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CBSE X
All India
MATHS PAPER 2014
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. If the height of a vertical pole is $\sqrt{3}$times the length of its shadow on the ground, then the angle of elevation of the Sun at that time is ______.
Apply appropriate trigonometric ratio to the sides of the right angled triangle.
1) Draw a figure representing the given data in the question.
2) Mark the angle of elevation of the Sun in the figure.
3) Find the tangent of the angle of elevation of the Sun.
4) Equate the value of the angle of elevation and find the value of Î¸.
Assume OA as the pole and OB is its shadow.
Let the length of the shadow be x.
Let Î¸ be the angle of elevation of the top of the pole from the ground.
Given that the height of the pole (h) = $\sqrt{3}$× length of its shadow
⇒h = $\sqrt{3}$x
In âˆ†OAB
tan Î¸ = $\frac{\mathrm{AO}}{\mathrm{OB}}$ = $\frac{h}{x}$ = $\frac{\sqrt{3}x}{x}$âˆš= $\sqrt{3}$
⇒ tan Î¸ = tan 60Â°
⇒ Î¸ = 60Â°
âˆ´ The angle of elevation of the Sun is 60Â°.
2. A bag contains cards numbered from 1 to 25. A card is drawn at random from the bag. The probability that the number on this card is divisible by both 2 and 3 is
Numbers which are divisible by 6 are possible favourable outcomes.
1) Write the number of cards as the number of possible outcomes.
2) List the possible favourable outcomes and assume it as E.
3) Find the probability of favourable outcomes which is the ratio of the number of favourable outcomes and the total number of possible outcomes.
Total number of cards = 25
Total number of possible outcomes = 25
Assume E as the event that the number on the card drawn is divisible by both 2 and 3.
âˆ´ E = {6, 12, 18, 24}
âˆ´ Number of possible outcomes favourable to E = 4
P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}E}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}$
P(E) = $\frac{4}{25}$.
3. Two different coins are tossed simultaneously. The probability of getting at least one head is ____.
Possible favourable outcomes should contain at least one head.
1) List the possible outcomes when two coins are tossed simultaneously.
2) List the favourable outcomes which contain atleast one head.
3) Find the ratio of the number of possible favourable outcomes and the total number of possible outcomes which is the required probability.
When two coins are tossed simultaneously, the possible outcomes are {(H, H), (H, T), (T, H), (T, T)}.
âˆ´ Total number of possible outcomes = 4
Assume the event of getting at least one head as 'E'.
Number of possible outcomes favourable to E is {(H, H), (H, T), (T, H)}.
âˆ´ Number of favourable outcomes = 3
P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}\mathrm{to}E}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$
P(E) = $\frac{3}{4}$.
4. Two concentric circles are of radii 5 cm and 3 cm. Length of the chord of the larger circle (in cm), which touches the smaller circle is _______.
Radius of the outer circle forms the hypotenuse of the right angled triangle formed by the tangent, radius of the inner circle and the radius of the outer circle.
1) Draw the figure representing the data given in the question.
2) Mark the radii at the point of contact of the the tangent with the inner circle.
3) Apply pythagoras theorem in the right angled triangle formed and find the length of the half of the tangent.
4) Double the length of the tangent to find the required length.
Assume 'O' as the center of the concentric circles.
Assume AB as the length of the chord of the larger circle touching the smaller circle at C.
Join O and C, O and B.
OC and OB are the radii of the smaller circle and the larger circle, respectively.
Therefore, AB is the tangent to the smaller circle.
The tangent at any point on a circle is perpendicular to the radius drawn at the point of contact.
âˆ´ OC âŠ¥ AB
In the right-angled triangle OCB
${\mathrm{OB}}^{2}$= ${\mathrm{OC}}^{2}$+ ${\mathrm{BC}}^{2}$(According to Pythagoras theorem)
⇒ ${\mathrm{BC}}^{2}$= (${5}^{2}$âˆ’${3}^{2})$
= (25 âˆ’ 9)
= 16 ${\mathrm{cm}}^{2}$
⇒ BC = 4 cm
But AB = 2(BC)
⇒ AB = 8 cm.
5. In figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S respectively. If AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm, find x.
Point of contact of the side of the quadrilateral with the circle is not the midpoint of the side.
1) Assume the sides of the quadrilateral as the tangents of the circle.
2) Find the lengths of the other parts of the tangent from the given part of the tangent.
3) Use the transitive property of equality and find the length of BP, the unknown tangent of the circle.
4) Add the values of AP and BP, and find the value of AB.
Given that
BC = 7 cm
CR = 3 cm
AS = 5 cm
AB = x cm
BQ = BP (Lengths of the tangents drawn from an external point to a circle are equal.)
AP = AS = 5 cm ------------ (1)
Similarly, CR = CQ = 3 cm
BQ = BC âˆ’ CQ and BC = 7 cm (Given)
⇒ BQ = (7 âˆ’ 3) cm = 4 cm
⇒ BQ = BP = 4 cm ------------ (2)
From (1) and (2), we get
AB = AP + PB = (5 + 4) cm = 9 cm
⇒ x = 9 cm.
6. The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is _______.
y-coordinate of a point on x-axis is zero and the x-coordinate of a point on y-axis is zero.
1) Draw the figure of the cartesian plane showing the triangle with the coordinates of the vertices of the triangle.
2) The triangle formed is a right angled triangle.
3) Find the length of the hypotenuse of the triangle using the Pythagoras theorem.
4) Find the sum of the sides of the triangle to find the perimeter of the triangle.
Plot the given coordinates i.e. O (0, 0), A (0, 4) and B (3, 0), on the Cartesian plane.
We can notice that the triangle AOB is a right-angled triangle with OB = 3 units and
OA = 4 units.
In the right angled triangle AOB,
${\mathrm{AB}}^{2}$= ${\mathrm{OA}}^{2}$+ ${\mathrm{OB}}^{2}$ (According to Pythagoras theorem)
⇒ ${\mathrm{AB}}^{2}$=${4}^{2}$ + ${3}^{2}$
= 16 + 9
= 25 sq. units
⇒ AB = 5 units
Perimeter of âˆ†AOB = OA + AB + BO
= 3 + 5 + 4 units
= 12 units.
7. A rectangular sheet of paper 40 cm × 22 cm, is rolled to form a hollow cylinder of height 40 cm. The radius of the cylinder (in cm) is _______.
Length of the cylinder is equal to the height of the cylinder. Therefore, assume that the rectangular sheet of paper is rolled about its breadth.
1) Draw the figure showing the data given in the question.
2) Find the area of the rectangular sheet of paper.
3) Assume the radius of the cylinder formed as 'r' cm.
4) Find the curved surface area of the cylinder.
5) Equate the curved surface area of the cylinder with the area of the rectangular sheet and find teh radius of the cylinder.
Area of the rectangular sheet = l x b = (40 × 22) cm2 = 880 cm2
Assume the radius of the cylinder as 'r'.
Given that height (h) of the cylinder = 40 cm
Area of the rectangular sheet is same as the curved surface area of the cylinder.
âˆ´ 2Ï€rh = Area of the rectangular sheet
⇒ 2 × $\frac{22}{7}$ × r × 40 = 880
⇒ r =$\frac{7}{2}$ = 3.5
âˆ´Radius of the cylinder is 3.5 cm.
8. The next term of the A.P. $\sqrt{7},\sqrt{28},\sqrt{63},$âˆš... is
Write the second term of the series as a mixed surd by prime factorizing the irrational number.
1) Write the second and the third terms of the series as mixed surds by prime factorizing the number.
2) Find the difference between the second the first term of the series which is the common difference of the sequence.
3) Find the next term of the sequence by adding the common difference to the third term of the sequence.
Given that $\sqrt{7}$, $\sqrt{28}$, $\sqrt{63}$.......... are in A.P.
$\sqrt{7}$, $\sqrt{28}$, $\sqrt{63}$..........
= $\sqrt{7}$, $\sqrt{(7\times 2\times 2)}$âˆš, $\sqrt{(7\times 3\times 3)}$ ............
= $\sqrt{(7\times 1\times 1)}$âˆš, $\sqrt{(7\times 2\times 2)}$, $\sqrt{(7\times 3\times 3)}$ ..............
= $\sqrt{7}$, 2$\sqrt{7}$, 3$\sqrt{7}$............
Given that the first term of the given AP is $\sqrt{7}$ and the common difference is (2$\sqrt{7}$âˆ’$\sqrt{7}$) = $\sqrt{7}$
So, a = d =âˆš$\sqrt{7}$
The next term of the sequence, i.e. the ${4}^{\mathrm{th}}$term = a + 3d
= 3$\sqrt{7}$+ $\sqrt{7}$ = 4$\sqrt{7}$= $\sqrt{7\times 4\times 4}$âˆš=âˆš$\sqrt{112}$.
9. In figure , XP and XQ are two tangents to the circle with centre O, drawn from an external point X. ARB is another tangent, touching the circle at R. Prove that XA + AR = XB + BR.
AP, AR, BQ and BR are the tangents from A and B.
1) XP and XQ are tangents from an external point X.
2) Tangents from an external point are equal.
3) A and B are external points from which AP, AR and BQ, BR are considered as tangents.
4) Use the property that the tangents from an external point are equal and prove the required proof.
From the figure, there is an external point X from where two tangents, XP and XQ, are drawn to the circle.
XP = XQ (The lengths of the tangents drawn from an external point to the circle are equal.)
Similarly,
AP = AR
BQ = BR
XP = XA + AP --------- (1)
XQ = XB + BQ --------- (2)
By substituting AP = AR in equation (1) and BQ = BR in equation (2), we get
XP = XA + AR
XQ = XB + BR
Since the tangents XP and XQ are equal, we get
XA + AR = XB + BR.
10. Prove that the tangents drawn at the ends of any diameter of a circle are parallel.
If two lines are perpendicular, angles on both sides at the point of intersection is a right angle.
1) Draw a figure showing the data given in the question.
2) A tangent at any point of a circle is perpendicular to the radius through the point of contact.
3) Angles on either side of the diameter at the points of intersection of the diameter and the tangent are right angles.
4) Alternate angles formed by the diameter as a transversal are equal.
5) Therefore, the tangents of the circle are parallel.
Assume AB as the diameter of a circle, with center O. The tangents PQ and RS are drawn at points A and B on the circle, respectively.
A tangent at any point of a circle is perpendicular to the radius through the point of contact.
âˆ´ OA âŠ¥ RS and OB âŠ¥ PQ
⇒ âˆ OAR = 90Â°
âˆ OAS = 90Â°
âˆ OBP = 90Â°
âˆ OBQ = 90Â°
âˆ OAR = âˆ OBQ and âˆ OAS = âˆ OBP
These angles are the pair of alternate interior angles.
The lines PQ and RS are parallel to each other as the alternate interior angles are equal.
Hence, proved.
11. Two different dice are rolled simultaneously. Find the probability that the sum of numbers appearing on the two dice is 10.
The outcome when two dice are rolled (4, 6) is not same as (6, 4).
1) Write the number that a dice could show when rolled.
2) List the possible outcomes when two dice are rolled.
3) Write the number of possible outcomes.
4) List the number of possible outcomes favourable to the given condition.
5) Find the ratio of the number of favourable outcomes to the total number of possible outcomes which is the required probability.
The numbers that two dice could show are 1, 2, 3, 4, 5, 6.
The possible outcomes when two dice are rolled are:
{(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),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
âˆ´ Number of possible outcomes = 6 × 6 = 36.
Assume 'E' as the event that the sum of numbers appearing on the two dice is 10.
The possible outcomes favourable to event, E are {(4, 6), (5, 5), (6, 4)}.
âˆ´ Number of favourable outcomes to event E = 3.
P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}E}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$ = $\frac{3}{36}$ = $\frac{1}{2}$.
12. In Figure , OABC is a quadrant of a circle of radius 7 cm. If OD = 4 cm, find the area of the shaded region. [Use Ï€ = $\frac{22}{7}$]
Angle at the center of the circle in the quadrant of a circle is 90Â°.
1) Find the area of the quadrant using the formula for the area of the sector.
2) The triangle which is unshaded is a right angled triangle whose base is the radius of the quadrant.
3) Find the area of the triangle.
4) Find the difference between the area of the quadrant and the area of the triangle to find the area of the shaded region.
Given that, radius of the circle = 7 cm
Area of the quadrant OABC = $\frac{1}{4}$ × Ï€${r}^{2}$
= $\frac{1}{4}$ × $\frac{22}{7}$ × 7 × 7 ${\mathrm{cm}}^{2}$= $\frac{77}{2}{\mathrm{cm}}^{2}$
Area of Î”ODC = $\frac{1}{2}$ × OD × OC
= ($\frac{1}{2}$ × 4 × 7) ${\mathrm{cm}}^{2}$ (Since OC is the radius of the circle)
= 14 ${\mathrm{cm}}^{2}$
Area of the shaded region = Area of the quadrant OABC âˆ’ Area of Î”ODC
Area of the shaded region = $\frac{77}{2}$âˆ’14 = $\frac{(77\xe2\u02c6\u201928)}{2}$= $\frac{49}{2}$ = 24.5${\mathrm{cm}}^{2}$
13. Solve for x: $\sqrt{3}{x}^{2}$âˆ’2$\sqrt{2}$xâˆ’2$\sqrt{3}$ = 0.
Use the factors of both the constant and the coefficient of ${x}^{2}.$
1) Split the middle term of the quadratic equation using the factors of the constant term and the leading term.
2) Find the factors by taking common in the first two terms and the last two terms.
3) Find the values of x by equating each of the terms to zero.
Given the quadratic equation as,
$\sqrt{3}$${x}^{2}$âˆ’2$\sqrt{2}$xâˆ’2$\sqrt{3}$ = 0
⇒ $\sqrt{3}{x}^{2}$âˆ’3$\sqrt{2}$x + $\sqrt{2}$xâˆ’2$\sqrt{3}$ = 0
⇒ $\sqrt{3}$x (xâˆ’$\sqrt{6}$) + $\sqrt{2}$(xâˆ’$\sqrt{6}$) = 0
⇒ ($\sqrt{3}$x +$\sqrt{2}$) (xâˆ’$\sqrt{6}$) = 0
⇒ x = $\xe2\u02c6\u2019\frac{\sqrt{2}}{\sqrt{3}}$ or x = $\sqrt{6}$
⇒ x = $\xe2\u02c6\u2019\sqrt{\frac{2}{3}}$ or x = $\sqrt{6}$âˆš
14. The sum of the first n terms of an AP is 5n âˆ’ ${n}^{2}$. Find the ${n}^{\mathrm{th}}$term of this A.P.
Sum of the first term of the A.P. is the first term of the A.P.
1) Write the formula to find the sum of the 'n' terms of an A.P.
2) Equate it to the given formula to find the 'n' terms of the A.P.
3) Simplify to get an equation.
4) Find the value of the first term by finding the sum of one term of AP using the given formula.
5) Substitute the value of 'a' in the equation obtained and simplify to find the formula for finding the ${n}^{\mathrm{th}}$term of the AP.
The sum of n terms of an A.P. is given by 5n -${n}^{2}$
But we know that ${S}_{n}$= $\frac{n}{2}$(a + ${t}_{n}$)
where a = first term ${T}_{n}$ = ${n}^{\mathrm{th}}$term
So, we have 5nâˆ’${n}^{2}$= $\frac{n}{2}$ ( a + ${t}_{n}$)
⇒ n(5âˆ’n) = $\frac{n}{2}$( a + ${t}_{n}$)
⇒ 10âˆ’2n = a + ${t}_{n}$ ...(1)
Now, we have sum of the first term ${S}_{1}$= a
⇒ a = (5(1)âˆ’${1}^{2}$) = 4
Substitute the value of a in equation (1), we get
10âˆ’2n = 4 + ${t}_{n}$
⇒ ${t}_{n}$= 6âˆ’2n .
Therefore, ${n}^{\mathrm{th}}$term of the A.P. is 6 âˆ’ 2n.
15. Points P, Q, R and S divide the line segment joining the points A (1, 2) and B (6, 7) in 5 equal parts. Find the coordinates of the points P, Q and R.
1) Ratio 2:3 is not equal to 3:2
2) Substitute the coordinates of points according the the section formula in order.
1) Draw a figure showing the line segment AB which is divided by four points P, Q, R and S into five equal parts.
2) Point P divides the line segment in the ratio of 1 : 4.
3) Use the section formula to find the coordinates of the point P.
4) Similarly the point Q divides the line segment in the ratio of 2:3 and the point R divides the line segment in the ratio of 3:2.
5) Use the section formula and find the coordinates of the points Q and R.
Given points P, Q, R and S divide the line segment AB formed by joining A(1, 2) and B (6, 7) into 5 equal parts.
âˆ´ AP = PQ = QR = RS = SB
First point P divides the line segment AB in the ratio of 1:4.
Coordinates of a point dividing the line segment joining the points (${x}_{1}$, ${y}_{1}$) and (${x}_{2}$, ${y}_{2}$) in the ratio of m:n are $\left(\frac{{\mathrm{mx}}_{2}+{\mathrm{nx}}_{1}}{m+n}\right)$, $\left(\frac{{\mathrm{my}}_{2}+{\mathrm{ny}}_{1}}{m+n}\right)$.
By applying section formula, the coordinates of point P:$(\frac{1\left(6\right)+4\left(1\right)}{1+4},\frac{1\left(7\right)+4\left(2\right)}{1+4})$= $(2,3)$
Similarly, the point Q divides the line segment AB in the ratio 2:3.
By applying section formula, the coordinates of point Q:
$(\frac{2\left(6\right)+3\left(1\right)}{2+3},\frac{2\left(7\right)+3\left(2\right)}{2+3})$= $(3,4)$
Similarly, the point R divides the line segment AB in the ratio 3:2.
By applying section formula, the coordinates of point R:
$(\frac{3\left(6\right)+2\left(1\right)}{3+2},\frac{3\left(7\right)+2\left(2\right)}{3+2})$= (4, 5).
16. In the figure, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region.
[Use Ï€ = 3.14]
Breadth of the rectangle is the hypotenuse of the right angled triangle and also the diameter of the semicircular region.
1) Find the hypotenuse of the right angled triangle using the Pythagoras property in the right angled triangle AED.
2) Hypotenuse of the right angled triangle is the breadth of the rectangle.
3) Find the area of the rectangle.
4) Breadth of the rectangle is the diameter of the semicircular region.
5) Find the area of the semicircular region.
6) Subtract the area of the right angled triangle from the sum of the areas of the rectangle and the area of the semicircular region to find the area of the shaded region of the figure.
AED is an right angle triangle,
${\mathrm{AD}}^{2}$= ${\mathrm{AE}}^{2}$+${\mathrm{ED}}^{2}$.
⇒ ${\mathrm{AD}}^{2}$ = (${9}^{2}$+ ${12}^{2}$)
= (81 + 144) $$
= 225 ${\mathrm{cm}}^{2}$
⇒ AD = 15 cm
Area of the rectangular region ABCD
= AB × AD
= (20 × 15)
= 300 ${\mathrm{cm}}^{2}$
Area of âˆ†AED
= $\frac{1}{2}$× AE × DE = ($\frac{1}{2}$ × 9 × 12) ${\mathrm{cm}}^{2}$ = 54 ${\mathrm{cm}}^{2}$
In a rectangle
AD = BC = 15 cm
Since, BC is the diameter of the circle, radius of the circle = 15 cm
Area of the semi-circle = $\frac{1}{2}$ × Ï€ × ${r}^{2}$
= ($\frac{1}{2}$ × 3.14 × $\frac{15}{2}$ × $\frac{15}{2}$) = 88.3125 ${\mathrm{cm}}^{2}$
Area of the shaded region = Area of the rectangle + Area of the semi-circle âˆ’ Area of the triangle
= (300 + 88.3125 âˆ’ 54) ${\mathrm{cm}}^{2}$
= 334.3125 ${\mathrm{cm}}^{2}$.
17. In the figure, ABCD is a quadrant of a circle of radius 28 cm and a semi circle BEC is drawn with BC as diameter. Find the area of the shaded region. [Use Ï€ = $\frac{22}{7}$]
Angle in the quadrant of a circle is a right angle. So, BC is the hypotenuse of the right angled triangle.
1) Find the area of the quadrant of the circle ABDC.
2) Find the area of the right angled triangle ABC.
3) Find the hypotenuse of the right angled triangle ABC using the Pythagoras property.
4) Subtract the area of the right angled triangle from the area of the quadrant of the circle to find the area of the segment BDC.
5) Diameter of the semicircle is the hypotenuse of the right angled triangle.
6) Find the area of the semicircular region.
7) Subtract the area of the segment BDC from the area of the semicircular region to find the area of the shaded region of the figure.
Given that,
Radius (r) of the circle = AB = AC = 28 cm
Area of quadrant ABDC
= $\frac{1}{4}$ × Ï€ ×${r}^{2}$ = ($\frac{1}{4}$ × $\frac{22}{7}$ × 28 × 28)
= 616${\mathrm{cm}}^{2}$
Area of âˆ†ABC
= $\frac{1}{2}$ × AC × AB = ($\frac{1}{2}$ × 28 × 28) ${\mathrm{cm}}^{2}$.
= 392 ${\mathrm{cm}}^{2}$
Area of segment BDC = Area of quadrant ABDC âˆ’ Area of âˆ†ABC
= (616 âˆ’ 392)
= 224 ${\mathrm{cm}}^{2}$ ----------- (1)
In a right-angled triangle BAC
${\mathrm{BC}}^{2}$=${\mathrm{BA}}^{2}$ + ${\mathrm{AC}}^{2}$(By Pythagoras theorem)
⇒${\mathrm{BC}}^{2}$ = (${28}^{2}$+ ${28}^{2}$)
⇒${\mathrm{BC}}^{2}$ = 28 × 28 × 2
⇒ BC = 28$\sqrt{2}$âˆšcm
In the figure BC is the diameter of the semi-circle BEC.
d = 2r
Radius = $\frac{28\sqrt{2}}{2}$cm = 14$\sqrt{2}$cm
Area of semi-circular region BEC
= $\frac{1}{2}$ × Ï€ × ${r}^{2}$= ( $\frac{1}{2}$ × $\frac{22}{7}$ × 14$\sqrt{2}$ × 14$\sqrt{2}$) $$
= ($\frac{1}{2}$ × $\frac{22}{7}$ × 14 × 14 × 2) = 616 ${\mathrm{cm}}^{2}$
Area of the shaded portion = Area of semi-circular region BEC âˆ’ Area of segment BDC
= 616$$âˆ’224 $$
= 392 ${\mathrm{cm}}^{2}$.
18. A 5 m wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used at the rate of Rs 25 per metre.
[Use Ï€ = $\frac{22}{7}$]
Rectangular cloth is not used to make the base of the cone. A conical tent has a curved surface area only.
1) Find the radius of the base of the conical tent from the diameter.
2) Find the slant height of the conical tent.
3) Assume the length of the rectangular cloth as 'l'.
4) Equate the area of the rectangular cloth to the curved surface area of the conical tent and find the length of the cloth required.
5) Find the total cost of the cloth required from the given rate.
Given that
Diameter of the conical tent = 14 m
âˆ´ Radius of the conical tent (r) = $\frac{d}{2}$ = 7 m
Height of the conical tent (h) = 24 m
Slant height of the cone (l) =âˆš$\sqrt{{r}^{2}+{h}^{2}}$
= $\sqrt{{7}^{2}+{24}^{2}}$
= $\sqrt{625}$
= 25 m
Curved surface area of the cone = Ï€rl
= ($\frac{22}{7}$ × 7 × 25) = 550 ${m}^{2}$
But Area of the cloth = Curved surface area of the cone
Length of the cloth × Width of the cloth = Curved surface area of the conical tent
⇒ Length of the cloth = $\frac{\mathrm{Curved}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{conical}\mathrm{tent}}{\mathrm{Width}\mathrm{of}\mathrm{the}\mathrm{cloth}}$
= $\frac{550}{5}$ = 110 m
Rate of 1 metre of cloth = Rs 25
Total cost of 110 m of cloth = Rs 25 × 110 = Rs 2750.
19. A girl empties a cylindrical bucket, full of sand, of base radius 18 cm and height 32 cm, on the floor to form a conical heap of sand. If the height of this conical heap is 24 cm, then find its slant height correct up to one place of decimal.
Radius of the cylindrical bucket and the radius of the conical heap are no the same.
1) Find the volume of the cylindrical bucket using the given measurements.
2) Assume the radius of the conical heap as 'R'.
3) Find volume of the conical heap.
4) Equate the volume of the conical heap to the volume of the cylindrical bucket and find the radius of the conical heap.
5) Find the slant height of the conical heap using the Pythagoras property.
Given that
Radius (r) of the cylindrical bucket = 18 cm
Height (h) of the cylindrical bucket = 32 cm
Volume of the cylindrical bucket = Ï€${r}^{2}h$
= (Ï€ × 18 × 18 × 32) ${\mathrm{cm}}^{3}$
Assume the radius of the conical heap as R.
Height (H) of the conical heap = 24 cm
Volume of the conical heap = $\frac{1}{3}$Ï€${R}^{2}H$ = ($\frac{1}{3}$Ï€${R}^{2}$ × 24) cm3 = 8Ï€${R}^{2}{\mathrm{cm}}^{3}.$
Volume of the sand in the bucket and the volume of conical heap are equal.
⇒Volume of the cylindrical bucket = Volume of the conical heap
⇒ Ï€ × 18 × 18 × 32 = 8 × Ï€ × ${R}^{2}$
⇒ ${R}^{2}$= $\frac{(18\times 18\times 32)}{8}$ = 18 × 18 × 4 = 1296.
⇒ R = 36 cm
Assume the slant height of the conical heap as l.
${l}^{2}$= ${R}^{2}$+ ${H}^{2}$
= ${36}^{2}$+ ${24}^{2}$
= 1296 + 576 = 1872
⇒ l = 43.3
âˆ´ Slant height (l) of the conical heap is 43.3 cm.
20. The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the ${28}^{\mathrm{th}}$ term of this A.P.
Use the formula to find the sum of the terms of the A.P. involving the first and the last term of the sequence.
1) Use the formula to find the ${n}^{\mathrm{th}}$term of the A.P. and find the ${7}^{\mathrm{th}}$, ${8}^{\mathrm{th}}$, ${14}^{\mathrm{th}}$terms of the A.P.
2) Find the sum of first seven terms of the A.P. using the formula to find the sum of the terms of the A.P. involving the first and the last terms.
3) Last term of the A.P. is considered as the ${7}^{\mathrm{th}}$term and mark the equation obtained as (1).
4) Find the sum of next seven terms of the A.P. using the formula to find the sum of the terms of the A.P. involving the first and the last terms.
5) First term of the A.P. is considered as the ${8}^{\mathrm{th}}$term and the last term is considered as ${14}^{\mathrm{th}}$ term and mark the equation obtained as (2).
6) Solve the equations obtained and find the values of a and d.
7) Find the value of the ${28}^{\mathrm{th}}$term using the values of a and d.
Assume the first term of the A.P. as a and the common difference as d.
${n}^{\mathrm{th}}$ term of an A.P. is ${t}_{n}$ = a + (n âˆ’ 1) d.
But ${7}^{\mathrm{th}}$term of the A.P. = a + 6d
${8}^{\mathrm{th}}$ term of the A.P. = a + 7d
${14}^{\mathrm{th}}$term of the A.P. = a + 13d
Assume the sum of the first seven terms of the A.P. as ${S}_{1\xe2\u02c6\u20197}$.
${S}_{1\xe2\u02c6\u20197}$= $\frac{n}{2}$(First term + Seventh term)
⇒ ${S}_{1\xe2\u02c6\u20197}$= $\frac{7}{2}$(a + a + 6d)
⇒ 63 = 7(a + 3d)
⇒ 9 = a + 3d ------------- (i)
Assume the sum of the next seven terms as ${S}_{8\xe2\u02c6\u201914}$
âˆ´ ${S}_{8\xe2\u02c6\u201914}$ = $\frac{n}{2}$(Eighth term + Fourteenth term)
${S}_{8\xe2\u02c6\u201914}$= $\frac{7}{2}$(a + 7d + a + 13d)
⇒161 = 7 (a + 10d)
⇒ 23 = a + 10d ------------- (ii)
Solving equation (i) and (ii), we get
7d = 14
⇒ d = 2
Substitute the value of d in equation (i), we get
a = 9 âˆ’ 6 = 3
${28}^{\mathrm{th}}$term of the A.P. = a + 27d = 3 + 27×2 = 57
âˆ´ ${28}^{\mathrm{th}}$term of the given A.P. is 57.
21. Two ships are approaching a light-house from opposite directions. The angles of depression of the two ships from the top of the light-house are 30Â° and 45Â°. If the distance between the two ships is 100 m, find the height of the light-house. [Use $\sqrt{3}$= 1.732]
Apply appropriate trigonometric ratio of the angles of depression.
1) Draw a figure representing the data given in the figure.
2) Complete two right angled triangles on either side of the light house.
3) Mark the alternate angles of the angles of depression.
4) Find the tangent of the angle of depression 30Â° and find the distance between the first ship and the light-house.
5) Find the tangent of the angle of depression 45Â° and find the height of the light-house.
Consider h be the height of the light house.
Assume the distance of one of the ships from the light house as x metres.
So, the distance of the other ship from the light house = 100 - x metres.
In the right-angled triangle AOD
tan 30Â° = $\frac{\mathrm{OD}}{\mathrm{AD}}$= $\frac{h}{x}$
⇒$\frac{1}{\sqrt{3}}$= $\frac{h}{x}$
⇒ x = $\sqrt{3}$âˆšh ------------- (1)
In a right-angled triangle BDO
tan 45Â° = $\frac{\mathrm{OD}}{\mathrm{BD}}$ = $\frac{h}{(100\xe2\u02c6\u2019x)}$
⇒1 =$\frac{h}{(100\xe2\u02c6\u2019x)}$
⇒ x + h = 100
By substituting x = $\sqrt{3}$h
$\sqrt{3}$âˆšh + h = 100
⇒ h ($\sqrt{3}$âˆš+ 1) =100
⇒ h = $\frac{100}{(\sqrt{3}+1)}$
By rationalizing the denominator of the fraction,
⇒ h =$[\frac{100}{(\sqrt{3}+1)}$×$\frac{(\sqrt{3}\xe2\u02c6\u20191)}{(\sqrt{3}\xe2\u02c6\u20191)}]$
⇒ h = [$\frac{100}{2}$] ($\sqrt{3}$âˆ’1)
⇒ h = 36.6 (approximately)
Therefore, the height of the light house is approximately 36.6 m
22. If 2 is a root of the quadratic equation 3${x}^{2}$+ px âˆ’ 8 = 0 and the quadratic equation 4${x}^{2}$âˆ’2px + k = 0 has equal roots, find the value of k.
Consider the coefficients of the terms in the quadratic equation along with their sign.
1) One of the roots of the quadratic equation with coefficient of x as 'a' is given as 2. Find the value of p as the root satisfies the equation.
2) Second equation has real and equal roots, so the discriminant of the equation is equal to zero.
3) Equate the discriminant of the second equation to zero and find the value of k in terms of p.
4) Substitute the value of p and find the value of k.
Given that 2 is a root of the quadratic equation 3${x}^{2}$+ px âˆ’ 8 = 0.
⇒ 3$({2)}^{2}$+ p(2) âˆ’ 8 = 0
⇒12 + 2p âˆ’ 8 = 0
⇒p = âˆ’2
Given that the equation 4${x}^{2}$âˆ’2px + k = 0 has equal roots.
⇒Discriminant of the quadratic equation is equal to 0
⇒${(\xe2\u02c6\u20192p)}^{2}$âˆ’4(4)(k) = 0
⇒16k = 4${p}^{2}$
⇒ k = $\frac{{p}^{2}}{4}$
Substituting the value of p, we get
k = $\frac{{(\xe2\u02c6\u20192)}^{2}}{4}$= $\frac{4}{4}$ = 1
âˆ´ The value of k is 1.
23. Construct a triangle PQR, in which PQ = 6 cm, QR = 7 cm and PR = 8 cm. Then construct another triangle whose sides are $\frac{4}{5}$ times the corresponding sides of âˆ†PQR.
Locate five points on PY such that they are equidistant.
1) Construct a triangle PQR using the measurements of three sides.
2) Draw a ray PY making an acute angle with PQ.
3) Mark 5 points on PY and join the fifth point to Q.
4) Draw a line ${P}_{4}$Q' parallel to ${P}_{5}$Q from ${P}_{4}$.
5) Draw a line Q'R' parallel to QR.
6) PQ'R' is the required triangle.
Steps of construction:
Step 1: Draw a ray PX.
Step 2: PQ = 6 cm as the radius, draw an arc from point P intersecting PX at Q.
Step 3: Q as the centre and radius equal to 7 cm, draw an arc.
Step 4: P as the centre and radius equal to 8 cm, draw another arc, cutting the previously drawn arc at R.
Step 5: Join PR and QR.
Therefore, PQR is the required triangle.
Step 6: Draw any ray PY making an acute angle with PQ.
Step 7: Now locate 5 points ${P}_{1}$, ${P}_{2}$, ${P}_{3}$, ${P}_{4}$and ${P}_{5}$ on PY so that P${P}_{1}$=${P}_{1}{P}_{2}$= ${P}_{2}{P}_{3}$= ${P}_{3}{P}_{4}$= ${P}_{4}{P}_{5}$.
Step 8: Join ${P}_{5}$Q and draw a line through ${P}_{4}$, which is parallel to ${P}_{5}$Q, to intersect PQ at Q'.
Step 9: Draw a line through Q', which is parallel to line QR, to intersect PR at R'.
Then we get, PQ'R' is the required triangle.
24. Find the value(s) of p for which the points (p + 1, 2p âˆ’ 2), (p âˆ’1, p) and (p âˆ’ 3, 2p âˆ’ 6) are collinear.
Substitute the coordinates of the points in the formula for the area of the triangle in the proper order.
1) The given points are collinear. So the area of the triangle formed by the points is zero.
2) Substitute the coordinates of the points in the formula for the area of the triangle and equate to zero.
3) Simplify the equation to find the value of the variable in the coordinates.
If the given points are collinear, the area of the triangle formed by the points must be equal to 0.
⇒ $\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})$] = 0
Comparing the given points with (${x}_{1}$,${y}_{1}$) (${x}_{2}$,${y}_{2}$) and (${x}_{3}$,${y}_{3}$),
${x}_{1}$ = p + 1, ${y}_{1}$ = 2pâˆ’2, ${x}_{2}$= pâˆ’1, ${y}_{2}$= p and ${x}_{3}$= pâˆ’3, ${y}_{3}$= 2pâˆ’6.
âˆ´ $\frac{1}{2}$[(p + 1){pâˆ’(2pâˆ’6)} + (pâˆ’1){(2pâˆ’6)âˆ’(2pâˆ’2)} + (pâˆ’3){(2pâˆ’2)âˆ’p}] = 0
⇒ (p + 1)(âˆ’p + 6) + (pâˆ’1)(âˆ’4) + (pâˆ’3)(pâˆ’2) = 0.
⇒ âˆ’4p + 16 = 0 ⇒
⇒ 4p = 16
⇒ p = 4
So that, the value of p is 4.
25. The mid-point P of the line segment joining the points A (âˆ’10, 4) and B (âˆ’2, 0) lies on the line segment joining the points C (âˆ’9, âˆ’4) and D (âˆ’4, y). Find the ratio in which P divides CD. Also find the value of y.
Assume the ratio in which P divides CD as k:1.
1) Find the coordinates of the mid-point of the line segment AB.
2) Assume the ratio in which the point P divides the line segment CD as k:1.
3) Find the coordiantes of the point P using the section formula.
4) Equate the x-coordinates of the point P and find the value of k.
5) Substitute the value of k in the equation which equates the y-coordinates and find the value of y.
Given that P is the mid-point of the line segment AB.
âˆ´ Coordinates of point P = $(\frac{\xe2\u02c6\u201910\xe2\u02c6\u20192}{2},\frac{4+0}{2})$= $\left(\xe2\u02c6\u2019\mathrm{6,2}\right)$
Then P (âˆ’6, 2) divide CD internally in the ratio k: 1.
Applying the section formula, we get
(-6 , 2) = $(\frac{k(\xe2\u02c6\u20194)+\left(1\right)(\xe2\u02c6\u20199)}{k+1},\frac{k\left(y\right)+\left(1\right)(\xe2\u02c6\u20194)}{k+1})$
(-6,2) = $(\frac{\xe2\u02c6\u20199\xe2\u02c6\u20194k}{k+1},\frac{\xe2\u02c6\u20194+\mathrm{ky}}{k+1})$
Equating on both sides
⇒ $\frac{\xe2\u02c6\u20199\xe2\u02c6\u20194k}{k+1}$= âˆ’6
⇒ âˆ’9âˆ’4k = âˆ’6kâˆ’6
⇒ k = $\frac{3}{2}$
⇒ $\frac{k}{1}$ = $\frac{3}{2}$
⇒k :1 = 3:2
Now, $\frac{\xe2\u02c6\u20194+\mathrm{ky}}{k+1}$= 2
Putting the value of k in the above equation, we get
$\frac{\xe2\u02c6\u20194+\frac{3}{2}y}{\frac{3}{2}+1}$ = 2
⇒ $\frac{\xe2\u02c6\u20198+3y}{5}$ = 2
⇒ âˆ’8+3y = 10
⇒ y = 6
âˆ´ The ratio in which P divides CD is 3:2 and the value of y is 6.
26. A quadrilateral is drawn to circumscribe a circle. Prove that the sums of opposite sides are equal.
Arrange the tangents of the circle according to the sides of the quadrilateral.
1) Draw a figure representing the data given in the question.
2) Write each side of the quadrilateral as sum of two parts.
3) Write the tangents from each vertex of the quadrilateral and they are equal.
4) Add one pair of opposite sides of the quadrilateral and equate to the sum of their parts.
5) Replace the equivalent tangents of the parts of the opposite sides considered.
6) Rearrange the parts of the sides and prove that the sums of the opposite sides of a quadrilateral are equal.
Consider ABCD as the quadrilateral circumscribing the circle.
Let E, F, G and H be the points of contact of the quadrilateral to the circle.
To Prove: AB + DC = AD + BC
Proof:
AB = AE + EB
AD = AH + HD
DC = DG + GC
BC = BF + FC
We know that AE = AH (Tangents drawn from an external point to the circle are equal.)
Similarly, BE = BF, DH = DG, CG = CF
Now, we have:
AB + DC = AE + EB + DG + GC
= AH + BF + DH + CF
= (AH + DH) + (BF + CF)
= AD + BC
⇒ AB + DC = AD + BC
âˆ´ If a quadrilateral is drawn to circumscribe a circle, the sums of opposite sides are equal.
Hence, it is proved.
27. The angle of elevation of the top of a chimney from the foot of a tower is 60Â° and the angle of depression of the foot of the chimney from the top of the tower is 30Â°. If the height of the tower is 40 m, find the height of the chimney. According to pollution control norms, the minimum height of a smoke emitting chimney should be 100 m. State if the height of the above mentioned chimney meets the pollution norms. What value is discussed in this question?
Apply appropriate trigonometric ratio of the angle of elevation and the angle of depression.
1) Draw a figure representing the data given in the question.
2) Form right angled triangles involving the angle of elevation and the angle of depression.
3) Find the tangent of the angle of depression and find the distance between the tower and the chimney.
4) Find the tangent of the angle of elevation and find the height of the chimney.
Let AB be the tower and CD be the chimney.
Given that height of the tower = AB = 40 m
In the right-angled âˆ†ABC,
tan 30Â° = $\frac{\mathrm{AB}}{\mathrm{BC}}$
⇒$\frac{40m}{\mathrm{BC}}$= $\frac{1}{\sqrt{3}}$
⇒ BC = 40$\sqrt{3}$âˆšm ..................(i)
In the right-angled âˆ†BCD,
tan 60Â° = $\frac{\mathrm{CD}}{\mathrm{BC}}$
⇒ $\frac{\mathrm{CD}}{\mathrm{BC}}$ = $\sqrt{3}$
⇒ CD = BC×$\sqrt{3}$
⇒ CD = 40$\sqrt{3}$×$\sqrt{3}$m [Using ( i)]
⇒ CD = 120 mâˆ´The height of the chimney is 120 m.
But the height of the chimney meets the pollution norms.
In this question, the conservation of environmental air pollution has been shown.
28. A hemispherical depression is cut out from one face of a cubical block of side 7 cm, such that the diameter of the hemisphere is equal to the edge of the cube. Find the surface area of the remaining solid. [Use Ï€ = $\frac{22}{7}$] .
Subtract the area of the circular portion from one of the faces of the cubical block.
1) Find the radius of the hemispherical depression from the diameter.
2) Find the total surface area of the cubical block, area of the circular region on one of the faces and the surface area of the hemispherical depression.
3) Find the surface area of the remaining solid by subtracting the area of the circular region from the sum of the surface areas of the cubical block and the surface area of the hemispherical depression.
In a cubical block
Edge of the cube, a = 7 cm
In the hemisphere
Diameter of the hemisphere = Length of a side of the cube = 7 cm
âˆ´ Radius(r) of the hemisphere =$\frac{7}{2}$ cm
Surface area of the remaining solid = Surface area of the cube âˆ’ Surface area of the circle on one of the faces of the cube + Surface area of the hemisphere
= 6${a}^{2}$âˆ’Ï€${r}^{2}$+ 2Ï€${r}^{2}$= 6${a}^{2}$+ Ï€${r}^{2}$
= (6 ×${7}^{2}$+$\frac{22}{7}$×$\frac{7}{2}$×$\frac{7}{2}$) c${m}^{2}$
= (294 + 38.5) ${\mathrm{cm}}^{2}$
= 332.5${\mathrm{cm}}^{2}$
âˆ´ Surface area of the remaining solid is 332.5 ${\mathrm{cm}}^{2}$
29. If ${S}_{n}$denotes the sum of the first n terms of an A.P., prove that${S}_{30}$ = 3 (${S}_{20}$ âˆ’ ${S}_{10}$).
Use the formula to find the sum of the terms of the A.P. involving a, d and n.
1) Use the formula to find the sum of the n terms of an A.P. and find the sum of 30 terms to obtain an equation.
2) Use the formula to find the sum of n terms of an A.P. and substitute the corresponding equations in RHS.
3) Simplify and prove that RHS and LHS are equal.
In an A.P , the sum of n$$terms is${S}_{n}$= $\frac{n}{2}$[2a+ (nâˆ’1) d].
where,
a = First term
d = Common difference
n = Number of terms in an A.P.
We have to prove that, ${S}_{30}$= 3(${S}_{20}$âˆ’${S}_{10}$).
Take R.H.S., 3(${S}_{20}$âˆ’${S}_{10}$)
= 3 [$\frac{20}{2}${2a + (20âˆ’1)d} âˆ’ $\frac{10}{2}${2a + (10âˆ’1)d} ]
= 3[10 (2a + 19d) âˆ’ 5(2a + 9d)]
= 3[20a + 190dâˆ’10aâˆ’45d]
= 3[10a + 145d]
= 15[2a + 29d]
Now take L.H.S., we get
${S}_{30}$= $\frac{30}{2}$[2a + (30âˆ’1)d]${S}_{30}$= 15[2a + 29d]
âˆ´ L.H.S. =R.H.S.Hence, proved.
30. A metallic bucket, open at the top, of height 24 cm is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are 7 cm and 14 cm respectively. Find
(i) the volume of water which can completely fill the bucket.
(ii) the area of the metal sheet used to make the bucket.
[Use Ï€ = $\frac{22}{7}$]
Surface area of the bucket includes the curved surface area and the area of the base.
1) Find the slant height of the bucket from the vertical height and radii using the Pythagoras property.
2) Find the volume of the bucket using the dimensions obtained.
3) Find the surface area of the bucket by finding the sum of the curved surface area and the area of the base of the bucket.
Upper radius of the bucket (R) = 14 cm
Lower radius of the bucket (r) = 7 cm
Height of the bucket (h) = 24 cm
Slant height of the bucket (l) = $\sqrt{{h}^{2}+{\left(R\xe2\u02c6\u2019r\right)}^{2}}$
= $\sqrt{{24}^{2}+(14\xe2\u02c6\u20197{)}^{2})}$âˆšcm
= $\sqrt{576+49}$cm
= $\sqrt{625}$cm= 25 cm
(i) We know that the volume of a frustum of a cone = V =$\frac{1}{3}$Ï€(${R}^{2}$+ Rr +${r}^{2}$)h
⇒V = $\frac{1}{3}$Ï€(${14}^{2}$+ 14 × 7 + ${7}^{2}$)}24 ${\mathrm{cm}}^{3}$
= ($\frac{1}{3}$ × $\frac{22}{7}$ × 343 × 24) ${\mathrm{cm}}^{3}$
= 8624 ${\mathrm{cm}}^{3}$
âˆ´the volume of water that can completely fill the bucket is 8624 ${\mathrm{cm}}^{3}$.
(ii) To measure the area of the metal sheet used in making the bucket, we have to find the surface area of the bucket.
Surface area of the bucket = Curved surface area of the frustum + Curved surface area of the circular base
= Ï€(R + r)l + Ï€${r}^{2}$
= [$\frac{22}{7}$ × (14 + 7) × 25 ] + [$\frac{22}{7}$ × ${7}^{2}$] ${\mathrm{cm}}^{2}$
= (1650 + 154) ${\mathrm{cm}}^{2}$
= 1804 ${\mathrm{cm}}^{2}$
âˆ´ The area of the metallic sheet used to make the bucket is 1804 ${\mathrm{cm}}^{2}$.
31. The sum of the squares of two consecutive even numbers is 340. Find the numbers.
Find two pairs of even numbers from two roots of the equation.
1) Assume the consecutive even numbers as 2n and (2n + 2).
2) Equate the sum of the assumed numbers to 340.
3) Simplify to obtain a quadratic equation.
4) Solve the quadratic equation and obtain two roots.
5) Find two pairs of even numbers from two roots of the equation.
Let the two consecutive even numbers be 2n and 2n + 2.
Given that sum of the squares of two consecutive even numbers is 340.
(2n${)}^{2}$+ (2n + 2${)}^{2}$= 340
⇒ 4${n}^{2}$ + 4${n}^{2}$+ 8n + 4 = 340
⇒8${n}^{2}$+ 8nâˆ’336 = 0⇒ 8(${n}^{2}$+ nâˆ’42) = 0
⇒${n}^{2}$+ 7nâˆ’6nâˆ’42 = 0
⇒ n(n+7)âˆ’6(n+7) = 0
⇒ (n + 7)(nâˆ’6) = 0
⇒ n = 6 or n = âˆ’7
Now, for n = 6, we get
First number = 2n = 2 × 6 = 12
Other number = 2n + 2 = 2 × 6 + 2 = 14
And, for n = âˆ’7, we get
First number = 2n = 2 × âˆ’7 = âˆ’14
Other number = 2n + 2 = 2 × âˆ’7 + 2 = âˆ’12
âˆ´ The two numbers are 12, 14 or âˆ’14, âˆ’12.
32. 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 Q on the tangent AB and draw a line segment joining the centre O and the point Q.
3) Let the line segment intersect the circle at R.
4) OQ is written as the sum of two parts OR and RQ.
5) OQ is greater than OR. So, OQ is greater than OP since OP and OR are radii of the circle.
6) OP is shorter than OQ.
7) OP is shorter than any other segment joining the centre of the circle and a point on the tangent.
8) This proves that OP, the radius at the point of contact of the tangent is perpendicular to the tangent.
In the figure AB be a tangent to the circle with centre O and touch the circle at point P.
To prove: OPâŠ¥AB
Construction: Mark a point Q on AB.
Join O and Q.
Let OQ intersect the circle at point R.
Proof:
In the above figure, we observe that any line segment joining the point O to a point on AB is larger than OP.
So, to prove that OPâŠ¥AB, it is sufficient to prove that OP is shorter than another segment joining O to any point on AB.In the given figure,
OP = OR (Radii of the same circle)
Now, OQ = OR + RQ
⇒ OQ > OR
⇒ OQ >OP (âˆµOP = OR)⇒ OP < OQ
As Q is any point, OP is shorter than any other line segment joining O to any point of AB.
âˆ´ OP âŠ¥ AB
âˆ´ The tangent at any point of a circle is perpendicular to the radius through the point of contact.
33. A dice is rolled twice. Find the probability that
(i) 5 will not come up either time.
(ii) 5 will come up exactly one time.
Do not include the outcome (5, 5) in finding the probability of 5 showing up exactly one time.
1) List the possible outcomes when two dice are rolled.
2) Find the number of possible outcomes.
3) List the outcomes in which 5 will not come up either time.
4) Find the number of favourable outcomes.
5) Find the probability by find the ratio of the favourable outcomes to the total number of possible outcomes.
6) List the outcomes where 5 comes up exactly one time.
7) Find the probability by finding the ratio of the favourable outcomes to the total number of possible outcomes.
In a die has 6 faces numbered 1,2,3,4,5 and 6.
Number of possible outcomes on rolling the dice twice are
{(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),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Total number of outcomes = 36.
(i) Favourable outcomes for the event that 5 will not show up either time are
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6)}
Total number of favourable outcomes = 25
âˆ´ Probability of not getting 5 either time = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favorable}\mathrm{out}\mathrm{comes}}{\mathrm{Total}\mathrm{Number}\mathrm{of}\mathrm{outcomes}}$ = $\frac{25}{36}$
(ii) Favourable outcomes for the event that 5 will show up exactly one time are
{(1, 5), (2, 5), (3,5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 5)}Total number of favourable outcomes = 10
âˆ´ Probability of getting 5 such that it will come up exactly one time
= $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}$ = $\frac{10}{36}$
34. Solve for x:
3$\left(\frac{3x\xe2\u02c6\u20191}{2x+3}\right)$âˆ’2 $\left(\frac{2x+3}{3x\xe2\u02c6\u20191}\right)$ = 5; xâ‰ $\frac{1}{3}$, âˆ’$\frac{3}{2}$.
Find two values for x from two values of y.
1) Assume the fraction of the first term as 'y'.
2) Fraction in the second term is the reciprocal of the first fraction.
3) Substitute in the equation and simplify by taking LCM.
4) Obtain a quadratic equation and find the values of y.
5) Substitute the values of y and simplify to find the values of x.
Given 3$\left(\frac{3x\xe2\u02c6\u20191}{2x+3}\right)$-2$\left(\frac{2x+3}{3x\xe2\u02c6\u20191}\right)$= 5
Assume $\frac{3x\xe2\u02c6\u20191}{2x+3}$= y --------(1)
âˆ´ 3y - 2$\times \frac{1}{y}$ = 5
⇒3${y}^{2}$- 2 = 5y
⇒3${y}^{2}$- 5y - 2 = 0
⇒3${y}^{2}$- 6y+ y -2 = 0
⇒3y(y -2) + 1(y - 2) = 0
⇒(y - 2)( 3y +1) = 0
⇒y = 2 or y = -$\frac{1}{3}$
Substitute the values of y in equation (1).
If y = 2 then,
$\frac{3x\xe2\u02c6\u20191}{2x+3}$ = 2
⇒3x -1 = 4x + 6
âˆ´ x = -7
If y = - $\frac{1}{3}$,
$\frac{3x\xe2\u02c6\u20191}{2x+3}$= -$\frac{1}{3}$
⇒3(3x-1) = -(2x+3)
⇒9x -3 = -2x - 3
⇒ 11x = 0
âˆ´ x = 0
âˆ´ The solutions of x are 0 and -7.
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