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CBSE X
Delhi
MATHS PAPER 2010
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 sum of first p terms of an Î‘.P. is ${\mathrm{ap}}^{2}+\mathrm{bp}$, find its common difference.
1) ${S}_{2}$is the sum of the first and second terms of the A.P.
2) Difference between ${S}_{2}$and ${S}_{1}$is the second term of the series.
1) Find the sum of first term and the sum of two terms.
2) Find the difference between ${S}_{2}$and ${S}_{1}$ which is the second term of the series.
3) Subtract the first term from the second term of the series which is the common difference of the series.
Let the first term of the arithmetic progression (A.P.) be 'a' and the common difference be 'd'.
According to the question, the sum of an A.P. with p terms is given as ${\mathrm{ap}}^{2}+\mathrm{bp}$.
Sum of the series with one term:
âˆ´ ${S}_{1}=a{\left(1\right)}^{2}+b\left(1\right)=a+b$
Sum of first two terms of the series:
${S}_{2}=a{\left(2\right)}^{2}+b\left(2\right)=4a+2b$
${S}_{1}$is the first term of the A.P.
${S}_{2}$is the sum of twice the first term and the common difference.
Therefore, the first term 'a' = (a + b) - (1)
Sum of the first two terms is a + (a + d) = (4a + 2b) - (2)
Subtracting the linear equation (1) from (2).
Multiply equation (1) by 2 and subtract from equation 2 to find the common difference 'd'.
(2a + d) - (2a) = 4a + 2b - 2a - 2b = 2a.
âˆ´d = 2a
Therefore, the common difference of the A.P. is 2a.
2. In the given fig. , S and T are points on the sides PQ and PR, respectively of Î”PQR, such that PT = 2 cm, TR = 4 cm and ST is parallel to QR. Find the ratio of the areas of Î”PST and Î”PQR.
1) Choose the correct pair of triangles to prove similar.
2) Choose the correct pair of corresponding sides.
1) Prove that â–³PQR and â–³PST are similar according to AA similarity rule.
2) Write the ratio of areas of the similar triangles as the ratio of the squares of the corresponding sides PT and PR.
Consider two triangles âˆ†PQR and âˆ†PST.
âˆ QPR = âˆ SPT as P is a common angle for both the triangles.
Also âˆ PQR = âˆ PST, as they are corresponding angles.
âˆ´ âˆ†PQR ~ âˆ†PST by AA similarity rule.
Ratio of areas of the two similar triangles is equal to the ratio of the squares of the corresponding sides.
âˆ´ $\frac{(\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{PQR})}{(\mathrm{Area}\mathrm{of}\xe2\u2013\xb3\mathrm{PST})}=\frac{{\left(\mathrm{PT}\right)}^{2}}{{\left(\mathrm{PR}\right)}^{2}}$
=$\frac{{\left(\mathrm{PT}\right)}^{2}}{{\left(\mathrm{PT}+\mathrm{TR}\right)}^{2}}$
= $\frac{{\left(2\right)}^{2}}{{\left(2+4\right)}^{2}}$= $\frac{4}{36}$= $\frac{1}{9}$
Therefore, the ratio of the areas of two triangles is 1 : 9.
3. In the given figure, Î”AHK is similar to Î”ABC. If AK = 10 cm, BC = 3.5 cm and HK = 7 cm, find AC.
Choose the correct pair of corresponding sides.
The given triangles are similar so, the corresponding sides are proportions.
Write the ratio of corresponding sides and simplify to find the value of AC.
Given that âˆ†ABC is similar to âˆ†AKH.
Corresponding sides of similar triangles are proportional to each other.
Therefore,
$\frac{\mathrm{BC}}{\mathrm{HK}}$= $\frac{\mathrm{AC}}{\mathrm{AK}}$ ⇒$\frac{(3.5\mathrm{cm})}{(7.5\mathrm{cm})}$ = $\frac{\mathrm{AC}}{(10\mathrm{cm})}$
âˆ´AC = 5 cm
4. If α, Î² are the zeroes of a polynomial, such that α + Î² = 6 and αÎ² = 4, then write the polynomial.
Write the standard form of the quadratic polynomial with proper signs.
Substitute the values of the sum of the zeroes and the product of the zeroes in the standard form of the quadratic polynomial.
Find the polynomial when the value of the real number is 1.
A polynomial can be represented by r[${x}^{2}$âˆ’(sum of zeroes) x + (product of zeroes)] where r is a real number.
Sum of zeroes = α + Î² = 6
Product of zeroes = αÎ² = 4
Substituting the values for sum of zeroes and product of zeroes, the polynomial is R (${x}^{2}$âˆ’6x + 4).
When r = 1, the polynomial becomes ${x}^{2}$âˆ’6x + 4.
Hence, the required polynomial is ${x}^{2}\xe2\u02c6\u20196x+4.$
5. Has the rational number $\frac{441}{({2}^{2}.{5}^{7}.{7}^{2})}$ a terminating or a non-terminating decimal representation?
Denominator of the fraction should be in exponential form with only 2 and 5 as bases.
1) Write the numerator in exponential form.
2) Simplify the fraction.
3) The denominator has an exponential form with only 2 and 5 as bases.
4) The rational number has the terminating decimal expansion.
Any rational number in its simplest form has a terminating decimal expression, if the denominator of the rational number can be expressed in the form of ${2}^{n}{5}^{m}$, where n and m are non negative integers.
Express the rational number $\frac{441}{({2}^{2}{5}^{7}{7}^{2})}$in its simplest form.
$\frac{441}{({2}^{2}{5}^{7}{7}^{2})}=\frac{{3}^{2}.{7}^{2}}{({2}^{2}.{5}^{7}.{7}^{2})}$= $\frac{{3}^{2}}{({2}^{2}.{5}^{7})}$
Simplified form of the rational number has the denominator in the form of ${2}^{n}{5}^{m}$.
So, the given rational number has a terminating decimal expansion.
6. If cosec Î¸ = 2x and cot Î¸ = $\frac{2}{x}$, find the value of 2(${x}^{2}$-$\frac{1}{{x}^{2}}$)
Square the values of squares of x and $\frac{1}{x}$within the bracket.
1) Find the values of x and $\frac{1}{x}$.
2) Substitute the values of x and $\frac{1}{x}$in the given expression.
3) Simplify and apply the related trigonometric identity to find the final value of the expression.
cosec Î¸ = 2x and ⇒ x = $\frac{\mathrm{cos}\mathrm{ec}\mathrm{\xce\xb8}}{2}$
Similarly, cot Î¸ = $\frac{2}{x}$ and hence $\frac{\mathrm{cot}\mathrm{\xce\xb8}}{2}$= $\frac{1}{x}$
Substituting the values of x and $\frac{1}{x}$ in the given expression:
2 $({x}^{2}\xe2\u02c6\u2019\frac{1}{{x}^{2}})$
= 2$(\frac{{\mathrm{cos}\mathrm{ec}}^{2}\mathrm{\xce\xb8}}{4}\xe2\u02c6\u2019\frac{{\mathrm{cot}}^{2}\mathrm{\xce\xb8}}{4})$= $\frac{2}{4}$(${\mathrm{cos}\mathrm{ec}}^{2}\mathrm{\xce\xb8}$ - ${\mathrm{cot}}^{2}\mathrm{\xce\xb8}$)
But, ${\mathrm{cos}\mathrm{ec}}^{2}\mathrm{\xce\xb8}$ - ${\mathrm{cot}}^{2}\mathrm{\xce\xb8}$ = 1
Therefore, the expression becomes $\frac{2}{4}$ × 1 = $\frac{1}{2}$.
7. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting a red face card.
Consider the exact number of red face cards in the pack of cards.
1) Number of red face cards in the deck is 6.
2) Find the probability of getting a red face card out of 52 cards of the pack.
Total number of cards in the pack = 52
Number of red face cards in the pack = 6
Therefore, probability of getting a red face card =$\frac{6}{52}$ = $\frac{3}{26}$
8. The slant height of frustum of a cone is 4 cm and the perimeters (circumferences) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
[Use Ï€ = $\frac{22}{7}$]
1) Write the correct formula to find the CSA of the frustum.
2) Avoid calculation mistakes in simplification.
1) Substitute the values of circumferences of the bases of the frustum of the cone and the slant height in the formula and simplify.
Let l represent the slant height of a cone. It is given that l = 4 cm
${r}_{1}$ and ${r}_{2}$ are the radii of the circular ends of the frustum of the cone with 'h' as its height.
Given that the perimeters or circumferences of the circular ends of the frustum of the cone are 18 cm and 6 cm.
âˆ´ 2Ï€${r}_{1}$ = 18 cm and 2Ï€${r}_{2}$ = 6 cm
âˆ´ Curved surface area of the frustum =$\frac{1}{2}\times $$(\mathrm{Sum}\mathrm{of}\mathrm{circumferences}\mathrm{of}\mathrm{bases})\times (\mathrm{Slant}\mathrm{height})$
âˆ´ Curved surface area
=$\frac{1}{2}$ (18 + 6) × 4
= $\frac{1}{2}\times 24\times 4$
= 48${\mathrm{cm}}^{2}$
Thus, the curved surface area of the frustum of the cone is 48 ${\mathrm{cm}}^{2}$.
9. If A(1, 2), B(4, 3) and C(6, 6) are the three vertices of the parallelogram ABCD, find the coordinates of the fourth vertex D.
Substitute the x-coordinates and y-coordinates of the points exactly in the midpoint formula.
1) AC and BD are the diagonals of the parallelogram.
2) Diagonals of a parallelogram bisect each other. So the midpoint of AC and midpoint of BD coincide.
3) Assume the coordinates of D as (x, y)
4) Find the coordinates of the midpoints of AC and BD using the midpoint formula and equate x and y coordinates.
5) Simplify the linear equations to find the coordinates of the fourth vertex D.
Let us assume that the coordinates of the fourth vertex D be (x, y).
We know that ABCD is a parallelogram and hence the diagonals AC and BD bisect each other.
âˆ´ Mid point of AC = Mid-point of BD
⇒ $(\frac{4+x}{2},\frac{3+y}{2})$ = $(\frac{1+6}{2},\frac{2+6}{2})$ = $(\frac{7}{2}$, $\frac{8}{2})$
âˆ´ 4 + x = 7, 3 + y = 8
⇒ x = 3 and y = 5
So, the coordinates of the fourth vertex are D(3, 5).
10. If P(2, p) is the mid-point of the line segment joining the points A(6, âˆ’ 5) and B(âˆ’ 2, 11), find the value of p.
Substitute the x-coordinates and y-coordinates of the points exactly in the midpoint formula.
1) Find the midpoint of the line segment AB using the midpoint formula.
2) Equate the midpoint to the coordinates of the given midpoint.
3) Equate y-coordinate to find the value of p.
P(2, p) is the midpoint of the line segment joining the points A (6, âˆ’5) and B (âˆ’2, 11).
Midpoint of the line segment joining the points ${(x}_{1},{y}_{1})$âˆ´$({x}_{2},{y}_{2})$= $(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2})$
âˆ´(2, p) = $(\frac{6+(\xe2\u02c6\u20192)}{2}$, $\frac{\xe2\u02c6\u20195+11}{2})$
⇒ (2, p) = (2, 3)
⇒ p = 3
16. Prove that (2$\sqrt{3}$-1) is an irrational number.
1) To prove that the given number is an irrational number, firstly prove that $\sqrt{3}$is an irrational number.
1) Prove that $\sqrt{3}$is an irrational number by disproving that it is a rational number.
2) Assume that 2$\sqrt{3}$-1 is a rational number and equate it to $\frac{a}{b}$.
3) Find the value of $\sqrt{3}$from the equation which can be a rational number.
4) Therefore, 2$\sqrt{3}$-1 is an irrational number by contradiction.
In order to prove that (2$\sqrt{3}$-1) is an irrational number we first prove that $\sqrt{3}$âˆšis irrational.
Let us consider that $\sqrt{3}$ is a rational number.
If we can find two integers a, b (b â‰ 0) such that $\sqrt{3}$= $\frac{a}{b}$, then $\sqrt{3}$ is rational.
Both a and b should be co-prime. i.e., they should have a common factor as 1.
Therefore,
a = $\sqrt{3}$b
${a}^{2}$= 3${b}^{2}$
It is evident that ${a}^{2}$ is divisible by 3. Hence, a is also divisible by 3.
Let a = 3m, where m is an integer.
${\left(3m\right)}^{2}={3b}^{2}$
${b}^{2}$= ${3m}^{2}$
From the above equation it is evident that ${b}^{2}$ is divisible by 3. Hence, b is divisible by 3.
From this analysis, where both numbers are divisible by 3, we can conclude that both a and b have 3 as common factor other than 1.
This does not satisfy the condition that both a and b are co-primes.
This implies that a and b have 3 as a common factor.
Hence, $\sqrt{3}$ is an irrational number.
Similarly, assume that (2$\sqrt{3}$-1) is a rational number.
Then, we can find two integers a, b (b â‰ 0) such that (2$\sqrt{3}\xe2\u02c6\u20191)$ = $\frac{a}{b}$.
⇒$\sqrt{3}$ = $\frac{1}{2}$($\frac{a}{b}$ -1).
Since a and b are integers, $\frac{1}{2}$($\frac{a}{b}$ - 1) is also rational.
Therefore, $\sqrt{3}$âˆšshould be rational.
This contradicts the fact that $\sqrt{3}$ is irrational. Hence, the assumption that "2$\sqrt{3}\xe2\u02c6\u20191$is rational" is false. Hence, (2$\sqrt{3}$- 1) is irrational.
17. In figure, ABC is a right triangle, right angled at C and D is the mid-point of BC. Prove that ${\mathrm{AB}}^{2}=4{\mathrm{AD}}^{2}\xe2\u02c6\u20193{\mathrm{AC}}^{2}$.
1) Apply Pythagoras theorem in triangle ACD in the first case.
2) Avoid calculation mistakes in taking the LCM of the equation.
1) D is the mid-point of BC. Therefore, equate both the parts obtained to half of BC.
2) Apply Pythagoras theorem in triangle ACD.
3) Substitute CD by half of BC and simplify to get an equation.
4) Apply Pythagoras theorem in triangle ABC.
5) Substitute the value of BC from the equation obtained above.
6) Simplify to get the required proof.
It is given that in Î”ABC, D is the mid-point of side BC.
âˆ´ BD = CD = $\frac{1}{2}$BC
In Î”ACD,
${\mathrm{AD}}^{2}={\mathrm{AC}}^{2}+{\mathrm{CD}}^{2}$ (Pythagoras theorem)
⇒ ${\mathrm{AD}}^{2}={\mathrm{AC}}^{2}+{(\frac{1}{2}\mathrm{BC})}^{2}$
⇒ ${\mathrm{AD}}^{2}={\mathrm{AC}}^{2}+\frac{1}{4}{\mathrm{BC}}^{2}$
⇒ ${4\mathrm{AD}}^{2}=4{\mathrm{AC}}^{2}+{\mathrm{BC}}^{2}$
⇒ ${\mathrm{BC}}^{2}=4{\mathrm{AD}}^{2}\xe2\u02c6\u20194{\mathrm{AC}}^{2}$--------- (1)
In Î”ABC, according to Pythagoras theorem,
${\mathrm{AB}}^{2}={\mathrm{AC}}^{2}+{\mathrm{BC}}^{2}$
⇒ ${\mathrm{AB}}^{2}={\mathrm{AC}}^{2}+(4{\mathrm{AD}}^{2}\xe2\u02c6\u20194{\mathrm{AC}}^{2})$ [from equation (1)]
âˆ´ ${\mathrm{AB}}^{2}=4{\mathrm{AD}}^{2}\xe2\u02c6\u20193{\mathrm{AC}}^{2}$
Hence Proved.
18. Prove the following
$\frac{\mathrm{tan}A}{(1\xe2\u02c6\u2019\mathrm{cot}A)}$ + $\frac{\mathrm{cot}A}{(1\xe2\u02c6\u2019\mathrm{tan}A)}$= 1+ tan A + cot â¡A
OR
Prove the following
(coses A - cos A)(sec A - cos A) = $\frac{1}{(\mathrm{tan}A+\mathrm{cot}A)}$
1) Avoid calculation mistakes in finding the LCM of the fractions with trigonometric ratios.
2) Use the appropriate formula for the cube of the binomial of the trigonometric ratios.
OR
1) Avoid calculation mistakes in finding the LCM of the fractions with trigonometric ratios.
2) Use the appropriate trigonometric identity to simplify the equation.
1) Write the reciprocals of cotangent and simplify by finding the LCM of the fractions.
2) Apply the formula for the cube of the binomial of the trigonometric ratios.
3) Simplify the fraction of the trigonometric ratios and prove that it is equal to RHS.
OR
1) Write the reciprocals of the trigonometric ratios cosecant, secant and cotangent.
2) Simplify by taking LCM of the fractions.
3) Apply the trigonometric identities and simplify to prove that LHS is equal to RHS.
Consider LHS of the equation
$\frac{\mathrm{tan}A}{(1\xe2\u02c6\u2019\frac{1}{\mathrm{tan}A})}$ + $\frac{\left(\frac{1}{\mathrm{tan}A}\right)}{1\xe2\u02c6\u2019\mathrm{tan}A}$
= $\frac{\frac{\mathrm{tan}A}{(\mathrm{tan}A\xe2\u02c6\u20191)}}{\mathrm{tan}A}$ + $\frac{1}{\mathrm{tan}A(1\xe2\u02c6\u2019\mathrm{tan}A)}$
= $\frac{{\mathrm{tan}}^{2}A}{(\mathrm{tan}A\xe2\u02c6\u20191)}$ - $\frac{1}{\mathrm{tan}A(\mathrm{tan}A\xe2\u02c6\u20191)}$
= $\frac{{(\mathrm{tan}}^{3}A\xe2\u02c6\u20191)}{\mathrm{tan}A(\mathrm{tan}A\xe2\u02c6\u20191)}$
= $\frac{(\mathrm{tan}A\xe2\u02c6\u20191)({\mathrm{tan}}^{2}A+\mathrm{tan}A+1)}{\mathrm{tan}A(\mathrm{tan}A\xe2\u02c6\u20191)}$
= $\frac{{(\mathrm{tan}}^{2}A+\mathrm{tan}A+1)}{\mathrm{tan}A}$
= tan A + 1 + $\frac{1}{\mathrm{tan}A}$
= 1 + tan A + $\frac{1}{\mathrm{tan}A}$
= R.H.S
Hence proved.
OR
Prove the following:
(cosec A - cos A)(sec A - cos A) = $\frac{1}{(\mathrm{tan}A+\mathrm{cot}A)}$
$(\frac{1}{\mathrm{sin}A}\xe2\u02c6\u2019\mathrm{sin}A)(\frac{1}{\mathrm{cos}A}\xe2\u02c6\u2019\mathrm{cos}A)=\frac{1}{(\mathrm{tan}A+\mathrm{cot}A)}$
$\left(\frac{1\xe2\u02c6\u2019{\mathrm{sin}}^{2}A}{\mathrm{sin}A}\right)$$\left(\frac{1\xe2\u02c6\u2019{\mathrm{cos}}^{2}A}{\mathrm{cos}A}\right)=\frac{1}{(\mathrm{tan}A+\mathrm{cot}A)}$
sin A cos A = $\frac{1}{(\frac{\mathrm{sin}A}{\mathrm{cos}A}+\frac{\mathrm{cos}A}{\mathrm{sin}A})}$
sin A cos A = $\frac{\mathrm{cos}A\mathrm{sin}A}{{\mathrm{sin}}^{2}A+{\mathrm{cos}}^{2}A}$
Since, ${\mathrm{sin}}^{2}A+{\mathrm{cos}}^{2}A$ = 1
sin A cos A = cos A sin A
Hence, LHS = RHS.
19. In an A.P., the sum of first ten terms is âˆ’150 and the sum of its next ten terms is âˆ’550. Find the A.P.
Substitute the value of n as 20 to find the sum of the first 20 terms of the series.
1) Find the sum of the first ten terms of the A.P., equate it to -150 and simplify. Consider it as equation (1).
2) Add the sum of the first ten terms and the sum of the next ten terms and consider it as the sum of the first twenty terms.
3) Find the sum of the first twenty terms of the A.P. equate it to -550 and simplify. Consider it as equation (2).
4) Solve equations (1) and (2) to find the values of the first term (a) and the common difference of the A.P.
5) Find the A.P. using the first term and the common difference.
Let 'a' be the first term of the A.P. and let 'd' be the common difference.
Given that the sum of first ten terms is - 150.
Sum of n terms ${S}_{n}=\frac{n}{2}$[2a + (n - 1)d]
Therefore, ${S}_{10}$= 5[2a + (10 - 1)d] = -150
2a + 9d = - 30 ------ (1)
Given that the sum of the next ten terms is -550.
So the sum of the first ten terms and the second ten terms =
$\frac{20}{2}$[2a + (20 - 1)d] = -150 + (-550) = -700
⇒ 2a + 19d = âˆ’70 ----- (2)
Solving linear equations (1) and (2)
d = - 4 and a = 3
So, the first term of the A.P. is 3 and the common difference is - 4.
Hence, the A.P. is 3, -1, -5, -9, -13 â€¦
20. The sum of numerator and denominator of a fraction is 3 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes $\frac{1}{2}$. Find the fraction.
Subtract 1 from both the numerator and the denominator of the fraction.
1) Assume the numerator and the denominator of the fraction as x and y.
2) Apply the first condition given in the question and consider the equation formed as the first equation.
3) Apply the second condition given in the question and consider the equation formed as the second equation.
4) Solve equations to find the values of x and y and the fraction.
Let the numerator of the fraction be x and denominator of the fraction be y. Hence the fraction can be represented by $\frac{x}{y}$. Given that the sum of the numerator and denominator of this fraction is 3 less than twice the denominator.
âˆ´ x + y = 2y âˆ’ 3
⇒ x + y âˆ’ 2y = âˆ’3
⇒ x âˆ’ y = âˆ’3 â€¦ (1)
The second part of the question says that if the numerator and denominator is decreased by 1, then the fraction becomes $\frac{1}{2}$.
Subtracting equation (1) from equation (2):
x = 1 âˆ’ (âˆ’ 3)
⇒ x = 4
Substituting the value of x in equation (1):
4 âˆ’ y = âˆ’ 3
⇒ y = 4 + 3 = 7
Hence, the fraction is $\frac{4}{7}.$
11. If $\sqrt{5}$ and -$\sqrt{5}$ are two zeroes of the polynomial ${x}^{3}+3{x}^{2}\xe2\u02c6\u20195x\xe2\u02c6\u201915$, find its third zero.
1) Apply the algebraic formula to find the product of the factors of the polynomial.
1) Find the factors of the polynomial from its zeroes.
2) Find the product of the factors.
3) Divide the polynomial by the product of the factors.
4) The quotient is the third factor of the polynomial.
5) Find the third zero of the polynomial from its third factor.
$\sqrt{5}$ and -$\sqrt{5}$ are two zeroes of the polynomial f(x) = ${x}^{3}+3{x}^{2}\xe2\u02c6\u20195x\xe2\u02c6\u201915$.
So, (x - $\sqrt{5}$) and (x +âˆš$\sqrt{5}$) are two factors of the polynomial f(x).
If the polynomial f(x) is divided by the the product of two factors, then we get the third factor.
$\frac{{(x}^{3}+3{x}^{2}\xe2\u02c6\u20195x\xe2\u02c6\u201915)}{({x}^{2}\xe2\u02c6\u20195)}$ = (x + 3)
âˆ´( x + 3) is the third factor from which we get x = - 3 as the third zero.
12. If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.
1) Group the tangents of the circle according to the sides of the parallelogram.
2) A parallelogram with all the four equal sides is a rhombus.
1) Equate the opposite sides of the parallelogram.
2) Equate the tangents of the circle from four external points.
3) Add the equations of tangents according to the sides of the parallelogram.
4) Prove that the adjacent sides of the parallelogram are equal.
5) If the adjacent sides of the parallelogram are equal, then all the four sides of the parallelogram are equal.
6) If all the four sides of the parallelogram are equal, then it is a square.
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 a circle are equal. If A, B, C and D are the external points,
âˆ´ DR = DS â€¦ (3)
CR = CQ â€¦ (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 tangents, 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 get
2AB = 2BC
âˆ´ AB = BC
âˆ´ AB = BC = CD = DA
Therefore, ABCD is a parallelogram that has all the sides of equal length.
Hence, the parallelogram ABCD is also a rhombus.
13. Without using trigonometric tables, find the value of the following expression:
$\frac{\mathrm{sec}(90\xc2\xb0\xe2\u02c6\u2019\mathrm{\xce\xb8}).\mathrm{cos}\mathrm{ec}\mathrm{\xce\xb8}\xe2\u02c6\u2019\mathrm{tan}(90\xc2\xb0\xe2\u02c6\u2019\mathrm{\xce\xb8}).\mathrm{cot}\mathrm{\xce\xb8}+{\mathrm{cos}}^{2}25\xc2\xb0+{\mathrm{cos}}^{2}65\xc2\xb0}{3\mathrm{tan}27\xc2\xb0.\mathrm{tan}63\xc2\xb0}$
OR
Find the value of cosec 30Â°, geometrically.
1) Use appropriate identities for the trigonometric ratios of complementary angles.
2) Use appropriate trigonometric identities.
OR
1) Draw a perpendicular from the vertex to the base of an equilateral triangle.
2) Prove that the triangles formed are congruent.
1) Use the trigonometric ratios of complementary angles and substitute.
2) Apply the trigonometric identities and simplify.
OR
1) Consider an equilateral triangle and draw a perpendicular from the vertex to the base.
2) Prove that the triangles are congruent according to RHS congruency rule.
3) Compare the corresponding parts of the congruent triangles and prove that the the perpendicular bisects the base and the vertical angle.
4) Apply secant to the vertical angles of the right angled triangle formed and find the value of secant 30Â°.
$\frac{\mathrm{sec}(90\xc2\xb0\xe2\u02c6\u2019\mathrm{\xce\xb8}).\mathrm{cos}\mathrm{ec}\mathrm{\xce\xb8}\xe2\u02c6\u2019\mathrm{tan}(90\xc2\xb0\xe2\u02c6\u2019\mathrm{\xce\xb8}).\mathrm{cot}\mathrm{\xce\xb8}+{\mathrm{cos}}^{2}25\xc2\xb0+{\mathrm{cos}}^{2}65\xc2\xb0}{3\mathrm{tan}27\xc2\xb0.\mathrm{tan}63\xc2\xb0}$
= $\frac{\mathrm{cos}\mathrm{ec}\mathrm{\xce\xb8}.\mathrm{cos}\mathrm{ec}\mathrm{\xce\xb8}\xe2\u02c6\u2019\mathrm{cot}\mathrm{\xce\xb8}.\mathrm{cot}\mathrm{\xce\xb8}+{\mathrm{sin}}^{2}65\xc2\xb0+{\mathrm{cos}}^{2}65\xc2\xb0}{3\mathrm{tan}27\xc2\xb0.\mathrm{tan}63\xc2\xb0}$
= [âˆµsec (90Â° - Î¸) = cosec Î¸, tan (90Â° - Î¸) = cot Î¸, also ${\mathrm{cos}}^{2}25={\mathrm{cos}}^{2}(90\xc2\xb0\xe2\u02c6\u201965\xc2\xb0)={\mathrm{sin}}^{2}65\xc2\xb0$]
= $\frac{{\mathrm{cos}\mathrm{ec}}^{2}\mathrm{\xce\xb8}\xe2\u02c6\u2019{\mathrm{cot}}^{2}\mathrm{\xce\xb8}+{\mathrm{sin}}^{2}65\xc2\xb0+{\mathrm{cos}}^{2}65\xc2\xb0}{3\mathrm{tan}27\xc2\xb0.\mathrm{cot}27\xc2\xb0}$
= $\frac{(1+1)}{3}$ = $\frac{2}{3}$
OR
Find the value of cosec 30Â°, geometrically.
Consider an equilateral triangle ABC.
Each angle of an equilateral triangle is 60Â°.
Therefore, âˆ A = âˆ B = âˆ C = 60Â°
Let a perpendicular AD be drawn from A to side BC.
Now, in Î”ABD and Î”ACD
âˆ ADB = âˆ ADC (each of them are right angles)
AB = AC (â–³ABC is an equilateral triangle)
AD = AD (Common side)
Î”ABD â‰… Î”ACD (By RHS congruency criterion)
âˆ´ BD = DC (CPCT)
âˆ BAD = âˆ CAD = 30Âº
Let the length of AB be 2a.
Therefore AB = BC = CD = 2a
BD = half of BC = a
We know that
cosec 30Â° = $\frac{1}{(\mathrm{sin}30\xc2\xb0)}$ = 2
Therefore, sin 30Â° = $\frac{\mathrm{BD}}{\mathrm{AB}}$ = $\frac{a}{2a}$= $\frac{1}{2}$.
14. Find the value of k for which the following pair of linear equations has infinitely many solutions: 2x + 3y = 7; (k âˆ’ 1) x + (k + 2)y = 3k.
1) Substitute the exact coefficients of the linear equations in the condition of the coefficients for infinite solutions.
1) Compare the given linear equations to the standard form of the linear equations.
2) Substitute the coefficients of the equation in the conditions for the infinite solutions of the linear equations.
3) Simplify to find the value of the unknown term.
The linear equations can be rearranged and written as 2x + 3y âˆ’ 7 = 0â€¦ (1)
and (k âˆ’ 1)x + (k + 2)y âˆ’ 3k = 0 â€¦ (2)
A pair of linear equations represented by ${a}_{1}x+{b}_{1}y+{c}_{1}=0$ and ${a}_{2}x+{b}_{2}y+{c}_{2}=0$ has infinite solutions, if $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}.$
From equations 1 and 2,
$\frac{2}{(k\xe2\u02c6\u20191)}$ = $\frac{3}{(k+2)}$= $\frac{(\xe2\u02c6\u20197)}{(\xe2\u02c6\u20193k)}$
⇒ 6k = 7(k âˆ’ 1)
⇒ 6k = 7k âˆ’ 7
⇒ 7 = 7k âˆ’ 6k
⇒ k = 7
Hence, the value of k is 7.
15. In an A.P., first term is 2, the last term is 29 and sum of the terms is 155. Find the common difference of the A.P.
1) Use the appropriate formula for the sum of the terms of the A.P.
2) Substitute the correct values of the terms in the ${n}^{\mathrm{th}}$term of the AP.
1) Substitute the values of the given terms in the formula for the sum of n terms of the A.P.
2) Find the number of terms of the A.P.
3) Find the common difference of the AP using the formula for the ${n}^{\mathrm{th}}$term of the A.P.
Let the first term of the A.P. be A and the common difference be d.
From the question we know that the first term, A = 2 and last term L = 29.
Sum of n terms of an A.P. is given by the equation ${S}_{n}=\frac{n}{2}(A+L)$
155 = $\frac{n}{2}$(2+29)
155 = $\frac{n}{2}$(31)
Solving the equation, we get n = 10.
${n}^{\mathrm{th}}$ term of an A.P. is given by ${A}_{n}$= A + (n âˆ’ 1) d.
âˆ´ 29 = 2 + (10 âˆ’ 1) × d
âˆ´ 9d = 29 âˆ’ 2 = 27
âˆ´ d = 3
Thus, the common difference of the A.P. is 3.
21. Construct a triangle PQR in which QR = 6 cm, âˆ Q = 60Â° and âˆ R = 45Â°. Construct another triangle similar to Î”PQR such that its sides are $\frac{5}{6}$ of the corresponding sides of Î”PQR.
1) Ray drawn on the other side of vertex P should make an acute angle with the base.
2) Draw lines at the fifth division which is exactly parallel to the line drawn at the sixth division.
1) Construct a triangle ABC with the given measurements.
2) Draw a ray on the other side of the vertex making an acute angle with QR.
3) Mark 6 equal divisions on the ray.
4) Join the sixth division to the other vertex of the base.
5) Draw a line from the fifth division which is parallel to the line drawn in the previous step.
6) Draw a line parallel to the third side from the point where the previous line intersects the base.
7) The required triangle is obtained.
Step 1:
Draw a Î”PQR with side QR = 6 cm, âˆ Q = 60Â° and âˆ R = 45Â°.
Step 2:
Draw a ray QX at an acute angle with QR on the opposite side of vertex P.
Step 3:
Mark 6 points (since 6 is greater between 5 and 6) ${Q}_{1},{Q}_{2},{Q}_{3},{Q}_{4},{Q}_{5},\mathrm{and}{Q}_{6}$on QX.
Step 4:
Join ${Q}_{6}R$. Draw a line through ${Q}_{5}$parallel to ${Q}_{6}R$ such that it intersects QR at R'.
Step 5:
Draw a line parallel to PR from R'such that it intersects QP at P'.
Thus, Î”P'QR' is the required triangle.
22. Cards bearing numbers 1, 3, 5, ..., 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing
(i) a prime number less than 15.
(ii) a number divisible by 3 and 5.
1) Find the prime numbers less than 15
2) FInd the numbers which are divisible by both 3 and 5.
1) Find the total number of cards with odd numbers less than 35.
2) Find the total number of primes less than 15.
3) Find the probability of getting a card with prime numbers less than 15 out of the total number of cards with odd numbers less than 35.
4) Find the total number of cards bearing a number which is divisible by both 3 and 5 i.e. divisible by 15.
5) Find the probability of getting a card bearing a number which is divisible by 15 out of the total number of cards.
Cards used are 1, 3, 5, â€¦ 35
That means there are a total of 18 cards on the table.
(i) Prime numbers less than 15 are 3, 5, 7, 11 and 13.
Therefore total number of prime numbers less than 15 is 5.
Hence the required probability is $\frac{5}{18}.$
(ii) Numbers divisible by 3 and 5 are 15 and 30. But 30 is not a valid card number. âˆ´
âˆ´ Probability of getting a number divisible by 3 and 5 = $\frac{1}{18}$.
Thus, the probability of getting a card bearing a number divisible by 3 and 5 is $\frac{1}{18}$.
23. If the point P (m, 3) lies on the line segment joining the points A ($\xe2\u02c6\u2019\frac{2}{5}$, 6 ) and B (2, 8), find the value of m.
1) Substitute the values of the appropriate coordinates in the formula for the area of the triangle.
1) The points A, P and B lie on a straight line.
2) So, the area of the triangle formed by the points is zero.
3) Find the area of the triangle and equate it to zero to find the value of m.
The points A, P and B are on a straight line and hence collinear.
Area of a triangle with vertices A(${x}_{1},{y}_{1}$), B(${x}_{2},{y}_{2})$and C(${x}_{3},{y}_{3}$) =
$\frac{1}{2}$[${x}_{1}({y}_{2}\xe2\u02c6\u2019{y}_{3})+{{x}_{2}({y}_{3}\xe2\u02c6\u2019{y}_{1})+{x}_{3}({y}_{1}\xe2\u02c6\u2019{y}_{2})]}_{}$
Since all the three points are on a straight line, the area of the triangle formed by the points is 0.
Therefore,
$\frac{1}{2}[\frac{(\xe2\u02c6\u20192)}{5}(3\xe2\u02c6\u20198)+m(8\xe2\u02c6\u20196)+2(6\xe2\u02c6\u20193\left)\right]=0$
$[\frac{(\xe2\u02c6\u20192)}{5}(\xe2\u02c6\u20195)+m(\xe2\u02c6\u20192)+2(3\left)\right]=0$
2 + 2m + 6 = 0
2m = âˆ’ 8
m = âˆ’ 4.
24. Point P divides the line segment joining the points A (2, 1) and B (5, âˆ’8) such that $\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{1}{3}$. If P lies on the line 2x âˆ’ y + k = 0, find the value of k.
1) Consider the fraction as the ratio in which the point P divides the line segment AB.
2) Coordinates of P should satisfy the linear equation 2x - y + k = 0 as it lies on the line.
1) The point P divides the line segment joining the points A and B in the ratio of 1:3.
2) Use section formula and find the coordinates of the point P.
3) Coordinates of the point P should satisfy the linear equation 2x - y + k = 0 as it lies on the line.
4) Find the value of k by substituting the coordinates of the point P for x and y in the equation.
Since the ratio of $\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{1}{3},$ assume that the point P divides the line AB in the ratio 1:2
Coordinates of the point P that divides the line joining A(${x}_{1},{y}_{1}),$B(${x}_{2},{y}_{2}$) in the ratio of m:n are
$[\frac{m{x}_{2}+n{x}_{1}}{m+n},\frac{m{y}_{2}+n{y}_{1}}{m+n}]$
Therefore,
⇒$[\frac{(1\times 5+2\times 2)}{(1+2)},\frac{(1\times \xe2\u02c6\u20198+2\times 1)}{(1+2)}]$
= (3, - 2)
Since the point lies on the line 2x âˆ’ y + k = 0, substituting the values for x and y we get k = -8.
25. In figure, the boundary of shaded region consists of four semicircular arcs, two smallest being equal. If diameter of the largest is 14 cm and that of the smallest is 3.5 cm, calculate the area of the shaded region. [Use Ï€ = $\frac{22}{7}$]
Or
Find the area of shaded region in figure 5, if AC = 24 cm, BC = 10 cm and O is the centre of the circle. [Use Ï€ = 3.14]
1) Calculate the correct radius of the shaded portion on the other side of the semicircular region.
OR
1) Hypotenuse of the right angled triangle is the diameter of the circle
1) Find the radius of all the semicircular regions of the figure.
2) Find the area of the bigger semicircle and the area of the middle shaded semicircle.
3) Find the area of the unshaded smaller semi circles.
4) Subtract the sum of the areas of the smaller semicircles from the sum of the areas of the shaded semicircles.
OR
1) Find the hypotenuse of the right angled triangle using the Pythgoras theorem.
2) Hypotenuse of the right angled triangle is the diameter of the circle.
3) Find the area of the right angled triangle and the area of the semicircle.
4) Subtract the area of the right angled triangle from the area of the semicircular region to find the area of the shaded region of the circle.
Part 1
Let ${r}_{1}$ the radius of the largest shaded semicircle, ${r}_{2}$be the radius of the smallest unshaded semicircle and ${r}_{3}$be the radius of the shaded semicircle with diameter BD.
As AE = 14 cm and AB = DE = 3.5 cm,
Therefore, ${r}_{2}$= $\frac{3.5}{2}$ cm
Also, ${r}_{3}$= ${r}_{1}\xe2\u02c6\u20192{r}_{2}$= 7 - 2 × $\frac{3.5}{2}$ = 3.5 cm
The area of the shaded region = (Area of the semicircle with radius ${r}_{1}$) + (Area of the semicircle with radius ${r}_{3}$âˆ’(2 × Area of semicircle with radius ${r}_{2}$)
$\frac{1}{2}\mathrm{\xcf\u20ac}{\left({r}_{1}\right)}^{2}+\frac{1}{2}\mathrm{\xcf\u20ac}{\left({r}_{3}\right)}^{2}\xe2\u02c6\u20192\times \frac{1}{2}\mathrm{\xcf\u20ac}{\left({r}_{2}\right)}^{2}$
= $\frac{1}{2}$Ï€{${\left({r}_{1}\right)}^{2}$ + ${\left({r}_{3}\right)}^{2}$ - 2 ${\left({r}_{2}\right)}^{2}$}
=$\frac{1}{2}\times \frac{22}{7}\{{\left(7\right)}^{2}+{\left(3.5\right)}^{2}\xe2\u02c6\u20192\times {\left(\frac{3.5}{2}\right)}^{2}$
=$\frac{11}{7}\{49+12.25\xe2\u02c6\u2019\frac{12.25}{2}\}$
= $\frac{11}{7}$ × 55.125 = 86.625
Thus, the area of the shaded region is 86.625 ${\mathrm{cm}}^{2}$.
Or
In the given figure, OACB forms a semicircle.
âˆ´ âˆ ACB = 90Â°
Therefore, Î”ABC is a right-angled triangle.
According to Pythagoras theorem,
${\mathrm{AB}}^{2}={\mathrm{AC}}^{2}+{\mathrm{BC}}^{2}$
⇒ ${\mathrm{AB}}^{2}$= ${(24)}^{2}+{\left(10\right)}^{2}$
⇒ ${\left(\mathrm{AB}\right)}^{2}$= ${\left(576\right)}^{2}+{\left(100\right)}^{2}$
⇒ ${\left(\mathrm{AB}\right)}^{2}=676{\mathrm{cm}}^{2}$
⇒ AB = $\sqrt{676}$
âˆ´ AB = 26 cm
⇒ OA = $\frac{26}{2}$ = 13 cm
The area of the shaded region is given by: Area of semicircle âˆ’ Area of Î”ABC
The shaded area = $\left[\frac{1}{2}{\mathrm{\xcf\u20ac}\left(\mathrm{OA}\right)}^{2}\right]\xe2\u02c6\u2019[\frac{1}{2}\times \mathrm{AC}\times \mathrm{BC}]$
Calculating the shaded area by substituting OA = 13 cm, AC = 24 cm and BC = 10 cm
We get shaded area = 145.57 ${\mathrm{cm}}^{2}$
Thus, the area of the shaded region is 145.57 ${\mathrm{cm}}^{2}$.
26. A milk container is made of metal sheet in the shape of frustum of a cone whose volume is 10459 $\frac{3}{7}{\mathrm{cm}}^{3}$. The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs. 1.40 per square centimetre. Use Ï€ = $\frac{22}{7}.$
Or
A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of base of the cone is 21 cm and its volume is $\frac{2}{3}$ of the volume of the hemisphere, calculate the height of the cone and the surface area of the toy. Ï€ = $\frac{22}{7}$.
1) Assume that the milk container in the shape of a frustum of frustum of a cone is open at the smaller end.
2) Surface area of the toy does not include the base area of the cone and the area of the flat surface of the hemisphere.
1) Find the height of the milk container from the volume using the formula for the volume of the frustum of a cone.
2) Find the curved surface area of the frustum of the cone and the area of the base.
3) Find the sum of both the areas to find the area of the metal sheet required to make the milk container.
4) Find the cost of the sheet required by multiplying the rate of the sheet with the area of the sheet.
OR
1) Find the height of the cone from the relation between the volume of the cone and the volume of the hemisphere.
2) Find the lateral surface area of the conical part and the surface area of the hemispherical part of the toy.
3) Sum of both the areas is the surface area of the toy.
The shape of the milk container as given in the figure.
Given that the volume of the container, V = 10459 $\frac{3}{7}{\mathrm{cm}}^{3}$= $\frac{73216}{7}{\mathrm{cm}}^{3}$
The radii of the lower and upper circular ends are ${r}_{1}$ = 20 cm, ${r}_{2}$= 8 cm.
Let the height of the container be h.
Volume of a frustum of a cone, V= $\frac{1}{3}\mathrm{\xcf\u20ach}({{r}_{1}}^{2}+{{r}_{2}}^{2}+{r}_{1}{r}_{2})$
The volume (V) of a frustum of cone is given by:
$\frac{73216}{7}$ = $\frac{1}{3}\mathrm{\xcf\u20ach}({20}^{2}+{20}^{2}+{7}^{2}+20\times 7)$
On calculating, we get h = 16 cm
To find the metal sheet or the surface area of the cone, find the sum of the curved surface area of the frustum of the cone and its base area.
Total surface area = Ï€l (${r}_{1}+{r}_{2})$
The metal sheet required for making the container is equal to the sum of the curved surface area of the frustum of the cone and its base area.
âˆ´Total surface area of the container where l is the slant height. = Ï€l (${r}_{1}$+${r}_{2}$) + Ï€${r}^{2}$
l =$\sqrt{{h}^{2}+{({r}_{1}\xe2\u02c6\u2019{r}_{2})}^{2}}$
Substituting h = 16 cm, ${r}_{1}$= 20 cm and ${r}_{2}$= 8 cm, we get l = 20 cm
Therefore total surface area = Ï€l (${r}_{1}+{r}_{2})$+ Ï€${{r}_{2}}^{2}$
(${r}_{1}+{r}_{2})+\mathrm{\xcf\u20ac}{{r}_{2}}^{2}$= Ï€ × 20 (8 + 20) ++ Ï€ ${\left(8\right)}^{2}$= $\frac{22}{7}\times 624{\mathrm{cm}}^{2}$
Cost of 1 ${\mathrm{cm}}^{2}$of sheet metal is about Rs. 1.40
Hence the total cost is $\frac{22}{7}\times 624{\mathrm{cm}}^{2}$×1.4 = Rs.2745.60
OR
Part 2
A picture of the toy is illustrated as given above. Let the height of the cone of the hemisphere be h cm.
Here both the radius of the cone and the hemisphere are same and is equal to 21 cm.
Volume of a cone = $\frac{1}{3}$${r}^{2}$ h
Similarly, volume of a hemisphere = $\frac{1}{2}$× $\frac{4}{3}$ × Ï€ ${r}^{3}$
Given that,
Volume of the cone = $\frac{2}{3}$ × Volume of the hemisphere
Therefore,
$\frac{1}{3}$ × Ï€ ${r}^{2}$h = $\frac{2}{3}$ × $\frac{1}{2}$ × $\frac{4}{3}$ × Ï€ ${r}^{3}$
Substituting for r and solving for h we get h = 28 cm.
The total surface area of the toy = Surface area of the cone + Surface area of the hemisphere.
= Ï€rl + 2 Ï€${r}^{2}$
Here l is the slant height. l =âˆš$\sqrt{({h}^{2}+{r}^{2})}$âˆš
Total surface area = Ï€r $\sqrt{({h}^{2}+{r}^{2})}$ + 2 Ï€${r}^{2}$
Substituting the values of h and r, total surface area = $\frac{22}{7}\times 21\times \sqrt{({28)}^{2}+{\left(21\right)}^{2}}+2\times \frac{22}{7}\times {\left(21\right)}^{2}$ = 5082 ${\mathrm{cm}}^{2}.$
27. Prove that, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Using the above, prove the following:
Point D is the mid-point of the side BC of a right triangle ABC, right angled at C. Prove that 4${\mathrm{AD}}^{2}=4{\mathrm{AC}}^{2}$+ ${\mathrm{BC}}^{2}$
1) Compare the corresponding angles of the triangles to prove that they are similar.
2) Consider the correct corresponding sides of the similar triangles.
1) Draw BDâŠ¥AC.
2) Prove that triangle ABC is similar to triangle BDC.
3) Obtain an equation by equating the ratio of the corresponding sides.
4) Prove that triangle ABC is similar to triangle ADB.
5) Obtain an equation by equating the ratio of the corresponding sides.
6) Multiply both the equations and simplify to prove that the square of the hypotenuse is equal to the sum of the squares of the other two sides.
7) Draw the triangle ABC with D as the midpoint of BC and angle C as right angle.
8) Apply Pythagoras theorem which we proved above in triangle ACD.
9) Substitute CD as $\left(\frac{\mathrm{BC}}{2}\right)$ and simplify to get the required proof.
Consider a â–³ABC that is right angled at D.
To prove : ${\mathrm{AC}}^{2}={\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}$
Construction: A perpendicular BD is dropped on the side AC to meet AC at D.
Proof: In Î”ABC and Î”BDC.
âˆ ACB = âˆ DCB (Common angle of two triangles)
âˆ ABC = âˆ BDC = 90Â° (Given)
âˆ´ Î”ABC âˆ¼ Î”BDC (AA criteria)
Corresponding sides of similar triangles are proportional.
âˆ´ $\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{BC}}{\mathrm{DC}}$
⇒ ${\mathrm{BC}}^{2}=\mathrm{AC}\times \mathrm{DC}$ ----- (1)
Similarly let us consider the triangles, Î”ABC and Î”ADB.
âˆ BAC = âˆ BAD (Common angle for the two triangles)
âˆ ABC = âˆ ADB = 90Â° (Given)
âˆ´Î”ABC âˆ¼ Î”ADB ( AA criteria)
Corresponding sides of similar triangles are proportional.
âˆ´$\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AB}}$
⇒ ${\mathrm{AB}}^{2}=\mathrm{AC}\times \mathrm{AD}$----- (2)
Adding equation (1) and (2):
${\mathrm{BC}}^{2}+{\mathrm{AB}}^{2}=(\mathrm{AC}\times \mathrm{DC})+(\mathrm{AC}\times \mathrm{AD})$
⇒ ${\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}$ = AC × (DC + AD)
⇒ ${\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}$= AC × AC [Since AC = DC + AD]
⇒ ${\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}={\mathrm{AC}}^{2}$
Hence, in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Proving that ${4\mathrm{AD}}^{2}=4{\mathrm{AC}}^{2}+{\mathrm{BC}}^{2}$:
Using the theorem proved above in Î”ACD
${\mathrm{AD}}^{2}={\mathrm{AC}}^{2}+{\mathrm{CD}}^{2}$
⇒ ${\mathrm{AD}}^{2}={\mathrm{AC}}^{2}+{\left(\frac{\mathrm{BC}}{2}\right)}^{2}$[Since D is the mid-point of BC]
⇒ ${\mathrm{AD}}^{2}={\mathrm{AC}}^{2}+\frac{{\left(\mathrm{BC}\right)}^{2}}{4}$
⇒ ${4\mathrm{AD}}^{2}=4{\mathrm{AC}}^{2}+{\mathrm{BC}}^{2}$
Hence proved.
28. Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46, find the integers.
Or
The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers.
1) Use quadratic formula to find the roots of the quadratic equation.
2) Consider the positive integer root as one of the integers to be found.
OR
1) Use quadratic formula to find the roots of the quadratic equation.
2) Consider the correct pair of smaller and larger numbers.
1) Assume the consecutive numbers as x, x + 1 and x + 2.
2) Apply the given conditions on the assumed numbers to get a quadratic equation.
3) Solve the quadratic equation and find the roots.
4) Find the numbers using the appropriate root of the quadratic equation.
OR
1) Assume the smaller number as x and find the larger number according to the given condition.
2) Apply the given condition on the assumed numbers to get a quadratic equation.
3) Solve the quadratic equation and find the roots.
4) Find the appropriate pair of numbers using the roots of the equation.
Assume three consecutive positive integers as x, x + 1 and x + 2.
According to the given condition:
${x}^{2}+(x+1)(x+2)=46$
⇒ ${x}^{2}+{x}^{2}+3x+2=46$
⇒ ${2x}^{2}+3x\xe2\u02c6\u201944=0$
The roots of a quadratic equation ${\mathrm{ax}}^{2}+{\mathrm{bx}}^{2}+c=0$ are $\frac{(\xe2\u02c6\u2019b\xc2\pm \sqrt{{(b}^{2}\xe2\u02c6\u20194\mathrm{ac})})}{2a}$.
Finding the roots of the ${2x}^{2}+3x\xe2\u02c6\u201944=0:$
$x=\frac{(\xe2\u02c6\u20193)\xc2\pm \sqrt{({3)}^{2}\xe2\u02c6\u20194(2\left)\right(\xe2\u02c6\u201944)}}{2\left(2\right)}$
= $\frac{(\xe2\u02c6\u20193+19)}{4}\mathrm{or}\frac{(\xe2\u02c6\u20193\xe2\u02c6\u201919)}{4}$
= 4 or $\frac{(\xe2\u02c6\u201922)}{4}$
It is given that x is a positive integer and hence we consider the value of x as 4. Hence the three numbers are x, x + 1 and x + 2 i.e., 4, 5 and 6.
OR
Let the smaller number be x.
According to the question the larger number is (2x âˆ’ 5)
According to the given condition:
${(2x\xe2\u02c6\u20195)}^{2}\xe2\u02c6\u2019{x}^{2}=88$
⇒ ${4x}^{2}+25\xe2\u02c6\u201920x\xe2\u02c6\u2019{x}^{2}=88$
⇒ ${3x}^{2}\xe2\u02c6\u201920x\xe2\u02c6\u201963=0$
Roots of a quadratic equation ${\mathrm{ax}}^{2}+\mathrm{bx}+c=0$are $\frac{(\xe2\u02c6\u2019b)\xc2\pm \sqrt{({b}^{2}\xe2\u02c6\u20194\mathrm{ac})}}{2a}$.
âˆ´ Solving for the equation ${3x}^{2}\xe2\u02c6\u201920x\xe2\u02c6\u201963=0$
$x=\frac{\xe2\u02c6\u2019(\xe2\u02c6\u201920)\xc2\pm \sqrt{{\left(\xe2\u02c6\u201920\right)}^{2}\xe2\u02c6\u20194\left(3\right)(\xe2\u02c6\u201963)}}{2\left(3\right)}$
= $\frac{(20+34)}{6},\mathrm{or}\frac{(20\xe2\u02c6\u201934)}{6}$
= 9 or $\xe2\u02c6\u2019\frac{7}{3}$
When x = 9, the smaller number is 9, and the larger number is 2(9) âˆ’ 5 = 13
When x = $\xe2\u02c6\u2019\frac{7}{3}$, the smaller number is $\xe2\u02c6\u2019\frac{7}{3}$ and the larger number is
x = $2(\xe2\u02c6\u2019\frac{7}{3})\xe2\u02c6\u20195=\xe2\u02c6\u2019\frac{29}{3}$
The two numbers are inconsistent as the smaller number according to our analysis is actually bigger than the calculated smaller number as $\xe2\u02c6\u2019\frac{29}{3}<\xe2\u02c6\u2019\frac{7}{3}$,
âˆ´ The second root is not considered and the required numbers are 9 and 13.
29. From the top of a 7 m high building, the angle of elevation of the top of a tower is 60Â° and the angle of depression of the foot of the tower is 30Â°. Find the height of the tower.
1) Find the correct alternate angle of the angle of the angle of depression.
2) Apply appropriate trigonometric ratio.
1) Draw the figure showing the building, tower and the angles of elevation and depression.
2) Consider the alternate angle of the angle of depression.
3) Apply tangent to the alternate angle of the angle of the depression and find the distance between the tower and the building.
4) Apply tangent to the angle of the elevation and find the height of the tower.
AB is the building and A is the viewing point. CD is the tower with C as the top of the tower.
AB = DE = 7 m
Also, BD = AE
In Î”ABD:
tan 30Â° = $\frac{\mathrm{AB}}{\mathrm{BD}}$
BD = 7$\sqrt{3}$âˆš
In Î”ACE:
tan 60Â° = $\frac{\mathrm{CE}}{\mathrm{AE}}$
CE = 7 $\sqrt{3}$×âˆš$\sqrt{3}$
CE = 21 m
âˆ´ CD = CE + AB = (21 + 7) = 28 m
Hence, the height of the tower is 28 m.
30. Find the mean, mode and median of the following frequency distribution:
1) Find the correct model class of the frequency distribution.
2) Find the correct median class of the frequency distribution.
3) Find the mean using the mid values of the classes.
1) Find the mid value of the classes.
2) Find the products of the frequencies and the corresponding mid values of the classes.
3) Find the sum of the frequencies and the sum of the products of the frequencies and the mid values.
4) Divide the sum of the products of the frequencies and the mid values by the sum of the frequencies to find the mean of the data.
5) Find the median class of the frequency distribution.
6) Use the formula to find the median of the frequency distribution.
7) Find the model class of the data.
8) Use the formula to find the mode of the data.
Frequency distribution table for the given data:
Mean = $\frac{(\xe2\u02c6\u2018{f}_{i}{x}_{i})}{\left({f}_{i}\right)}=\frac{2850}{80}=35.625$
âˆ´Mean = 35.625
Since the maximum class frequency is 20, the modal class is 30âˆ’40.
The formula for finding mode is
$l+\frac{({f}_{1}\xe2\u02c6\u2019{f}_{0})-}{({2f}_{1}\xe2\u02c6\u2019{f}_{0}\xe2\u02c6\u2019{f}_{2})-}\times h$
l = 30, h = 10, ${f}_{1}$= 20, ${f}_{0}$ = 15, ${f}_{2}$= 12
Substituting for variables
30 + $\frac{(20\xe2\u02c6\u201915)}{(2\times 20\xe2\u02c6\u201915\xe2\u02c6\u201912)}\times 10=33.85$
âˆ´Mode = 33.85
The cumulative frequency distribution table:
n = 80
$\frac{n}{2}$ value lies in the class 30âˆ’40.
âˆ´ Median class is 30âˆ’40.
l = 30, cf = 30, f = 20, h = 10
Median = $l+\frac{({f}_{1}\xe2\u02c6\u2019{f}_{0})}{f}\times h$
= 30 + $\frac{(40\xe2\u02c6\u201930)}{20}\times 10=35$
âˆ´ Median = 35
Hence, from the calculations we get that the mean, mode and median of the given frequency distribution as 35.625, 33.85 and 35 respectively.
Section A | Section B | Section C | Section D |
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