CBSE Class 10 - Maths

Use the factors of both the constant term and the coefficient of $x^2$ to spit the middle term of the equation.

1) Find the product of the constant term and the coefficient of $x^2.$

2) Use the factor pair of the product which gives the middle term on subtraction.

3) Find the factors of the equation by taking common in the first two terms and the last two terms.

4) Find the values of x be equating the each factor to zero.

Given quadratic equation is

4$\sqrt{3}{x}^2$+ 5x - 2$\sqrt{3}$= 0

⇒4$\sqrt{3}{x}^2$+ 8x - 3x - 2$\sqrt{3}$= 0

⇒4x($\sqrt{3}x$ + 2) - $\sqrt{3}$( $\sqrt{3}x$ + 2) = 0

⇒(4x - $\sqrt{3})$($\sqrt{3}x$ + 2) = 0

∴ x = $\frac{\sqrt{3}}{4}$ or x = - $\frac{2}{\sqrt{3}}$

Write the data using the inequalities and simplify.

1) Numbers which are divisible by 7 are in the form of 7n.

2) Write the condition using the inequalities.

3) Divide all the terms of the inequality by 7.

4) Write the fractions in the form of mixed fractions.

5) Find the difference between the integral part of the mixed numbers to find the number of natural numbers which are divisible by 7.

Let three digit numbers from 100 to 999.

All the three-digit natural numbers that are divisible by 7 will be of the form 7n.

∴ 100 ≤ 7n ≤ 999

$\frac{100}{7}$≤n ≤$\frac{999}{7}$

14$\frac{2}{7}$≤n ≤ 142$\frac{5}{7}$

As n is an integer, number of 3-digit natural numbers that are divisible by 7 are 142 − 14 i.e. 128.

Length of CF i.e. r can be found by subtracting the sum of p and q from the sum of p, q and r.

1) Tangents drawn from the vertices of the triangle are equal.

2) Assume the lengths of the equal tangents from three sides as p, q and r.

3) Each side of the triangle is the sum of two different tangents.

4) Find the sum of the sides and equate it to the perimeter of the triangle.

5) Substitute the values of p, q and r and simplify to find the sum of p, q and r.

6) Find the length of p, q and r by subtracting the sum of two of them from the sum of all the three.

7) Lengths of p, q and r gives the required lengths.

Lengths of sides of the triangle are given as: AB = 12 cm, BC = 8 cm and AC = 10 cm.

Assume AD = AF = p cm, BD = BE = q cm and CE = CF = r cm

(Tangents drawn from an external point to the circle are equal )

⇒2(p + q + r) = AB + BC + AC = 30 cm

⇒(p + q + r) = 15 cm

⇒AB = AD + DB = p + q = 12 cm

∴ r = CF = 15 - 12 = 3 cm.

⇒ AC = AF + FC = p + r = 10 cm

∴q = BE = 15 - 10 = 5 cm.

⇒ BC = BE + EC = q + r = 8 cm

∴ p = AD = 15 − 8 = 7 cm.

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.

Length of the tangents drawn to a circle from an exterior point are equal.

∴ 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 (Since, AB = DC and AD = BC)

∴ AB = BC = DC = AD.

∴ ABCD is a rhombus.

Hence, it is proved.

Consider the cards that are neither a queen nor a king of both the colours in the pack of playing cards.

1) Find the number of cards in the pack of playing cards that are either a king or a queen.

2) Total number of cards is 52. So, the number of cards that are neither a king nor a queen is 44.

3) Note the total number of possible outcomes as 52 and the number of favourable outcomes as 44.

4) Ratio of the number of favourable outcomes and the total number of outcomes gives the required probability.

Let E be the event of that the drawn card is neither a king nor a queen.

Total number of possible outcomes = 52.

Total number of cards that are king or queen in the pack of playing cards = 4 + 4 = 8.

∴ Number of cards that are neither a king nor a queen = 52 − 8 = 44.

Total number of favourable out comes = 44.

∴ Required probability = P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$ = $\frac{44}{52}$ = $\frac{11}{13}$

∴ The probability that the drawn card is neither a king nor a queen is $\frac{11}{13}$.

Circles of the maximum area has the diameter equal to the width of the rectangle and half of the length.

1) The circle of the maximum area has the diameter equal to the width of the rectangle.

2) Find the radius of the circle from the diameter.

3) Find the area of two circles.

4) Subtract the area of two circles from the area of the rectangle to get the area of the remaining card board.

Dimension of the rectangular card board = 14 cm × 7 cm

Two circular pieces of equal radii and maximum area touching each other are cut from the rectangular card board, therefore, the diameter of each circular piece is $\frac{14}{2}$ = 7 cm.

Radius of each circular piece = $\frac{d}{2}$=$\frac{7}{2}$ cm

∴ Sum of area of two circular pieces = 2×π ×${\left(\frac{7}{2}\right)}^2$= 2×$\frac{22}{7}$×$\frac{49}{4}$ = 77 ${\mathrm{cm}}^2$.

Area of the remaining card board = Area of the card board − Area of two circular pieces.

= (14 × 7) − (77)$$= 98−77 = 21 ${\mathrm{cm}}^2$.

∴The area of the remaining card board is 21 ${\mathrm{cm}}^2$.

Consider the coefficients of the quadratic equation along with their signs.

1) Roots of the quadratic equation are equal, so the discriminant of the equation is zero.

2) Write the coefficients of the quadratic equation and find the discriminant.

3) Equate the value of the discriminant to zero and find simplify to find the value of k.

Given quadratic equation is kx(x − 2) + 6 = 0.

The equation can be simplified as k$x^2$−2kx + 6 = 0.

If the roots of a quadratic equation are equal, it’s discriminant is equal to zero. i.e. D = 0.

⇒ $b^2$−4ac = 0, where a = k, b = −2k and c = 6.

⇒ 4$k^2$−24k = 0

⇒ 4k(k− 6 ) = 0

⇒ k = 0 or k = 6

But k cannot be 0, so the value of k is 6.

Consider the last term of the A.P. along with its sign.

1) Find the common difference of the A.P.

2) Equate the n th term to the last term of the A.P. and substitute the values.

3) Simplify to find the value of n which gives the number of terms of the A.P.

4) Find the sum of all the terms of the A.P. using the given values and the value of n.

The terms AP is given by 18, 15$\frac{1}{2}$, 13, ..........., - 49$\frac{1}{2}$.

First term a = 18, common difference, d = 15$\frac{1}{2}$ - 18 = -2$\frac{1}{2}$ and the last term of the A.P. = - 49$\frac{1}{2}$

Assume that A.P. has n terms.

∴$a_n$= a + (n − 1) d

⇒- 49$\frac{1}{2}$ = 18 - ( n -1) × 2$\frac{1}{2}$

⇒- $\frac{99}{2}$ = 18 - $\frac{5}{2}$( n - 1)

⇒- 99 = 36 - 5( n -1)

⇒5( n -1) = 135

⇒5n = 140

∴ n = $\frac{140}{5}$= 28

∴The given A.P. has 28 terms.

Sum of all the terms ($S_n$):

$S_n$= $\frac{n}{2}$[ 2a + ( n -1) d ]

$S_{28}$= $\frac{28}{2}$$[2\times 18+(28-1)\times -(\frac{5}{2})]$

$S_{28}$= 14 $[36-\frac{135}{2}]$

$S_{28}$= - 441

∴The sum of all the terms of the A.P. is = - 441.

Locate three points on AX such that they are equidistant.

1) Construct a triangle ABC using the measurements of three sides. 2) Draw a ray AX making an acute angle with AB.

3) Mark 3 points on AX and join the third point to B.

4) Draw a line $A_2$B' parallel to $A_3$B from $A_2$.

5) Draw a line B'C' parallel to BC.

6) AB'C' is the required triangle.

Step1:

Draw a line segment AB = 4 cm. Consider the point A as centre, draw an arc of 5 cm radius.

Similarly, consider the point B as its centre, draw an arc of 6 cm radius.

Both the arcs will intersect each other at C and the required triangle ABC is formed.

Step 2:

Draw a ray AX making an acute angle with the line segment AB on the opposite side of the vertex C.

Step 3:

Locate 3 points $A_1$, $A_2$, $A_3$on line AX such that A$A_1$ = $A_1{A}_2$= $A_2{A}_3$.

Step 4:

Join B and $A_3$i.e. B$A_3$and draw a line through $A_2$parallel to B$A_3$ to intersect AB at point B'.

Step 5:

Draw a line segment through B' parallel to BC to intersect AC at C'.

∴ΔAB'C' is the required triangle.

Find the tangent of the alternate angle of the angle of the depression.

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

2) Draw a horizontal line at the top of the first pole.

3) Mark the alternate angle of the angle of depression of the top of the first pole from the top of the second pole.

4) Assume the height of the first pole as 'h'.

5) Obtain the measures of the sides of the right angled triangle involving the alternate angle of the angle of depression.

6) Find the tangent of the alternate angle of the angle of depression. Substitute the values and simplify to find the height of the first pole.

Assume AB and CD as two poles and CD = 24 m.

Given that the angle of depression of the top of the pole AB from the top of the pole CD is 30° and the distance between the two poles is 15 m i.e. BD = 15 m

∴ ∠CAL = 30° and BD = 15 m.

Required to find the height of the pole AB.

Let the height of the pole AB be 'h' metres.

AL = BD = 15 m and AB = LD = h

∴CL = CD - LD = 24 - h

In the right angled triangle ACL,

tan ∠CAL = $\frac{\mathrm{Opposite}\mathrm{side}}{\mathrm{Adjacent}\mathrm{side}}$= $\frac{\mathrm{CL}}{\mathrm{AL}}$

tan 30° = $\frac{24-h}{15}$

$\frac{1}{\sqrt{3}}$= $\frac{24-h}{15}$

24 - h = $\frac{15}{\sqrt{3}}$

24 - h = 5 $\sqrt{3}$

h = 24 - (5 × 1.732)

h = 15.34

∴ Height of the pole (AB) = h = 15.34 m.

Angle opposite to the longest side i.e. hypotenuse is the right angle in a right angled triangle.

1) Assume the three given points as A, B and C.

2) Find the distances between each pair of points using the distance formula.

3) Note that the lengths of two sides are equal and the triangle formed by the points is an isosceles triangle.

4) Note that the sum of the squares of two sides is equal to the square of the other side.

5) The sides obey the Pythagoras property so, they form a right angled triangle.

6) The angle opposite to the longest side i.e. the hypotenuse is the right angle.

7) So, the triangle formed by the given points is an isosceles right angled triangle.

The given points are A (7, 10), B (− 2, 5) and C (3, − 4).

By Using distance formula, we have

AB = $\sqrt{{(7+2)}^2+{(10-5)}^2}$= $\sqrt{81+25}$= $\sqrt{106}$

BC = $\sqrt{{(-2-3)}^2+{(5+4)}^2}$= $\sqrt{25+81}$= $\sqrt{106}$

CA = $\sqrt{{(7-3)}^2+{(10+4)}^2}$= $\sqrt{16+196}$= $\sqrt{212}$

AB = BC ⇒ ΔABC is an isosceles triangle.

${\mathrm{AB}}^2$+ ${\mathrm{BC}}^2$= 106 + 106 = 212 = ${\mathrm{AC}}^2$

∴ ΔABC is a right angled triangle right angled at ∠B.

Therefore, ΔABC is an isosceles right angled triangle.

∴ The points (7, 10), (− 2, 5) and (3, − 4) are the vertices of an isosceles right triangle.

Equate the y-coordinates of the point to find the value of k.

1) Assume the ratio in which the point on y-axis divides the line segment joining the given points as k:1.

2) Find the coordinates of the point dividing using the section formula.

3) Every point on the y-axis is in the form of (0, y).

4) Equate the x-coordinates of the points and simplify to find the value of k.

5) Find the ratio in which the point on x-axis divides using the value of k.

6) Substitute the value of k in the fraction obtained by using the section formula.

7) Simplify to find the coordinates of the point on y-axis.

Let a point on y-axis divide the line segment joining the points (−4, −6) and (10, 12) in the ratio k : 1.

The point of the intersection be (0, y).

So, by putting the values of coordinates and the ratio in the section formula, we get

$(\frac{10k+(-4)}{k+1},\frac{12k+(-6)}{k+1})$ = (0, y)

∴$\frac{10k-4}{k+1}$= 0 ⇒10k - 4 = 0

⇒ k = $\frac{4}{10}$= $\frac{2}{5}$

∴ y = $\frac{12k-6}{k+1}$= $\frac{12\times \frac{2}{5}-6}{\frac{2}{5}+1}$= $\frac{\frac{24-30}{5}}{\frac{2+5}{5}}$ = - $\frac{6}{7}$

∴ y-axis divides the line segment joining the given points in the ratio of 2 : 5 and the point of intersection is$\left(0,-\frac{6}{7}\right)$.

Radius of the bigger circle is the diameter of the smaller circle and the altitude of the right angled triangle.

1) Radius of the bigger circle is the diameter of the smaller circle.

2) Find the radius of the smaller circle.

3) Radius of the bigger circle is also the altitude of the right angled triangle corresponding to the side BC.

4) Subtract the sum of the areas of the smaller circle and the area of the right angled triangle from the area of the bigger circle to find the area of the shaded region.

Given that AB and CD are the diameters of a circle with centre O.

∴ OA = OB = OC = OD = 7 cm (Radius of the circle)

Area of the shaded region

= Area of the circle with diameter OB + (Area of the semi-circle ACDA − Area of ΔACD)

= π$r^2$ + ( $\frac{1}{2}$π$r^2$ -$\frac{1}{2}$ x CD x OA)

= $\frac{22}{7}$×${\left(\frac{7}{2}\right)}^2$+$(\frac{1}{2}\times \frac{22}{7}\times 49-\frac{1}{2}\times 14\times 7)$

= ($\frac{77}{2}$+ 77 - 49) ${\mathrm{cm}}^2$

= 66.5 ${\mathrm{cm}}^2$

∴The area of the shaded region is 66.5${\mathrm{cm}}^2$.

1) Find the sum of the inner and the outer surface areas of the vessel.

2) Ignore the thickness of the vessel.

1) Find the radius of the bowl and the cylinder from the diameter.

2) Find the height of the cylindrical part of the vessel.

3) Find two times of the sum of the curved surface area of the cylinder and the surface area of the hemisphere.

4) Simplify to find the total surface area of the vessel.

Assume the radius and the height of the cylinder as r cm and h cm respectively.

The diameter of the hemispherical bowl = 14 cm

∴ Radius of the hemispherical bowl = Radius of the cylinder = r = $\frac{14}{2}$ cm = 7 cm

Total height of the vessel = 13 cm

∴ Height of the cylinder = Total height of the vessel − Radius of the hemispherical bowl

= 13− 7 = 6 cm

Total surface area of the vessel = 2 (Curved surface area of the cylinder + Surface area of the hemisphere)

= 2(2πrh + 2π $r^2$)

= 4πr(h + r)

= 4 × $\frac{22}{7}$ × 7 × (6 + 7)

= 1144 ${\mathrm{cm}}^2$.

∴ Total surface area of the vessel is 1144 ${\mathrm{cm}}^2$.

Radius of the hemisphere and the cylinder are equal.

1) Find the radius of the hemisphere from the diameter.

2) Subtract two time of the volume of the hemisphere from the volume of the cylinder to find the volume of the wood in the toy.

3) Substitute the values in the formula and simplify to find the total volume.

Given that the height of the cylinder, h = 10 cm

And radius of the cylinder = Radius of each hemisphere = r = 3.5 cm

Volume of wood in the toy = Volume of the cylinder − 2 × Volume of each hemisphere.

= (π $r^2$h) − (2 × $\frac{2}{3}$ × π $r^3$)

= [ $\frac{22}{7}$× ${\left(3.5\right)}^2$×10] −[ $\frac{4}{3}$× $\frac{22}{7}$× ${\left(3.5\right)}^3$]

= 385 − $\frac{539}{3}$

= $\frac{616}{3}$

= 205.33 ${\mathrm{cm}}^3$(Approximately)

∴ The volume of the wood in the toy is approximately 205.33 ${\mathrm{cm}}^3$.

Units of the area of the sector are ${\mathrm{cm}}^2$.

1) Note the angle subtended by the arc at the centre of the circle and the radius of the circle.

2) Write the formula to find the length of the arc of the circle.

3) Substitute the values and simplify to find the result.

4) Write the formula to find the area of the sector formed by the arc.

5) Substitute the values and simplify to find the result.

Given that the radius of the circle = 21 cm.

Measure of the angle subtended by the arc at the centre = 60°.

(i) Length of the arc

l = $\frac{\theta }{360°}$ × 2πr

where r = 21 cm and θ = 60°

= $\frac{60°}{360°}$ × 2 × $\frac{22}{7}$ × 21

= 22 cm.

(ii) Area of the sector formed by the arc

A = $\frac{\theta }{360°}$×π$r^2$

r = 21 cm and θ = 60°

= $\frac{60°}{360°}$ × $\frac{22}{7}$ × ${\left(21\right)}^2$

= 231${\mathrm{cm}}^2$.

Factorise the equation by rearranging the terms of the equation.

1) Rearrange the fractions of the equation such that the terms with variable 'x'

2) Find the LCM of the fractions and simplify.

3) Find the factors of the quadratic equation obtained by taking common.

4) Equate each factor to zero and simplify to find the values of x.

The given equation is

$\frac{1}{2a+b+2x}=\frac{1}{2a}$ + $\frac{1}{b}$ + $\frac{1}{2x}$

⇒$\frac{1}{2a+b+2x}-\frac{1}{2x}=\frac{1}{2a}$ + $\frac{1}{b}$$$

⇒$\frac{2x-2a-b-2x}{2x(2a+b+2x)}$ = $\frac{b+2a}{2\mathrm{ab}}$

⇒$\frac{-(2a+b)}{2x(2a+b+2x)}$ = $\frac{2a+b}{2\mathrm{ab}}$

⇒$\frac{-1}{x(2a+b+2x)}$ = $\frac{1}{\mathrm{ab}}$

⇒2$x^2$+ 2ax + bx + ab = 0

⇒2x(x + a) + b(x +a) = 0

⇒(x +a)(2x +b) = 0

⇒x + a = 0 or 2x + b = 0

⇒ x = −a or x = −$\frac{b}{2}$.

1) Substitute the value of either x or y obtained from equation (1) in equation (2).

2) Factorise the quadratic equation using the factors of the constant.

3) Do not consider the negative values as the lengths.

1) Assume the lengths of sides of the square as x and y where x>y.

2) Find the sum of the areas of the squares and obtain equation (1).

3) Find the difference between the perimeters of the squares and obtain equation (2).

4) Find the value of x in terms of y from equation (2)

5) Substitute the value of x in equation (1) and simplify to get a quadratic equation.

6) Solve the quadratic equation by splitting the middle term and find the value of y.

7) Find the value of x using equation (2). x and y are the required lengths of sides of the squares.

Assume that the sides of the two squares be x cm and y cm where x > y.

The areas of the squares are $x^2$ and $y^2$and their perimeters are 4x and 4y.

Given that, $x^2$+ $y^2$= 400 … (1)

and 4x − 4y = 16

⇒ 4(x − y) = 16 ⇒ x - y = 4 … (2)

⇒ x = y + 4

Substituting the value of x in eqn (1), we get ${(y+4)}^2$+ $y^2$= 400

⇒ $y^2$+16 + 8y +$y^2$ = 400

⇒2$y^2$+ 8y + 16 = 400

⇒ $y^2$+ 4y - 192 = 0

⇒ $y^2$ + 16y −12y - 192 = 0

⇒ y(y + 16) −12(y + 16) = 0

⇒ (y + 16) (y −12) = 0

⇒ y = −16 or y = 12

Since the value of y cannot be negative so the value of y = 12.

Value of x = y + 4 = 12 + 4 = 16.

∴The sides of the two squares are 16 cm and 12 cm.

Use the formula to find the sum of n terms with n and d terms.

1) Find the sum of the first 7 terms of the A.P. and obtain equation (1)

2) Find the sum of the first 17 terms of the A.P. and obtain equation (2)

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

4) Substitute the values of a and d in the formula to find the sum of n terms of an A.P. and simplify.

Given that, the sum of first 7 terms of an A.P. is 49 i.e.$S_7$ = 49 and the sum of first 17terms of an AP is 289 i.e$S_{17}$ = 289.

$S_n$=$\frac{n}{2}$[2a + (n - 1)d ]

⇒$S_7$=$\frac{7}{2}$[ 2a + 6d ]

⇒49 = $\frac{7}{2}$[2a + 6d]

⇒2a + 6d = 14

⇒ a + 3d = 7 ------------ (1)

Similarly,

$S_{17}$= $\frac{17}{2}$[2a +( 17 - 1) d ]

289 = $\frac{17}{2}$[2a + 16d ]

⇒ a + 8d = 17 ------------ (2)

Solving equations (1) and equation (2), we get 5d = 10.

∴ d = 2

∴ If d = 2 then, a = 7 − 3d = 1.

$S_n$= $\frac{n}{2}$[2a + (n - 1)d]

$S_n$= $\frac{n}{2}$[ 2 + ( n - 1)2 ]

$S_n$= $\frac{n}{2}$[ 2n]

∴ $S_n$= $n^2$.

∴The sum of n terms of the A.P. is $n^2$.

The shortest distance between two points is the perpendicular distance.

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

2) Mark any point B on the tangent 'l' and draw a line segment joining the centre O and the point B.

3) Let the line segment intersect the circle at C.

4) OB is written as the sum of two parts OC and CB.

5) OB is greater than OC. So, OB is greater than OA since OA and OC are radii of the circle.

6) OA is shorter than OB.

7) OA is shorter than any other segment joining the centre of the circle and a point on the tangent.

8) This proves that OA, the radius at the point of contact of the tangent is perpendicular to the tangent.

Given:In a circle C (0, r) and a tangent l touches the circle at point A.

To prove: OA ⊥ l

Construction: Consider a point B on l, other than A, on the tangent l . Join OB. Let OB intersect the circle in C.

Proof: Among all line segments joining the point O to a point on l, the perpendicular is shortest to l.

OA = OC (Radius of the circle)

Now, OB = OC + BC.

∴ OB > OC

⇒ OB > OA

⇒ OA < OB

As B is any point on the tangent l, OA is shorter than any other line segment joining O to any point on l.

There fore, OA ⊥ l.

Hence, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Prove that the corresponding sides of the triangles are equal to prove that they are congruent.

Sum of the angles of a straight line is 180°.

1) Join O and C.

2) Compare the triangles OAD and OCD and prove that they are congruent.

3) Angles DOA and COD are equal according to CPCT.

4) Similarly prove that the triangles BOE and COE are congruent.

5) Angles BOE and EOC are equal according to CPCT.

6) Equate the sum of the angles on the diameter AOB and equate to 180°.

7) Simplify using the equality of angles and prove that the sum of ∠COD and ∠COE is a right angle.

8) Sum of ∠COD and ∠COE is∠DOE.

Given: l and m are two tangents to the circle with centre O parallel to each other and touching the circle at A and B respectively. DE is another tangent at the point C, which intersects l at D and m at E.

Required to prove: ∠DOE = 90°

Construction: Join OC.

Proof:

Consider ΔODA and ΔODC,

OA = OC (Radii of the same circle)

AD = DC (Length of tangents drawn from an external point to a circle are equal)

DO = OD (Common side)

ΔODA ≅ ΔODC (SSS congruence property)

∴ ∠DOA = ∠COD …........ (i) (C.P.C.T)

Similarly, we can prove that ΔOEB ≅ ΔOEC

∠EOB = ∠COE … (ii)

AOB is a diameter of the circle which is a straight line and the angle on a straight line is a straight angle.

∴ ∠DOA + ∠COD + ∠COE + ∠EOB = 180º.

From (i) and (ii), we get,

2∠COD + 2 ∠COE = 180º

⇒ ∠COD + ∠COE = 90º

⇒ ∠DOE = 90°

Hence, it is proved.

Apply appropriate trigonometric ratio to the angles of the elevation.

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

2) Complete the right angled triangles involving each of the angles of elevation.

3) Find the cotangent of the angle of elevation of the tower and simplify to find the distance between the building and the tower.

4) Find the tangent of the angle of elevation of the building and simplify to find the height of the building.

Assume AB as the building and CD as the tower.

Suppose the height of the building AB as 'h' m.

Given that, ∠ACB = 30°, ∠CBD = 60° and CD = 60 m.

ΔBCD is a right angled triangle then,

cot 60° = $\frac{\mathrm{BC}}{\mathrm{CD}}$ ⇒ BC = CD cot 60°

⇒ BC = 60 × $\frac{1}{\sqrt{3}}$= 20 $\sqrt{3}$m

ΔACB is a right angled triangle then,

tan 30° = $\frac{\mathrm{AB}}{\mathrm{BC}}$ ⇒ AB = BC tan 30°

⇒ h = 20 $\sqrt{3}$ × $\frac{1}{\sqrt{3}}$= 20m

∴ The height of the building is 20 m.

Find the total number of persons who are extremely honest or kind.

1) Note the total number of persons in the group which is considered as the total number of possible outcomes.

2) Assume the number of persons who are extremely patient as $E_1.$

3) Note the number of persons favourable to $E_1$.

4) Assume the number of persons who are extremely extremely kind or honest as $E_2$.

5) Find the number of persons who are extremely kind or honest.

6) Note the number of persons favourable to $E_2$.

7) Find the ratio of the number of outcomes favourable to $E_1$and the total number of possible outcomes to get the probability of the person selected is extremely patient.

8) Find ratio of the number of outcomes favourable to $E_2$and the total number of possible outcomes to get the probability of the person selected is extremely kind or honest.

Given that the group consists of 12 persons.

∴Total number of possible outcomes = 12

Assume $E_1$as the event of selecting persons who are extremely patient.

∴ Number of favourable outcomes to $E_1$is 3.

Assume $E_2$as the event of selecting persons who are extremely kind or honest.

Number of persons who are extremely honest is 6.

Number of persons who are extremely kind = 12 − (6 + 3) = 3

∴ Number of favourable outcomes to$E_2$ = 6 + 3 = 9.

(i) Probability of selecting a person who is extremely patient is P($E_1$).

P($E_1$) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{E}_1}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$ = $\frac{3}{12}$ = $\frac{1}{4}$.

(ii) Probability of selecting a person who is extremely kind or honest is

P ($E_2$) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{E}_2}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$ = $\frac{9}{12}$ = $\frac{3}{4}$.

Find the area of one of the triangle formed by the diagonals of the parallelogram and multiply it by 2.

1) Assume the coordinates of the fourth vertex as (x, y).

2) Equate the midpoints of the diagonals of the parallelogram as the diagonals of a parallelogram bisect each other.

3) Simplify to find the coordinates of the fourth vertex.

4) Area of a quadrilateral is equal to twice the area of the triangles formed by the diagonal of the parallelogram.

5) Find the area of one of the triangles using the area of the triangle formula.

6) Multiply the area of the triangle by 2 to obtain the area of the parallelogram.

Given that three vertices of the parallelogram ABCD are A (3, −4), B (−1, −3) and C (−6, 2).

Assume that the coordinates of the vertex D be (x, y).

In a parallelogram, the diagonals bisect each other.

∴Mid point of AC = Mid point of BD

⇒$(\frac{3-6}{2},\frac{-4+2}{2})$= $(\frac{-1+x}{2},\frac{-3+y}{2})$

⇒$(\frac{-3}{2},\frac{-2}{2})$= $(\frac{-1+x}{2},\frac{-3+y}{2})$

⇒$\frac{-3}{2}$= $\frac{-1+x}{2}$, $\frac{-2}{2}$= $\frac{-3+y}{2}$

⇒ x = −2, y = 1.

So that, the coordinates of the fourth vertex D is (−2, 1).

Now, area of parallelogram ABCD

= area of triangle ABC + area of triangle ACD

= 2 × area of triangle ABC [Diagonal divides the parallelogram into two triangles of equal area]

The area of a triangle whose vertices are ($x_1$, $y_1$), ($x_2$, $y_2$) and ($x_3$, $y_3$) is $\frac{1}{2}$[$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$]

Area of triangle ABC

= $\frac{1}{2}$|[3(-3 - 2) + (-1)(2 - (-4)) + (-6)( -4 -(-3) )]|

= $\frac{1}{2}$| [-15 -6 + 6]| = $\frac{\mid -15\mid }{2}$square units.

∴Area of the triangle ABC = $\frac{15}{2}$ square units (Area of the triangle cannot be negative)

∴ Area of the parallelogram ABCD = 2 × $\frac{15}{2}$ = 15 square units.

1) Convert the diameter of the cylindrical pipe into metres.

2) Find the speed of the water in metres per minute.

1) Find the radius of the circular end of the pipe in metres.

2) Find the area of the cross section of the pipe in terms of π.

3) Find the speed of the water in metres per minute.

4) Multiply the speed of the water by the area of the cross section of the pipe to get the volume of the water flown in one minute.

5) Find the volume of the water flown through the pipe in 30 minutes.

6) Assume the level of the water raised in the tank as 'h'.

7) Equate the volume of the water flown through the pie in 30 minutes and the volume of the water filled in the tank.

8) Simplify to find the rise in level of water in the tank by the water filled in half an hour.

Given that, internal diameter of circular end of pipe = 2 cm

∴ Radius ($r_1$) of circular end of pipe = $\frac{2}{100\times 2}$ m = 0.01 m

Area of cross-section of the circular end of pipe = π${r_1}^2$= π${\left(0.01\right)}^2$= 0.0001π $m^2$

Speed of water = 0.4 m/s = 0.4 ×60 = 24 metre / min.

Volume of water that flows through the pipe in 1 minute = 24 × 0.0001 π$m^3$= 0.0024π $m^3$

Volume of water that flows through the pipe in 30 minutes = 30 × 0.0024 π$m^3$ = 0.072π $m^3$

Radius ($r_2$) of base of cylindrical tank = 40 cm = 0.4 m

Let the rise in level of the water in the cylindrical tank filled in 30 minutes be h m.

Volume of water filled in tank in 30 minutes = Volume of water flowed in 30 minutes from the pipe.

∴ π ×${\left(r_2\right)}^2$×h = 0.072π

⇒ ${\left(0.4\right)}^2$×h = 0.072

⇒ 0.16 × h = 0.072

⇒h = $\frac{0.072}{0.16}$

⇒ h = 0.45 m = 45 cm

∴ The rise in level of water in the tank in half an hour is 45 cm.

Consider the area of the base of the bucket only as it is open at the top.

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

2) Find the slant height of the bucket in the shape of frustum of a cone using the Pythagoras theorem.

3) Area of metal sheet required to make the bucket is equal to the sum of the curved surface area of the frustum of the cone and the area of the base of the bucket.

4) Substitute the values in the formula and simplify to get the final answer.

5) Multiply the area of the sheet required by the cost of the sheet to find the total cost of the sheet.

Given that diameter of upper end of bucket = 30 cm

∴ Radius ( $r_1$) of upper end of bucket = 15 cm

Diameter of lower end of bucket = 10 cm

∴ Radius ( $r_2$) of lower end of bucket = 5 cm

Height (h) of bucket = 24 cm

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

= $\sqrt{{(15-5)}^2+{24}^2}$ = $\sqrt{100+576}$

= $\sqrt{676}$= 26 cm

Area of metal sheet required to make the bucket = π( $r_1$+ $r_2{)}^2$l

+ π ${r_2}^2$

= π (15 + 5) 26 + π( $5^2$)

= 520 π + 25 π = 545 π ${\mathrm{cm}}^2$

Cost of 100 ${\mathrm{cm}}^2$metal sheet = Rs 10

Cost of 545π ${\mathrm{cm}}^2$metal sheet = Rs $\frac{545\times 3.14\times 10}{100}$ = Rs.171.13

∴Total cost of metal sheet used to make the bucket is Rs 171.13