CBSE Class 10 - Maths

AP, AR, BQ and BR are the tangents from A and B.

1) XP and XQ are tangents from an external point X.

2) Tangents from an external point are equal.

3) A and B are external points from which AP, AR and BQ, BR are considered as tangents.

4) Use the property that the tangents from an external point are equal and prove the required proof.

From the figure, there is an external point X from where two tangents, XP and XQ, are drawn to the circle.

XP = XQ (The lengths of the tangents drawn from an external point to the circle are equal.)

Similarly,

AP = AR

BQ = BR

XP = XA + AP --------- (1)

XQ = XB + BQ --------- (2)

By substituting AP = AR in equation (1) and BQ = BR in equation (2), we get

XP = XA + AR

XQ = XB + BR

Since the tangents XP and XQ are equal, we get

XA + AR = XB + BR.

If two lines are perpendicular, angles on both sides at the point of intersection is a right angle.

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

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

3) Angles on either side of the diameter at the points of intersection of the diameter and the tangent are right angles.

4) Alternate angles formed by the diameter as a transversal are equal.

5) Therefore, the tangents of the circle are parallel.

Assume AB as the diameter of a circle, with center O. The tangents PQ and RS are drawn at points A and B on the circle, respectively.

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

∴ OA ⊥ RS and OB ⊥ PQ

⇒ ∠OAR = 90°

∠OAS = 90°

∠OBP = 90°

∠OBQ = 90°

∠OAR = ∠OBQ and ∠OAS = ∠OBP

These angles are the pair of alternate interior angles.

The lines PQ and RS are parallel to each other as the alternate interior angles are equal.

Hence, proved.

The outcome when two dice are rolled (4, 6) is not same as (6, 4).

1) Write the number that a dice could show when rolled.

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

3) Write the number of possible outcomes.

4) List the number of possible outcomes favourable to the given condition.

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

The numbers that two dice could show are 1, 2, 3, 4, 5, 6.

The possible outcomes when two dice are rolled are:

{(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)}

∴ Number of possible outcomes = 6 × 6 = 36.

Assume 'E' as the event that the sum of numbers appearing on the two dice is 10.

The possible outcomes favourable to event, E are {(4, 6), (5, 5), (6, 4)}.

∴ Number of favourable outcomes to event E = 3.

P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}E}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$ = $\frac{3}{36}$ = $\frac{1}{2}$.

Angle at the center of the circle in the quadrant of a circle is 90°.

1) Find the area of the quadrant using the formula for the area of the sector.

2) The triangle which is unshaded is a right angled triangle whose base is the radius of the quadrant.

3) Find the area of the triangle.

4) Find the difference between the area of the quadrant and the area of the triangle to find the area of the shaded region.

Given that, radius of the circle = 7 cm

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

= $\frac{1}{4}$ × $\frac{22}{7}$ × 7 × 7 ${\mathrm{cm}}^2$= $\frac{77}{2}{\mathrm{cm}}^2$

Area of ΔODC = $\frac{1}{2}$ × OD × OC

= ($\frac{1}{2}$ × 4 × 7) ${\mathrm{cm}}^2$ (Since OC is the radius of the circle)

= 14 ${\mathrm{cm}}^2$

Area of the shaded region = Area of the quadrant OABC − Area of ΔODC

Area of the shaded region = $\frac{77}{2}$−14 = $\frac{(77-28)}{2}$= $\frac{49}{2}$ = 24.5${\mathrm{cm}}^2$

Use the factors of both the constant and the coefficient of $x^2.$

1) Split the middle term of the quadratic equation using the factors of the constant term and the leading term.

2) Find the factors by taking common in the first two terms and the last two terms.

3) Find the values of x by equating each of the terms to zero.

Given the quadratic equation as,

$\sqrt{3}$$x^2$−2$\sqrt{2}$x−2$\sqrt{3}$ = 0

⇒ $\sqrt{3}{x}^2$−3$\sqrt{2}$x + $\sqrt{2}$x−2$\sqrt{3}$ = 0

⇒ $\sqrt{3}$x (x−$\sqrt{6}$) + $\sqrt{2}$(x−$\sqrt{6}$) = 0

⇒ ($\sqrt{3}$x +$\sqrt{2}$) (x−$\sqrt{6}$) = 0

⇒ x = $-\frac{\sqrt{2}}{\sqrt{3}}$ or x = $\sqrt{6}$

⇒ x = $-\sqrt{\frac{2}{3}}$ or x = $\sqrt{6}$√

Sum of the first term of the A.P. is the first term of the A.P.

1) Write the formula to find the sum of the 'n' terms of an A.P.

2) Equate it to the given formula to find the 'n' terms of the A.P.

3) Simplify to get an equation.

4) Find the value of the first term by finding the sum of one term of AP using the given formula.

5) Substitute the value of 'a' in the equation obtained and simplify to find the formula for finding the $n^{\mathrm{th}}$term of the AP.

The sum of n terms of an A.P. is given by 5n -$n^2$

But we know that $S_n$= $\frac{n}{2}$(a + $t_n$)

where a = first term $T_n$ = $n^{\mathrm{th}}$term

So, we have 5n−$n^2$= $\frac{n}{2}$ ( a + $t_n$)

⇒ n(5−n) = $\frac{n}{2}$( a + $t_n$)

⇒ 10−2n = a + $t_n$ ...(1)

Now, we have sum of the first term $S_1$= a

⇒ a = (5(1)−$1^2$) = 4

Substitute the value of a in equation (1), we get

10−2n = 4 + $t_n$

⇒ $t_n$= 6−2n .

Therefore, $n^{\mathrm{th}}$term of the A.P. is 6 − 2n.

1) Ratio 2:3 is not equal to 3:2

2) Substitute the coordinates of points according the the section formula in order.

1) Draw a figure showing the line segment AB which is divided by four points P, Q, R and S into five equal parts.

2) Point P divides the line segment in the ratio of 1 : 4.

3) Use the section formula to find the coordinates of the point P.

4) Similarly the point Q divides the line segment in the ratio of 2:3 and the point R divides the line segment in the ratio of 3:2.

5) Use the section formula and find the coordinates of the points Q and R.

Given points P, Q, R and S divide the line segment AB formed by joining A(1, 2) and B (6, 7) into 5 equal parts.

∴ AP = PQ = QR = RS = SB

First point P divides the line segment AB in the ratio of 1:4.

Coordinates of a point dividing the line segment joining the points ( $x_1$, $y_1$) and ( $x_2$, $y_2$) in the ratio of m:n are $\left(\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{m+n}\right)$, $\left(\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{m+n}\right)$.

By applying section formula, the coordinates of point P: $(\frac{1\left(6\right)+4\left(1\right)}{1+4},\frac{1\left(7\right)+4\left(2\right)}{1+4})$= $(2,3)$

Similarly, the point Q divides the line segment AB in the ratio 2:3.

By applying section formula, the coordinates of point Q:

$(\frac{2\left(6\right)+3\left(1\right)}{2+3},\frac{2\left(7\right)+3\left(2\right)}{2+3})$= $(3,4)$

Similarly, the point R divides the line segment AB in the ratio 3:2.

By applying section formula, the coordinates of point R:

$(\frac{3\left(6\right)+2\left(1\right)}{3+2},\frac{3\left(7\right)+2\left(2\right)}{3+2})$= (4, 5).

Breadth of the rectangle is the hypotenuse of the right angled triangle and also the diameter of the semicircular region.

1) Find the hypotenuse of the right angled triangle using the Pythagoras property in the right angled triangle AED.

2) Hypotenuse of the right angled triangle is the breadth of the rectangle.

3) Find the area of the rectangle.

4) Breadth of the rectangle is the diameter of the semicircular region.

5) Find the area of the semicircular region.

6) Subtract the area of the right angled triangle from the sum of the areas of the rectangle and the area of the semicircular region to find the area of the shaded region of the figure.

AED is an right angle triangle,

${\mathrm{AD}}^2$= ${\mathrm{AE}}^2$+${\mathrm{ED}}^2$.

⇒ ${\mathrm{AD}}^2$ = ($9^2$+ ${12}^2$)

= (81 + 144) $$

= 225 ${\mathrm{cm}}^2$

⇒ AD = 15 cm

Area of the rectangular region ABCD

= AB × AD

= (20 × 15)

= 300 ${\mathrm{cm}}^2$

Area of ∆AED

= $\frac{1}{2}$× AE × DE = ($\frac{1}{2}$ × 9 × 12) ${\mathrm{cm}}^2$ = 54 ${\mathrm{cm}}^2$

In a rectangle

AD = BC = 15 cm

Since, BC is the diameter of the circle, radius of the circle = 15 cm

Area of the semi-circle = $\frac{1}{2}$ × π × $r^2$

= ($\frac{1}{2}$ × 3.14 × $\frac{15}{2}$ × $\frac{15}{2}$) = 88.3125 ${\mathrm{cm}}^2$

Area of the shaded region = Area of the rectangle + Area of the semi-circle − Area of the triangle

= (300 + 88.3125 − 54) ${\mathrm{cm}}^2$

= 334.3125 ${\mathrm{cm}}^2$.

Angle in the quadrant of a circle is a right angle. So, BC is the hypotenuse of the right angled triangle.

1) Find the area of the quadrant of the circle ABDC.

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

3) Find the hypotenuse of the right angled triangle ABC using the Pythagoras property.

4) Subtract the area of the right angled triangle from the area of the quadrant of the circle to find the area of the segment BDC.

5) Diameter of the semicircle is the hypotenuse of the right angled triangle.

6) Find the area of the semicircular region.

7) Subtract the area of the segment BDC from the area of the semicircular region to find the area of the shaded region of the figure.

Given that,

Radius (r) of the circle = AB = AC = 28 cm

Area of quadrant ABDC

= $\frac{1}{4}$ × π ×$r^2$ = ($\frac{1}{4}$ × $\frac{22}{7}$ × 28 × 28)

= 616${\mathrm{cm}}^2$

Area of ∆ABC

= $\frac{1}{2}$ × AC × AB = ($\frac{1}{2}$ × 28 × 28) ${\mathrm{cm}}^2$.

= 392 ${\mathrm{cm}}^2$

Area of segment BDC = Area of quadrant ABDC − Area of ∆ABC

= (616 − 392)

= 224 ${\mathrm{cm}}^2$ ----------- (1)

In a right-angled triangle BAC

${\mathrm{BC}}^2$=${\mathrm{BA}}^2$ + ${\mathrm{AC}}^2$(By Pythagoras theorem)

⇒${\mathrm{BC}}^2$ = (${28}^2$+ ${28}^2$)

⇒${\mathrm{BC}}^2$ = 28 × 28 × 2

⇒ BC = 28$\sqrt{2}$√cm

In the figure BC is the diameter of the semi-circle BEC.

d = 2r

Radius = $\frac{28\sqrt{2}}{2}$cm = 14$\sqrt{2}$cm

Area of semi-circular region BEC

= $\frac{1}{2}$ × π × $r^2$= ( $\frac{1}{2}$ × $\frac{22}{7}$ × 14$\sqrt{2}$ × 14$\sqrt{2}$) $$

= ($\frac{1}{2}$ × $\frac{22}{7}$ × 14 × 14 × 2) = 616 ${\mathrm{cm}}^2$

Area of the shaded portion = Area of semi-circular region BEC − Area of segment BDC

= 616$$−224 $$

= 392 ${\mathrm{cm}}^2$.

Rectangular cloth is not used to make the base of the cone. A conical tent has a curved surface area only.

1) Find the radius of the base of the conical tent from the diameter.

2) Find the slant height of the conical tent.

3) Assume the length of the rectangular cloth as 'l'.

4) Equate the area of the rectangular cloth to the curved surface area of the conical tent and find the length of the cloth required.

5) Find the total cost of the cloth required from the given rate.

Given that

Diameter of the conical tent = 14 m

∴ Radius of the conical tent (r) = $\frac{d}{2}$ = 7 m

Height of the conical tent (h) = 24 m

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

= $\sqrt{7^2+{24}^2}$

= $\sqrt{625}$

= 25 m

Curved surface area of the cone = πrl

= ( $\frac{22}{7}$ × 7 × 25) = 550 $m^2$

But Area of the cloth = Curved surface area of the cone

Length of the cloth × Width of the cloth = Curved surface area of the conical tent

⇒ Length of the cloth = $\frac{\mathrm{Curved}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{conical}\mathrm{tent}}{\mathrm{Width}\mathrm{of}\mathrm{the}\mathrm{cloth}}$

= $\frac{550}{5}$ = 110 m

Rate of 1 metre of cloth = Rs 25

Total cost of 110 m of cloth = Rs 25 × 110 = Rs 2750.

Radius of the cylindrical bucket and the radius of the conical heap are no the same.

1) Find the volume of the cylindrical bucket using the given measurements.

2) Assume the radius of the conical heap as 'R'.

3) Find volume of the conical heap.

4) Equate the volume of the conical heap to the volume of the cylindrical bucket and find the radius of the conical heap.

5) Find the slant height of the conical heap using the Pythagoras property.

Given that

Radius (r) of the cylindrical bucket = 18 cm

Height (h) of the cylindrical bucket = 32 cm

Volume of the cylindrical bucket = π $r^{2}h$

= (π × 18 × 18 × 32) ${\mathrm{cm}}^3$

Assume the radius of the conical heap as R.

Height (H) of the conical heap = 24 cm

Volume of the conical heap = $\frac{1}{3}$π $R^{2}H$ = ( $\frac{1}{3}$π $R^2$ × 24) cm3 = 8π $R^2{\mathrm{cm}}^3.$

Volume of the sand in the bucket and the volume of conical heap are equal.

⇒Volume of the cylindrical bucket = Volume of the conical heap

⇒ π × 18 × 18 × 32 = 8 × π × $R^2$

⇒ $R^2$= $\frac{(18\times 18\times 32)}{8}$ = 18 × 18 × 4 = 1296.

⇒ R = 36 cm

Assume the slant height of the conical heap as l.

$l^2$= $R^2$+ $H^2$

= ${36}^2$+ ${24}^2$

= 1296 + 576 = 1872

⇒ l = 43.3

∴ Slant height (l) of the conical heap is 43.3 cm.

Use the formula to find the sum of the terms of the A.P. involving the first and the last term of the sequence.

1) Use the formula to find the $n^{\mathrm{th}}$term of the A.P. and find the $7^{\mathrm{th}}$, $8^{\mathrm{th}}$, ${14}^{\mathrm{th}}$terms of the A.P.

2) Find the sum of first seven terms of the A.P. using the formula to find the sum of the terms of the A.P. involving the first and the last terms.

3) Last term of the A.P. is considered as the $7^{\mathrm{th}}$term and mark the equation obtained as (1).

4) Find the sum of next seven terms of the A.P. using the formula to find the sum of the terms of the A.P. involving the first and the last terms.

5) First term of the A.P. is considered as the $8^{\mathrm{th}}$term and the last term is considered as ${14}^{\mathrm{th}}$ term and mark the equation obtained as (2).

6) Solve the equations obtained and find the values of a and d.

7) Find the value of the ${28}^{\mathrm{th}}$term using the values of a and d.

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

$n^{\mathrm{th}}$ term of an A.P. is $t_n$ = a + (n − 1) d.

But $7^{\mathrm{th}}$term of the A.P. = a + 6d

$8^{\mathrm{th}}$ term of the A.P. = a + 7d

${14}^{\mathrm{th}}$term of the A.P. = a + 13d

Assume the sum of the first seven terms of the A.P. as $S_{1-7}$.

$S_{1-7}$= $\frac{n}{2}$(First term + Seventh term)

⇒ $S_{1-7}$= $\frac{7}{2}$(a + a + 6d)

⇒ 63 = 7(a + 3d)

⇒ 9 = a + 3d ------------- (i)

Assume the sum of the next seven terms as $S_{8-14}$

∴ $S_{8-14}$ = $\frac{n}{2}$(Eighth term + Fourteenth term)

$S_{8-14}$= $\frac{7}{2}$(a + 7d + a + 13d)

⇒161 = 7 (a + 10d)

⇒ 23 = a + 10d ------------- (ii)

Solving equation (i) and (ii), we get

7d = 14

⇒ d = 2

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

a = 9 − 6 = 3

${28}^{\mathrm{th}}$term of the A.P. = a + 27d = 3 + 27×2 = 57

∴ ${28}^{\mathrm{th}}$term of the given A.P. is 57.

Apply appropriate trigonometric ratio of the angles of depression.

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

2) Complete two right angled triangles on either side of the light house.

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

4) Find the tangent of the angle of depression 30° and find the distance between the first ship and the light-house.

5) Find the tangent of the angle of depression 45° and find the height of the light-house.

Consider h be the height of the light house.

Assume the distance of one of the ships from the light house as x metres.

So, the distance of the other ship from the light house = 100 - x metres.

In the right-angled triangle AOD

tan 30° = $\frac{\mathrm{OD}}{\mathrm{AD}}$= $\frac{h}{x}$

⇒ $\frac{1}{\sqrt{3}}$= $\frac{h}{x}$

⇒ x = $\sqrt{3}$√h ------------- (1)

In a right-angled triangle BDO

tan 45° = $\frac{\mathrm{OD}}{\mathrm{BD}}$ = $\frac{h}{(100-x)}$

⇒1 = $\frac{h}{(100-x)}$

⇒ x + h = 100

By substituting x = $\sqrt{3}$h

$\sqrt{3}$√h + h = 100

⇒ h ( $\sqrt{3}$√+ 1) =100

⇒ h = $\frac{100}{(\sqrt{3}+1)}$

By rationalizing the denominator of the fraction,

⇒ h = $[\frac{100}{(\sqrt{3}+1)}$× $\frac{(\sqrt{3}-1)}{(\sqrt{3}-1)}]$

⇒ h = [ $\frac{100}{2}$] ( $\sqrt{3}$−1)

⇒ h = 36.6 (approximately)

Therefore, the height of the light house is approximately 36.6 m

Consider the coefficients of the terms in the quadratic equation along with their sign.

1) One of the roots of the quadratic equation with coefficient of x as 'a' is given as 2. Find the value of p as the root satisfies the equation.

2) Second equation has real and equal roots, so the discriminant of the equation is equal to zero.

3) Equate the discriminant of the second equation to zero and find the value of k in terms of p.

4) Substitute the value of p and find the value of k.

Given that 2 is a root of the quadratic equation 3$x^2$+ px − 8 = 0.

⇒ 3$({2)}^2$+ p(2) − 8 = 0

⇒12 + 2p − 8 = 0

⇒p = −2

Given that the equation 4$x^2$−2px + k = 0 has equal roots.

⇒Discriminant of the quadratic equation is equal to 0

⇒${(-2p)}^2$−4(4)(k) = 0

⇒16k = 4$p^2$

⇒ k = $\frac{p^2}{4}$

Substituting the value of p, we get

k = $\frac{{(-2)}^2}{4}$= $\frac{4}{4}$ = 1

∴ The value of k is 1.

Locate five points on PY such that they are equidistant.

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

2) Draw a ray PY making an acute angle with PQ.

3) Mark 5 points on PY and join the fifth point to Q.

4) Draw a line $P_4$Q' parallel to $P_5$Q from $P_4$.

5) Draw a line Q'R' parallel to QR.

6) PQ'R' is the required triangle.

Steps of construction:

Step 1: Draw a ray PX.

Step 2: PQ = 6 cm as the radius, draw an arc from point P intersecting PX at Q.

Step 3: Q as the centre and radius equal to 7 cm, draw an arc.

Step 4: P as the centre and radius equal to 8 cm, draw another arc, cutting the previously drawn arc at R.

Step 5: Join PR and QR.

Therefore, PQR is the required triangle.

Step 6: Draw any ray PY making an acute angle with PQ.

Step 7: Now locate 5 points $P_1$, $P_2$, $P_3$, $P_4$and $P_5$ on PY so that P$P_1$=$P_1{P}_2$= $P_2{P}_3$= $P_3{P}_4$= $P_4{P}_5$.

Step 8: Join $P_5$Q and draw a line through $P_4$, which is parallel to $P_5$Q, to intersect PQ at Q'.

Step 9: Draw a line through Q', which is parallel to line QR, to intersect PR at R'.

Then we get, PQ'R' is the required triangle.

Substitute the coordinates of the points in the formula for the area of the triangle in the proper order.

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

2) Substitute the coordinates of the points in the formula for the area of the triangle and equate to zero.

3) Simplify the equation to find the value of the variable in the coordinates.

If the given points are collinear, the area of the triangle formed by the points must be equal to 0.

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

Comparing the given points with ($x_1$,$y_1$) ($x_2$,$y_2$) and ($x_3$,$y_3$),

$x_1$ = p + 1, $y_1$ = 2p−2, $x_2$= p−1, $y_2$= p and $x_3$= p−3, $y_3$= 2p−6.

∴ $\frac{1}{2}$[(p + 1){p−(2p−6)} + (p−1){(2p−6)−(2p−2)} + (p−3){(2p−2)−p}] = 0

⇒ (p + 1)(−p + 6) + (p−1)(−4) + (p−3)(p−2) = 0.

⇒ −4p + 16 = 0 ⇒

⇒ 4p = 16

⇒ p = 4

So that, the value of p is 4.

Assume the ratio in which P divides CD as k:1.

1) Find the coordinates of the mid-point of the line segment AB.

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

3) Find the coordiantes of the point P using the section formula.

4) Equate the x-coordinates of the point P and find the value of k.

5) Substitute the value of k in the equation which equates the y-coordinates and find the value of y.

Given that P is the mid-point of the line segment AB.

∴ Coordinates of point P = $(\frac{-10-2}{2},\frac{4+0}{2})$= $\left(-\mathrm{6,2}\right)$

Then P (−6, 2) divide CD internally in the ratio k: 1.

Applying the section formula, we get

(-6 , 2) = $(\frac{k(-4)+\left(1\right)(-9)}{k+1},\frac{k\left(y\right)+\left(1\right)(-4)}{k+1})$

(-6,2) = $(\frac{-9-4k}{k+1},\frac{-4+\mathrm{ky}}{k+1})$

Equating on both sides

⇒ $\frac{-9-4k}{k+1}$= −6

⇒ −9−4k = −6k−6

⇒ k = $\frac{3}{2}$

⇒ $\frac{k}{1}$ = $\frac{3}{2}$

⇒k :1 = 3:2

Now, $\frac{-4+\mathrm{ky}}{k+1}$= 2

Putting the value of k in the above equation, we get

$\frac{-4+\frac{3}{2}y}{\frac{3}{2}+1}$ = 2

⇒ $\frac{-8+3y}{5}$ = 2

⇒ −8+3y = 10

⇒ y = 6

∴ The ratio in which P divides CD is 3:2 and the value of y is 6.

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) 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.

Consider ABCD as the quadrilateral circumscribing the circle.

Let E, F, G and H be the points of contact of the quadrilateral to the circle.

To Prove: AB + DC = AD + BC

Proof:

AB = AE + EB

AD = AH + HD

DC = DG + GC

BC = BF + FC

We know that AE = AH (Tangents drawn from an external point to the circle are equal.)

Similarly, BE = BF, DH = DG, CG = CF

Now, we have:

AB + DC = AE + EB + DG + GC

= AH + BF + DH + CF

= (AH + DH) + (BF + CF)

= AD + BC

⇒ AB + DC = AD + BC

∴ If a quadrilateral is drawn to circumscribe a circle, the sums of opposite sides are equal.

Hence, it is proved.

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) Find the tangent of the angle of depression and find the distance between the tower and the chimney.

4) Find the tangent of the angle of elevation and find the height of the chimney.

Let AB be the tower and CD be the chimney.

Given that height of the tower = AB = 40 m

In the right-angled ∆ABC,

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

⇒ $\frac{40m}{\mathrm{BC}}$= $\frac{1}{\sqrt{3}}$

⇒ BC = 40 $\sqrt{3}$√m ..................(i)

In the right-angled ∆BCD,

tan 60° = $\frac{\mathrm{CD}}{\mathrm{BC}}$

⇒ $\frac{\mathrm{CD}}{\mathrm{BC}}$ = $\sqrt{3}$

⇒ CD = BC× $\sqrt{3}$

⇒ CD = 40 $\sqrt{3}$× $\sqrt{3}$m [Using ( i)]

∴The height of the chimney is 120 m.

But the height of the chimney meets the pollution norms.

In this question, the conservation of environmental air pollution has been shown.

Subtract the area of the circular portion from one of the faces of the cubical block.

1) Find the radius of the hemispherical depression from the diameter.

2) Find the total surface area of the cubical block, area of the circular region on one of the faces and the surface area of the hemispherical depression.

3) Find the surface area of the remaining solid by subtracting the area of the circular region from the sum of the surface areas of the cubical block and the surface area of the hemispherical depression.

In a cubical block

Edge of the cube, a = 7 cm

In the hemisphere

Diameter of the hemisphere = Length of a side of the cube = 7 cm

∴ Radius(r) of the hemisphere = $\frac{7}{2}$ cm

Surface area of the remaining solid = Surface area of the cube − Surface area of the circle on one of the faces of the cube + Surface area of the hemisphere

= 6 $a^2$+ π $r^2$

= (6 × $7^2$+ $\frac{22}{7}$× $\frac{7}{2}$× $\frac{7}{2}$) c $m^2$

= (294 + 38.5) ${\mathrm{cm}}^2$

= 332.5 ${\mathrm{cm}}^2$

∴ Surface area of the remaining solid is 332.5 ${\mathrm{cm}}^2$

Use the formula to find the sum of the terms of the A.P. involving a, d and n.

1) Use the formula to find the sum of the n terms of an A.P. and find the sum of 30 terms to obtain an equation.

2) Use the formula to find the sum of n terms of an A.P. and substitute the corresponding equations in RHS.

3) Simplify and prove that RHS and LHS are equal.

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

where,

a = First term

d = Common difference

n = Number of terms in an A.P.

We have to prove that, $S_{30}$= 3($S_{20}$−$S_{10}$).

Take R.H.S., 3($S_{20}$−$S_{10}$)

= 3 [$\frac{20}{2}${2a + (20−1)d} − $\frac{10}{2}${2a + (10−1)d} ]

= 3[10 (2a + 19d) − 5(2a + 9d)]

= 3[20a + 190d−10a−45d]

= 3[10a + 145d]

= 15[2a + 29d]

Now take L.H.S., we get

$S_{30}$= 15[2a + 29d]

Hence, proved.

Surface area of the bucket includes the curved surface area and the area of the base.

1) Find the slant height of the bucket from the vertical height and radii using the Pythagoras property.

2) Find the volume of the bucket using the dimensions obtained.

3) Find the surface area of the bucket by finding the sum of the curved surface area and the area of the base of the bucket.

Upper radius of the bucket (R) = 14 cm

Lower radius of the bucket (r) = 7 cm

Height of the bucket (h) = 24 cm

Slant height of the bucket (l) = $\sqrt{h^2+{\left(R-r\right)}^2}$

= $\sqrt{{24}^2+(14-7{)}^2)}$√cm

= $\sqrt{576+49}$cm

= 25 cm

(i) We know that the volume of a frustum of a cone = V = $\frac{1}{3}$π( $R^2$+ Rr + $r^2$)h

⇒V = $\frac{1}{3}$π( ${14}^2$+ 14 × 7 + $7^2$)}24 ${\mathrm{cm}}^3$

= ( $\frac{1}{3}$ × $\frac{22}{7}$ × 343 × 24) ${\mathrm{cm}}^3$

= 8624 ${\mathrm{cm}}^3$

∴the volume of water that can completely fill the bucket is 8624 ${\mathrm{cm}}^3$.

(ii) To measure the area of the metal sheet used in making the bucket, we have to find the surface area of the bucket.

Surface area of the bucket = Curved surface area of the frustum + Curved surface area of the circular base

= π(R + r)l + π $r^2$

= [ $\frac{22}{7}$ × (14 + 7) × 25 ] + [ $\frac{22}{7}$ × $7^2$] ${\mathrm{cm}}^2$

= (1650 + 154) ${\mathrm{cm}}^2$

= 1804 ${\mathrm{cm}}^2$

∴ The area of the metallic sheet used to make the bucket is 1804 ${\mathrm{cm}}^2$.

Find two pairs of even numbers from two roots of the equation.

1) Assume the consecutive even numbers as 2n and (2n + 2).

2) Equate the sum of the assumed numbers to 340.

3) Simplify to obtain a quadratic equation.

4) Solve the quadratic equation and obtain two roots.

5) Find two pairs of even numbers from two roots of the equation.

Let the two consecutive even numbers be 2n and 2n + 2.

Given that sum of the squares of two consecutive even numbers is 340.

(2n${)}^2$+ (2n + 2${)}^2$= 340

⇒ 4$n^2$ + 4$n^2$+ 8n + 4 = 340

⇒ 8($n^2$+ n−42) = 0

⇒$n^2$+ 7n−6n−42 = 0

⇒ n(n+7)−6(n+7) = 0

⇒ (n + 7)(n−6) = 0

⇒ n = 6 or n = −7

Now, for n = 6, we get

First number = 2n = 2 × 6 = 12

Other number = 2n + 2 = 2 × 6 + 2 = 14

And, for n = −7, we get

First number = 2n = 2 × −7 = −14

Other number = 2n + 2 = 2 × −7 + 2 = −12

∴ The two numbers are 12, 14 or −14, −12.

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 Q on the tangent AB and draw a line segment joining the centre O and the point Q.

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

4) OQ is written as the sum of two parts OR and RQ.

5) OQ is greater than OR. So, OQ is greater than OP since OP and OR are radii of the circle.

6) OP is shorter than OQ.

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

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

In the figure AB be a tangent to the circle with centre O and touch the circle at point P.

To prove: OP⊥AB

Construction: Mark a point Q on AB.

Join O and Q.

Let OQ intersect the circle at point R.

Proof:

In the above figure, we observe that any line segment joining the point O to a point on AB is larger than OP.

In the given figure,

OP = OR (Radii of the same circle)

Now, OQ = OR + RQ

⇒ OQ > OR

⇒ OP < OQ

As Q is any point, OP is shorter than any other line segment joining O to any point of AB.

∴ OP ⊥ AB

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

Do not include the outcome (5, 5) in finding the probability of 5 showing up exactly one time.

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

2) Find the number of possible outcomes.

3) List the outcomes in which 5 will not come up either time.

4) Find the number of favourable outcomes.

5) Find the probability by find the ratio of the favourable outcomes to the total number of possible outcomes.

6) List the outcomes where 5 comes up exactly one time.

7) Find the probability by finding the ratio of the favourable outcomes to the total number of possible outcomes.

In a die has 6 faces numbered 1,2,3,4,5 and 6.

Number of possible outcomes on rolling the dice twice are

(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)}

Total number of outcomes = 36.

(i) Favourable outcomes for the event that 5 will not show up either time are

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6)}

Total number of favourable outcomes = 25

∴ Probability of not getting 5 either time = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favorable}\mathrm{out}\mathrm{comes}}{\mathrm{Total}\mathrm{Number}\mathrm{of}\mathrm{outcomes}}$ = $\frac{25}{36}$

(ii) Favourable outcomes for the event that 5 will show up exactly one time are

Total number of favourable outcomes = 10

∴ Probability of getting 5 such that it will come up exactly one time

= $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}$ = $\frac{10}{36}$

Find two values for x from two values of y.

1) Assume the fraction of the first term as 'y'.

2) Fraction in the second term is the reciprocal of the first fraction.

3) Substitute in the equation and simplify by taking LCM.

4) Obtain a quadratic equation and find the values of y.

5) Substitute the values of y and simplify to find the values of x.

Given 3$\left(\frac{3x-1}{2x+3}\right)$-2$\left(\frac{2x+3}{3x-1}\right)$= 5

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

∴ 3y - 2$\times \frac{1}{y}$ = 5

⇒3$y^2$- 2 = 5y

⇒3$y^2$- 5y - 2 = 0

⇒3$y^2$- 6y+ y -2 = 0

⇒3y(y -2) + 1(y - 2) = 0

⇒(y - 2)( 3y +1) = 0

⇒y = 2 or y = -$\frac{1}{3}$

Substitute the values of y in equation (1).

If y = 2 then,

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

⇒3x -1 = 4x + 6

∴ x = -7

If y = - $\frac{1}{3}$,

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

⇒3(3x-1) = -(2x+3)

⇒9x -3 = -2x - 3

⇒ 11x = 0

∴ x = 0

∴ The solutions of x are 0 and -7.