆ’1âˆ’4)]
= $\frac{1}{2}$[(4 × 8) + (3 × âˆ’3) âˆ’2 (âˆ’5)]
= $\frac{1}{2}$[32 âˆ’9 + 10]
= $\frac{33}{2}$
= 16.5 sq units
Hence, area of (ABCD) = area of (Î”ABD) + area of (Î”CDB) = (11.5 + 16.5) sq units = 28 sq units.
OR
ABC is the triangle with A (2, 1), B (4, 3) and C (2, 5) as the vertices of the triangle.
Assume that P, Q and R are the mid-points of sides AB, BC and CA respectively of Î”ABC.
Coordinates of the midpoints of the points (${x}_{1}$, ${y}_{1}$) and (${x}_{2}$, ${y}_{2}$) = $(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2})$
Coordinates of P = $(\frac{4+2}{2},\frac{3+1}{2})$ = (3, 2)
Coordinates of Q = $(\frac{4+2}{2},\frac{3+5}{2})$ = (3, 4)
Coordinates of R = $(\frac{2+2}{2},\frac{5+1}{2})$ = (2, 3)
Area of the triangle with vertices, (${x}_{1}$ , ${y}_{1}$ ), (${x}_{2}$, ${y}_{2}$) 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 Î”PQR
= $\frac{1}{2}\left[3\right(4\xe2\u02c6\u20193)+3(3\xe2\u02c6\u20192)+2(2\xe2\u02c6\u20194\left)\right]$
= $\frac{1}{2}\left[3\right(1)+3(1)+2(\xe2\u02c6\u20192\left)\right]$
= $\frac{1}{2}[3+3\xe2\u02c6\u20194]$
= $\frac{2}{2}$
= 1 sq unit
Therefore, the area of the triangle with given vertices is 1 sq unit.
27. From the top of a vertical tower, the angles of depression of two cars, in the same straight line with the base of the tower, at an instant are found to be 45Â° and 60Â°. If the cars are 100 m apart and are on the same side of the tower, find the height of the tower. [Use $\sqrt{3}$= 1.73]
Find the tangent of the alternate angles of the angles of depression.
1) Draw a figure representing the data given in the question.
2) Mark the alternate angles of the angles of depression.
3) Assume the height of the tower as 'h' m and the distance between the second boat and the tower as 'x' m.
4) Find the tangent of the alternate angle of the angle of depression of the second car.
5) Substitute the values of the sides and find the value of OA in terms of 'h'.
6) Find the tangent of the alternate angle of the angle of depression of the first car.
7) Substitute the values of the sides and find the value of OB in terms of 'h'.
8) Equate the difference between OA and OB to 100.
9) SImplify and find the value of h.
The illustration for the question is given below
Tower is represented by OP. Point A and Point B are the two positions of the cars.
Assume the height of the tower as h i.e., OP = h.
âˆ XPB = 45Â° and âˆ XPA = 60Â°(Given)
⇒ âˆ PBO = âˆ XPB = 45Â° and âˆ PAO = âˆ XPA = 60Â° (Alternate interior angles)
Consider Î”OPA:
tan 60Â° = $\frac{h}{\mathrm{OA}}$
⇒$\sqrt{3}$= $\frac{h}{\mathrm{OA}}$
⇒OA = $\frac{h}{\sqrt{3}}$
Consider Î”OPB:
tan 45Â° = $\frac{h}{\mathrm{OB}}$
⇒1 = $\frac{h}{\mathrm{OB}}$
⇒OB = h
The distance between the two cars is 100 m (Given)
Therefore, AB = 100 m
⇒ OB âˆ’ OA = 100
⇒h âˆ’$\frac{h}{\sqrt{3}}$ = 100
⇒ $\frac{\sqrt{3}\xe2\u02c6\u20191}{\sqrt{3}}h$ = 100
⇒h = $\frac{100\sqrt{3}}{\sqrt{3}\xe2\u02c6\u20191}$ = $\frac{100\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}\xe2\u02c6\u20191)(\sqrt{3}+1)}$ = $\frac{100(3+\sqrt{3)}}{2}$ = 50 × (3 + 1.73) = 236.5 m
Hence, the height of the tower = 236.5 m.
28. Two dice are rolled once. Find the probability of getting such numbers on the two dice, whose product is 12.
OR
A box contains 80 discs which are numbered from 1 to 80. If one disc is drawn at random from the box, find the probability that it bears a perfect square number.
Consider the factors of 12 between 1 and 6 to find the favourable outcomes.
OR
Consider the perfect squares between 1 and 80 including 1 to find the favourable outcomes.
1) Note the number of possible outcomes when two dice are rolled.
2) Assume the event of getting the numbers whose product is 12 as E.
3) List the outcomes favourable to E.
4) Find the number of favourable outcomes.
5) Ratio of number of favourable outcomes and the total number of outcomes is the required probability.
OR
1) Number of discs in the box is the total number of possible outcomes.
2) Note the total number of possible outcomes of the event.
3) Selecting a disc with a perfect square number is the favourable outcome.
4) List the favourable outcomes and note the number of favourable outcomes.
5) Ratio of number of favourable outcomes and the total number of possible outcomes is the required probability.
List out the sample space of the given experiment
S = {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)}
Therefore, n (S) = 36
Assume the event of getting the numbers whose product is 12 as E.
E = {(2, 6), (3, 4), (4, 3), (6, 2)}
Therefore, n (E) = 4
Hence, the required probability = $\frac{n\left(E\right)}{n\left(S\right)}$= $\frac{4}{36}$= $\frac{1}{9}$.
OR
Discs are numbered from 1 to 80.
So, the total number of possible outcomes = 80.
We know that the perfect squares less than 80 are 1, 4, 9, 16, 25, 36, 49 and 64.
The total number of perfect squares from 1 to 80 is 8
So, the total number of favourable outcomes = 8
Hence, the required probability = $\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$= $\frac{8}{80}$= $\frac{1}{10}$.
29. 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.
In the illustration above, O is the centre of the circle and AB is the tangent to the circle at point P.
To prove that OP âŠ¥ AB.
Consider Q any point on AB other than P.
Draw OQ to meet the circle at point R.
The shortest line segment joining O to a point on AB is perpendicular to it.
So, to prove that OP âŠ¥ AB, it is sufficient to prove that OP is shorter than any other segment joining O to any point on AB.
OP = OR [Radii to the circle]
OQ = OR + RQ
⇒ OQ > OR
⇒ OQ > OP [Since OP = OR]
⇒ OP < OQ
Hence, OP is shorter than any other segment joining O to any point on AB.
Therefore, OP âŠ¥ AB.
30. The first and the last terms of an A.P. are 8 and 350 respectively. If its common difference is 9, how many terms are there and what is their sum?
OR
How many multiples of 4 lie between 10 and 250? Also find their sum.
'n' value of the last term of the A.P. gives the number of terms of the A.P.
OR
Multiples of 4 form an A.P. with a common difference of 4.
1) Note the first and last terms of the A.P. and the common difference.
2) Equate the ${n}^{\mathrm{th}}$term of an A.P. to the last term of the A.P.
3) Substitute the values of a and d and simplify to find the value of n.
4) 'n' value of the last term of the A.P. gives the number of terms of the A.P.
OR
1) List the multiples of 4 between 10 and 250.
2) Note the first and the last terms of the A.P. and the common difference.
3) Equate the ${n}^{\mathrm{th}}$term of an A.P. to the last term of the A.P.
4) Substitute the values of a and d and simplify to find the value of n.
5) 'n' value of the last term of the A.P. gives the number of terms of the A.P.
Let the first term of the A.P. be a and common difference be d.
a = 8 , d = 9 (Given)
Consider the last term as n th term.
l = ${a}_{n}$= 350
⇒ a + (n âˆ’ 1) d = 350
⇒ 8 + (n âˆ’ 1) × 9 = 350
⇒ (n âˆ’ 1) × 9 = 342
⇒(n - 1) = $\frac{342}{9}$= 38
⇒n = 38 + 1 = 39
So, there are 39 terms in the given A.P.
Sum of 39 terms = ${S}_{39}$= $\frac{39}{2}$(a + ${a}_{39}$)
= $\frac{39}{2}$ x (8 + 350)
= $\frac{39}{2}$x 358
= 6981
OR
Multiples of 4 between 10 and 250 in the increasing order are 12, 16, 20 â€¦ 248.
The series forms an A.P. with first term, a = 12 and the common difference, d = 4.
If there are n terms in this A.P.
We get that ${t}_{n}$= 248
⇒ a + (n âˆ’ 1) d = 248
⇒ 12 + (n âˆ’ 1) × 4 = 248
⇒ (n âˆ’ 1) × 4 = 236
⇒(n - 1) = $\frac{236}{4}$ = 59
⇒ n = 59 + 1 = 60
So, there are 60 terms in the A.P.
Sum of all the 60 terms = $\frac{60}{2}$(12 + 248) = 30 x 260 = 7800.
31. A train travels 180 km at a uniform speed. If the speed had been 9 km/hour more, it would have taken 1 hour less for the same journey. Find the speed of the train.
OR
Find the roots of the equation.
$\frac{1}{2x\xe2\u02c6\u20193}+\frac{1}{x+5}=1,x\xe2\u2030\frac{3}{2},5$
Simplify the equation representing the difference between the timings by finding the LCM of the denominators.
OR
Solve the quadratic equation obtained by using the quadratic formula.
1) Assume the speed of the train as 'x' km/hr.
2) Find the time taken to travel 180 km.
3) Find the increased speed of the train and the time taken with the increased speed.
4) Find the difference between the timings and equate it to the given time.
5) Solve the quadratic equation obtained by splitting the middle term.
6) Positive root of the equation gives the speed of the train.
OR
1) Simplify the equation by finding the LCM of the denominators.
2) Obtain the quadratic equation by cross multiplication.
3) Find the roots of the quadratic equation by using the quadratic formula.
Assume the speed of the train as x km/h.
Time taken by the train to travel the distance of 180 km = $\frac{180}{x}$ hr
If the speed of the train had been 9 km/h more, then its speed would have been (x + 9) km/hr.
Time taken by the train to travel 180 km = $\frac{180}{x+9}$ hr
Given that, $\frac{180}{x}\xe2\u02c6\u2019\frac{180}{x+9}=1$
⇒$\frac{180(x+9)\xe2\u02c6\u2019180x}{x(x+9)}=1$
⇒$\frac{1620}{{x}^{2}+9x}=1$
⇒${x}^{2}$+ 9x = 1620
⇒${x}^{2}$+ 9x - 1620 = 0
⇒(x + 45)(x - 36) = 0
⇒x = - 45 or x = 36
As the speed of a train cannot be negative, x = 36.
Therefore, the speed of the train is 36 km/h.
OR
$\frac{1}{2x\xe2\u02c6\u20193}+\frac{1}{x+5}=1$
⇒$\frac{(x\xe2\u02c6\u20195)+(2x\xe2\u02c6\u20193)}{(2x\xe2\u02c6\u20193)(x\xe2\u02c6\u20195)}=1$
⇒3x âˆ’ 8 = (2x âˆ’3)(x âˆ’5)
⇒3x âˆ’ 8 = 2${x}^{2}$âˆ’10x âˆ’3x + 15
⇒3x âˆ’ 8 = 2${x}^{2}$âˆ’13x + 15
⇒2${x}^{2}$âˆ’16x + 23 = 0
âˆ´ x = $\frac{\xe2\u02c6\u2019(\xe2\u02c6\u201916)\xc2\pm \sqrt{{(\xe2\u02c6\u201916)}^{2}\xe2\u02c6\u20194\times 2\times 23}}{2\times 2}$
= $\frac{16\xc2\pm \sqrt{256\xe2\u02c6\u2019184}}{4}$
= $\frac{16\xc2\pm \sqrt{72}}{4}$
= $\frac{16\xc2\pm 6\sqrt{2}}{4}$
= $\frac{2(8\xc2\pm 3\sqrt{2})}{4}$
= $\frac{1}{2}$(8 + 3$\sqrt{2}$) or $\frac{1}{2}$(8 âˆ’3$\sqrt{2}$)
âˆ´x =$\frac{1}{2}$(8 + 3$\sqrt{2}$) or $\frac{1}{2}$(8 âˆ’3$\sqrt{2}$)
32. In Figure 6, three circles each of radius 3.5 cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these three circles (shaded region). [Use Ï€ = $\frac{22}{7}$]
Length of the side of the triangle formed by joining the centres of the circles is equal to twice the radius of the circles.
1) Mark the centres of the circles as A, B and C. And the points where the circles are touching each other as P, Q and R.
2) Triangle ABC is an equilateral triangle as the lengths of the sides are equal.
3) Find the area of the triangle using the formula.
4) Areas of sectors within the circles are equal.
5) Find the area of the sector taking the angle made by the arc at the centre as 60Â°.
6) Subtract the area of three sectors from the area of the triangle to find the area between the circles.
Let A, B and C be the centres of the circles.
Given that radius of each circle, r = 3.5 cm
So, each side of Î”ABC = 3.5 + 3.5 = 7 cm.
ABC is an equilateral triangle as the sides of triangle ABC are of equal length.
Therefore, âˆ A = âˆ B = âˆ C = 60Â°
Area of the shaded region = Area of Î”ABC âˆ’ (sum of areas of sectors APR, BPQ and CQR)
= $\left[\frac{\sqrt{3}}{4}{a}^{2}\right]\xe2\u02c6\u2019[3\times \frac{\mathrm{\xce\xb8}}{360\xc2\xb0}\times \mathrm{\xcf\u20ac}{r}^{2}]$
= $[\frac{1.732}{4}\times {\left(7\right)}^{2}]\xe2\u02c6\u2019[3\times \frac{60\xc2\xb0}{360\xc2\xb0}\times \frac{22}{7}\times 3.5\times 3.5]$
= 21.217 - 19.25
= 1.967
â‰ˆ1.97 ${\mathrm{cm}}^{2}$.
33. Water is flowing at the rate of 15 km/hour through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm?
1) Convert the rate of flow of water into metres/sec.
2) Convert the diameter of the pipe and rise in level of water in the pond into metres.
1) Convert the rate of flow of water into metres/sec. 2) Convert the diameter of the pipe and rise in level of water in the pond into metres.
3) Find the volume of the water flowing through the pipe in 1 sec.
4) Find the volume of the pond.
5) Divide the volume of the pond by the volume of the water flowing through the pipe in 1 second.
6) Simplify and and find the time taken to rise the level of water in the pond by 21 cm.
Speed of the water flowing through the pipe = 15 km/h = $\frac{15000m}{3600\mathrm{sec}}$= $\frac{25}{6}$m/sec
Radius of the pipe = $\frac{14}{2}\mathrm{cm}=7\mathrm{cm}=\frac{7}{100}m$
Length of the cuboidal pond is 50 m and the breadth of the cuboidal pond is 44 m.
Rise in level of water in the pond = 21 cm = $\frac{21}{100}$m
Therefore, the time (in sec) taken by the pipe to fill the pond
= $\frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{pond}}{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{water}\mathrm{flowing}\mathrm{through}\mathrm{the}\mathrm{pipe}\mathrm{in}1\mathrm{sec}}$
= $\frac{50\times 44\times \frac{21}{100}}{\frac{22}{7}\times \frac{7}{100}\times \frac{7}{100}\times \frac{25}{6}}$
= 7200 sec
= $\frac{7200}{3600}$ hrs
= 2 hours
So, the time taken by the pipe to fill the pond is 2 hours.
34. The angle of elevation of the top of a vertical tower from a point on the ground is 60Â°. From another point 10 m vertically above the first, its angle of elevation is 30Â°. Find the height of the tower.
Find the tangent of the angles of elevation of the top of the tower from two points.
1) Draw a figure to represent the data given in the question.
2) Assume the height of the tower CD as h.
3) Draw a horizontal line at B, and note the length of DE as (hâˆ’10) m.
4) Find the tangent of the angle of elevation of the top of the tower from a point on the ground.
5) Substitute the values and simplify to find the length of AC in terms of h.
6) Find the tangent of the angle of elevation of the top of the tower from a height of 10 m.\
7) Substitute the values and simplify to find the length of BE.
8) Equate the values of AC and BE, and find the height of the tower.
Draw a figure to represent the data given in the question.
CD is the tower with height as h m.
Angles of elevation of point C from points A and B are 60Â° and 30Â° respectively.
Therefore, AB = CE = 10 m.
And DE = CD âˆ’ CE = (h âˆ’ 10) m
Consider the right-angled triangle ACD:
tan 60Â° = $\frac{\mathrm{CD}}{\mathrm{AC}}$
⇒$\sqrt{3}$= $\frac{h}{\mathrm{AC}}$m
⇒AC = $\frac{h}{\sqrt{3}}$m
Consider the right-angled triangle BDE:
tan 30Â° = $\frac{\mathrm{DE}}{\mathrm{BE}}$
⇒$\frac{1}{\sqrt{3}}$= $\frac{(h\xe2\u02c6\u201910)}{\mathrm{BE}}$m
⇒BE = (h - 10) $\sqrt{3}$m
Since, AC = BE
⇒$\frac{h}{\sqrt{3}}$= (h - 10) $\sqrt{3}$
⇒h = 3(h - 10)
⇒- 2h = - 30
⇒h = 15
Therefore, the height of the tower is 15 m.
Section A | Section B | Section C | Section D |