CBSE Class 10 - Maths

1) Consider the correct values of a, b and c from the given quadratic equation.

2) Value of c in the given quadratic equation is negative.

1) Compare the given quadratic equation to a$x^2$+ bx + c = 0 and write the values of a, b and c.

2) Write the quadratic formula to find the roots of a quadratic equation.

3) Substitute the values of a, b and c in the formula and simplify to get the solutions of the quadratic equation.

Given quadratic equation

2$x^2$ + ax −$a^2$ = 0

Comparing the given equation with the standard quadratic equation a$x^2$+ bx + c = 0,

a = 2, b = a and c = −$a^2$

Roots of the equation, x = $\frac{-b\pm \sqrt{b^2-4\mathrm{ac}}}{2a}$

⇒ x = $\frac{-a\pm \sqrt{(a^2-4\times 2\times {(-a)}^2)}}{2\times 2}$

⇒ x = $\frac{-a\pm \sqrt{9a^2}}{4}$

⇒ x = $\frac{\left(-a\pm 3a\right)}{4}$

⇒ x = $\frac{\left(-a+3a\right)}{4}$ = $\frac{a}{2}$ or x = $\frac{(-a-3a)}{4}$ = −a.

∴ The solutions of the given quadratic equation are x = $\frac{a}{2}$ or −a.

1) Use the formula to find the sum of the terms of an A.P. including the last term.

1) Use the formula to find the n th term of an A.P. and obtain an equation by substituting the values of a and $T_n$.

2) Use the formula to find the sum of n terms of A.P. and find the value of n.

3) Substitute the value of n in the equation obtained in step 1 and simplify to find the value of d.

Assume 'a' as the first term and 'd' as the common difference between the terms of an A.P.

Given that

a = 5

$T_n$ = 45

$S_n$= 400

$n^{\mathrm{th}}$ term of an A.P.:

$T_n$ = a + (n − 1)d

⇒ 45 = 5 + (n − 1)d

⇒ 40 = (n − 1)d ------------ (1)

Sum of n terms of an A.P.:

$S_n$ =$\frac{n}{2}$(a + $T_n$)

⇒ 400 = $\frac{n}{2}$(5 + 45)

⇒ $\frac{400}{50}$= $\frac{n}{2}$

⇒ n = 2 × 8

⇒ n = 16

Substituting the value of 'n' equation (1), we get

40 = (16 − 1) d

⇒ 40 = (15) d

⇒ d = $\frac{40}{15}$ = $\frac{8}{3}$.

Therefore, the common difference between the terms of the A.P. is $\frac{8}{3}$.

1) If the sum of two adjacent angles is 180°, then the line formed by the non-common arms of the angles is a straight line.

1) Draw two tangents to the circle with centre 'O' parallel to each other.

2) Draw a line through the center parallel to the tangents.

3) Join the centre of the circle with the points of contact of the tangents.

4) Prove that the angle between the lines from the centres are right angles using the property of interior angles on the same side of the transversal.

5) If the sum of two adjacent angles is a right angle then the line formed by the non-common arms of the angles is a straight line.

6) Hence, the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.

Assume XBY and PCQ as two parallel tangents to a circle with centre O.

Join OB and OC.

Draw OA∥XY

Therefore XB ∥ AO

⇒ ∠XBO + ∠AOB = 180° (Sum of interior angles on the same side of a transversal is 180°)

But, ∠XBO = 90° (A tangent to a circle is perpendicular to the radius through the point of contact)

⇒ 90° + ∠AOB = 180°

⇒ ∠AOB = 180° − 90° = 90°

Similarly, ∠AOC = 90°

∴ ∠AOB + ∠AOC = 90° + 90° = 180°

∴ BOC is a straight line passing through O.

∴ The line segment joining the points of contact of two parallel tangents of a circle passes through its centre.

Use the appropriate trigonometric ratio in proving the relation.

1) Draw the figure representing the data given in the question.

2) Join PO.

3) Angle between the radius and the tangent is a right angle. Angle between the tangent and the line joining the centre and the external point ∠OPQ measures 60°.

4) Apply cosine to the angle OPQ.

5) Simplify and find the relation between the tangent PO and PQ.

Draw the circle with centre O from an external point P and two tangents PQ and PR.

Radius of a circle is perpendicular to the tangent at the point of contact.

∴∠OQP = 90°

The tangents drawn to a circle from an external point make equal angles with the segment, joining the centre to that point.

∴∠QPO = 60°

Consider ∆OPQ,

Cos 60° = $\frac{\mathrm{PQ}}{\mathrm{OP}}$

⇒ $\frac{1}{2}$ = $\frac{\mathrm{PQ}}{\mathrm{PO}}$

∴ 2PQ = PO.

An outcome with atleast one of the faces as tails is a favorable outcome.

1) List all the possible outcomes when two coins are tossed.

2) Write the total number of possible outcomes.

3) List all the favourable outcomes which contains at least one tail.

4) Write the total number of favourable outcomes.

5) Find the ratio of the number of favourable outcomes and the number of possible outcomes which is the required probability.

Rahim tosses two coins simultaneously.

There the number of possible outcomes {HH, HT, TH, TT}.

Total number of outcomes = 4

Out of these getting at least one tail on tossing the two coins = {HT, TH, TT}

Number of favourable outcomes of getting at least one tail = 3

∴ Probability of getting at least one tail on tossing the two coins = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}$ = $\frac{3}{4}$

1) Angle made by a quadrant of a circle at the centre of the circle is a right angle.

2) Diagonal of the square is the hypotenuse of the right angled triangle which is also the radius of the quadrant.

1) Draw the figure representing the data given in the figure.

2) Find the length of the diagonal of the square using the Pythagoras theorem in the right angled triangle.

3) Hypotenuse of the triangle is the radius of the quadrant of the circle.

4) Find the area of the quadrant of the circle and the area of the square.

5) Find the difference between the area of the quadrant and the area of the square which gives the area of the shaded portion.

Construction: join O and B i.e. OB.

In a ΔOAB is an right angle triangle then

${\mathrm{OB}}^2$ = ${\mathrm{OA}}^2$ + ${\mathrm{AB}}^2$ = ${20}^2$ + ${20}^2$ = 2 × ${20}^2$

⇒ OB = 20 $\sqrt{2}$

Radius of the circle, r = 20$\sqrt{2}$ √cm

Area of quadrant OPBQ = $\frac{\theta }{360°}$×π$r^2$

= $\frac{90°}{360°}$× 3.14 × (20$\sqrt{2}$${)}^2$ $$

= $\frac{1}{4}$ × 3.14 ×800 $$

= 628 ${\mathrm{cm}}^2$.

Area of square OABC = ${\left(\mathrm{Side}\right)}^2$= (20${)}^2$= 400$$

∴ Area of the shaded region = Area of quadrant OPBQ − Area of square OABC

= (628 − 400) ${\mathrm{cm}}^2$

= 228 ${\mathrm{cm}}^2$.

Write the quadratic equation in lowest terms before solving.

1) Simplify on both sides of the equation by taking LCM.

2) Cross multiply the equation.

3) Simplify to obtain a quadratic equation.

4) Solve the quadratic equation by splitting the middle term.

Given equation is

$\frac{4}{x}$ - 3 = $\frac{5}{2x+3}$ ; x≠0,−$\frac{3}{2}$

$\frac{4}{x}$ - 3 = $\frac{5}{2x+3}$

$\Rightarrow \frac{4-3x}{x}$ = $\frac{5}{2x+3}$

⇒ (4−3x)(2x+3) = 5x

⇒ - 6$x^2$+ 8x−9x + 12 = 5x

⇒ 6$x^2$+ 6x−12 = 0

⇒$x^2$ + x−2 = 0

⇒ $x^2$ + 2x−x−2 = 0

⇒ (x+2)(x−1) = 0

⇒ x + 2 = 0 or x−1 = 0

⇒ x = −2 or x = 1

∴ The solutions of the given equation are − 2 and 1.

Solve the equations involving fractions by taking LCM.

1) Find the seventh term of the A.P. using the formula to find the n th term of an AP and equate it to the given value.

2) Obtain equation (1).

3) Find the ninth term of the A.P. using the formula to find the n th term of an A.P. and equate to the given value.

4) Obtain equation (2).

5) Find the values of 'a' and 'd' by solving the equations.

6) Find the ${63}^{\mathrm{rd}}$term of the A.P. using the formula to find the n th term of an A.P.

Assume 'a' as the first term and 'd' as the common difference of the given A.P.

Given that

$a_7$= $\frac{1}{9}$

$a_9$ = $\frac{1}{7}$

$t_n$= a + (n - 1)d

$a_7$ = a + (7 − 1) d = $\frac{1}{9}$

⇒ a + 6d = $\frac{1}{9}$ .............. (i)

$a_9$= a + (9 − 1)d = $\frac{1}{7}$

⇒ a + 8d = $\frac{1}{7}$ ..............(ii)

Solving equations (i) and (ii), we get:

2d = $\frac{2}{63}$

⇒ d = $\frac{1}{63}$

Substituting the value of d in equation (i), we get

a + $(6\times \frac{1}{63})$ = $\frac{1}{9}$

a = $\frac{1}{63}$

∴ $a_{63}$= a + (63 − 1)d = $\frac{1}{63}$+ 62$\left(\frac{1}{63}\right)$ = $\frac{63}{63}$ = 1

∴ The ${63}^{\mathrm{rd}}$term of the given A.P. is 1.

The following are the steps to construct the figure.

Step 1:

Draw a line BC of length 8 cm.

Step 2:

Draw BX perpendicular to BC.

Step 3:

Make a mark of an arc at the distance of 6 cm on BX. Mark it as A.

Step 4:

Join A and C. There fore ∆ABC is the required triangle.

Step 5:

B as the centre, draw an arc on AC.

Step 6:

Draw the bisector of this arc and join it with B. Thus, BD is perpendicular to AC.

Step 7:

Now, draw the perpendicular bisector of BD and CD. Take the point of intersection as O.

Step 8:

With O as the centre and OB as the radius, draw a circle passing through points B, C and D.

Step 9:

Join A and O and bisect it. Let P be the midpoint of AO.

Step 10:

Take point P as the centre and PO as radius, draw a circle which will intersect the circle at point B and G. Join A and G.

∴ AB and AG are the required tangents to the circle from A.

Simplify the equation equating the distances by squaring on both sides.

1) Equate the distances between the points A to B and A to C.

2) Substitute the values of the coordinates in the distance formula.

3) Simplify the equation to find the value of the unknown coordinate.

4) Use the distance formula to find the length of AB.

The given points are A (0, 2), B (3, p) and C (p, 5).

Given that the distance from A to B is equal to the distance from A to C.

∴ AB = AC

⇒${\mathrm{AB}}^2$ =${\mathrm{AC}}^2$

⇒${\left(3-0\right)}^2$ + ${\left(p-2\right)}^2$ = ${\left(p-0\right)}^2$ + ${\left(5-2\right)}^2$

⇒ 9 + $p^2$ + 4 − 4p = $p^2$ + 9

⇒ 4 − 4p = 0

⇒ 4p = 4

⇒ p = 1

Hence, the value of p is 1.

Length of AB = $\sqrt{{\left(3-0\right)}^2+{\left(1-2\right)}^2}$ = $\sqrt{3^2+{\left(-1\right)}^2}$√= $\sqrt{9+1}$ = $\sqrt{10}$ units.

1) Find the alternate angles of the angles of depression.

2) Use the appropriate trigonometric ratio of the angles of depression in each triangle .

1) Draw the figure representing the data given in the question.

2) Mark the alternate angles of the angles of depression in the right angled triangles on either side of the light house.

3) Find the tangent of the angle of depression of 45° in the first triangle and find the distance between the first ship and the light house.

4) Find the tangent of the angle of depression of 60° in the second triangle and simplify to find the distance between the ships.

Consider d be the distance between the two ships. Suppose that distance of one of the ships from the light house is x metres, then the distance of the other ship from the light house is (d−x) metres.

Consider the right angled triangle ADO,

Tan 45° = $\frac{\mathrm{OD}}{\mathrm{AD}}$ = $\frac{200}{x}$

⇒ 1 = $\frac{200}{x}$

⇒ x = 200 ................(i)

Consider the right angled triangle BDO,

Tan 60° = $\frac{\mathrm{OD}}{\mathrm{BD}}$ = $\frac{200}{\left(d-x\right)}$

⇒ $\sqrt{3}$√= $\frac{200}{\left(d-x\right)}$

⇒ d−x = $\frac{200}{\sqrt{3}}$

Substituting x = 200, we have

d−200 = 200d = $\frac{200}{\sqrt{3}}$

d = $\frac{200}{\sqrt{3}}$+ 200

⇒ d = 200( $\sqrt{3}$ + $\frac{1}{\sqrt{3}}$)

⇒ d = 200 ×1.58

⇒ d = 316 (approximately.)

∴ The distance between two ships is approximately 316 m.

1) Substitute the correct coordinates of the points in the formula to find the area of the triangle.

1) The given points are collinear. Therefore, the area of the triangle formed by these points is zero.

2) Substitute the values of the coordinates in the formula for the area and equate it to zero.

3) Simplify to find an equation.

4) Solve the given equation and the equation obtained to find the values of a and b.

The points A (−2, 1), B (a, b) and C (4, −1) are collinear.

Therefore, the area of the triangle ABC formed by the points is 0.

If the three points are ($x_1$, $y_1$), ($x_2$, $y_2$) and ($x_3$, $y_3$),

⇒$\frac{1}{2}$[$x_1$($y_2$−$y_3$) + $x_2$($y_3$−$y_1$) + $x_3$($y_1$−$y_2$)] = 0

Comparing with the given coordinates, $x_1$ = -2, $y_1$ = 1 , $x_2$ = a, $y_2$ = b and $x_3$ = 4, $y_3$ = −1.

∴ $\frac{1}{2}$[−2 (b + 1) + a (−1−1) + 4(1−b)] = 0

⇒ −2b−2−2a + 4 − 4b = 0

⇒ 2a + 6b = 2

⇒ a + 3b = 1 -------------- (i)

Given that a−b = 1 -------- (ii)

Solving equations (i) and (ii), we get 4b = 0 ⇒b = 0

Putting b = 0 in equation (ii), we get a = 1

∴ The values of a and b are 1 and 0 respectively.

Altitudes of the equilateral triangle intersect at the centre of the incircle.

1) Draw the figure representing the data given in the question.

2) Draw the altitudes of the equilateral triangle which intersect at the centre of the circle.

3) Equate the sum of the areas of the isosceles triangles into which the triangle is divided and equate it to the area of the equilateral triangle.

4) Simplify to find the radius of the circle.

5) Find the difference between the area of the equilateral triangle and the area of the circle to find the area of the shaded region.

Given that ABC is an equilateral triangle of side 12 cm.

Construction:

Join O and A, O and B, and O and C.

P, Q, R are the points on BC, CA and AB respectively then,

OP⊥BC

OQ⊥AC

OR⊥AB

Assume the radius of the circle as r cm.

Area of ∆AOB + Area of ∆BOC + Area of ∆AOC = Area of ∆ABC

⇒( $\frac{1}{2}$× AB × OR) + ($\frac{1}{2}$× BC × OP) + ($\frac{1}{2}$× AC × OQ) = $\frac{\sqrt{3}}{4}$× ${\left(\mathrm{side}\right)}^2$.

⇒ ($\frac{1}{2}$× 12 × r) + ($\frac{1}{2}$× 12 × r) + ($\frac{1}{2}$× 12 × r) = √$\frac{\sqrt{3}}{4}$× ${\left(12\right)}^2$

⇒ 3 × $\frac{1}{2}$× 12 × r =$\frac{\sqrt{3}}{4}$× 12 × 12

⇒ r = 2$\sqrt{3}$ = 2 × 1.73 = 3.46

Hence, the radius of the inscribed circle is 3.46 cm.

Area of the shaded region = Area of ∆ABC − Area of the inscribed circle

= [$\frac{\sqrt{3}}{4}$×${\left(12\right)}^2$ − π${\left(2\sqrt{3}\right)}^2$]

= [36$\sqrt{3}$√−12π] $$

= [36 × 1.73 − 12 × 3.14] $$

= [62.28 − 37.68] $$

= 24.6 ${\mathrm{cm}}^2$

∴ The area of the shaded region is 24.6 ${\mathrm{cm}}^2$

Unshaded semicircle is included in the shaded semicircle.

1) Find the radius of the semicircles.

2) Find the areas of the semicircles using the formula.

3) Subtract the area of the unshaded semicircle from the sum of the areas of the shaded semicircles.

Find the radius of the semicircles from the given diameters.

Radius = $\frac{1}{2}\times \mathrm{Diameter}$

Radius of semicircle PSR = $\frac{1}{2}$ × 10 = 5 cm

Radius of semicircle RTQ = $\frac{1}{2}$ × 3 = 1.5 cm

Radius of semicircle PAQ = $\frac{1}{2}$ × 7 = 3.5 cm

Perimeter of the shaded region = (Circumference of semicircle PSR) + (Circumference of semicircle RTQ) - (Circumference of semicircle PAQ)

= [$\frac{1}{2}$× 2π(5)] + [$\frac{1}{2}$× 2π(1.5)] - [$\frac{1}{2}$× 2π(3.5)]

= π (5 + 1.5 - 3.5)

= 3.14 × 3

= 9.42 cm

1) Convert the inner radius of the pipe into metres.

2) Convert the rate of flow of water through the pipe into metres per hour.

1) Find the radius and the of the pipe in metres.

2) Assume the time taken to fill the tank by the pipe as 't' hours.

3) Convert the rate of flow of water through the pipe into metres per hour.

4) Equate the volume of the water flown thrown the pipe with the volume of the cylindrical tank.

5) Find the value of 't', by simplifying the equation.

Measurements of the cylindrical tank:

Diameter = 10 m

Radius, R = 5 m

Depth, H = 2 m

Inner radius of the pipe = r = $\frac{20}{2}$ cm = 10 cm = $\frac{1}{10}$ m

Rate of flow of water = v = 4 km/h = 4000 m/h [1 km = 1000 m]

Consider the time taken to fill the tank as 't'.

Water flown through the pipe in t hours is equal to the volume of the tank.

∴ π$r^2$× v × t = π$R^2$H

⇒ ${\left(\frac{1}{10}\right)}^2$× 4000 × t = ${\left(5\right)}^2$× 2

⇒ t = 25 × 2 × $\frac{100}{4000}$ = $\frac{5}{4}$ = 1$\frac{1}{4}$

∴ The time taken to fill the tank is 1$\frac{1}{4}$hours.

1) Angle at the vertex of t he cone is bisected by the vertical height of the cone.

2) Length of the wire to be drawn is considered as the height of the cylindrical wire.

1) Draw the figure representing the data given in the question.

2) Draw a plane cutting the cone at the middle of the vertical height.

3) Find the dimensions of the frustum of the cone obtained by using the given dimensions.

4) Find the volume of the frustum of the cone.

5) Assume the length of the wire which is drawn from the frustum of the cone as 'l'.

6) Equate the volume of the wire to the volume of the frustum of the cone.

7) Simplify and find the value of 'l'.

Consider ACB be the cone whose vertical angle ∠ACB = 60°.

Let R and x be the radii of the lower and upper end of the frustum.

But height of the cone, OC = H = 20 cm

Height CP = h = 10 cm

Let us consider P as the mid-point of OC.

After cutting the cone into two parts through P,

OP = $\frac{20}{2}$ = 10 cm

But ∠ACO and ∠OCB = ( ½) × 60° = 30°.

After cutting the cone CQS from cone CBA, the remaining solid obtained is a frustum.

But in a triangle CPQ

Tan 30° = $\frac{x}{10}$

⇒ $\frac{1}{\sqrt{3}}$ = $\frac{x}{10}$ ⇒ x = $\frac{10}{\sqrt{3}}$ cm

In a triangle COB

Tan 30° = $\frac{R}{\mathrm{CO}}$

⇒ $\frac{1}{\sqrt{3}}$ = $\frac{R}{20}$

⇒ R = $\frac{20}{\sqrt{3}}$ cm

Volume of the frustum, V = $\frac{1}{3}$ π ( $R^2$H − $x^2$h)

⇒ V = $\frac{1}{3}$π ( $\frac{20}{\sqrt{3}}{)}^2$× 20 − ${\left(\frac{10}{\sqrt{3}}\right)}^2\times$10

= $\frac{1}{3}$ π $(\frac{8000}{3}$− $\frac{1000}{3})$

= $\frac{1}{3}$ π ( $\frac{7000}{3}$)

= $\frac{1}{9}$ π × 7000

= $\frac{7000}{9\pi }$

The volumes of the frustum of the cone is equal to the volume of the wire formed.

π × ${\left(\frac{1}{24}\right)}^2$× l = $\frac{7000}{9\pi }$ [Volume of wire = π $r^2$h]

⇒ l = ( $\frac{7000}{9}$) × 24 × 24 ⇒

⇒ l = 448000 cm = 4480 m

∴ The length of the wire is 4480 m.

If one natural number is greater than the other, then the reciprocal of the number is smaller than the reciprocal of the other.

1) Assume the numbers as x and y.

2) Find the difference between the numbers, equate it to 5 and obtain equation (1).

3) Find the difference between the reciprocals of the numbers, equate it to $\frac{1}{10}$and obtain equation (2).

4) Find the value of x from equation (1) and substitute in equation (2).

5) Obtain a quadratic equation and solve the equation to find the value of y.

6) Find the value of x using the value of x. Thus the required natural numbers are found.

Let us assume the two natural numbers as x and y such that x > y.

Given that difference between the two natural numbers = 5 ⇒ x - y = 5 ------ (i)

Difference of their reciprocals = $\frac{1}{10}$⇒$\frac{1}{y}$ - $\frac{1}{x}$= $\frac{1}{10}$

⇒$\frac{x-y}{\mathrm{xy}}$= $\frac{1}{10}$

⇒$\frac{5}{\mathrm{xy}}$= $\frac{1}{10}$

xy = 50 .............(ii)

Substituting the value of x from equation (i) in equation (ii), we get

y(y + 5) = 50.

⇒$y^2$+ 5y - 50 = 0

⇒$y^2$+ 10y - 5y - 50 = 0

⇒ y(y + 10) - 5 ( y + 10) = 0

⇒(y - 5)( y + 10) = 0

⇒ y = 5 or -10.

Here y is a natural number.

∴ y = 5

Other natural number x = y + 5 = 5 + 5 = 10

∴ The two natural numbers are 5 and 10.

Prove that the triangles are congruent according to the RHS congruence criterion.

1) Draw the figure of a circle with two tangents from an external point.

2) Draw radii at the points of contact of the tangents.

3) Draw a line from the centre of the circle to the external point.

4) Compare both the triangles formed and prove that they are congruent according to RHS congruency criterion.

5) The two tangents are equal as the corresponding parts of the congruent triangles are equal.

Let AP and BP be the two tangents to the circle with centre O.

Required to prove: AP = BP

Proof: Consider ∆OAP and ∆OBP:

OA = OB (radii of the same circle)

∠OAP = ∠OBP = 90° (The radius of a circle drawn at the point of contact is perpendicular to the tangent.)

OP = OP (common side)

∴ ∆AOP ≅ ∆BOP (According to R.H.S. congruence criterion)

∴ AP = BP (Corresponding parts of congruent triangles)

∴ The length of the tangents drawn from an external point to a circle is equal.

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.

5) Find the difference between the heights of the building and the tower.

Consider AB is the building and CD is the tower.

∆ABD is a right angle triangle ⇒ $\frac{\mathrm{AB}}{\mathrm{BD}}$= Tan 60°

⇒ $\frac{60}{\mathrm{BD}}$= $\sqrt{3}$

⇒ BD = $\frac{60}{\sqrt{3}}$

⇒ BD = 20 $\sqrt{3}$

Consider the right angled triangle ACE

$\frac{\mathrm{CE}}{\mathrm{AE}}$= Tan 30°

⇒ $\frac{\mathrm{CE}}{\mathrm{BD}}$= $\frac{1}{\sqrt{3}}$ ( ∴ AE = BD )

⇒ CE = $\frac{20\sqrt{3}}{\sqrt{3}}$= 20.

Height of the tower, CD = CE + ED = CE + AB = 20 + 60 = 80 m.

Difference between the heights of the tower and the building = 80 − 60 = 20 m

Distance between the tower and the building = BD = 20 $\sqrt{3}$√m.

1) 1 and 49 are also included in the list of favourable outcomes of odd numbers.

2) 1 and 49 are also included in the list of favourable outcomes of perfect squares.

1) Find the number of possible outcomes.

2) Find the number of favourable outcomes in each case.

3) Probability of each event is equal to the ratio of the number of favourable outcomes and the number of possible outcomes.

(i)

Total number of cards in a bag = 49.

Total number of possible outcomes = 49

The odd numbers from 1 to 49 are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47 and 49.

∴Total number of favourable outcomes = 25

∴ Probability = $\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}$ = $\frac{25}{49}$

(ii)

Total number of possible outcomes = 49

The multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40 and 45.

Total number of favourable outcomes = 9

∴ Probability = $\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}$ = $\frac{9}{49}$

(iii)

Total number of outcomes = 49

The numbers 1, 4, 9, 16, 25, 36 and 49 are perfect squares.

Total number of favourable outcomes = 7

∴ Probability = $\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}$ = $\frac{7}{49}$ = $\frac{1}{49}$.

(iv)

Total number of outcomes = 49

We know that 2 is only one even prime number.

Total number of favourable outcomes = 1

∴ Probability = $\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}$ = $\frac{1}{49}$.

1) Assume ratio in which the point P divides the line segment AB as k:1.

2) Find the value of k first and then the unknown coordinate of the point dividing.

1) Assume the ratio in which the point P divides AB as k:1.

2) Find the coordinates of the point P and equate them to the given coordinates.

3) Find the value of the k, by equating the y-coordinates.

4) Find the value of x by substituting the value of k in the equation equating the x-coordinates.

5) Find the ratio from the value of k.

Given that the point P (x, 2) divide the line segment joining the points A (12, 5) and B (4, −3) in the ratio k : 1.

∴ The coordinates of P are using section formula$(\frac{4k+12}{k+1},\frac{-3k+5}{k+1})$

Now, the coordinates of P are (x, 2).

∴ $\frac{4k+12}{k+1}$ = x and $\frac{-3k+5}{k+1}$= 2

⇒ -3k + 5 = 2( k + 1)

⇒5k = 3

⇒k = $\frac{3}{5}$

Substituting k = $\frac{3}{5}$in $\frac{4k+12}{k+1}$ = x, we get

⇒x = $\frac{4\times \frac{3}{5}+12}{\frac{3}{5}+1}$

⇒x = $\frac{12+60}{3+5}$

⇒x = $\frac{72}{8}$

⇒x = 9.

∴ The value of x is 9.

Also, the point P divides the line segment joining the points A(12, 5) and (4, −3) in the ratio $\frac{3}{5}$ : 1 is 3 : 5.

Consider both the values of k to find two equal roots of the equation.

1) Equate the given quadratic equation to ${\mathrm{ax}}^2+\mathrm{bx}+c=0$ and find the values of a, b and c.

2) Substitute the value of a, b and c in the discriminant $b^2-4\mathrm{ac}$ and equate it to zero as the roots of the equation are equal.

3) Solve the quadratic equation obtained to find two value for k.

4) Use each value of k to find two equal roots of the equation.

Given that the quadratic equation (k + 4)$x^2$+ ( k + 1)x + 1 = 0 has equal roots ⇒Discriminant of the equation is zero.

If ${\mathrm{ax}}^2+\mathrm{bx}+c=0$ is the quadratic equation, the discriminant $b^2-4\mathrm{ac}=0$

a = (k + 1), b = (k + 1), c = 1

⇒${(k+1)}^2$- 4 × (k + 4) × 1 = 0

⇒$k^2$+ 2k + 1 - 4k - 16 = 0

⇒$k^2$- 2k -15 = 0

⇒$k^2$-5k + 3k - 15 = 0

⇒(k - 5)( k + 3) = 0

⇒k - 5 = 0 or k + 3 = 0.

⇒k = 5 or -3 .

Thus the values of k are 5 and -3.

For k = 5 :

( k + 4)$x^2$+ ( k + 1)x + 1 = 0

⇒9$x^2$+ 6x + 1 = 0

⇒${(3x+1)}^2$= 0

⇒ x = $\frac{-1}{3}$,$\frac{-1}{3}$

For k = -3 :

(k + 4)$x^2$+ ( k +1)x + 1 = 0

⇒ $x^2$- 2x + 1 = 0

⇒${(x-1)}^2$= 0

∴ x = 1, 1

Therefore, the equal root of the given quadratic equation is either 1 or $-\frac{1}{3}$.

${15}^{\mathrm{th}}$ term from the last is the the ${36}^{\mathrm{th}}$term from the first.

1) Use the formula to find the sum of the n terms of an A.P., substitute the value of n as 10 and obtain equation (1).

2) ${15}^{\mathrm{th}}$term from the last term is the ${36}^{\mathrm{th}}$term from the first. Use the formula to find the sum of the n terms of an A.P., substitute the value of n as 15, value of a as the${36}^{\mathrm{th}}$term and obtain equation (2).

3) Solve equation (1) and (2) and find the values of a and d.

Consider a and d as the first term and the common difference of an A.P. respectively.

$n^{\mathrm{th}}$term of an A.P., $a_n$= a + ( n - 1)d

Sum of n terms of an A.P., $S_n$= $\frac{n}{2}$[2a + (n - 1)d]

Given that the sum of the first 10 terms is 210.

⇒$\frac{10}{2}$[2a + 9d ] = 210

⇒5[2a + 9d] = 210

⇒2a + 9d = 42 ----------- (1)

${15}^{\mathrm{th}}$term from the last = ( 50 - 15 + 1${)}^{\mathrm{th}}$= ${36}^{\mathrm{th}}$ term from the beginning

⇒$a_{36}$= a + 35d

Sum of the last 15 terms = $\frac{15}{2}$[2$a_{36}$+ ( 15 - 1)d ] = 2565

⇒$\frac{15}{2}$[ 2(a + 35d) + 14d ] = 2565

⇒ 15 [ a + 35d + 7d ] = 2565

⇒a + 42d = 171 ----------(2)

From (1) and (2), we have d = 4 and a = 3.

Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199.

Rearrange the lengths of tangents to form the sides of the rhombus.

1) Draw the figure showing a parallelogram circumscribing a circle.

2) Tangents to a circle from an external point are equal. So, write the tangents from each vertex of the parallelogram and equate the corresponding tangents.

3) Add the tangents such that they form the sides of the parallelogram.

4) Simplify the equation using the property of the parallelogram and prove that it is a rhombus.

Given: Draw ABCD a parallelogram circumscribing a circle with centre O.

Required to prove: ABCD is a rhombus.

The tangents drawn to a circle from an exterior point are equal in length.

∴ AP = AS, BP = BQ, CR = CQ and DR = DS.

AP + BP + CR + DR = AS + BQ + CQ + DS

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

∴ AB + CD = AD + BC or 2AB = 2BC (Opposite sides of the parallelogram are equal. So, AB = DC and AD = BC)

∴ AB = BC = DC = AD.

∴ ABCD is a rhombus.

Hence, it is proved.

Convert the diameter of the ball into radius.

1) Find the radius of the balls dropped in the vessel.

2) Assume the number of balls as 'n'.

3) Equate the volume of the water spilled out of the vessel to the total volume of the spherical balls.

4) Substitute the given values in the equation and simplify to find the value of 'n'.

Given that height (h) of the conical vessel = 11 cm

Radius ( $r_1$) of the conical vessel = 2.5 cm

Radius ( $r_2$) of the metallic spherical balls = $\frac{0.5}{2}$ = 0.25 cm

Assume the number of spherical balls that were dropped in the vessel as 'n'.

Volume of the water spilled out of the vessel = Total volume of the spherical balls dropped

$\frac{2}{5}$ × Volume of the cone = n × Volume of each spherical ball

⇒ $\frac{2}{5}$× $\frac{1}{3}$π ${r_1}^2$h = n × $\frac{4}{3}$π ${r_2}^3$

⇒ ${r_1}^2$h = n × 10 ${r_2}^3$

⇒ ${\left(2.5\right)}^2$× 11 = n × 10 × ${\left(0.25\right)}^3$

⇒ 68.75 = 0.15625 n

⇒n = 440.

∴The number of spherical balls that were dropped in the vessel is 440.

Sushant made the arrangement so that the water that flows out, irrigates the flower beds.

This shows the importance of conservation of water.

1) Base of the conical cavity and the base of the solid cylinder coincide each other. So, consider only one base of the cylinder.

1) Draw the figure representing the given data.

2) Find the slant height of the cone using the radius and the vertical height.

3) Surface area of the remaining solid includes the curved surface area of the cylinder and the curved surface area of the cone and the base area of the cylinder.

This figure shows that the required cylinder and the conical cavity.

Given that

Height of the conical part (h) = Height of the cylindrical part (h) = 2.8 cm

Diameter of the cylindrical part = Diameter of the conical part = 4.2 cm

∴ Radius of the cylindrical part (r) = Radius of the conical part (r) = 2.1 cm

Slant height of the conical part (l) = $\sqrt{r^2+h^2}$

l = $\sqrt{{\left(2.1\right)}^2+{\left(2.8\right)}^2}$

l = $\sqrt{4.41+7.84}$

l = $\sqrt{12.25}$

l = 3.5 cm

Total surface area of the remaining solid =

Curved surface area of the cylindrical part + Curved surface area of the conical part + Area of the cylindrical base.

= 2πrh + πrl + π $r^2$

= (2× $\frac{22}{7}$×2.1×2.8) + ( $\frac{22}{7}$×2.1×3.5) + ( $\frac{22}{7}$×2.1×2.1)

= (36.96 + 23.1 + 13.86) $$

= 73.92 ${\mathrm{cm}}^2$

∴ The total surface area of the remaining solid is 73.92 ${\mathrm{cm}}^2$.