CBSE Class 10 - Maths

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 $\sqrt{3}$√is irrational.

Let us consider that $\sqrt{3}$ is a rational number.

If we can find two integers a, b (b ≠ 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 ≠ 0) such that (2$\sqrt{3}-1)$ = $\frac{a}{b}$.

⇒$\sqrt{3}$ = $\frac{1}{2}$($\frac{a}{b}$ -1).

Since a and b are integers, $\frac{1}{2}$($\frac{a}{b}$ - 1) is also rational.

Therefore, $\sqrt{3}$√should be rational.

This contradicts the fact that $\sqrt{3}$ is irrational. Hence, the assumption that "2$\sqrt{3}-1$is rational" is false. Hence, (2$\sqrt{3}$- 1) is irrational.

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 ΔABC, D is the mid-point of side BC.

∴ BD = CD = $\frac{1}{2}$BC

In ΔACD,

${\mathrm{AD}}^2={\mathrm{AC}}^2+{\mathrm{CD}}^2$ (Pythagoras theorem)

⇒ ${\mathrm{AD}}^2={\mathrm{AC}}^2+{(\frac{1}{2}\mathrm{BC})}^2$

⇒ ${\mathrm{AD}}^2={\mathrm{AC}}^2+\frac{1}{4}{\mathrm{BC}}^2$

⇒ ${4\mathrm{AD}}^2=4{\mathrm{AC}}^2+{\mathrm{BC}}^2$

⇒ ${\mathrm{BC}}^2=4{\mathrm{AD}}^2-4{\mathrm{AC}}^2$--------- (1)

In ΔABC, according to Pythagoras theorem,

${\mathrm{AB}}^2={\mathrm{AC}}^2+{\mathrm{BC}}^2$

⇒ ${\mathrm{AB}}^2={\mathrm{AC}}^2+(4{\mathrm{AD}}^2-4{\mathrm{AC}}^2)$ [from equation (1)]

∴ ${\mathrm{AB}}^2=4{\mathrm{AD}}^2-3{\mathrm{AC}}^2$

Hence Proved.

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

$\frac{\tan A}{(1-\frac{1}{\tan A})}$ + $\frac{\left(\frac{1}{\tan A}\right)}{1-\tan A}$

= $\frac{\frac{\tan A}{(\tan A-1)}}{\tan A}$ + $\frac{1}{\tan A(1-\tan A)}$

= $\frac{{\tan }^{2}A}{(\tan A-1)}$ - $\frac{1}{\tan A(\tan A-1)}$

= $\frac{{(\tan }^{3}A-1)}{\tan A(\tan A-1)}$

= $\frac{(\tan A-1)({\tan }^{2}A+\tan A+1)}{\tan A(\tan A-1)}$

= $\frac{{(\tan }^{2}A+\tan A+1)}{\tan A}$

= tan A + 1 + $\frac{1}{\tan A}$

= 1 + tan A + $\frac{1}{\tan A}$

= R.H.S

Hence proved.

OR

Prove the following:

(cosec A - cos A)(sec A - cos A) = $\frac{1}{(\tan A+\cot A)}$

$(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A)=\frac{1}{(\tan A+\cot A)}$

$\left(\frac{1-{\sin }^{2}A}{\sin A}\right)$$\left(\frac{1-{\cos }^{2}A}{\cos A}\right)=\frac{1}{(\tan A+\cot A)}$

sin A cos A = $\frac{1}{(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A})}$

sin A cos A = $\frac{\cos A\sin A}{{\sin }^{2}A+{\cos }^{2}A}$

Since, ${\sin }^{2}A+{\cos }^{2}A$ = 1

sin A cos A = cos A sin A

Hence, LHS = RHS.

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 = −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 …

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.

∴ x + y = 2y − 3

⇒ x + y − 2y = −3

⇒ x − y = −3 … (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 − (− 3)

⇒ x = 4

Substituting the value of x in equation (1):

4 − y = − 3

⇒ y = 4 + 3 = 7

Hence, the fraction is $\frac{4}{7}.$

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) = $x^3+3x^2-5x-15$.

So, (x - $\sqrt{5}$) and (x +√$\sqrt{5}$) 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.

$\frac{{(x}^3+3x^2-5x-15)}{(x^2-5)}$ = (x + 3)

∴( x + 3) is the third factor from which we get x = - 3 as the third zero.

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 … (1)

BC = AD … (2)

Lengths of tangents drawn from an external point to a circle are equal. If A, B, C and D are the external points,

∴ DR = DS … (3)

CR = CQ … (4)

BP = BQ … (5)

AP = AS … (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)

∴ CD + AB = AD + BC

From equations (1) and (2), we get

2AB = 2BC

∴ AB = BC

∴ 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.

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 30°.

$\frac{\sec (90°-\theta ).\cos \mathrm{ec}\theta -\tan (90°-\theta ).\cot \theta +{\cos }^{2}25°+{\cos }^{2}65°}{3\tan 27°.\tan 63°}$

= $\frac{\cos \mathrm{ec}\theta .\cos \mathrm{ec}\theta -\cot \theta .\cot \theta +{\sin }^{2}65°+{\cos }^{2}65°}{3\tan 27°.\tan 63°}$

= [∵sec (90° - θ) = cosec θ, tan (90° - θ) = cot θ, also ${\cos }^{2}25={\cos }^2(90°-65°)={\sin }^{2}65°$]

= $\frac{{\cos \mathrm{ec}}^2\theta -{\cot }^2\theta +{\sin }^{2}65°+{\cos }^{2}65°}{3\tan 27°.\cot 27°}$

= $\frac{(1+1)}{3}$ = $\frac{2}{3}$

OR

Find the value of cosec 30°, geometrically.

Consider an equilateral triangle ABC.

Each angle of an equilateral triangle is 60°.

Therefore, ∠A = ∠B = ∠C = 60°

Let a perpendicular AD be drawn from A to side BC.

Now, in ΔABD and ΔACD

∠ADB = ∠ADC (each of them are right angles)

AB = AC (△ABC is an equilateral triangle)

AD = AD (Common side)

ΔABD ≅ ΔACD (By RHS congruency criterion)

∴ BD = DC (CPCT)

∠BAD = ∠CAD = 30º

Let the length of AB be 2a.

Therefore AB = BC = CD = 2a

BD = half of BC = a

We know that

cosec 30° = $\frac{1}{(\sin 30°)}$ = 2

Therefore, sin 30° = $\frac{\mathrm{BD}}{\mathrm{AB}}$ = $\frac{a}{2a}$= $\frac{1}{2}$.

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 − 7 = 0… (1)

and (k − 1)x + (k + 2)y − 3k = 0 … (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,

$\frac{2}{(k-1)}$ = $\frac{3}{(k+2)}$= $\frac{(-7)}{(-3k)}$

⇒ 6k = 7(k − 1)

⇒ 6k = 7k − 7

⇒ 7 = 7k − 6k

⇒ k = 7

Hence, the value of k is 7.

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 − 1) d.

∴ 29 = 2 + (10 − 1) × d

∴ 9d = 29 − 2 = 27

∴ d = 3

Thus, the common difference of the A.P. is 3.

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 ΔPQR with side QR = 6 cm, ∠Q = 60° and ∠R = 45°.

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, ΔP'QR' is the required triangle.

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, … 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. ∴

∴ 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 $\frac{1}{18}$.

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($x_3,y_3$) =

$\frac{1}{2}$[$x_1(y_2-y_3)+{x_2(y_3-y_1)+x_3(y_1-y_2)]}_{}$

Since all the three points are on a straight line, the area of the triangle formed by the points is 0.

Therefore,

$\frac{1}{2}[\frac{(-2)}{5}(3-8)+m(8-6)+2(6-3\left)\right]=0$

$[\frac{(-2)}{5}(-5)+m(-2)+2(3\left)\right]=0$

2 + 2m + 6 = 0

2m = − 8

m = − 4.

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,

⇒$[\frac{(1\times 5+2\times 2)}{(1+2)},\frac{(1\times -8+2\times 1)}{(1+2)}]$

= (3, - 2)

Since the point lies on the line 2x − y + k = 0, substituting the values for x and y we get k = -8.

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, $r_3$= $r_1-2r_2$= 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 $r_3$−(2 × Area of semicircle with radius $r_2$)

$\frac{1}{2}\pi {\left(r_1\right)}^2+\frac{1}{2}\pi {\left(r_3\right)}^2-2\times \frac{1}{2}\pi {\left(r_2\right)}^2$

= $\frac{1}{2}$π{${\left(r_1\right)}^2$ + ${\left(r_3\right)}^2$ - 2 ${\left(r_2\right)}^2$}

=$\frac{1}{2}\times \frac{22}{7}\{{\left(7\right)}^2+{\left(3.5\right)}^2-2\times {\left(\frac{3.5}{2}\right)}^2$

=$\frac{11}{7}\{49+12.25-\frac{12.25}{2}\}$

= $\frac{11}{7}$ × 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.

∴ ∠ACB = 90°

Therefore, ΔABC 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 = $\sqrt{676}$

∴ AB = 26 cm

⇒ OA = $\frac{26}{2}$ = 13 cm

The area of the shaded region is given by: Area of semicircle − Area of ΔABC

The shaded area = $\left[\frac{1}{2}{\pi \left(\mathrm{OA}\right)}^2\right]-[\frac{1}{2}\times \mathrm{AC}\times \mathrm{BC}]$

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 ${\mathrm{cm}}^2$.

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= $\frac{1}{3}\mathrm{\pi h}({r_1}^2+{r_2}^2+r_1{r}_2)$

The volume (V) of a frustum of cone is given by:

$\frac{73216}{7}$ = $\frac{1}{3}\mathrm{\pi h}({20}^2+{20}^2+7^2+20\times 7)$

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 = πl ( $r_1+r_2)$

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.

∴Total surface area of the container where l is the slant height. = πl ( $r_1$+ $r_2$) + π $r^2$

l = $\sqrt{h^2+{(r_1-r_2)}^2}$

Substituting h = 16 cm, $r_1$= 20 cm and $r_2$= 8 cm, we get l = 20 cm

Therefore total surface area = πl ( $r_1+r_2)$+ π ${r_2}^2$

( $r_1+r_2)+\pi {r_2}^2$= π × 20 (8 + 20) ++ π ${\left(8\right)}^2$= $\frac{22}{7}\times 624{\mathrm{cm}}^2$

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 = $\frac{1}{2}$× $\frac{4}{3}$ × π $r^3$

Given that,

Volume of the cone = $\frac{2}{3}$ × Volume of the hemisphere

Therefore,

$\frac{1}{3}$ × π $r^2$h = $\frac{2}{3}$ × $\frac{1}{2}$ × $\frac{4}{3}$ × π $r^3$

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.

= πrl + 2 π $r^2$

Here l is the slant height. l =√ $\sqrt{(h^2+r^2)}$√

Total surface area = πr $\sqrt{(h^2+r^2)}$ + 2 π $r^2$

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 ${\mathrm{cm}}^2.$

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 BD⊥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 △ABC 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 ΔABC and ΔBDC.

∠ACB = ∠DCB (Common angle of two triangles)

∠ABC = ∠BDC = 90° (Given)

∴ ΔABC ∼ ΔBDC (AA criteria)

Corresponding sides of similar triangles are proportional.

∴ $\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{BC}}{\mathrm{DC}}$

⇒ ${\mathrm{BC}}^2=\mathrm{AC}\times \mathrm{DC}$ ----- (1)

Similarly let us consider the triangles, ΔABC and ΔADB.

∠BAC = ∠BAD (Common angle for the two triangles)

∠ABC = ∠ADB = 90° (Given)

∴ΔABC ∼ ΔADB ( AA criteria)

Corresponding sides of similar triangles are proportional.

∴ $\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AB}}$

⇒ ${\mathrm{AB}}^2=\mathrm{AC}\times \mathrm{AD}$----- (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 ΔACD

${\mathrm{AD}}^2={\mathrm{AC}}^2+{\mathrm{CD}}^2$

⇒ ${\mathrm{AD}}^2={\mathrm{AC}}^2+{\left(\frac{\mathrm{BC}}{2}\right)}^2$[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.

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:

$x^2+(x+1)(x+2)=46$

⇒ $x^2+x^2+3x+2=46$

⇒ ${2x}^2+3x-44=0$

The roots of a quadratic equation ${\mathrm{ax}}^2+{\mathrm{bx}}^2+c=0$ are $\frac{(-b\pm \sqrt{{(b}^2-4\mathrm{ac})})}{2a}$.

Finding the roots of the ${2x}^2+3x-44=0:$

$x=\frac{(-3)\pm \sqrt{({3)}^2-4(2\left)\right(-44)}}{2\left(2\right)}$

= $\frac{(-3+19)}{4}\mathrm{or}\frac{(-3-19)}{4}$

= 4 or $\frac{(-22)}{4}$

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 − 5)

According to the given condition:

${(2x-5)}^2-x^2=88$

⇒ ${4x}^2+25-20x-x^2=88$

⇒ ${3x}^2-20x-63=0$

Roots of a quadratic equation ${\mathrm{ax}}^2+\mathrm{bx}+c=0$are $\frac{(-b)\pm \sqrt{(b^2-4\mathrm{ac})}}{2a}$.

∴ Solving for the equation ${3x}^2-20x-63=0$

$x=\frac{-(-20)\pm \sqrt{{\left(-20\right)}^2-4\left(3\right)(-63)}}{2\left(3\right)}$

= $\frac{(20+34)}{6},\mathrm{or}\frac{(20-34)}{6}$

= 9 or $-\frac{7}{3}$

When x = 9, the smaller number is 9, and the larger number is 2(9) − 5 = 13

When x = $-\frac{7}{3}$, the smaller number is $-\frac{7}{3}$ and the larger number is

x = $2(-\frac{7}{3})-5=-\frac{29}{3}$

The two numbers are inconsistent as the smaller number according to our analysis is actually bigger than the calculated smaller number as $-\frac{29}{3}<-\frac{7}{3}$,

∴ The second root is not considered and the required numbers are 9 and 13.

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 ΔABD:

tan 30° = $\frac{\mathrm{AB}}{\mathrm{BD}}$

BD = 7 $\sqrt{3}$√

In ΔACE:

tan 60° = $\frac{\mathrm{CE}}{\mathrm{AE}}$

CE = 7 $\sqrt{3}$×√ $\sqrt{3}$

CE = 21 m

∴ CD = CE + AB = (21 + 7) = 28 m

Hence, the height of the tower is 28 m.

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 = $\frac{(\sum f_i{x}_i)}{\left(f_i\right)}=\frac{2850}{80}=35.625$

∴Mean = 35.625

Since the maximum class frequency is 20, the modal class is 30−40.

The formula for finding mode is

$l+\frac{(f_1-f_0)-}{({2f}_1-f_0-f_2)-}\times h$

l = 30, h = 10, $f_1$= 20, $f_0$ = 15, $f_2$= 12

Substituting for variables

30 + $\frac{(20-15)}{(2\times 20-15-12)}\times 10=33.85$

∴Mode = 33.85

The cumulative frequency distribution table:

n = 80

$\frac{n}{2}$ value lies in the class 30−40.

∴ Median class is 30−40.

l = 30, cf = 30, f = 20, h = 10

Median = $l+\frac{(f_1-f_0)}{f}\times h$

= 30 + $\frac{(40-30)}{20}\times 10=35$

∴ 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.