CBSE Class 10 - Maths

Substitute the values of the coordinates in the formula of the discriminant along with their signs.

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

2) Equate the discriminant of the quadratic equation to zero.

3) Substitute the values and simplify to find the value of k.

Given quadratic equation is$x^2$−4kx + k = 0

Given equation has equal roots. So, the discriminant of the equation, D = 0.

Discriminant of the equation a$x^2$+ bx + c = 0 is $b^2$-4ac.

⇒$b^2$−4ac = 0

⇒${\left(-4k\right)}^2$−4$\left(1\right)\left(k\right)$= 0

⇒16$k^2$−4k = 0

⇒4k( 4k−1) = 0

⇒4k = 0 or 4k−1 = 0

⇒ k = 0 or k = $\frac{1}{4}$

Therefore, the values of k are 0 or$\frac{1}{4}$ for which the given equation will have equal roots.

n value of the last term gives the number of terms of the A.P.

1) List the three digit numbers which are divisible by 11.

2) Note the first and the last term of the series.

3) Equate the last term of the series to the $n^{\mathrm{th}}$term of an A.P. and find the value of n.

4) Find the sum of n terms of the A.P. using the values obtained.

The smallest and the largest three digit natural numbers, that are divisible by 11 are 110 and 990 respectively.

So, the three digit numbers which are divisible by 11 are 110, 121, 132, …, 990.

The numbers form an A.P. with first term, a = 110 and the common difference, d = 11.

Assume that there are n terms in the sequence.

Therefore,$a_n$ = 990

$a_n$= a + ( n - 1) d = 990

⇒110 + ( n - 1)11 = 990

⇒110 + 11n - 11 = 990

⇒11n = 990 - 99

⇒11n = 891

⇒ n = 81.

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

= $\frac{81}{2}$[ 2$\left(10\right)$+ ( 81 -1)11]

= $\frac{81}{2}$[ 220 + 880 ]

= $\frac{81}{2}$[ 1100 ]

= 44550.

∴The sum of all three digit numbers which are multiples of 11 is 44550.

OP is the hypotenuse in both the triangles OAP and OBP.

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

2) Join OA, OB and OP.

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

4) So, the triangles OAP and OBP are the right angled triangles.

5) Apply Pythagoras theorem in triangle OAP and find the length of OP.

6) Apply Pythagoras theorem in triangle OBP and find the length of OB.

Given : OA = 8 cm, OB = 5 cm and AP = 15 cm

To find: BP

Construction: Join O and P i.e OP.

Now OA ⊥AP and OB ⊥BP [ Tangent to a circle is perpendicular to the radius through the point of contact ]

⇒ ∠OAP = ∠OBP = 90°

On applying Pythagoras theorem in ΔOAP, we get

${\left(\mathrm{OP}\right)}^2$= ${\left(\mathrm{OA}\right)}^2$+ ${\left(\mathrm{AP}\right)}^2$

⇒ ${\left(\mathrm{OP}\right)}^2$ = ${\left(8\right)}^2$+ ${\left(15\right)}^2$

⇒ ${\left(\mathrm{OP}\right)}^2$= 64 + 225

⇒ ${\left(\mathrm{OP}\right)}^2$ = 289

⇒ OP = 17

∴The length of OP is 17 cm.

Now applying Pythagoras theorem in ΔOBP, we get

${\mathrm{OP}}^2$= ${\mathrm{OB}}^2$+ ${\mathrm{BP}}^2$

⇒ ${\left(17\right)}^2$= ${\left(5\right)}^2$+ ${\left(\mathrm{BP}\right)}^2$

⇒ 289 = 25 + ${\left(\mathrm{BP}\right)}^2$

⇒ ${\left(\mathrm{BP}\right)}^2$= 289 − 25

⇒ ${\left(\mathrm{BP}\right)}^2$= 264

⇒ BP = 16.25 cm (approximately)

∴The length of BP is 16.25 cm.

If BP = PC then, P is the midpoint of BC.

OR

If OC bisects AB then, AC = BC.

1) Tangents drawn from an external point to a circle are equal. So, equate the tangents drawn from the vertices of the triangle.

2) Equate AB and AC of the triangle.

3) Write the parts of AB and AC.

4) Replace the equal tangents of the circle and simplify to prove that BP and CP are equal and hence P bisects BC.

OR

1) Join OC which is a radius at the point of contact of the tangent to the inner circle.

2) The radius at the point of contact of a tangent at any point on the circle is perpendicular to the tangent.

3) Tangent of the inner circle is also the chord of the outer circle.

4) Perpendicular from the centre to the chord bisects the chord.

Given: In isosceles triangle ABC with AB = AC, circumscribing a circle.

To prove: P bisects BC

Proof: AR and AQ are the tangents drawn to the circle from an external point A.

∴ AR = AQ (Tangents drawn from an external point to the circle are equal)

Similarly, BR = BP and CP = CQ.

Given that in ΔABC, AB = AC.

⇒ AR + RB = AQ + QC

⇒ BR = QC (Since AR = AQ)

⇒ BP = CP (Since BR = BP and CP = CQ)

⇒ P bisects BC

Hence, it is proved.

OR

Given: Two concentric circles $C_1$ and $C_2$ with centre O, and AB is the chord of $C_1$ touching $C_2$ at C.

To prove: AC = CB

Construction: Join OC.

Proof: AB is the chord of $C_1$ touching $C_2$ at C, then AB is the tangent to the circle $C_2$ at C with OC as radius.

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

∴ OC ⊥ AB

But AB as the chord of the circle $C_1$. So, OC ⊥ AB.

∴ OC is the bisector of the chord AB.

Therefore, AC = CB (Perpendicular from the centre to the chord bisects the chord).

Consider the volume of the hemisphere and radius in the form of a fraction.

1) Convert the volume of the hemisphere into an improper fraction.

2) Assume the radius of the hemisphere as 'r'.

3) Find the volume of the hemisphere and equate it to the given value.

4) Simplify and find the radius of the hemisphere.

5) Find the curved surface area of the hemisphere using the radius obtained.

Given that volume of hemisphere = 2425$\frac{1}{2}{\mathrm{cm}}^3$= $\frac{4851}{2}{\mathrm{cm}}^3$

Consider the radius of the hemisphere as ‘r’ cm.

Volume of a hemisphere = $\frac{2}{3}$π$r^3$

⇒$\frac{2}{3}$π$r^3$= $\frac{4851}{2}$

⇒$\frac{2}{3}$ × $\frac{22}{7}\times r^3$= $\frac{4851}{2}$

⇒$r^3$= $\frac{4851\times 3\times 7}{2\times 2\times 22}$

⇒$r^3$= $\frac{441\times 21}{2\times 2\times 2}$

⇒$r^3$= $\frac{21\times 21\times 21}{2\times 2\times 2}$

⇒r = $\sqrt[3]{\frac{21\times 21\times 21}{2\times 2\times 2}}$

⇒r = $\frac{21}{2}$ cm ----------- (1)

∴Curved surface area of a hemisphere = 2π$r^2$

⇒2 × $\frac{22}{7}$ × ${\left(\frac{21}{2}\right)}^2$

= 2 × $\frac{22}{7}$ × $\frac{21\times 21}{2\times 2}$

= 693 ${\mathrm{cm}}^2$.

Radius of the quadrant of the circle is equal to the side of the square.

1) Find the area of the square using the given length of the side.

2) Radius of the quadrant of the circle is equal to the side of the square.

3) Find the area of the quadrant of the circle.

4) Subtract the area of the quadrant of the circle from the area of the square to find the area of the shaded region of the square.

Given that OABC is a square of side 7 cm.

∴ Area of square OABC = ${\left(7\right)}^2$= 49${\mathrm{cm}}^2$

Given that OAPC is a quadrant of circle with centre O.

∴Radius of the quadrant of the circle = OA = 7 cm

∴ Area of quadrant of the circle = $\frac{1}{4}$π$r^2$

= $\frac{1}{4}$ × $\frac{22}{7}$ × $7^2$

= $\frac{1}{2}$ × 11 × 7

= $\frac{77}{2}{\mathrm{cm}}^2$

Area of the shaded region = Area of the square − Area of the quadrant of the circle.

= ( 49 - $\frac{77}{2}$)

=$\frac{(98-77)}{2}$

= $\frac{21}{2}$

= 10.5${\mathrm{cm}}^2$

∴ Area of the shaded region is 10.5${\mathrm{cm}}^2$

Find the distance between the points by substituting the coordinates in the distance formula in correct order.

1) Point A is equidistant from B and C. So, equate the distances AB and AC.

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

3) Simplify and find the value of p.

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

Given that A is equidistant from the points B and C.

∴ AB = AC

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

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

Squaring on both sides, we get

⇒$p^2$−4p + 13 =$p^2$+ 9

⇒ −4p = − 4

⇒ p = 1.

Number selected up to 50 must be a multiple of both 3 and 4 i.e. a multiple of 12.

1) List the natural numbers from 1 to 50.

2) Note the number of possible outcomes.

3) List the numbers which are divisible by 3 and 4 between 1 and 50.

4) Note the number of favourable outcomes.

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

The first fifty natural numbers are 1, 2, 3,..........50.

Total number of possible outcomes = 50

Multiples of 3 and 4 which are less than or equal to 50 are:

12, 24, 36, 48

Number of favourable outcomes = 4

Let E be the favourable outcome.

P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{4}{50}$

= $\frac{2}{25}$

Thus, the probability of the number selected is a multiple of 3 and 4 is $\frac{2}{25}$.

Split the middle term using the constant term and the coefficient of $x^2$.

OR

Consider the coefficients of the terms along with their sign.

1) Split the middle term of the quadratic equation using the factors of the constant term and the coefficient of $x^2$.

2) Factorise the equation by taking common in the first two and the last two terms.

3) Equate each of the factors to zero and find the value of x.

OR

1) Equate the given quadratic equation to the standard form of the quadratic equation and find the values of a, b and c.

2) Find the value of the discriminant by using the coefficients.

3) Find the roots of the quadratic equation by substituting the given values in the formula to find the roots of the quadratic equation.

The given quadratic equation is 4$x^2$ − 4ax + ($a^2$−$b^2$) = 0

⇒ 4$x^2$ − 4ax + ( a - b)( a + b) = 0

⇒ 4$x^2$ −( a - b)( a + b) = 0

⇒ 2x [ 2x - ( a - b) ] - ( a + b)[ 2x - ( a - b) ] = 0

⇒ [2x - ( a - b)] [ 2x - ( a + b) ] = 0

⇒ 2x = ( a - b) or 2x = ( a + b)

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

∴ The solution of the given quadratic equation is given by x = $\frac{a-b}{2}$ or x = $\frac{a+b}{2}$

OR

Given quadratic equation is 3$x^2$-2$\sqrt{6}$x + 2 = 0.

Comparing coefficients with the quadratic equation a$x^2$+ bx + c = 0, we have a = 3 , b = - 2$\sqrt{6}$ and c = 2.

Discriminant of the given quadratic equation,

D = $b^2$-4ac = ${\left(2\sqrt{6}\right)}^2$-4$\left(3\right)\left(2\right)=24-24=0$

Roots of a quadratic equation a$x^2$+ bx + c = 0 are $\frac{-b\pm \sqrt{D}}{2a}$.

x = $\frac{-\left(-2\sqrt{6}\right)\pm \sqrt{0}}{2\times 3}$

x = $\frac{2\sqrt{6}}{6}$

x = $\frac{\sqrt{6}}{3}$

Hence, the solution of the given quadratic equation is x = $\frac{\sqrt{6}}{3}$.

1) Lengths of tangents drawn from an external point to the circle are equal.2) Group the tangents according to the sides of the parallelogram.

OR

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

1) In a parallelogram ABCD , the circle touch the sides AB, BC, CD and AD at P, Q, R and S respectively.

2) Since, ABCD is a parallelogram then AB = CD and BC = AD.

3) Lengths of tangents drawn from an external point to the circle are equal.

4) Adding all these and grouping the lengths of these tangents according to the sides of the parallelogram.

5) Therefore, ABCD is a parallelogram that has sides of equal length.

6) Hence, the parallelogram ABCD is also a rhombus.

OR

1) Draw a quadrilateral ABCD circumscribing a circle touching the circle at P, Q, R and S.

2) Draw lines joining the vertices of the quadrilateral to the centre of the circle.

3) Draw perpendiculars OP, OQ, OR and OS on the sides AB, BC, CD and AD of the quadrilateral.

4) Prove that the triangles OAP and OAS are congruent and hence the angles ∠POA and ∠AOS are equal.

5) Similarly prove that the angles in the pairs of adjacent right angled triangles are equal.

6) Add the equations of angles and regroup the angles.

7) Simplify and prove that the angles subtended by the opposite sides of the quadrilateral are supplementary.

Given that, ABCD is a parallelogram.

AB = CD … (1)

BC = AD … (2)

We know that

DR = DS (Tangents to the circle from point D)

CR = CQ (Tangents to the circle from point C)

BP = BQ (Tangents to the circle from point B)

AP = AS (Tangents to the circle from point A)

Adding all these equations, we get

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

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

CD + AB = AD + BC

On substituting the values of equations (1) and (2) in this equation, we get

2AB = 2BC

AB = BC … (3)

Comparing equations (1), (2), and (3), we obtain

AB = BC = CD = DA

∴ABCD is a rhombus.

OR

Consider ABCD a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R and S.

Let us join the vertices of the quadrilateral ABCD to the centre of the circle.

In ΔOAP and ΔOAS,

AP = AS (Tangents from the same point)

OP = OS (Radii of the same circle)

OA = OA (Common side)

ΔOAP ≅ ΔOAS (SSS congruence criterion)

Therefore, A ↔ A, P ↔ S, O ↔ O

And thus, ∠POA = ∠AOS

∠1 = ∠8

Similarly,

∠2 = ∠3

∠4 = ∠5

∠6 = ∠7

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360º

(∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º

2∠1 + 2∠2 + 2∠5 + 2∠6 = 360º

2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º

(∠1 + ∠2) + (∠5 + ∠6) = 180º

∠AOB + ∠COD = 180º

Similarly, we can prove that ∠BOC + ∠DOA = 180º

∴ Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Locate five points on BX such that they are equidistant.

1) Construct a triangle ABC using the measurements of three sides.

2) Draw a ray BX making an acute angle with BC.

3) Mark 5 points on BX and join the fifth point to C.

4) Draw a line $B_3$C' parallel to $B_5$C from $B_3$.

5) Draw a line C'A' parallel to CA.

6) A'BC' is the required triangle.

Given that

BC = 6 cm, AC = 8 cm

The triangle to be formed is to be right angled triangle.

Steps of construction:

Step 1: Draw a line segment, BC = 6 cm.

Step 2: Draw a ray CN at C making an angle of 90°.

Step 3: From C as centre, 8 cm as the radius draw an arc at CN intersecting it at A and Join AB.

Step 4: ABC is the triangle whose similar triangle is to be drawn.

Step 5: Draw any ray BX making an acute angle with BC on the opposite side of the vertex A.

Step 6: Locate 5 (Greater of 3 and 5 in $\frac{3}{5}$) points $B_1$, $B_2$,$B_3$,$B_4$and$B_5$on BX such that B$B_1$= $B_1{B}_2$= $B_2{B}_3$=$B_3{B}_4$= $B_4{B}_5$.

Step 7: Join $B_5$C and draw a line through$B_3$ (Smaller of 3 and 5 in $\frac{3}{5}$) parallel to $B_5$C to intersect BC at C’.

Step 8: Draw a line through C’parallel to CA to intersect BA at A’.

Step 9: A’BC’ is the required similar triangle whose sides are $\frac{3}{5}$ times the corresponding sides of ΔABC.

Angle subtended by both the arcs at the centre is 30°.

1) Shaded region lies between the sectors formed by the arcs.

2) Difference between the areas of the sectors formed by the arcs of the concentric circles gives the area of the shaded region.

3) Angle subtended by both the arcs at the centre is 30°.

4) substitute the values and simplify to find the area of the shaded region.

Let PQ and AB are the arcs of two concentric circles of radii 7 cm and 3.5 cm respectively.

Assume $r_1$and$r_2$as the radii of the outer and the inner circle respectively.

Suppose the angle subtended by the arcs at the centre O as θ.

Then$r_1$= 7 cm, $r_2$= 3.5 cm and θ = 30°

Area of the shaded region = Area of sector OPQ − Area of sector OAB

= $\frac{\theta }{360°}$× π${r_1}^2$−$\frac{\theta }{360}$ × π${r_2}^2$

= $\frac{\theta }{360°}$π(${r_1}^2$−${r_2}^2$)

= $\frac{30°}{360°}$ × $\frac{22}{7}$ × ($7^2$−${\left(3.5\right)}^2$)

= $\frac{1}{12}$ × $\frac{22}{7}$ × 36.75

= 9.265 ${\mathrm{cm}}^2$

∴ The area of the shaded region is 9.625 ${\mathrm{cm}}^2$.

Consider one base area of the cylinder to find the total surface area of the remaining solid.

OR

Equate the volumes of the cylinder and the conical heap to find the radius of the conical heap.

1) Find the radius of the cylindrical part of the solid cylinder.

2) Find the slant height of the conical part using the radius and the height of the cylinder.

3) Total surface area of the remaining solid is the sum of the curved surface areas of the cylindrical part and the conical part, and the base area of the cylindrical part.

4) Substitute the values and simplify to find the total surface area of the remaining solid.

OR

1) Assume the radii of the cylindrical bucket and the conical heap as $r_1$and $r_2$.

2) Equate the volumes of the sand in the form of a cylinder and the conical heap.

3) Substitute the values and simplify to find the radius of the conical heap.

4) Find the slant height of the conical heap of sand using the vertical height and the radius.

Given that, height (h) of cylindrical part = height (h) of the conical part = 7 cm

Diameter of the cylindrical part = 12 cm

∴Radius (r) of the cylindrical part = $\frac{12}{2}$= 6 cm

∴ Radius of conical part = 6 cm

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

= $\sqrt{6^2+7^2}$

= $\sqrt{36+49}$

= $\sqrt{85}$

= 9.22 cm ( approximately)

Total surface area of the remaining solid

= CSA of cylindrical part + CSA of conical part + Base area of the circular part

= 2πrh + πrl + π $r^2$

= (2 × $\frac{22}{2}$ × 6 × 7) + ( $\frac{22}{7}\times$6 × 9.22) + ( $\frac{22}{7}$ × 6 × 6)

= (264 + 173.86 + 113.14) ${\mathrm{cm}}^2$

= 551 ${\mathrm{cm}}^2$

OR

Height ( $h_1$) of the cylindrical bucket = 32 cm

Radius ( $r_1$) of circular end of the bucket = 18 cm

Height ( $h_2$) of the conical heap = 24 cm

Let the radius of the circular end of conical heap be $r_2$.

The volume of sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.

Volume of sand in the cylindrical bucket = Volume of sand in conical heap

⇒π ${r_1}^2{h}_1$= $\frac{1}{3}\pi {r_2}^2{h}_2$

⇒π $\times {18}^2\times$32 = $\frac{1}{3}$π × ${r_2}^2$× 24

⇒ ${r_2}^2$= $\frac{3\times {18}^2\times 32}{24}$= ${18}^2$× 4

⇒ $r_2$ = 18 ×2 = 36 cm

Slant height (l) = $\sqrt{{36}^2+{24}^2}$ = 12 $\sqrt{13}$cm

Hence, the radius and slant height of the conical heap are 36 cm and 12 $\sqrt{13}$cm respectively

Apply appropriate trigonometric ratio to the angles of depression of two ships from the top of the light house.

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

2) Assume 'h' as the height of the light house and 'x' as the distance between the foot of the light house and the first ship.

3) Mark the alternate angles of the angles of depression.

4) Find the tangent of the alternate angle of the angle of depression of the first ship and find the value of 'x' in terms of 'h'.

5) Find the tangent of the alternate angle of the angle of depression of the second ship and find the value of h.

6) Substitute the value of 'x' in the terms of h in the trigonometric ratio and simplify to find the height of the light house.

The given solution can be represented as,

Consider AB as the light house and the ships are at the points C and D.

Let h be the height of the light house and BC = x.

In a right angled ΔABC:

tan 45° = $\frac{\mathrm{AB}}{\mathrm{AC}}$

⇒1 = $\frac{h}{x}$

⇒x = h

In a right angled ΔABD,

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

⇒ tan 30° = $\frac{h}{x+200}$

⇒ tan 30° = $\frac{h}{h+200}$

⇒ $\frac{1}{\sqrt{3}}$ = $\frac{h}{h+200}$

⇒ $\sqrt{3}h$ - h = 200

⇒h( $\sqrt{3}$ - 1) = 200

⇒h = $\frac{200}{\sqrt{3}-1}$

⇒h = $\frac{200}{\sqrt{3}-1}$× $\frac{\sqrt{3}+1}{\sqrt{3}+1}$

⇒h = $\frac{200}{2}(\sqrt{3}+1)$

⇒h = 100 ( $\sqrt{3}$+ 1 )

∴The height of the light house is 100 ( $\sqrt{3}$+ 1) m.

Substitute the values of the coordinates according to the section formula in order.

1) The point P divides the line segment AB in the ratio of k:1.

2) Use the section formula and find the coordinates of the point P.

3) The point P lies on the line x + y = 0, so it should satisfy the equation.

4) Substitute the coordinates of the point P for x and y in the given equation.

5) Simplify to find the value of k.

The given points are A (3, −5) and B (− 4, 8).

Consider $x_1$= 3, $y_1$= −5, $x_2$= −4 and$y_2$= 8.

Since $\frac{\mathrm{AP}}{\mathrm{PB}}$ = $\frac{k}{1}$, the point P divides the line segment joining the points A and B in the ratio of k : 1. The coordinates of P can be found using the section formula .

$(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Put, m = k and n = 1

Coordinates of P = $(\frac{k\times -4+1\times 3}{k+1},\frac{k\times 8+1\times -5}{k+1})$

= $(\frac{-4k+3}{k+1},\frac{8k-5}{k+1})$

Given that, P lies on the line x + y = 0.

$\frac{-4k+3}{k+1}$+ $\frac{8k-5}{k+1}$= 0

$\frac{-4k+3+8k-5}{k+1}$= 0

4k - 2 = 0

k = $\frac{1}{2}$

∴ The required value of k is $\frac{1}{2}$.

Find two values of p, by equating the area to both positive and negative values of the given area.

1) Find the area of the triangle formed by the given points using the formula to find the area of the triangle.

2) Equate the expression obtained to both positive and negative values of the given area.

3) Simplify and find two values of p.

Given that vertices of a triangle are (1, −3), (4, p) and (−9, 7).

$x_1$= 1, $y_1$= -3 , $x_2$= 4 , $y_2$= p , $x_3$= -9 and $y_3$= 7.

Area of given triangle = $\frac{1}{2}$[$x_1$($y_2$- $y_3$) + $x_2$($y_3$- $y_1$) + $x_3$($y_1$- $y_2$)

= $\frac{1}{2}$[1$\left(p-7\right)$+ 4$\left(7+3\right)$+ $\left(-9\right)\left(-3-p\right)$]

= $\frac{1}{2}$[p - 7 + 40 + 27 + 9p ]

= $\frac{1}{2}$[ 10p + 60 ]

= 5( p + 6)

The expression may be positive or negative.

⇒5(p + 6) = 15 or 5(p + 6) = − 15

⇒ p + 6 = 3 or p + 6 = − 3

∴ p = − 3 or p = − 9.

1) Cards other than yellow includes red and blue cards.

2) Cards which are neither yellow nor blue are red.

1) Find the total number of cards of all the three colours.

2) Ratio of number of blue cards and the total number of cards gives the probability that the card drawn is a blue card.

3) Ratio of the sum of the number of red and blue cards, and the total number of cards gives the probability that the card drawn is not a yellow card.

4) Ratio of number of red cards and the total number of cards gives the probability that the card drawn is neither a yellow nor a blue card.

Number of red cards = 100

Number of yellow cards = 200

Number of blue cards = 50

Total number of cards = 100 + 200 + 50 = 350

(i)

P(A blue card) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{blue}\mathrm{card}s}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{card}s}$

= $\frac{50}{350}$

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

(ii)

P(Not a yellow card) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{card}s\mathrm{other}\mathrm{than}\mathrm{yellow}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{card}s}$

= $\frac{\mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{card}s+\mathrm{Number}\mathrm{of}\mathrm{blue}\mathrm{card}s}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{card}s}$

= $\frac{100+50}{350}$

= $\frac{150}{350}$

= $\frac{3}{7}$.

(iii)

P(Neither yellow nor a blue card) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{card}s\mathrm{that}\mathrm{are}\mathrm{neither}\mathrm{yellow}\mathrm{nor}\mathrm{blue}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{card}s}$

= $\frac{\mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{card}s}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{card}s}$

= $\frac{100}{350}$

= $\frac{2}{7}$.

Find the value of a in terms of d.

1) Write the equation representing the given condition.

2) Substitute the formulas to find the ${17}^{\mathrm{th}}$term and $8^{\mathrm{th}}$term in the given condition.

3) Simplify the equation and find the value of a in terms of d.

4) Find the ${11}^{\mathrm{th}}$term of the A.P. and equate it to the given value.

5) Substitute the value of a in terms of d.

6) Simplify to find the value of d.

7) Find the value of a using the value of d.

8) Find the $n^{\mathrm{th}}$term of the A.P. by substituting the values of a and d in the formula to find the $n^{\mathrm{th}}$term of an A.P.

Let a be the first term and d be the common difference of the given A.P.

According to the given condition,

${17}^{\mathrm{th}}$term = (2 ×$8^{\mathrm{th}}$ term) + 5

i.e.,$a_{17}$ = 2$a_8$ + 5

⇒a + (17−1)d = 2[a + (8− 1)d] + 5

⇒a + 16d = 2[a + 7d] + 5

⇒a + 16d = 2a + 14d + 5

⇒a = 2d − 5 ----------- (1)

But${11}^{\mathrm{th}}$term,$a_{11}$= 43

⇒a + ( 11 − 1)d = 43

⇒a + 10d = 43

⇒2d − 5 + 10 d = 43 [From (i)]

⇒12d = 48

⇒ d = 4

∴a = 2d−5 = (2 × 4) − 5 = 3

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

Substituting the values of a and d, we get

$a_n$= 3 + (n − 1) 4 = 3 + 4n − 4 = 4n - 1

∴$n^{\mathrm{th}}$ term of the given A.P. is 4n − 1.

Subtract the fractions by taking LCM.

1) Assume the number of purchased by the shopkeeper as x.

2) Find the original cost price of the book in terms of x.

3) Find the new cost price for the increased number of books.

4) Difference between the cost prices is Rs 1.

5) Write the equation representing the difference between the cost prices as Rs 1.

6) Simplify and solve the quadratic equation obtained to find the number of books.

OR

1) Assume the first number as x.

2) Find the second number as (9 - x)

3) Obtain a quadratic equation by applying the given condition to the numbers.

4) Solve the quadratic equation and find the assumed number.

5) Find the other number by using the sum of the numbers.

Consider the number of books purchased by the shopkeeper be x.

Cost price of x books = Rs 80

∴Original cost price of one book = Rs.$\frac{80}{x}$

If the shopkeeper had purchased 4 more books, then the number of books purchased by him would be (x + 4).

∴New cost price of one book = Rs $\frac{80}{(x+4)}$

Given that, original cost price of one book − New cost price of one book = Rs 1

⇒$\frac{80}{x}$ - $\frac{80}{(x+4)}$= 1

⇒$\frac{80(x+4)-80x}{x(x+4)}$= 1

⇒80x + 320 - 80x = x(x + 4)

⇒$x^2$+ 4x = 320

⇒$x^2$+ 4x - 320 = 0

⇒$x^2$+ 20x - 16x - 320 = 0

⇒x(x + 20) - 16( x + 20) = 0

⇒x - 16 = 0 or x + 20 = 0

⇒x = 16 or x = -20 ( Number of books cannot be negative)

∴ x = 16

∴ The number of books purchased by the shopkeeper is 16.

OR

Assume the first number as x.

Given that, the sum of the first number and the second number is 9.

∴ (x + Second number) = 9

⇒ Second number = 9 − x

Sum of their reciprocals is $\frac{1}{2}$.

∴ $\frac{1}{x}$ + $\frac{1}{9-x}$= $\frac{1}{2}$

⇒$\frac{9-x+x}{x(9-x)}$= $\frac{1}{2}$

⇒ 9 × 2 = 9x −$x^2$ ⇒

⇒$x^2$−9x + 18 = 0 ⇒

⇒ $x^2$−6x − 3x + 18 = 0

⇒ x(x − 6) − 3(x − 6) = 0 ⇒

⇒ (x − 3) (x − 6) = 0 ⇒

⇒ x − 3 = 0 or x − 6 = 0

⇒ x = 3 or x = 6

When x = 3, we have 9 − x = 9 − 3 = 6

When x = 6, we have 9 − x = 9 − 6 = 3

∴ The two numbers are 3 and 6.

Use the formula to find the sum of the n terms of the A.P. involving the first term and the common difference.

1) Assume the common difference of the A.P. as d.

2) Find the sum of the 14 terms of the A.P. by substituting the values of a and n and equate it to the given value.

3) Simplify the equation to find the value of d.

4) Find the ${25}^{\mathrm{th}}$term of the A.P. by using the values of a, n and d.

Given that first term, a = 10 and sum of first 14 terms is 1505.

Let the common difference of the A.P. be d.

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

The sum of first 14 terms of the A.P.:

$S_{14}$ = $\frac{14}{2}$[2×10 + (14 −1)d] = 1505

⇒7[20 + 13d] = 1505

⇒20 + 13d = $\frac{1505}{7}$

⇒20 + 13d = 215

⇒13d = 215 − 20

⇒13d = 195

⇒ d = 15.

∴${25}^{\mathrm{th}}$term of the A.P., $a_{25}$= 10 + (25 − 1) 15 = 10 + (24 × 15) = 370. [$a_n$= a + (n − 1)d]

∴ ${25}^{\mathrm{th}}$term of the A.P is 370.

The shortest distance between two points is the perpendicular distance.

OR

Arrange the tangents of the circle according to the sides of the quadrilateral.

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.

OR

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

2) Write each side of the quadrilateral as sum of two parts.

3) Write the tangents from each vertex of the quadrilateral and they are equal.

4) Add one pair of opposite sides of the quadrilateral and equate to the sum of their parts.

5) Replace the equivalent tangents of the parts of the opposite sides considered.

6) Rearrange the parts of the sides and prove that the sums of the opposite sides of a quadrilateral are equal.

Given: A circle of centre O and a tangent l at point A.

To prove: OA ⊥ l

Construction: Consider any point B other than A on the tangent l. Join O and B.

Suppose OB intersects the circle at C.

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

So, in order to prove OA ⊥ l, we need to prove that OA is shorter than OB.

OA = OC (Radius of same circle)

Now, OB = OC + BC

⇒ OB > OC

⇒ OB > OA

⇒ OA < OB

B is an arbitrary point on the tangent l.

Thus, OA is shorter than any other line segment joining O to any point on l.

∴ OA ⊥ l.

OR

Let the sides of the quadrilateral ABCD touch the circle at points P, Q, R and S as shown in the figure.

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

Therefore,

AP = AS ---------(1)

BP = BQ ---------(2)

CQ = CR -----------(3)

DR = DS ------------(4)

∴ AB + CD = (AP + BP) + (CR + DR)

= (AS + BQ) + (CQ + DS) [Using (1) ,(2),(3) and (4)]

= (AS + DS) + (BQ + CQ)

= AD + BC

∴ AB + CD = AD + BC

Hence, it is proved.

Radius and the height of the hemisphere are equal.

1) Find the height of the cone by subtracting the radius i.e. height of the hemisphere from the total height of the solid.

2) Find the volume of the hemisphere and the volume of the cone using the given dimensions.

3) Sum of the volumes of the hemisphere and the cone gives the volume of the solid.

Given that radius of cone = Radius of hemisphere = 3.5 cm

Total height of solid (OC) = 9.5 cm

⇒ OD + DC = 9.5

⇒ OD + 3.5 = 9.5 [DC (radius of hemisphere) = 3.5 cm]

⇒ OD = 6 cm

∴ Height of cone = OD = 6 cm

Volume of solid = Volume of cone + Volume of hemisphere

= ( $\frac{1}{3}$π $r^2$h) + ( $\frac{2}{3}$π $r^3$)

= $\frac{\pi r^2}{3}$[ h + 2r]

= $\frac{22}{7}\times \frac{3.5\times 3.5}{3}$× [ 6 + (2 × 3.5)]

= $\frac{22\times 0.5\times 3.5\times 13}{3}$

= $\frac{500.5}{3}$

= 166.83 $$

∴ The volume of the solid is 166.83 ${\mathrm{cm}}^3$.

Convert the volume of the water into${\mathrm{cm}}^3$.

1) Assume the height of the bucket in the form of frustum of a cone as h.

2) Find the volume of the bucket using the formula to find the volume of the frustum of the cone.

3) Convert the given volume of the water into${\mathrm{cm}}^3$.

4) Equate the volumes and simplify to find the value of h which gives the height of the bucket.

Consider the height of the bucket be h cm.

Let$r_1$and$r_2$ be the radii of the circular ends of the bucket.

Given that,$r_1$ = 28 cm and$r_2$= 21 cm

Capacity of bucket = 28.49 litres

∴Volume of the bucket = 28.49 × 1000${\mathrm{cm}}^3$ [1 litre = 1000 ${\mathrm{cm}}^3$]

⇒$\frac{1}{3}\mathrm{\pi h}({r_1}^2+{r_2}^2+r_1{r}_2)=28.49\times 1000$

⇒$\frac{1}{3}$ × $\frac{22}{7}$ × h × [ ${28}^2$+ ${21}^2$+ (28 × 21)] = 28490 $$

⇒$\frac{22}{21}$ × h × 1813 = 28490

⇒ h = $\frac{28940\times 21}{22\times 1813}$cm

⇒ h = 15 cm

∴ Height of the bucket is 15 cm.

Apply appropriate trigonometric ratio of the angle of elevation and the angle of depression.

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

2) Form right angled triangles involving the angle of elevation and the angle of depression.

3) Mark the alternate angle of the angle of depression.

4) Find the tangent of the alternate angle of the angle of depression and find the distance between the tower and the hill.

5) Find the tangent of the angle of elevation and find the height of the hill.

Consider AB is the hill and CD is the tower.

Angle of elevation of the hill at the foot of the tower is 60°, i.e., ∠ADB = 60°

The angle of depression of the foot of hill from the top of the tower is 30°, i.e., ∠CBD = 30°.

In a right angled triangle ΔCBD:

⇒ tan 30° = $\frac{\mathrm{CD}}{\mathrm{BD}}$

⇒ BD = $\frac{\mathrm{CD}}{\tan 30°}$

⇒ BD = $\frac{50}{\frac{1}{\sqrt{3}}}$ = 50 $\sqrt{3}$

In a right angled triangle ΔABD:

⇒tan 60° = $\frac{\mathrm{AB}}{\mathrm{BD}}$

⇒ AB = BD × tan 60°

⇒ AB = 50 $\sqrt{3}$ × $\sqrt{3}$

⇒ AB = 150 m

∴ The height of the hill is 150 m.