CBSE Class 10 - Maths

Radius of the larger circle is the diameter of the smaller circle.

OR

Angle at the centre of the circle in a quadrant is a right angle.

1) Find the lengths of the arc of the sectors around the shaded regions using the formula.

2) Find the sum of the lengths of all the arcs.

OR

1) Find the radius of the circle using the given circumference of the circle.

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

3) area of the quadrant of the circle is one fourth of the area of the circle.

To find the perimeter of the shaded region, add the lengths of APB, ARC, CQD and DSB

∴ Perimeter of the shaded region = Length of APB + Length of ARC + Length of CQD + Length of DSB

Length of APB = $\frac{1}{2}$ × 2π$(\frac{7}{2})$ = 11cm

Length of ARC = $\frac{1}{2}$ × 2π$\left(7\right)$= 22 cm

Length of CQD = $\frac{1}{2}$ × 2π$\left(\frac{7}{2}\right)$= 11cm

Length of DSB = $\frac{1}{2}$ × 2π$\left(7\right)$= 22 cm

Hence, perimeter of the shaded region = 11 + 22 + 11 + 22 = 66 cm

OR

Assume the radius of the circle is r.

Given the perimeter of the circle = 44 cm.

Circumference of a circle = 2πr

∴ 2πr = 44 cm

⇒2 × $\frac{22}{7}$ × r = 44

⇒r = 7 cm.

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

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

= $\frac{1}{4}$ × $\frac{22}{7}$ × 49 = 38.5

∴ Area of quadrant of the circle = 38.5 ${\mathrm{cm}}^2$

The perpendicular on a chord of a circle bisects the chord. So, PM = QM.

1) Assume O as the centre of the concentric circles, PQ be the tangent of the smaller circle and M be the point contact.

2) OM is perpendicular on PQ. So, PM = QM.

3) OM is also the radius of the smaller circle.

4) Apply Pythagoras theorem in triangle OPM and find the value of OP.

5) OP is the required radius of the larger circle (r).

Assume O as the centre of the two concentric circles.

Let PQ be the chord of the larger circle which touches the smaller circle and M be the point where the circle touches the tangent.

Given that, PQ = 48 cm.

OM = radius of the smaller circle = 7 cm

Assume the radius of the larger circle (OP) as r.

PQ is a tangent to the inner circle.

∴ OM ⊥ PQ

Hence, OM bisects PQ.

⇒PM = MQ = $\frac{48}{2}$= 24 cm.

Now, applying Pythagoras Theorem in ΔOPM,

${\mathrm{OP}}^2$= ${\mathrm{OM}}^2$ + ${\mathrm{PM}}^2$

⇒ ${\mathrm{OP}}^2$= $7^2$+ ${24}^2$

= (49 + 576) = 625 ${\mathrm{cm}}^2$= ${\left(25\right)}^2$

⇒ OP = 25 cm

∴ Radius of the larger circle = r = 25 cm.

Substitute the coordinates of the points in the distance formula according to the order.

1) Find the distance between the points P and Q using the distance formula.

2) Equate the distance to the given value.

3) Simplify the equation to obtain a quadratic equation.

4) Solve the quadratic equation and find the roots.

5) Roots of the equation are the values of x.

Given that the distance between the points P (x, 4) and Q (9, 10) is 10 units.

If the coordinates of two points are given by ($x_1$, $y_1$) and ($x_2$, $y_2$), then $x_1$= x, $y_1$= 4,$x_2$= 9,$y_2$= 10.

Using the distance formula,

d = $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

10 = $\sqrt{{\left(9-x\right)}^2+{\left(10-4\right)}^2}$

10 = $\sqrt{81+x^2-18x+36}$

On squaring both the sides,

100 =$x^2$−18x + 117

⇒ $x^2$−18x + 17 = 0

⇒$x^2$−17x − x + 17 = 0

⇒ x (x − 17) − 1(x − 17) = 0

⇒ (x − 1) (x − 17) = 0

⇒ x = 1, 17

Hence, the values of x are 1 and 17.

n value of every term in an A.P. is a natural number.

1) Find the common difference of the A.P. from the first term and the second term of the A.P.

2) Equate the n th term of the A.P. to − 150.

3) Substitute the values of a and d and find the value of n.

4) If the value of n is a whole number, then −150 is a term of the A.P.

The given A.P. is 17, 12, 7, 2, …

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

a = 17 and d = 12 − 17 = −5

For − 150 to be a term of the given A.P.,$a_n$ = −150, where n is a natural number

⇒a + ( n − 1)d = −150

⇒17 + ( n −1)( −5) = -150

⇒−5n + 5 = −167

⇒−5n = −172

⇒ n = $\frac{172}{5}$= 34.4

34.4 is not a natural number.

Therefore, −150 is not a term of the given A.P.

Breadth and the height of the new cuboid formed are equal to the side of the cube.

1) Find the length, breadth and height of the new cuboid formed.

2) Length of the cuboid is twice the length of the side of the cube. Breadth and the height of the cuboid formed are equal to the length of the side of the cube.

3) Find the total surface area of the resulting cuboid using the corresponding formula.

If two cubes of side 4cm. If we join the two cubes end to end, the resulting cuboid’s length (l) = 8 cm, breadth (b) = 4cm and height (h) = 4cm.

∴ Surface area of the resulting cuboid = 2 (lb + bh + lh)

= 2 (8 cm × 4 cm + 4 cm × 4 cm + 8 cm × 4 cm)

= 2 × (32 + 16 + 32) ${\mathrm{cm}}^2$

= 160 ${\mathrm{cm}}^2$

Therefore, the surface area of the cuboid formed = 160 ${\mathrm{cm}}^2$.

Mark points on ray AX such that they are equidistant.

1) Draw a line segment AB of length 6 cm.

2) Draw a ray AX at A making an acute angle with AB.

3) Mark 7 points (sum of 3 and 4) $A_1$, $A_2$..... $A_7$ on AX such that they are equidistant.

4) Join $A_7$ to B.

5) Draw a line at $A_3$, parallel to $A_7$B.

6) The line touches the line segment AB at P.

7) P is the required point which divides the line segment AB in the ratio of 3:4.

The point P which divides a line segment of length 6 cm in the ratio 3:4. can be marked as follows

(i) Draw a line segment AB of length 6 cm.

(ii) Draw a ray AX which makes an acute angle with AB

(iii) Mark 7 points on AX as $A_1$, $A_2$, $A_3$, $A_4$ … $A_7$ such that A $A_1$ = $A_1{A}_2$ = $A_2{A}_3$= $A_3{A}_4$ etc

(iv) Join B and $A_7$.

(v) With an angle equal to ∠A $A_7$B at $A_3$, draw a line parallel to the line B $A_7$ through the point $A_3$, which intersects AB at point P.

P is the point that divides the line segment AB of length 6 cm in the ratio of 3:4.

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

1) Write the given equation in the standard form of the quadratic equation.

2) Compare the coefficients of the equation with the coefficients of the standard quadratic equation and note the values.

3) Find the discriminant of the quadratic equation and equate it to zero.

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

The given quadratic equation = kx(3x − 10) + 25 = 0

⇒ 3k$x^2$ −10kx + 25 = 0

If a$x^2$ + bx + c = 0 is a quadratic equation, then in above equation, a = 3k, b = −10k, c = 25

If a quadratic equation has two equal roots then, the discriminant of the equation is 0.

i.e., $b^2$−4ac = 0

⇒ ${\left(-10k\right)}^2$−4 × 3k × 25 = 0

⇒ 100$k^2$ − 300k = 0

⇒ 100k (k − 3) = 0

⇒ k = 3

Therefore, when k = 3, the given equation will have two equal roots.

Favourable outcomes should not have two heads.

1) List the sample space when two coins are tossed.

2) Write the number of possible outcomes.

3) List the favourable outcomes which includes the outcomes with no heads and one heads.

4) Write the number of favourable outcomes.

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

Let S represent the set of possible outcomes.

If a coin is tossed two times, S = {HH, HT, TH, TT}

Therefore number of possible outcomes or n(S) = 4

From these, the favourable outcomes = HT, TH and TT.

Hence the number of favourable outcomes = 3

P( not more than one head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$= $\frac{3}{4}$

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

1) Join the points O and A, O and E, O and F, O and B, and O and C.

2) OD is perpendicular on BC as the radius at the point of contact of a tangent of a circle is perpendicular to the tangent.

3) Similarly OE and OF are perpendiculars on AC and AB respectively.

3) Area of triangle ABC is the sum of the areas of triangles OAB, OBC and OAC.

4) Assume the lengths AF and AE as x.

5) Find the lengths of the sides of the triangle in terms of x.

6) Equate the sum of the areas of the triangles OAB, OBC and OAC to the given value.

7) Simplify to find the value of x.

8) Find the lengths of sides AB and AC using the value of x.

O is the center of the circle.

Join O with the points A, B, C, E and F.

OD = OE = OF = 2 cm (radius of the circle)

BD = 4 cm and DC = 3 cm

∴ BC = BD + DC = 4 + 3 = 7 cm.

BF and BD are tangents to the circle from point B.

Hence, BF = BD = 4 cm

Similarly, CE and CD are tangents to the circle from point C.

Hence, CE = CD = 3 cm

To find the lengths of AB and AC:

AF and AE are tangents to the circle from point A.

Assume AF =AE = x cm.

AB = AF + BF = (4 + x) cm

AC = AE + CE = (3 + x) cm

Given that, area of ΔOBC + area of ΔOAB + area of ΔOAC = area of ΔABC

= ( $\frac{1}{2}$×BC×OD) + ( $\frac{1}{2}$×AB×OF) + ( $\frac{1}{2}$×AC×OE) = 21

⇒ [ $\frac{1}{2}$×7×2] + [ $\frac{1}{2}$×(4 + x)×2] + [ $\frac{1}{2}$×(3 + x)×2] = 21

⇒ $\frac{1}{2}$×2(7 + 4 + x + 3 + x) = 21

⇒ 14 + 2x = 21

⇒2x = 7

⇒x = 3.5

Therefore, AB = (4 + 3.5) cm = 7.5 cm and AC = (3 + 3.5) cm = 6.5 cm.

Consider the outcomes such as (1, 4) and (4, 1) as favourable outcomes.

OR

Consider the outcomes such as (HHT) and (THH) as possible outcomes.

1) List the possible outcomes when two dice are rolled.

2) Note the number of possible outcomes.

3) List the favourable outcomes in which the product of the numbers on the dice is a perfect square.

4) Consider the outcomes such as (1, 4) and (4, 1) as favourable outcomes.

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

OR

1) List the possible outcomes when a coin is tossed three times.

2) Note the number of possible outcomes.

3) Assume the outcomes with three heads and three tails as E.

4) List and note the number of outcomes favourable to E.

5) Find the ratio of the number of outcomes favourable to E and the total number of possible outcomes which gives the probability of E.

6) Find the probability of the outcomes which are not favouravle to E by subtracting the probability of E from 1.

7) The fraction obtained is the required probability.

Since each dice has numbers from 1 to 6, the set of outcomes or S is:

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

∴ n (S) = 36

Let E be the event of getting numbers on the two dices whose product forms a perfect square

E = {(1, 1), (1, 4), (2, 2), (3, 3), (4, 1), (4, 4), (5, 5), (6, 6)}

∴ n (E) = 8

P(E) = $\frac{n\left(E\right)}{n\left(S\right)}$= $\frac{8}{36}$ = $\frac{2}{9}$.

OR

The coin is tossed three times.

If S represents the set of possible outcomes, then

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Assume the event of getting three heads or three tails as E.

∴ E = {HHH, TTT}

∴ Probability of winning = P (E) = $\frac{n\left(E\right)}{n\left(S\right)}$= $\frac{2}{8}$= $\frac{1}{4}$

∴ Probability of losing = P(E) = 1 − P(E)

P(E ' ) = 1 − $\frac{1}{4}$= $\frac{3}{4}$

So, the probability of Hanif losing the game = $\frac{3}{4}$.

Square on both sides of an equation to remove the square root on both sides.

OR

The ratio in which the point Q divides the line segment PQ should have one term as 1.

1) Assume the three vertices of the equilateral triangle as A, B and C.

2) Assume the coordinates of the third vertex of the equilateral triangle as (x, y)

3) Find the distances between A and B, and B and C, and A and C.

4) Equate two of the distances at a time to find the values of x and y.

OR

1) Assume the ratio in which the point Q divides the line segment PR as a:1.

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

3) Equate the coordinates of the point Q with the given coordinates to find the value of a.

4) Use the value of a and find the value of k.

Assume A (3, 0) and B (6, 0) as the two given vertices of the equilateral triangle

and the coordinates of the third vertex as C (x, y).

ABC is an equilateral triangle,

⇒ AB = BC = CA (Sides of an equilateral triangle are equal)

⇒$\sqrt{{\left(6-3\right)}^2+{\left(0-0\right)}^2}=\sqrt{{\left(x-6\right)}^2+{\left(y-0\right)}^2}=\sqrt{{\left(x-3\right)}^2+{\left(y-0\right)}^2}$

⇒9 = ${\left(x-6\right)}^2$+ $y^2$ = ${\left(x-3\right)}^2$+ $y^2$

∴ ${\left(x-6\right)}^2$+ $y^2$ = ${\left(x-3\right)}^2$+ $y^2$

⇒−12x + 36 = −6x + 9

⇒−6x = −27

⇒ x = $\frac{9}{2}$

Using distance formula,

⇒${\left(y-0\right)}^2$+ ${\left(x-6\right)}^2$ = 9

⇒$y^2$+ ${(\frac{9}{2}-6)}^2$= 9

⇒$y^2$= 9 − $\frac{9}{4}$

⇒$y^2$= $\frac{27}{4}$

⇒y = ± $\sqrt{\frac{27}{4}}$= ± $\frac{3\sqrt{3}}{2}$

Therefore, the coordinates of the third vertex are $(\frac{9}{2},\frac{3\sqrt{3}}{2})$or $(\frac{9}{2},\frac{-3\sqrt{3}}{2}).$

OR

Points P, Q and R are to be collinear.

Let us assume that the point Q (7, k) divides the line segment joining P (5, 4) and R (9, −2) in the ratio a : 1

∴ Coordinates of the point Q are $(\frac{9a+5}{a+1},\frac{-2a+4}{a+1})$

∴$\frac{9a+5}{a+1}$ = 7 and k = $\frac{-2a+4}{a+1}$

⇒9a + 5 = 7a + 7

⇒2a = 2

⇒a = 1

So, k = $\frac{-2a+4}{a+1}$

⇒k = $\frac{-2+4}{1+1}$ = $\frac{2}{2}$= 1

So that the value of k = 1.

Consider the coefficients of the terms in the quadratic formula along with their signs.

1) Compare

Given quadratic equation is 2$\sqrt{3}{x}^2$−5x + $\sqrt{3}$= 0

Compare the quadratic equation with the standard form of a quadratic equation a$x^2$ + bx + c = 0,

a = 2$\sqrt{3}$, b = −5 , c = $\sqrt{3}$

The roots of the quadratic equation are x = $\frac{-b\pm \sqrt{b^2-4\mathrm{ac}}}{2a}$

x = $\frac{5\pm \sqrt{{(-5)}^2-4\times 2\sqrt{3}\times \sqrt{3}}}{2\times 2\sqrt{3}}$

x = $\frac{5\pm \sqrt{25-24}}{4\sqrt{3}}$

x = $\frac{5\pm \surd 1}{4\sqrt{3}}$

x = $\frac{5+1}{4\sqrt{3}}$or $\frac{5-1}{4\sqrt{3}}$

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

x = $\frac{\sqrt{3}}{2}$or $\frac{1}{\sqrt{3}}$

Hence, the roots of the given quadratic equation are $\frac{\sqrt{3}}{2}$and $\frac{1}{\sqrt{3}}$.

1) Middle term of the series when the number of terms is odd = ${\left(\frac{n+1}{2}\right)}^{\mathrm{th}}$term.

OR

Find the value of d from the given condition.

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

2) Equate the last term of the A.P. to the n th term of an A.P. and find the n value of the term.

3) Find the middle term of the total number of terms.

4) Find the middle term of the A.P. using the formula to find the n th term of an A.P.

OR

1) Find the fourth term of the A.P. using the formula to find the n th term of an A.P.

2) Equate it to the given value and obtain an equation.

3) Find the value of d from the given condition.

4) Substitute the value of d in the equation to find the value of a.

5) Find the A.P. using the values of a and d.

Given : A.P. = −6, −2, 2, …, 58.

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

a = −6

d = −2 − (−6) = − 2 + 6 = 4

If l is the last term, then l = 58

⇒ a + (n − 1) d = 58

⇒ − 6 + ( n − 1)4 = 58

⇒( n − 1)4 = 64

⇒ ( n − 1) = 16

⇒ n = 17

Middle term of the A.P. = ${\left(\frac{n+1}{2}\right)}^{\mathrm{th}}$term = ${\left(\frac{17+1}{2}\right)}^{\mathrm{th}}$term = $9^{\mathrm{th}}$ term

$a_9$= a + (9 − 1) d = −6 + 8 × 4 = 2−6 + 32 = 26.

Therefore, the middle term of the A.P. = 26.

OR

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

Given that$4^{\mathrm{th}}$term =18, i.e.$a_4$ = 18

⇒a + ( 4 − 1)d = 18

⇒a + 3d = 18 ------------------------- (1)

Also given that $a_{15}$−$a_9$= 30

⇒{a + 14d} − {a + 8d} = 30

⇒6d = 30

⇒ d = 5.

Substituting the value of d in (i)

a + 3 × 5 = 18

⇒ a + 15 = 18

⇒ a = 18 − 15

⇒ a = 3.

Hence, a = 3 and d = 5.

A.P. = 3, 3 + 5, 3 + (2 × 5), 3 + (3 × 5) …

= 3, 8, 13, 18, …

Subtract the area of the minor segment from the area of the circle to find the area of the major segment.

1) Find the area of the minor sector AOB.

2) Find the area of the right angled triangle AOB.

3) Subtract the area of the triangle from the area of the sector to find the area of the minor segment.

4) Find the area of the circle using the given radius.

5) Subtract the area of the minor segment from the area of the circle to find the area of the major segment APB.

OA = OB = radius of the circle = 35 cm

Area of the minor segment = Area of sector OAB − area of ΔAOB

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

= $\frac{90°}{360°}$ × $\frac{22}{7}$ × ${\left(35\right)}^2$

= $\frac{1}{4}$ × $\frac{22}{7}$ × 1225

= 962.5 ${\mathrm{cm}}^2$

Area of ΔAOB = $\frac{1}{2}$× OA × OB

= $\frac{1}{2}$ × 35 × 35$$

= 612.5 ${\mathrm{cm}}^2$.

Area of the minor segment = 962.5 − 612.5 = 350 ${\mathrm{cm}}^2$

Area of the major segment = Area of the circle − Area of the minor segment

Area of the circle = π$r^2$= $\frac{22}{7}$×${\left(35\right)}^2$= 3850 ${\mathrm{cm}}^2$

∴ Area of the major segment APB = 3850 − 350 = 3500 ${\mathrm{cm}}^2$

Find the tangent of the alternate angles of the angles of depression.

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

2) Mark the alternate angles of the angles of depressions.

3) Find the tangent of the angles of depression in each of the right angled triangles.

4) Find the horizontal distances between the foot of the tower and the cars.

5) Find the sum of the distances to find the distance between two cards.

Assume AB = the height of the tower = 100 m.

If A is the top of the tower, C and D are the positions of the two cars.

In right ΔACB,

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

⇒1 = $\frac{100m}{\mathrm{BC}}$

BC = 100 m

In right angled triangle ABD

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

⇒ $\frac{1}{\sqrt{3}}$ = $\frac{100}{\mathrm{BD}}$

Distance between two cars = CD = BC + BD

CD = 100 + 100 $\sqrt{3}$

= 100 + 100 × 1.73

= 100 + 173

= 273 m

Hence, the distance between two cars is 273 m.

Consider the positive value as the radius of the bucket.

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

2) Equate the volume to the given value of the volume of the bucket.

3) Solve the quadratic equation obtained by splitting the middle term using the factors of the constant.

4) Consider the positive value as the radius of the bucket.

Height of the bucket, h = 15 cm

Radius of one end of bucket, R = 14 cm

Radius of the other end of the bucket is r.

Given that volume of the bucket = 5390 ${\mathrm{cm}}^3$.

$\Rightarrow \frac{1}{3}$π($R^2$+ $r^2$+ Rr)h = 5390

⇒$\frac{1}{3}$×$\frac{22}{7}$× [${14}^2$+ $r^2$+ 14r] ×15 = 5390

⇒196 + $r^2$+ 14 r = $\frac{5390\times 7}{22\times 5}$= 343

⇒$r^2$+ 14r + 196 − 343 = 0

⇒$r^2$+ 14r - 147 = 0

⇒$r^2$+ 21r - 7r - 147 = 0

⇒r( r + 21) - 7 ( r + 21) = 0

⇒r - 7 = 0 or r + 21 = 0

⇒r = 7 or r = -21 ( radius cannot be negative )

Hence, the value of r = 7 cm.

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 segment $B_4$C' parallel to $B_5$C from $B_4$.

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

6) A'BC' is the required triangle.

Given that ∠B = 45° and ∠A = 105°.

But sum of the interior angles of a triangle is 180°.

Therefore, ∠A + ∠B + ∠C = 180°

105° + 45° + ∠C = 180°

∠C = 180° − 150°

∠C = 30°

Step 1: Draw a △ABC of side BC = 7 cm, ∠B = 45° ,∠C = 30°.

Step 2: Draw a ray BX making an acute angle with BC on the opposite side of the vertex A.

Step 3: Mark 5 points $B_1$, $B_2$, $B_3$, $B_4$, $B_5$on BX.

Step 4: Join $B_{5}C$. Draw a line segment through $B_3$ parallel to $B_{5}C$ intersecting BC at C '.

Step 5: Through C ', draw a line segment parallel to AC intersecting AB at A' . ΔA'BC' is the required triangle.

Substitute the coordinates of P, Q and R in the distance formula in the correct order.

1) Equate the distance between P and Q, and P and R using the distance formula.

2) Simplify and find the value of x.

3) Find the distance between P and Q.

Given that the point P (2, 4) is equidistant from Q (7, 0) and R (x, 9).

So, PQ = PR.

⇒$\sqrt{{\left(7-2\right)}^2+{\left(0-4\right)}^2}$= $\sqrt{{\left(x-2\right)}^2+{\left(9-4\right)}^2}$

⇒25 + 16 = ${\left(x-2\right)}^2$+ 25

⇒${\left(x-2\right)}^2$= 16

⇒x − 2 = ±4

⇒ x − 2 = 4 or x - 2 = −4

⇒ x = 6 or x = −2

Hence, the values of x are 6 and −2.

PQ = $\sqrt{{\left(7-2\right)}^2+{\left(0-4\right)}^2}$= $\sqrt{25+16}$= $\sqrt{41}$units

Therefore, PQ =$\sqrt{41}$ units.

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.

Assume that T is the external point to the circle and PT and QT are the tangents that are drawn from the external point T to the circle.

To prove that PT = TQ.

Construction : Join OT.

Proof : But a tangent to circle is perpendicular to the radius through the point of contact.

Hence ∠OPT = ∠OQT = 90°

In the ΔOPT and ΔOQT,

OT = OT (Common side)

OP = OQ ( Radius of the circle)

∠OPT = ∠OQT ( = 90°)

∴ ΔOPT≅ ΔOQT ( By RHS congruence criterion)

⇒ PT = TQ (CPCT)

Hence, the lengths of the tangents drawn from an external point to a circle are equal

Speed of the boat in upstream is equal to the speed of the boat minus the speed of the stream.

OR

Simplify the fractions by finding the LCM of the denominators.

1) Assume the speed of the stream as x km/hr.

2) Find the speed of the boat in upstream and down stream.

3) Find the time taken in upstream and downstream.

4) Find the difference between the timings and equate the difference to 1 hour.

5) Solve the quadratic equation obtained on simplification.

6) Split the middle term of the quadratic equation and factorise.

7) Find the roots of the quadratic equation and consider the positive value which gives the speed of the stream.

OR

1) Find the LCM of the denominators and simplify to obtain a quadratic equation.

2) Split the middle term of the quadratic equation using the factors of the constant term.

3) Find the factors of the equation and equate them to zero.

4) Find the roots of the equation.

Assume the speed of the stream as x km/hr.

If the speed of the boat while going upstream would be (20 - x) km/hr

Speed of the boat down stream will be (20 + x) km/hr

Time taken for the upstream journey = $\frac{\mathrm{Dis}\tan \mathrm{ce}}{\mathrm{Speed}}$ = $\frac{48}{\left(20-x\right)}$h

Time taken for the downstream journey = $\frac{\mathrm{Dis}\tan \mathrm{ce}}{\mathrm{Speed}}$ = $\frac{48}{\left(20+x\right)}$h

Given that

Time taken for upstream trip = Time taken for downstream trip + 1 hour

⇒$\frac{48}{\left(20-x\right)}$ - $\frac{48}{\left(20+x\right)}$= 1

⇒$\frac{960+48x-960+48x}{\left(20-x\right)\left(20+x\right)}$= 1

⇒$\frac{96x}{400-x^2}$ = 1

⇒ 400 - $x^2$ = 96x

⇒ $x^2$+ 96 x - 400 = 0

⇒$x^2$+ 100x - 4x - 400 = 0

⇒x( x + 100) - 4( x + 100) = 0

⇒x + 100 = 0 or x - 4 = 0

⇒ x = -100 or x = 4 ( speed cannot be negative )

Therefore, x = 4 .

Hence, the speed of the stream is 4 km/hr

OR

⇒$\frac{x-7-x-48}{(x+4)(x-7)}$= $\frac{11}{30}$

⇒$\frac{-11}{x^2-3x-28}$= $\frac{11}{30}$

⇒ $x^2$−3x − 28 = −30

⇒ $x^2$−3x + 2 = 0

⇒$x^2$−2x − x + 2 = 0

⇒x( x − 2) − 1( x − 2) = 0

⇒x − 1 = 0 or x − 2 = 0

⇒ x = 1 or x = 2.

The roots of the given equation are 1 and 2.

Find the values of a and d using the given data.

OR

Integers divisible by six are multiples of 6.

1) Find the sum of first 4 terms of the A.P. using the formula.

2) Equate the sum of the four terms to the given value and obtain equation 1.

3) Find the sum of first 14 terms of the A.P. using the formula.

4) Equate the sum of the fourteen terms to the given value and obtain equation 2.

5) Solve both the equations and find the values of a and d.

6) FInd the sum of n terms of an A.P. using the formula.

OR

1) List the positive integers which are divisible by 6.

2) The numbers form an A.P.

3) Note the values of a and d from the series.

4) Find the sum of 30 terms of the A.P. using the formula.

Assume that ‘a’ be the first term and ‘d’ be the common difference.

It is given that the sum of first four terms = 40

$S_4$= 40

⇒$\frac{4}{2}$[ 2a + ( 4 − 1) d ] = 40

⇒2a + 3d = 20 --------------(1)

The sum of first 14 terms is 280.

$S_{14}$= 280

⇒$\frac{14}{2}$[ 2a + ( 14 − 1)d ] = 280

⇒ 2a + 13d = 40 -------------(2)

Subtracting equation (1) from equation (2),

⇒2a + 13d − (2a + 3d) = 40 −20

⇒10d = 20

⇒d = 2

Substitute d = 2 in eqn (1) we get,

⇒2a + 3×2 = 20

⇒2a = 20 − 6 = 14

⇒a = $\frac{14}{2}$= 7.

Sum of the first n terms $S_n$= $\frac{n}{2}$[ 2a + ( n − 1)d]

= $\frac{n}{2}$[ 2(7) + ( n − 1)2 ]

= $\frac{n}{2}$[ 14 + 2n − 2 ]

= $\frac{n}{2}$[ 2n + 12]

= n( n + 6)

= $n^2$+ 6n

OR

The positive integers divisible by six are 6,12, 18, 24 …. 180 which is an A.P. with first term as 6 and common difference as 6.

Therefore, the sum of first 30 integers are 6 + 12 + ---------+ 180

= $\frac{30}{2}$[6 + 180] [ $S_n$= $\frac{n}{2}$[ a + l ] ]

= 15×186

= 2790 .

Find the tangent of the angles of elevation of the top and bottom of the tower.

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

2) Find the tangent of the angle of elevation of the foot of the tower and find the horizontal distance between the tower and the point of observation.

3) Find the tangent of the angle of elevation of the top of the tower and find the height of the tower from the ratio of sides.

From above figure,

P - point of observation.

AB - building of height 10 m

AC - transmission tower.

In ΔABP,

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

⇒ $\frac{1}{\sqrt{3}}$= $\frac{10}{\mathrm{BP}}$

⇒BP = 10 $\sqrt{3}$

In ΔCBP,

tan 60° = $\frac{\mathrm{CB}}{\mathrm{BP}}$

⇒ $\sqrt{3}$= $\frac{\mathrm{AB}+\mathrm{AC}}{10\sqrt{3}}$

⇒AB + AC = 30

⇒10 + AC = 30

⇒AC = 20 m

Therefore, height of the transmission tower is 20 m.

Areas of all the quadrants formed by the arcs are equal.

1) Find the radius of the arcs of the shaded region.

2) Arcs intersect at the midpoints of the sides of the square. So, the area formed by each arc is a quadrant of a circle.

3) Find the area of one quadrant using the formula to find the area of the sector.

4) Find the total area of all the four quadrants which gives the area of the non-shaded region of the square.

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

6) Subtract the total area of the quadrants from the area of the square to find the area of the shaded region of the square.

Given that length of each side of square is 14 cm.

P, Q, R and S are the midpoints of sides AB, BC, CD and DA respectively.

Radius of each arc = $\frac{14}{2}$ = 7 cm

∠SAP = ∠PBQ = ∠QCR = ∠RDS = 90°

Area of sector SAP = Area of sector PBQ = area of sector QCR = area of sector RDS

Area of sector APS = $\frac{90°}{360°}$ × $\frac{22}{7}$ × ${\left(7\right)}^2$

= $\frac{1}{4}$ × $\frac{22}{7}$ × 49

= 38.5 ${\mathrm{cm}}^2$.

Hence the area of the non-shaded region = 4 × Area of sector APS

= 4 × 38.5 ${\mathrm{cm}}^2$ = 154 ${\mathrm{cm}}^2$.

Area of square = $a^2$= ${14}^2$= 196.

Area of the shaded region = Area of square - Area of the non-shaded region

= 196 - 154 = 42 ${\mathrm{cm}}^2$.

Therefore, area of the shaded region is 42 ${\mathrm{cm}}^2$.

Radius of the base of the cone and the radius of the base of the cylinder are same.

1) Total surface area of the toy is the sum of the curved area of the cone, and the curved surface area of the cylinder, and the base area of the cylinder.

2) Write the respective formulas and substitute the values of r and h.

3) Substitute the value of slant height in terms of r and h.

4) Simplify and find the total surface area of the toy.

The toy represented as like this

For the cylindrical part

Height = EF = 21 cm

Diameter ( CD = $h_1$) = 40 cm

Radius $r_1$= 20 cm

For the conical part

Height (PF = $h_2$) = 15 cm

Radius of conical part ( $r_2$) = $r_1$= 20 cm.

Let $r_1$= $r_2$= r = 20 cm

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

= 2π $r_1{h}_1$+ π $r_2\sqrt{{r_2}^2+{h_2}^2}$+ π ${r_1}^2$

= 2πr $h_1$+ πr $\sqrt{r^2+{h_2}^2}$+ π $r^2$

= πr $(2h_1+\sqrt{r^2+{h_2}^2}+r)$

= 3.14×20 $\left(2\times 21+\sqrt{{\left(20\right)}^2+{\left(15\right)}^2}+20\right)$

= 62.8 ( 42 + $\sqrt{625}$+ 20)

= 62.8 ( 42 + 25 + 20 ) $$

= 62.8 × 87 $$

= 5463.6 ${\mathrm{cm}}^2$

Hence, the total surface area of the toy is 5463.6 ${\mathrm{cm}}^2$.