You can use same login credentials on LearnNext that of NextGurukul. Click on the below link to agree LearnNext terms and conditions and proceed.
CBSE X
Delhi
MATHS PAPER 2009
Time allowed: 180 minutes; Maximum Marks: 90
General Instructions: | |
1) | All questions are compulsory. |
2) | The question paper consists of thirty questions divided into 4 sections A, B, C and D. Section A comprises of ten questions of 01 mark each, Section B comprises of five questions of 02 marks each, Section C comprises ten questions of 03 marks each and Section D comprises of five questions of 06 marks each. |
3) | All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. |
4) | There is no overall choice. However, internal choice has been provided in one question of 02 marks each, three questions of 03 marks each and two questions of 06 marks each. You have to attempt only one of the alternatives in all such questions. |
5) | In question on construction, drawing should be near and exactly as per the given measurements. |
6) | Use of calculators is not permitted. |
Section A | Section B | Section C | Section D |
1. The decimal expansion of the rational number $\frac{43}{{2}^{4}{5}^{3}}$ will terminate after how many places of decimals?
1) Write the denominator of the rational number in the exponential form using prime factors with multiples of 10.
2) Verify the how many places of decimals will terminate.
1) If the denominator is of the form ${2}^{n}{5}^{m}$, where n and m are non-negative integers, then the rational number has a decimal expansion which terminates.
2) If the denominator is not of the form ${2}^{n}{5}^{m}$, where n and m are non-negative integers, then the rational number has a decimal expansion which is non-terminating repeating (recurring).
3)Write the denominator of the rational number in the exponential form using prime factors with multiples of 10.
4) Write the how many places of decimals will terminate.
5)Finally dividing the numerator by denominator to get the solution.
The expression $\frac{43}{{2}^{4}{5}^{3}}$can be written as
$\frac{43}{{2}^{4}{5}^{3}}$= $\frac{43}{\left[\right(2\times 5)\times (2\times 5)\times (2\times 5)\times 2]}$ = $\frac{43}{[1000\times 2]}$ = $\frac{0.043}{2}$= 0.0215
Hence the given expression will terminate after 4 decimal points.
2. For what value of k , âˆ’4 is a zero of the polynomial${x}^{2}$ âˆ’ x âˆ’ (2k + 2)?
1) Write the quadratic polynomial with correct signs.
2) Avoid calculation mistakes while substitute zeros of the polynomial.
1) Assume that polynomial is p(x).
2) If - 4 is a zero of the polynomial then p(- 4) = 0.
3) Substitute - 4 in quadratic polynomial to get k value.
Assume that polynomial p(x) =${x}^{2}$ âˆ’ x âˆ’ (2k + 2).
If (âˆ’4) is a zero of p(x), then p (âˆ’4) = 0.
⇒${(\xe2\u02c6\u20194)}^{2}$ âˆ’ (âˆ’4) âˆ’ 2k âˆ’ 2 = 0
⇒ 16 + 4 âˆ’ 2k âˆ’ 2 = 0
⇒ 18 âˆ’ 2k = 0
⇒2k = 18
⇒ k = 9
Hence k = 9.
3. For what value of p, 2p âˆ’ 1, 7 and 3p are three consecutive terms of an A.P.?
1) Write the correct formula to find the arithmetic mean of an A.P.
2) Substitute the given values of the terms in the formula correctly.
3) Avoid calculation mistakes.
1) Write the formula for arithmetic mean of three terms of an A.P.
2) Substitute the given values of the terms in the formula.
3) Simplify and find the value of k.
It is given that 2p âˆ’ 1, 7, and 3p are the three consecutive terms of an A.P.
The common difference is the difference between any two consecutive terms of an A.P.
The common difference is a non-changing value in an A.P.
âˆ´2nd term âˆ’ 1st term = 3rd term âˆ’ 2nd term
⇒ 7 âˆ’ (2p âˆ’ 1) = 3p âˆ’ 7
⇒ 7 âˆ’ 2p + 1 = 3p âˆ’ 7
⇒ 8 âˆ’ 2p = 3p âˆ’ 7
⇒ 15 = 5p
⇒p = $\frac{15}{5}$ =3
In that case if p = 3, then 5, 7, 9 are the three consecutive terms of an A.P.
2p âˆ’ 1, 7, and 3p are the three consecutive terms of an A.P for p = 3.
4. In the figure CP and CQ are tangents to a circle with centre O. ARB is another tangent touching the circle at R. If CP = 11 cm, and BC = 7 cm, then find the length of BR.
1) Compare the corresponding tangents to the circle.
2) Find the length of tangents using properties.
3) Choose the corresponding sides correctly to find the length of given tangent.
1) Given that CP and CQ are tangents from so that CP = CQ .
2) Using BC find the value of BQ with the help of CQ.
3)If BQ and BR are the tangents from the same point B, By using theorem BQ = BR.
4) There fore, to find length of the BR.
In the above figure given, CP and CQ are tangents from the same point C. Hence CP = CQ = 11 cm [ as CP = 11 cm]
âˆ´CQ = 11 cm
⇒ CB + BQ = 11 cm
⇒BQ + 7 cm = 11 cm [BC = 7 cm]
⇒ BQ = 11 cm âˆ’ 7 cm = 4 cm
If we take point B then the two tangents BQ and BR.
BR = BQ = 4 cm.
Hence, the length of BR is 4 cm.
5. In Fig. 2, âˆ M = âˆ N = 46Â°. Express x in terms of a, b and c where a, b and c are lengths, of LM, MN and NK respectively.
1) Comparing the similar triangles choose the correct ratio of corresponding sides.
1) Take two triangles Î”KPN and Î”KLM and compare it. 2) By AA similarity criterion of triangles Î”KPN âˆ¼ Î”KLM.
3) Therefore, ratio of corresponding sides are equal.
4) Substitute corresponding sides of values and simplify to get x value in terms of a,b,c.
Consider Î”KPN and Î”KLM, we have
âˆ KNP = âˆ KML = 46Â° [Given in the question]
âˆ NKP = âˆ MKL [Common angle of the two triangles]
Thus, Î”KPN âˆ¼ Î”KLM [by AA similarity criterion of triangles]
Therefore $\frac{\mathrm{KN}}{\mathrm{KM}}$= $\frac{\mathrm{PN}}{\mathrm{LM}}$
$\frac{c}{(c+b)}$= $\frac{x}{a}$
Therefore we get that x = $\frac{\mathrm{ac}}{(b+c)}$.
6. If sin Î¸ =$\frac{1}{3}$ , then find the value of (2${\mathrm{cot}}^{2}$Î¸ + 2).
1) Write the correct suitable identity.
2) Avoid calculation mistakes.
1) Given that sin Î¸.
2) Using identity find out cos Î¸ .
3)After that find out cot Î¸ using Trigonometric ratios.
4) Substitute cot Î¸ in (2${\mathrm{cot}}^{2}$Î¸ + 2).
Let us first find the value of cos â¡Î¸
cos Î¸ = âˆš(1-${\mathrm{sin}}^{2}$â¡Î¸ )= âˆš ((1- ($\frac{1}{3}{)}^{2}$) = âˆš (1 - $\frac{1}{9}$) = âˆš($\frac{8}{9}$) = $\frac{2\sqrt{2}}{3}$
âˆ´cot â¡Î¸ = $\frac{\mathrm{cos}\mathrm{\xce\xb8}}{\mathrm{sin}\mathrm{\xce\xb8}}$ = $\frac{\frac{2\sqrt{2}}{3}}{\frac{1}{3}}$= 2âˆš 2
2 ${\mathrm{cot}}^{2}\mathrm{\xce\xb8}$ + 2 = 2(${2\sqrt{2})}^{2}$+ 2 = 2×8 + 2 = 18.
7. Find the value of a so that the point (3, a) lies on the line represented by 2x âˆ’ 3y = 5.
1) Avoid calculation mistakes.
1) First substitute (3,a) in 2x âˆ’ 3y = 5
2) Simplify and to get the value of a.
It is given that the point (3, a) lies on the line represented by 2x âˆ’ 3y = 5. Therefore the point(3, a) satisfies the equation 2x âˆ’ 3y = 5.
âˆ´2 × 3 âˆ’ 3 × a = 5
⇒ 6 âˆ’ 3a = 5
⇒ 6 âˆ’ 5 = 3a
⇒ 1 = 3a
Therefore, a = 1/3.
8. A cylinder and a cone are of same base radius and of same height. Find the ratio of the volume of cylinder to that of the cone.
1) Write the correct formulas.
2) Check the given conditions.
1) Given that cylinder and a cone are of same base radius and of same height.
2)Write the Volume of cylinder and cone formulas.
3) Now find the ratio of Volume of cylinder and cone to get the solution.
Given that the cylinder and the cone are of same base radius and same height.
Hence,
Radius of cylinder = Radius of cone = r
Height of cylinder = Height of cone = h
âˆ´Volume of cylinder = Ï€${r}^{2}$ h
Volume of cone = $\frac{1}{3}$Ï€${r}^{2}$ h
Therefore,
$\frac{\mathrm{Volume}\mathrm{of}\mathrm{cylinder}}{\mathrm{Volume}\mathrm{of}\mathrm{cone}}$= ($\frac{\mathrm{\xcf\u20ac}{r}^{2}h}{\frac{1}{3}\mathrm{\xcf\u20ac}{r}^{2}h}$) = 3
So that ,the ratio between the volume of the cylinder to that of the cone is 3 : 1.
9. Find the distance between the points (-$\frac{8}{5}$, 2) and ($\frac{2}{5}$, 2).
1) Write the correct formula.
2) Avoid calculation mistakes.
1) Write the distance between two points formula.
2) Substitute given points in that formula to get the solution.
The distance between two points (${x}_{1}$, ${y}_{1}$) and (${x}_{2}$, ${y}_{2}$) is given by the equation âˆš((${X}_{2}$- ${X}_{1}{)}^{2}$+ (${Y}_{2}$- ${Y}_{1}$${)}^{2}$)
Substituting the values given into the equation we get AB = 2 units.
10. Write the median class of the following distribution :
1) Avoid calculation mistakes while writing the cumulative frequency.
2) Select correct median class of the given distribution.
1) Take another column and write the cumulative frequency.
2) Find total frequency n and find out $\frac{n}{2}$ value.
3) Now the median class of a set of data is that class whose cumulative frequency is greater than and also nearest to the value of $\frac{n}{2}$.
4) Hence the median class of the given distribution is 30âˆ’40.
Let us make a tabular column to represent the data that is given in the question.
Median class of a set of data is that class whose cumulative frequency is greater than and also nearest to the value of $\frac{n}{2}$.
From the table above n = 50. Therefore $\frac{n}{2}$= $\frac{50}{2}$= 25
The corresponding class is 30-40 from the table and hence the median class of the given distribution is 30âˆ’40.
11. If the polynomial 6${x}^{4}$+ 8${x}^{3}$+ 17${x}^{2}$+ 21x + 7 is divided by another polynomial 3${x}^{2}$ + 4x + 1 , the remainder comes out to be (ax+b), find a and b.
1) Find the reaminder by the long division method of polynomials.
2) Avoid calculation mistakes.
3) Equate both the terms of the remainder to ax + b.
1) Divide the given polynomial by 3${x}^{2}$-1 using long division method.
2) Equate the remainder to ax + b .
3)find the values of 'a' and 'b'.
Given polynomial 6${x}^{4}$+ 8${x}^{3}$+ 17${x}^{2}$+ 21x + 7 can be divided by 3${x}^{2}$ + 4x +1 as
The remainder obtained is x + 2.
Since it is given that the remainder is of the form ax + b.
Now, equating both remainders to get a = 1 and b = 2.
12. Find the value(s) of k for which the pair of linear equations kx + 3y = k âˆ’ 2 and 12x + ky = k has no solution.
1) Write the pair of linear equations are in correct order by comparing the standard form.
2) Find the ratio of corresponding coefficients of the equations.
3) Relate the exact condition for finding the relation between the pair of lines.
1) Write the coefficients of the equations by comparing the standard form of the linear equations.
2) Find the ratios of the corresponding coefficients if the equations have no solution.
3) By comparing the ratios of coefficients of the linear equations to find the value of k.
Write the given equations in the form of ${a}_{1}$x + ${b}_{1}$y + ${c}_{1}$ = 0 and ${a}_{2}$x + ${b}_{2}$y + ${c}_{2}$ = 0
These equations have no solution if $\frac{{a}_{1}}{{a}_{2}}$ = $\frac{{b}_{1}}{{b}_{2}}$ â‰ $\frac{{c}_{1}}{{c}_{2}}$.
Let us analyze the given equations and for that first rewrite it as,
kx + 3y âˆ’ (k âˆ’ 2) = 0, and 12x + ky âˆ’ k = 0
On comparing the given equations, we get
${a}_{1}$ = k, ${a}_{2}$ = 12, ${b}_{1}$ = 3, ${b}_{2}$= k, ${c}_{1}$ = âˆ’(k âˆ’ 2), ${c}_{2}$ = âˆ’k
Therefore $\frac{k}{12}$= $\frac{3}{k}$â‰ $\frac{\xe2\u02c6\u2019(k\xe2\u02c6\u20192)}{\xe2\u02c6\u2019k}$
On comparing first part of the equation $\frac{k}{12}$= $\frac{3}{k}$, we get ${k}^{2}$ = 36, which gives k = Â± 6.
So that , When k = 6 and âˆ’6, the given pair of equations will have no solutions.
13. If ${S}_{n}$is the sum of first n terms of an A.P. is given by ${S}_{n}$= 3${n}^{2}$- 4n, then find its ${n}^{\mathrm{th}}$ term.
1) Write the correct formula to find its ${n}^{\mathrm{th}}$ term, If the sum of first n terms of an A.P. is given.
1) Write the correct formula of ${n}^{\mathrm{th}}$ term, If the sum of first n terms of an A.P is given.
2) Substitute values and simply will get n th term of the given A.P.
Let the ${n}^{\mathrm{th}}$ term of the A.P. be ${a}_{n}$.
Since it is given that ${S}_{n}$ = 3${n}^{2}$âˆ’4n.
But ${S}_{n}$-${S}_{n\xe2\u02c6\u20191}$= ${a}_{n}$.
Therefore
⇒3${n}^{2}$- 4n - { 3(${n\xe2\u02c6\u20192)}^{2}$- 4( n - 1)} = ${a}_{n}$
⇒3${n}^{2}$- 4n - { 3(${n}^{2}$+ 1 - 2n) - 4(n- 1)} = ${a}_{n}$${a}_{n}$
⇒3${n}^{2}$- 4n - { 3(n2+ 1 - 2n) - 4(n- 1)} = ${a}_{n}$
⇒3${n}^{2}$- 4n - 3${n}^{2}$-7 + 10n = ${a}_{n}$
⇒ 6n - 7 = ${a}_{n}$
Therefore , the n th term of the given A.P. is (6n âˆ’ 7).
14. Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that âˆ APB = 2 âˆ OAB.
OR
Prove that the parallelogram circumscribing a circle is a rhombus.
1) Draw a line OB.
2) Observe OAPB is a quadrilateral.
or
1) Lengths of tangents drawn from an external point to the circle are equal.
2) Group the tangents according to the sides of the parallelogram.
1) Given that two tangents PA and PB are drawn to a circle with centre O and P is an external point.
2) join O and B i.e. OB.
3) The tangent at any point of a circle is perpendicular to the radius through the point of contact.
4) Observe OAPB is a quadrilateral. Therefore sum of the angles in a quadrilateral is 360 degrees.
5) Find âˆ AOB .
6) Observe OAB is a triangle and OA and OB are radii then âˆ OAB = âˆ OBA.
7) The sum of angles in a triangle is equal to 180 Â°.
8) Find âˆ APB = 2âˆ OAB by using âˆ AOB.
OR
1) In a parallelogram ABCD , the circle touch the sides AB, BC, CD and AD at P, Q, R and S respectively.
2) Since, ABCD is a parallelogram then AB = CD and BC = AD.
3) Lengths of tangents drawn from an external point to the circle are equal.
4) Adding all these and grouping these tangent lengths according to the sides of the parallelogram.
5) Therefore, ABCD is a parallelogram that has sides of equal length.
6) Hence, the parallelogram ABCD is also a rhombus.
To complete the quadrilateral by joining OB.
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
âˆ´âˆ OAP = âˆ OBP = 90Âº
Now,
âˆ OAP + âˆ APB + âˆ OBP + âˆ AOB = 360Âº [Angle sum property of quadrilaterals]
⇒ 90Âº + âˆ APB + 90Âº + âˆ AOB = 360Âº
⇒âˆ AOB = 360Âº âˆ’ 180Âº âˆ’ âˆ APB = 180Âº âˆ’ âˆ APB ---------------------- (1)
Now, in Î”OAB,
OA is equal to OB as both are radii.
⇒âˆ OAB = âˆ OBA [In an isosceles triangle, angles opposite to equal sides are equal]
The sum of angles in a triangle is equal to 180 Â°.
In Î”AOB,
âˆ OAB +âˆ OBA + âˆ AOB = 180Âº
⇒ 2âˆ OAB + âˆ AOB = 180Âº
⇒ 2âˆ OAB + (180Âº âˆ’ âˆ APB) = 180Âº [From (1)]
⇒ 2âˆ OAB = âˆ APB
⇒âˆ APB = 2âˆ OAB
Hence proved.
OR
Let the parallelogram be ABCD and let the circle touch the sides AB, BC, CD and AD at P, Q, R and S respectively.
Since, ABCD is a parallelogram.
AB = CD â€¦ (1)
BC = AD â€¦ (2)
Lengths of tangents drawn from an external point to the circle are equal. If A, B, C, D are the external points then,
DR = DS ----- (3)
CR = CQ ----- (4)
BP = BQ â€¦ (5)
AP = AS â€¦ (6)
On adding equations (3), (4), (5) and (6), we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS
Grouping these tangent lengths according to the sides of the parallelogram.
⇒ (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
âˆ´ CD + AB = AD + BC
From equations (1) and (2), we obtain
2AB = 2BC
âˆ´ AB = BC
âˆ´ AB = BC = CD = DA
Therefore, ABCD is a parallelogram that has sides of length are equal.
Hence, the parallelogram ABCD is also a rhombus.
15. Simplify: $\frac{{\mathrm{sin}}^{3}\mathrm{\xce\xb8}+{\mathrm{cos}}^{3}\mathrm{\xce\xb8}}{\mathrm{sin}\mathrm{\xce\xb8}+\mathrm{cos}\mathrm{\xce\xb8}}$ + sinÎ¸ cosÎ¸
1) Write the correct formula and apply the identity.
1) Apply the ${a}^{3}$+ ${b}^{3}$ formula.
2) Cancel the similar terms and simplify using identity to get required solution.
Given $\frac{{\mathrm{sin}}^{3}\mathrm{\xce\xb8}+{\mathrm{cos}}^{3}\mathrm{\xce\xb8}}{\mathrm{sin}\mathrm{\xce\xb8}+\mathrm{cos}\mathrm{\xce\xb8}}$+ sinÎ¸ cosÎ¸
Since${a}^{3}+{b}^{3}$ = ( a + b)(${a}^{2}+{b}^{2}$ - 2ab)
= $\frac{(\mathrm{sin}\mathrm{\xce\xb8}+\mathrm{cos}\mathrm{\xce\xb8})({\mathrm{sin}}^{2}\mathrm{\xce\xb8}+{\mathrm{cos}}^{2}\mathrm{\xce\xb8}\xe2\u02c6\u2019\mathrm{sin}\mathrm{\xce\xb8}\mathrm{cos}\mathrm{\xce\xb8})}{(\mathrm{sin}\mathrm{\xce\xb8}+\mathrm{cos}\mathrm{\xce\xb8})}$ + sin Î¸ cos Î¸
= ${(\mathrm{sin}}^{2}$ Î¸ +${\mathrm{cos}}^{2}$ Î¸ - sinÎ¸ cos Î¸) + sin Î¸ cos Î¸
= 1 - sinÎ¸ cosÎ¸ + sinÎ¸ cos Î¸ = 1
Hence, the value of the given expression is 1.
16. Prove that âˆš5 is an irrational number.
1) Substituting appropriate values properly.
1)Assume that âˆš5 is rational number.
2) To prove that we assume there are positive co-primes a and b such that âˆš5 = $\frac{a}{b}$.
3) This shows that 5 is a factor of ${a}^{2}$, which comes down to the fact that 5 is factor of a.
4) Let a = 5c. Similarly apply we get ${b}^{2}$ is a multiple of 5, or b is a multiple of 5.
5) Since both a and b are multiples of 5 or 5 is a factor of both a and b.
6) This is contradiction shows that our assumption incorrect .7) Therefore, âˆš5 is an irrational number.
For proving that âˆš5 is an irrational number try to prove the contrary.
Let us assume on the contrary that âˆš5 is rational. To prove that we assume there are positive co-primes a and b such that
âˆš5 = $\frac{a}{b}$
5 = $\frac{{a}^{2}}{{b}^{2}}$
$\Rightarrow {b}^{2}$x 5 = ${a}^{2}$
This shows that 5 is a factor of ${a}^{2}$, which comes down to the fact that 5 is factor of a. So that a = 5c for a positive integer c.
${\Rightarrow b}^{2}$x 5 = ${a}^{2}$
⇒${b}^{2}$x 5 =${5}^{2}$ x ${c}^{2}$
$\Rightarrow {b}^{2}$= 5 x ${c}^{2}$
So that ${b}^{2}$ is a multiple of 5, or b is a multiple of 5.
Since both a and b are multiples of 5 or 5 is a factor of both a and b.
This contradicts shows that a and b as co-primes, so our assumption incorrect .
Hence, âˆš5 is an irrational number.
17. Solve the following pair of equations:
$\frac{5}{(x\xe2\u02c6\u20191)}+\frac{1}{(y\xe2\u02c6\u20192)}$= 2
$\frac{6}{(x\xe2\u02c6\u20191)}\xe2\u02c6\u2019\frac{3}{(y\xe2\u02c6\u20192)}=1$
1) Assuming the given equations$\frac{1}{(x\xe2\u02c6\u20191)}=u$ and $\frac{1}{(y\xe2\u02c6\u20192)}$= v .
2) Avoid calculation mistakes.
3) Substitute the u and v values to get x and y values.
1) Let the equations be (1) and (2) .
2) Assuming the given equations$\frac{1}{(x\xe2\u02c6\u20191)}=u$ and $\frac{1}{(y\xe2\u02c6\u20192)}$= v .
3) We get (3) and (4) equations.
4) Solving (3) and (4) equations to get u and v values.
5) Substitute u and v values to get x and y values.
Let
$\frac{5}{(x\xe2\u02c6\u20191)}+\frac{1}{(y\xe2\u02c6\u20192)}=2$----------------(1)
$\frac{6}{(x\xe2\u02c6\u20191)}$- $\frac{3}{(y\xe2\u02c6\u20192)}$ = 1 ---------------- (2)
Assume that $\frac{1}{(x\xe2\u02c6\u20191)}=u$ and $\frac{1}{(y\xe2\u02c6\u20192)}$= v
So now the equations will become
5u + v = 2 â€¦......... (3)
6u âˆ’ 3v = 1 â€¦......... (4)
Solving equations (3) and (4) as linear equations we get that
u = $\frac{1}{3}$ and v = $\frac{1}{3}$
That means
$\frac{1}{(x\xe2\u02c6\u20191)}$= $\frac{1}{3}$ and $\frac{1}{(y\xe2\u02c6\u20192)}$= $\frac{1}{3}$
From which we get that x = 4 and y = 5 which is the solution for the given pair of equation.
18. The sum of 4th and 8th terms of an A.P. is 24 and sum of 6th and 10th terms is 44. Find the A.P.
1) Write the correct formula of ${n}^{\mathrm{th}}$ term of an A.P.
2) By solving equations avoid calculation mistakes.
3) Write the correct A.P. series by using difference value.
1) Let us assume that a is the first term and d is the common difference of the A.P
2) Write the ${n}^{\mathrm{th}}$term of an A.P.
3) Find ${4}^{\mathrm{th}}$, ${6}^{\mathrm{th}}$, ${8}^{\mathrm{th}}$and ${10}^{\mathrm{th}}$ terms.
4) Adding 4th and 8th term is 24 and 6th term and 10th term is 44, to get two equations .
5) Solving these equations to get a and d values.
6) Finally write the correct A.P. series by using difference value.
Let us assume that a is the first term of the A.P. and d is the common difference of the A.P.
It is known that the ${n}^{\mathrm{th}}$term of an A.P. is given by ${a}_{n}$ = a + (n -1)d.
The fourth term of the A.P. ${a}_{4}$ = a + (4 -1)d = a + 3d.
8th term, ${a}_{8}$ = a + (8 - 1)d = a + 7d
6th term, ${a}_{6}$ = a + (6 -1)d = a + 5d
10th term, ${a}_{10}$ = a + (10 - 1)d = a + 9d
It is given that ${a}_{4}$ + ${a}_{8}$= 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d= 12 ------------ (1)
And also
${a}_{6}$ + ${a}_{10}$ = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 ----------- (2)
On solving linear equations (1) and (2) we get,
d = 5 and a = -13
The A.P. is a, a + d, a + 2d â€¦
⇒ âˆ’13, âˆ’13 + 5, âˆ’13 + 10â€¦
Thus, the A.P. is âˆ’13, âˆ’8, âˆ’3â€¦
19. Construct a Î”ABC in which BC = 6.5 cm, AB = 4.5 cm and âˆ ABC = 60Â°. Construct a triangle similar to this triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the triangle ABC.
1) Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
2) Locate 4 points${B}_{1}$, ${B}_{2}$, ${B}_{3}$, and ${B}_{4}$ on line segment BX.
3) Join B, C and draw a line through ${B}_{3}$ such that it is parallel to ${B}_{4}$C and intersecting BC at C '.
4) Draw a line through C ' parallel to AC and intersecting AB at A'.
1) Draw a line segment BC of length 6.5 cm.
2) Draw an arc of any radius while taking B as the centre. Taking O as the centre, draw another arc to cut the previous arc at point O'.
3) Taking B as the centre, draw an arc of radius 4.5 cm intersecting the extended at point A. Join AC.
4) Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
5) Locate 4 points on line segment BX.
6) Join two lines which is parallel to AC and ${B}_{4}$C then to get a required similar triangle.
7) Hence the required construction will satisfy the condition.
In a triangle whose sides are $\frac{3}{4}$of the corresponding sides of Î”ABC and it can be drawn by following the given steps.
Step 1: Draw a line segment BC of length 6.5 cm.
Step 2: Draw an arc of any radius while taking B as the centre. Let it intersect line BC at point O. Taking O as the centre, draw another arc to cut the previous arc at point O'. Join BO', which is the ray making 60Â° with line BC.
Step 3: Taking B as the centre, draw an arc of radius 4.5 cm intersecting the extended line segment BO' at point A. Join AC. In Î”ABC, AB = 4.5 cm, BC = 6.5 cm, and âˆ ABC = 60Â°.
Step 4: Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
Step 5: Locate 4 points (as 4 is greater in 3 and 4) ${B}_{1}$, ${B}_{2}$, ${B}_{3}$, and ${B}_{4}$ on line segment BX.
Step 6: Join ${B}_{4}$C and draw a line through ${B}_{3}$ such that it is parallel to ${B}_{4}$C and intersecting BC at C '.
Step 7: Draw a line through C ' parallel to AC and intersecting AB at A'. Î”A'BC' is the required triangle.
This is possible because
A' B = $\frac{3}{4}$AB, BC ' = $\frac{3}{4}$BC, A' C ' = $\frac{3}{4}$AC
In Î”Aâ€²BCâ€² and Î”ABC,
âˆ Aâ€²Câ€²B = âˆ ACB ( they are corresponding angles)
âˆ Aâ€²BCâ€² = âˆ ABC ( common angle)
âˆ´ Î”Aâ€²BCâ€² âˆ¼ Î”ABC (by AA similarity criterion)
Therefore $\frac{(A\text{'}B)}{\mathrm{AB}}$= $\frac{\mathrm{BC}\text{'}}{\mathrm{BC}}$= $\frac{(A\text{'}C\text{'})}{\mathrm{AC}}$ ----------- (1)
In Î”B${B}_{3}$Câ€² and Î”B${B}_{4}$C,
âˆ ${B}_{3}$BCâ€² = âˆ ${B}_{4}$BC (Common angle)
âˆ B${B}_{3}$Câ€² = âˆ B${B}_{4}$C (Alternate interior angles)
âˆ´ Î”B${B}_{3}$Câ€² âˆ¼ Î”B${B}_{4}$C (byAA similarity criterion)
$\frac{\mathrm{BC}\text{'}}{\mathrm{BC}}$= $\frac{B{B}_{3}}{B{B}_{4}}$
$\frac{\mathrm{BC}\text{'}}{\mathrm{BC}}$ = $\frac{3}{4}$ ----------(2)
Therefore from equations (1) and (2), we obtain
$\frac{A\text{'}B}{\mathrm{AB}}$ = $\frac{\mathrm{BC}\text{'}}{\mathrm{BC}}$ = $\frac{A\text{'}C\text{'}}{\mathrm{AC}}$= $\frac{3}{4}$
Therefore A'B = $\frac{3}{4}$AB, BC ' = $\frac{3}{4}$ BC, A' C ' =$\frac{3}{4}$AC
Hence the required construction will satisfy the condition.
20. In Fig. 4, Î”ABC is right angled at C and DE âŠ¥ AB. Prove that Î”ABC âˆ¼ Î”ADE and hence find the lengths of AE and DE.
OR
In Fig, 5, DEFG is a square and âˆ BAC = 90Â°. Show that ${\mathrm{DE}}^{2}$ = BD × EC.
1) Comparing the similarity of traingles
2) Check the ratio corresponding sides of similar triangles are proportional.
3) Pythagoras theorem applied to Î”ABC.
or
1) Let âˆ DBG = α. In each triangle find the remaining angles using sum of angles in a triangle = 180Â°.
2) Comparing the AA similarity criterion.
3) Check the ratio corresponding sides of similar triangles are proportional.
1) Given that Î”ABC is right angled at C and DE âŠ¥ AB.
2) Comparing the similarity of traingles.
3) Since thecorresponding sides of similar triangles are proportional,4) Substitute thevalues and using Pythagoras theorem to find the side AB.
5) Using side AB to find DE and AE.
or
1) In a Î”ABC is right angled at A and DEFG is a square.
2) Find angles âˆ BDG and âˆ CEF .
3) Let âˆ DBG = α then find âˆ DGB .
4) âˆ´In a Î”ABC to find âˆ ECF and In a Î”FEC to find âˆ CFE.
â‡’By AA similarity criterion, Î”BDG âˆ¼ Î”FEC.
6) Corresponding sides of similar triangles are proportional then we get ${\mathrm{DE}}^{2}$= BD Ã— EC.
Consider Î”ABC and Î”ADE
âˆ ACB = âˆ AED (both 90Â°)
âˆ BAC = âˆ DAE (as common angle)
âˆ´ Î”ABC âˆ¼ Î”ADE (By AA similarity criterion)
Since the corresponding sides of similar triangles are proportional,
âˆ´$\frac{\mathrm{AB}}{\mathrm{AD}}$ =$\frac{\mathrm{BC}}{\mathrm{DE}}$ = $\frac{\mathrm{AC}}{\mathrm{AE}}$
âˆ´ $\frac{\mathrm{AB}}{\mathrm{AD}}$ = $\frac{\mathrm{BC}}{\mathrm{DE}}$= $\frac{(\mathrm{AD}+\mathrm{DC})}{\mathrm{AE}}$
â‡’$\frac{\mathrm{AB}}{3}$ = $\frac{12}{\mathrm{DE}}$= $\frac{(3+2)}{\mathrm{AE}}$
â‡’$\frac{\mathrm{AB}}{3}$ = $\frac{12}{\mathrm{DE}}$= $\frac{5}{\mathrm{AE}}$ ------------(1)
Pythagoras theorem applied to Î”ABC, we get
${\mathrm{AB}}^{2}$ = ${\mathrm{AC}}^{2}$ + ${\mathrm{BC}}^{2}$
â‡’ ${\mathrm{AB}}^{2}$ = 52 + 122 = 25 + 144 = 169
â‡’ AB = 13 cm
Substituting the value of AB in equation (1), we obtain
$\frac{13}{3}$= $\frac{12}{\mathrm{DE}}$ = $\frac{5}{\mathrm{AE}}$
$\frac{13}{3}$ = $\frac{12}{\mathrm{DE}}$
DE = 2.77 cm
Similarly AE = 1.15 cm
Hence DE and AE are approximately 2.77cm and 1.55cm.
OR
In a figure given that DEFG is a square.
âˆ´âˆ EDG = âˆ FED = 90Â°
Now, âˆ BDG + âˆ EDG = 180Â° (straight line)
â‡’âˆ BDG + 90Â° = 180Â°
â‡’âˆ BDG = 90Â°
Similarly, âˆ CEF = 90Â°
Let âˆ DBG = α
Sum of angles in a triangle = 180Â°
In a Î”BDG:
âˆ DGB + âˆ DBG + âˆ BDG = 180Â°
â‡’âˆ DGB = 90Â° âˆ’ α .
âˆ´ a Î”ABC:
âˆ ECF + âˆ ABC + âˆ BAC = 180Â° [âˆ ECF = âˆ BCA]
â‡’âˆ ECF = 90Â° âˆ’ α.
In a Î”FEC:
âˆ CEF + âˆ ECF + âˆ CFE = 180Â°
â‡’âˆ CFE = α
Now, in Î”BDG and Î”FEC:
âˆ BDG = âˆ FEC ( = 90Â°)
âˆ DBG = âˆ EFC ( = α)
By AA similarity criterion, Î”BDG âˆ¼ Î”FEC
As corresponding sides of similar triangles are proportional.
$\frac{\mathrm{BD}}{\mathrm{FE}}$ = $\frac{\mathrm{DG}}{\mathrm{EC}}$
$\frac{\mathrm{BD}}{\mathrm{DE}}$ = $\frac{\mathrm{DE}}{\mathrm{EC}}$ [In square DEFG, DE = DG = EF = GF]
${\mathrm{DE}}^{2}$ = BD Ã— EC
Hence proved.
21. Find the value of sin 30Â° geometrically.
OR
Without using trigonometrical tables, evaluate:
$\frac{\mathrm{cos}58\xc2\xb0}{\mathrm{sin}32\xc2\xb0}$+ $\frac{\mathrm{sin}22\xc2\xb0}{\mathrm{cos}68\xc2\xb0}$ - $\frac{\mathrm{cos}38\xc2\xb0\mathrm{cos}\mathrm{ec}52\xc2\xb0}{\mathrm{tan}18\xc2\xb0\mathrm{tan}35\xc2\xb0\mathrm{tan}60\xc2\xb0\mathrm{an}72\xc2\xb0\mathrm{tan}55\xc2\xb0}$
1) Draw a line AD which is perpendicular to side BC
2) Comparing two triangles.
3) Let the length of AB be 2a.
or
1) Using complimentary angles to find trigonometric ratios.
1) Consider an equilateral triangle ABC, each angle in an equilateral triangle is 60Â°.
2) Draw a line AD which is perpendicular to side BC.
3) Now Comparing two triangles.
4) By RHS congruency criterion to get Î”ABD â‰… Î”ACD.
5)âˆ´BD = DC ( by CPCT)âˆ BAD = âˆ CAD = 30Âº.
6) Let the length of AB be 2a. therefore AB = BC = CD = 2a. There fore, BD = half of BC = a.
7) To find Sin 30Â° using right angled triangle.
or
1) Using complimentary angles find the trigonometric ratios.
2) Substitute and simplify to get solution.
Consider an equilateral triangle ABC, each angle in an equilateral triangle is 60Â°.
Therefore, âˆ A = âˆ B = âˆ C = 60Â°
Draw a line AD which is perpendicular to side BC.
Now, in Î”ABD and Î”ACD
âˆ ADB = âˆ ADC (both are 90Â°)
AB = AC (ABC is an equilateral triangle)
AD = AD (Common line)
Î”ABD â‰… Î”ACD (By RHS congruency criterion)
âˆ´ BD = DC ( by CPCT)
âˆ BAD = âˆ CAD = 30Âº
Let the length of AB be 2a. therefore AB = BC = CD = 2a
BD = half of BC = a.
Therefore, Sin 30Â° =$\frac{\mathrm{BD}}{\mathrm{AB}}$ = $\frac{a}{2a}$= $\frac{1}{2}$.
OR
$\frac{\mathrm{cos}58\xc2\xb0}{\mathrm{sin}32\xc2\xb0}$+ $\frac{\mathrm{sin}22\xc2\xb0}{\mathrm{cos}68\xc2\xb0}$ - $\frac{\mathrm{cos}38\xc2\xb0\mathrm{cos}\mathrm{ec}52\xc2\xb0}{\mathrm{tan}18\xc2\xb0\mathrm{tan}35\xc2\xb0\mathrm{tan}60\xc2\xb0\mathrm{tan}72\xc2\xb0\mathrm{tan}55\xc2\xb0}$
$\frac{\mathrm{cos}(90\xc2\xb0\xe2\u02c6\u201932\xc2\xb0)}{\mathrm{sin}32\xc2\xb0}$+ $\frac{\mathrm{sin}(90\xc2\xb0\xe2\u02c6\u201968\xc2\xb0)}{\mathrm{cos}68\xc2\xb0}$- $\frac{\mathrm{cos}38\xc2\xb0\mathrm{cos}\mathrm{ec}(90\xc2\xb0\xe2\u02c6\u201938\xc2\xb0)}{\mathrm{tan}18\xc2\xb0\mathrm{tan}35\xc2\xb0\mathrm{tan}60\xc2\xb0\mathrm{tan}(90\xc2\xb0\xe2\u02c6\u201918\xc2\xb0)\mathrm{tan}(90\xc2\xb0\xe2\u02c6\u201935\xc2\xb0)}$
We know that sin (90Â° - Î¸) = cos Î¸, cos(90Â° - Î¸) =sin Î¸, cosec (90Â° - Î¸) = sec Î¸, tan (90Â° - Î¸) = cot Î¸
$\frac{\mathrm{sin}32\xc2\xb0}{\mathrm{sin}32\xc2\xb0}$+ $\frac{\mathrm{cos}68\xc2\xb0}{\mathrm{cos}68\xc2\xb0}$ - $\frac{\mathrm{cos}38\xc2\xb0\mathrm{sec}38\xc2\xb0}{\mathrm{tan}18\xc2\xb0\mathrm{tan}35\xc2\xb0\xe2\u02c6\u01613\mathrm{cot}18\xc2\xb0\mathrm{cot}35\xc2\xb0}$
⇒ (1+1) - $\frac{1}{\xe2\u02c6\u01613}$
⇒ $\frac{(2\xe2\u02c6\u01613\xe2\u02c6\u20191)}{\xe2\u02c6\u01613}$x $\frac{\xe2\u02c6\u01613}{\xe2\u02c6\u01613}$
⇒ $\frac{(6\xe2\u02c6\u2019\xe2\u02c6\u01613)}{3}$.
22. Find the point on y - axis which is equidistant from the points (5, âˆ’2) and (âˆ’3, 2)
OR
The line segment joining the points A (2, 1) and B (5, âˆ’8) is trisected at the points P and Q such that P is nearer to A. If P also lies on the line given by
2x âˆ’ y + k = 0, find the value of k.
1) Take correct coordinates of y - axis.
2) Write the correct formula for distance between two points.
OR
1) Write the correct section formula.
1) Assume that the coordinates of y - axis are (0, y).
2) Write the distance between two points.
3) Substitute in given condition.
4) Squaring on both sides to get required point.
OR
1) Let P, Q are two points of trisection of line segment AB.
2) The point P divides AB internally in the ratio 1:2.
3) Using section formula find the point P.
4) Substitute P in given line to get k value.
The required point is at equal distance from points (5, âˆ’2) and (âˆ’3, 2).
The point lies on the y-axis and hence its x-coordinate will be 0.
Assume that the coordinates are (0, y).
Distance between two points (${x}_{1}$, ${y}_{1}$) and (${x}_{2}$, ${y}_{2}$) is given byâˆš${{(x}_{2}\xe2\u02c6\u2019{x}_{1})}^{2}$ + ${{(y}_{2}\xe2\u02c6\u2019{y}_{1})}^{2}$ )
Distance between (0,y) and (5, -2) = âˆš(${(5\xe2\u02c6\u20190)}^{2}+({\xe2\u02c6\u20192\xe2\u02c6\u2019y)}^{2}$).
And the distance between (0, y) and (âˆ’3, 2) =âˆš(${(\xe2\u02c6\u20193\xe2\u02c6\u20190)}^{2}+({2\xe2\u02c6\u20190)}^{2}$)
Given that âˆš(${(5\xe2\u02c6\u20190)}^{2}$+ ${(\xe2\u02c6\u20192\xe2\u02c6\u2019y)}^{2}$) = âˆš(${(\xe2\u02c6\u20193\xe2\u02c6\u20190)}^{2}$+ ${(2\xe2\u02c6\u2019y)}^{2}$) squaring on both sides
25 + 4 + ${y}^{2}$+ 4y = 9 + 4 + ${y}^{2}$ - 4y
8y = -16
y = - 2
Hence the point is (0,-2)
OR
Let P (${x}_{1}$, ${y}_{1}$) and Q (${x}_{2}$, ${y}_{2}$) be the points of trisection of line segment AB, i.e., AP = PQ = QB
Hence the point P divides AB internally in the ratio 1:2.
By using section formula
(${x}_{1},{y}_{1})$= [$\frac{(1\times 5+2\times 2)}{(1+2)}$, $\frac{(1\times \xe2\u02c6\u20198+2\times 1)}{(1+2)}$]
${x}_{1}$= 3 and ${y}_{1}$ = -2
Therefore the coordinates are P(3,-2)
It is given that P(3,-2) on 2x âˆ’ y + k = 0
⇒2(3) - (-2) + k = 0
k = -8
Hence k = - 8
23. If P (x, y) is any point on the line joining the points A (a, 0) and B (0, b), then show that
$\frac{x}{a}$+ $\frac{y}{b}$= 1.
1) Write the correct formula.
2) Avoid calculation mistakes.
1) Given that three points A (a, 0), P (x, y), and C (0, b) lie on the same line, hence the area of the triangle is zero.
2) Substitute three points convert into line intercept form.
It is given that the three points A (a, 0), P (x, y), and C (0, b) lie on the same line, hence the area of the triangle formed by these three points is zero.
⇒$\frac{1}{2}$[a (y - b) + x ( b - 0) + 0 (0 - y) ] = 0
⇒[a ( y - b) + x ( b - 0) + 0 ( 0 - y) ] = 0
⇒ay - ab + xb = 0
⇒$\frac{\mathrm{ay}}{\mathrm{ab}}$ + $\frac{\mathrm{xb}}{\mathrm{ab}}$= 1
$\frac{x}{a}$ + $\frac{y}{b}$ = 1
Hence Proved.
24. In Fig. 6, PQ = 24 cm, PR = 7 cm and O is the centre of the circle. Find the area of shaded region (take Ï€ = 3.14)
1) Î”PQR represents a right-angled triangle. Find QR.
2) Diameter convert into radius.
2) Area of the shaded region = Area of semicircle âˆ’ Area of Î”PQR
1) Given that PQ = 24 cm, PR = 7 cm and O is the centre of the circle then QOR is diameter of the circle.
2) The angle in a semicircle is a right angle.Therefore, Î”PQR represents a right-angled triangle.
3) Apply pythagoras theorem to find QR.
4) Find Area of semicircle and Area of triangle.
5) Finally to find area of the shaded region.
Given that QOR is the diameter of the circle.
The angle in a semicircle is a right angle.
âˆ´âˆ RPQ = 90Â°
Therefore, Î”PQR represents a right-angled triangle where PR = 7 cm and PQ = 24 cm.
When Pythagoras theorem is applied to right Î”PQR, we obtain
${\mathrm{QR}}^{2}$ = ${\mathrm{PQ}}^{2}$ + ${\mathrm{PR}}^{2}$ = ${24}^{2}$ + ${7}^{2}$ = (576 + 49) = 625 ${\mathrm{cm}}^{2}$
QR = âˆš625 = 25 cm
Area of semicircle = $\frac{1}{2}$ Ï€${r}^{2}$,where r = $\frac{25}{2}$ cm = 12.5 cm
Area of semicircle = $\frac{1}{2}$×3.14×12.5×12.5 ${\mathrm{cm}}^{2}$= 245.3125 ${\mathrm{cm}}^{2}$
Area of Î”PQR = $\frac{1}{2}$× Base × Height = $\frac{1}{2}$×7×24 ${\mathrm{cm}}^{2}$= 7 × 12 ${\mathrm{cm}}^{2}$ = 84 ${\mathrm{cm}}^{2}$.
Hence , area of the shaded region = Area of semicircle âˆ’ Area of Î”PQR
= 245.3125 âˆ’ 84 = 161.3125 ${\mathrm{cm}}^{2}$.
25. The king, queen and jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of (i) heart (ii) queen (iii) clubs.
1) Out of 52 cards only king, queen and jack of clubs are removed.
2) Total$\mathrm{number}\mathrm{of}\mathrm{remaining}\mathrm{card}$s are 49.
1) Remove king, queen and jack of clubs in a pack of 52 cards.Remaining 49 cards left and in that 13 are hearts, 13 are spades, 13 are diamonds and 10 are clubs.
2) Applying the formula for i) Probability of getting a heart ii) Probability of getting a queen iii) Probability of getting a club card,
When we remove the King, Queen and Jack of Clubs we have about 49 cards left and in that 13 are hearts, 13 are spades, 13 are diamonds and 10 are clubs. In this
Probability of getting a heart = $\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{hearts}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{remaining}\mathrm{card}s}$= $\frac{13}{49}$
Probability of getting a queen = $\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{Queens}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{remaining}\mathrm{card}s}$= $\frac{3}{49}$
Probability of getting a club card = $\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{remaining}\mathrm{club}\mathrm{card}s}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{remaining}\mathrm{card}s}$ = $\frac{10}{49}$.
26. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
OR
Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
1) Write the two consecutive odd numbers perfectly.
2) Avoid calculation mistakes.
or
1) Consider two conditions when both cars travel in the same direction and when both cars travel in opposite directions.
1) Let the speed of car A be u and the car B be v.
2) Consider two conditions when both cars travel in the same direction and when both cars travel in opposite directions.
3) Using distance = time Ã— speed formula to get two equations.
4) Solving two equations to get speed of two cars.
We can represent the two odd numbers as (2n âˆ’ 1) and (2n + 1), where n is a natural number.
It is given that (2n âˆ’ 1${)}^{2}$ + (2n + 1${)}^{2}$ = 394
⇒ 4${n}^{2}$ + 1 âˆ’ 4n + 4${n}^{2}$+ 1 + 4n = 394
⇒ 8${n}^{2}$ + 2 = 394
⇒ 8${n}^{2}$ = 392
⇒ ${n}^{2}$ = 49
⇒ n = 7
⇒ 2n âˆ’ 1 = 2 (7) âˆ’ 1 = 13
⇒ 2n + 1 = 2 (7) + 1 = 15
Therefore, the odd numbers are 13 and 15.
OR
Consider the speed of car A be u and the car B be v.
Condition 1 : when both cars travel in the same direction.
Let the cars meet at point C in 5 hours.
Car A travels distance AC, whereas car B travels distance BC.
Distance = time × speed
⇒ AC = 5 × u
⇒BC = 5 × v
⇒AC âˆ’ BC = 100
⇒ 5u âˆ’ 5v = 100
⇒ u âˆ’ v = 20 â€¦ (1)
Condition 2 : When both cars travel in opposite directions
Let both cars meet at point D.
Car A will travel AD, while car B will travel BD when they meet.
Distance = time × speed
⇒ AD = 1 × u
⇒ BD = 1 × v
AD + BD = 100
⇒ u + v = 100 ----------((2)
On adding equations (1) and (2), we obtain
2u = 120
⇒ u = 60
From equation (2), we obtain
60 + v = 100
⇒ v = 40
Hence, the speed of car A is 60 km/h and the car B is 40 km/h.
27. Prove that, if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Using the above result, do the following:
In Fig. 7, DE || BC and BD = CE. Prove that Î”ABC is an isosceles triangle.
1) Draw a perpendicular lines from T and S to PS and PT i.e. TU and SV respectively and also join QT and RS.
2) Observe that height of triangles will take out side and find area of a triangle.
3) Compare the ratio of similar triangles.
In a triangle PQR as shown below a line ST drawn parallel to the base QR. ST is intersecting the sides PQ and PR at points S and T.
A line ST divides PQ into two parts PS and SQ, similarly ST divides PR into two parts PT and TR.Prove that $\frac{\mathrm{PS}}{\mathrm{SQ}}$= $\frac{\mathrm{PT}}{\mathrm{TR}}$.
Draw a perpendicular lines from T and S to PS and PT i.e. TU and SV respectively and also join QT and RS.
Area or a triangle = $\frac{1}{2}$×base ×height
Area of âˆ† PST = $\frac{1}{2}$×PS ×TU
If PT as base and SV as height, so that
Area of âˆ† PST= $\frac{1}{2}$×PT ×SV
In the similar way
Area of âˆ† QST = $\frac{1}{2}$×QS ×TU
Area of âˆ† RST = $\frac{1}{2}$×SV ×TR
$\frac{\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{PST}}{\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{QST}}$ = $\frac{\frac{1}{2}\times \mathrm{PS}\times \mathrm{TU}}{\frac{1}{2}\times \mathrm{QS}\times \mathrm{TU}}$
Therefore
$\frac{\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{PST}}{\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{QST}}$= $\frac{\mathrm{PS}}{\mathrm{QS}}$ ------------------------- (1)
$\frac{\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{PST}}{\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{RST}}$=$\frac{\frac{1}{2}\times \mathrm{PT}\times \mathrm{SV}}{\frac{1}{2}\times \mathrm{TR}\times \mathrm{SV}}$
$\frac{\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{PST}}{\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{RST}}$ = $\frac{\mathrm{PT}}{\mathrm{TR}}$ ----------------------- (2)
In the diagram given above Î”QST and Î”RST are drawn on the same base i.e., ST and between the same parallel lines i.e., ST and QR.
âˆ´area of (Î”QST) = area of (Î”RST)------------------------- (3)
From equations (1), (2), and (3), we obtain
$\frac{\mathrm{PS}}{\mathrm{QS}}$= $\frac{\mathrm{PT}}{\mathrm{TR}}$.
Hence proved.
In Î”ABC a line DE parallel to its base (i.e., BC) is drawn such that it intersects sides AB and BC at points D and E respectively.
From the above result
$\frac{\mathrm{AD}}{\mathrm{BD}}$= $\frac{\mathrm{AE}}{\mathrm{CE}}$
It is given that BD = CE ------------------------------- (4)
Therefore AD = AE ------------------------------- (5)
Lets add equations (4) and (5), we obtain
BD + AD = CE + AE
⇒ AB = AC
Hence, Î”ABC is an isosceles triangle.
28. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30Â°, which is approaching the foot of the tower with a uniform speed. Six seconds later the angle of depression of the car is found to be 60Â°. Find the time taken by the car to reach the foot of the tower from this point.
1) Write the angles in correct order.
2) Since the car is moving in uniform speed. Take the ratio will be constant.
1) Using angles find the distance between foot of the tower and car starting point.
2) Since the car is moving in uniform speed then Speed =$$$\frac{\mathrm{dis}\mathrm{tan}\mathrm{ce}}{\mathrm{speed}}$
3) Since ratio will be constant because of uniform speed.
4) Hence to get time taken by the car to reach the tower.
Consider that AB is the tower.
The car was originally at point C.
It travelled for 6 sec and reached point D and hence CD is the distance travelled.
The angle of depression changes from âˆ XAC to âˆ XAD.
âˆ ACB = âˆ XAC = 30Â° (Alternate interior angles)
And, âˆ ADB = âˆ XAD = 60Â° (Alternate interior angles)
In Î”ABD
Tan 60Â° = $\frac{\mathrm{AB}}{\mathrm{BD}}$
$\frac{\mathrm{AB}}{\mathrm{BD}}$ = âˆš3 --------------------------- (1)
In Î”ABC,
Tan 30Â° = $\frac{\mathrm{AB}}{\mathrm{BC}}$
$\frac{\mathrm{AB}}{\mathrm{BC}}$= $\frac{1}{\sqrt{3}}$------------------------ (2)
Divide (1) by (2)
$\frac{\mathrm{BC}}{\mathrm{BD}}$= 3
$\frac{(\mathrm{BD}+\mathrm{DC})}{\mathrm{BD}}$= 3
1+ $\frac{\mathrm{DC}}{\mathrm{BD}}$= 3
$\frac{\mathrm{DC}}{\mathrm{BD}}$= 2
BD = $\frac{\mathrm{DC}}{2}$---------------------------(3)
Since the car is moving in uniform speed
Speed = $\frac{\mathrm{dis}\mathrm{tan}\mathrm{ce}}{\mathrm{time}}$
This ratio will be constant because of uniform speed.
$\frac{\mathrm{DC}}{6}$= $\frac{\mathrm{DC}}{2t}$
T = 3 sec
Hence the time taken by the car to reach the tower = 3 seconds.
29. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and of base radius 6 cm, is hollowed out. Find the volume of the remaining solid correct to two places of decimals. Also find the total surface area of the remaining solid. (take Ï€ = 3.1416)
OR
In Fig. 8, ABC is a right triangle right angled at A. Find the area of shaded region if AB = 6 cm, BC = 10 cm and O is the centre of the incircle of Î”ABC.
(take Ï€ = 3.14)
1) Write the correct formulas.
2) Avoid calculation mistakes.
or
1) The tangents drawn from an external point to the circle are equal.
1) Obtain the volume of the remaining solid from subtracting the volume of conical cavity from the volume of cylinder.
2) Find slant height .
3) By adding curved surface area of cylinder , area of top face and curved surface area of cone to get the total surface area of remaining solid.
or
1) By applying Pythagoras theorem to find remaining side.
2) The circle touches all sides then the radius of the centre of the circle is perpendicular to the tangent through the point of contact.
3) Find the radius if the tangents drawn from an external point to the circle are equal.
4) Area of shaded region from subtracting the area of circle from the area of triangle.
Since the conical cavity is hollowed out from the cylinder, we can obtain the volume of the remaining solid from subtracting the volume of conical cavity from the volume of cylinder
i.e;Volume of the remaining solid = Volume of cylinder âˆ’ Volume of conical cavity
Volume of the remaining solid = Ï€${r}^{2}$h - $\frac{1}{3}$Ï€${r}^{2}$h
=$\frac{2}{3}$Ï€${r}^{2}$h
= $\frac{2}{3}$×3.1416×(6${)}^{2}$× 8
Volume of the remaining solid = 603.1872 ${\mathrm{cm}}^{3}$.
The total surface area of remaining solid = Curved surface area of cylinder + area of top face + curved surface area of cone
= 2Ï€rh + Ï€r2 + Ï€rl
= Ï€r (2h + r + l)
We know that the slant height (l) of the cone is given byâˆš(${r}^{2}$+ ${h}^{2})$.
âˆ´ Slant height of conical cavity =âˆš(${6}^{2}$+ ${8}^{2}$) = 10 cm
Hence, total surface area of the solid = 3.1416 × (6 cm) (2 × 8 cm + 6 cm + 10 cm)
= 3.1416 × (6 cm) (16 cm + 6 cm + 10 cm)
= 3.1416 × 6 cm × 32 cm
= 603.1872 ${\mathrm{cm}}^{2}$
Therefore the total surface area of the remaining solid is approximately 603.18 ${\mathrm{cm}}^{2}$.
OR
In Î”ABC, applying Pythagoras theorem we obtain
${\mathrm{BC}}^{2}$ = ${\mathrm{AC}}^{2}$ + ${\mathrm{AB}}^{2}$
⇒ ${\mathrm{AC}}^{2}$ = ${\mathrm{BC}}^{2}$âˆ’${\mathrm{AB}}^{2}$= (${10\mathrm{cm})}^{2}$ âˆ’ ${(6\mathrm{cm})}^{2}$
⇒ ${\mathrm{AC}}^{2}$ = 64 ${\mathrm{cm}}^{2}$
⇒ AC = 8 cm
Let us assume the radius of the incircle to be â€˜râ€™.
The circle touches side AB of the triangle at P, side AC at Q, and side BC at R.
The radius from the centre of the circle is perpendicular to the tangent through the point of contact.
âˆ´ OPâŠ¥AB, OQâŠ¥AC, and ORâŠ¥BC
The tangents drawn from an external point to the circle are equal.
âˆ´ AP = AQ (Tangents from point A)
BP = BR (Tangents from point B)
CR = CQ (Tangents from point C)
Now, in quadrilateral OPAQ, AQ = AP and âˆ AQO = âˆ APO = âˆ PAQ = 90Â°
Therefore, OPAQ is a square.
âˆ´ OP = AQ = AP = OQ = r
âˆ´ AB = 6 cm âˆ’ r
⇒ BR = 6 cm âˆ’ r
⇒CQ = 8 cm âˆ’ r
⇒ CR = 8 âˆ’ r
Now, BC = BR + CR
⇒ 10 cm = 6 cm âˆ’ r + 8 cm âˆ’ r
⇒ 10 cm = 14 cm âˆ’ 2r
⇒ r = 2 cm
Area of shaded region = Area of Î”ABC âˆ’ Area of circle
=$\frac{1}{2}$×AB×AC - Ï€${r}^{2}$
= $\frac{1}{2}$×8×6 - Ï€${2}^{2}$= 24 - 12.56 = 11.44 ${\mathrm{cm}}^{2}$.
Hence, the area of the shaded region is 11.44 cm2.
30. The following table gives the daily income of 50 workers of a factory:
Find the Mean, Mode and Median of the above data.
1) Take correct assumed mean.
2) In the mode class interval depends on corresponds to the maximum frequency .
3) In the median class interval depends on number of observations.
1) Find the Cumulative frequency and deviations.
2) Substitute all values in appropriate formulas of Mean, Mode and Median to get the solution.
Let us represent the assumed mean by â€˜aâ€™ and assume it as 150.
The table for the given data can be drawn as
Mean = a + $\frac{\mathrm{\xce\pounds}{f}_{i}{d}_{i}}{{f}_{i}}$
Therefore Mean = 150 + $\frac{(\xe2\u02c6\u2019240)}{50}$= 150 + (- 4.8) = 145.2
The class that corresponds to the maximum frequency of 14 is 120 -140.
Modal class = 120 - 140
Lower limit (l) = 120
Class size (h) = 140 âˆ’ 120 = 20
Frequency of modal class (${f}_{1}$) = 14
Frequency of class preceding the modal class (${f}_{0}$) = 12
Frequency of class succeeding the modal class (${f}_{2}$) = 8
Mode = l + $\frac{({f}_{1}\xe2\u02c6\u2019{f}_{0})}{(2{f}_{1}\xe2\u02c6\u2019{f}_{0}\xe2\u02c6\u2019{f}_{2})}$× h
Therefore, Mode = 120 + $\frac{(14\xe2\u02c6\u201912)}{(2\times 14\xe2\u02c6\u201912\xe2\u02c6\u20198)}$× 20
Mode = 120 + $\frac{2}{8}$×20
= 120 + 5
= 125.
Hence, mode of the given data is 125.
Number of observations (n) = 50
$\frac{n}{2}$= 25.
This value lies in class interval 120âˆ’140.
Therefore, the median class is 120âˆ’140.
Lower limit of median class (l) = 120
Cumulative frequency of class preceding the median class (c.f) = 12
Frequency of median class (f) = 14
Median of the data = l + $\frac{\frac{n}{2}\xe2\u02c6\u2019c.f}{f}$× h
Therefore median = 120 + $\frac{(125\xe2\u02c6\u201912)}{14}$× 20
= 120 + $\frac{130}{7}$ = 120 + 18.57 = 138.57
Hence, median = 138.57
Section A | Section B | Section C | Section D |
Nice! You have completed the lesson level test. To further perfect your understanding on chapter name , take this chapter level test.