CBSE Class 10 - Maths

Substitute the coefficients of the terms of the equation in the discriminant along with their sign.

1) Simplify and write the given quadratic equation in the standard form.

2) The roots of the given equation are equal. So, the discriminant of the equation is equal to zero.

3) Substitute the values of the coefficients in the formula for finding the discriminant and equate to zero.

4) Simplify and find two values of p.

The quadratic equation px(x − 2) + 6 = 0 can be simplified as p$x^2$−2px + 6 = 0.

Given that the roots of the given quadratic equation are equal.

Therefore, its discriminant is equal to 0.

∴ D = 0.

⇒$b^2$−4ac = 0

⇒ ${\left(-2p\right)}^2$−4 × p × 6 = 0 [where a = p, b = −2p, c = 6]

⇒ 4$p^2$−24p = 0

⇒ 4p (p − 6) = 0

⇒ 4p = 0 or p − 6 = 0

⇒ p = 0 or p = 6

Therefore, p = 0 or 6.

Two-digit numbers are from 10 to 99.

1) List the two digit numbers divisible by 3.

2) Numbers form A.P. with the first term as 12, last term as 99 and the common difference as 3.

3) Equate the last term of the series 99 to $n^{\mathrm{th}}$term of an A.P. and find the value of n.

4) Value of n gives the number of terms which are divisible by 3.

Two-digit numbers which are divisible by 3 are 12, 15, 18, 21 ........... 99.

These numbers form an A.P. with first term (a) = 12 and common difference (d) = 3.

To find the number of terms, assume 99 as the $n^{\mathrm{th}}$term.

⇒ $t_n$= 99

⇒ a + (n − 1)d = 99

⇒ 12 + (n − 1)3 = 99

⇒ 12 + 3n − 3 = 99

⇒ 3n + 9 = 99

⇒ 3n = 99 − 9

⇒ 3n = 90

⇒ n = $\frac{90}{3}$= 30 .

Hence, the number of two-digit numbers which are divisible by 3 is 30.

Area of the triangle ABC is equal to the sum of the areas of the triangles OAB, OBC and OCA.

1) Find the area of the right angled triangle using the formula for the area of the triangle.

2) Join OA, OB and OC.

3) Let OM, ON and OP be the radii at the points of contact of the tangents of the circle AC, BC and AB.

4) Find the length of AC using the Pythagoras theorem.

5) Radii of the circle are perpendicular to the tangents i.e. sides of the triangle.

6) Equate the area of the triangle ABC to the sum of the areas of OAB, OBC and OCB.

7) Substitute the values in the formula for the area of the triangle and simplify to find the value of radius of the circle.

The given triangle ABC is a right angled triangle such with ∠B = 90° , BC = 6 cm, AB = 8 cm.

O is given as the centre of the circumscribed circle and r is the radius of the circle.

Consider A, B and C as external points to the circle.

AP and AM will be tangents to the circle from A

BP and BN will be tangents to the circle from B

CN and CM will be tangents to the circle from C.

OP = ON = OM = r (radius of the circle)

For the right-angled triangle, ABC

Area of ΔABC = $\frac{1}{2}$× 6 × 8 = 24 ${\mathrm{cm}}^2$

Using Pythagoras theorem,

${\mathrm{CA}}^2$= ${\mathrm{AB}}^2$+ ${\mathrm{BC}}^2$

⇒ ${\mathrm{CA}}^2$ = $8^2$ + $6^2$

⇒ ${\mathrm{CA}}^2$= 100

⇒ CA = 10 cm

Area of ∆ABC = (Area of ∆OAB) + (Area of ∆OBC) + (Area of ∆OCA)

24 = ( $\frac{1}{2}$ × r × AB) + ( $\frac{1}{2}$ × r × BC) + ( $\frac{1}{2}$ × r × CA)

24 = $\frac{1}{2}r(\mathrm{AB}+\mathrm{BC}+\mathrm{CA})$

⇒r = $\frac{(2\times 24)}{(\mathrm{AB}+\mathrm{BC}+\mathrm{CA})}$

⇒r = $\frac{\left(48\right)}{(8+6+10)}$= $\frac{48}{24}$

⇒r = 2 cm.

Hence r = 2 cm.

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.

Let us assume that AB is the diameter of the circle.

Draw a tangent PQ at point B and tangent RS at point A of the circle.

Draw radius to these tangents, it will be perpendicular to the tangents.

Hence we get OA ⊥ RS and OB ⊥ PQ

∠OAR = 90º

∠OAS = 90º

∠OBP = 90º

∠OBQ = 90º

From the figure, it can be observe that

∠OAR = ∠OBQ (Since they are alternate interior angles)

∠OAS = ∠OBP (Since they are alternate interior angles)

As the alternate interior angles are equal, it is proved that the lines PQ and RS will be parallel.

Angle in a quadrant is a right angle. Area of the the shaded region is obtained by subtracting the area of four quadrants and the area of the circle from the area of the square.

OR

Diameter of the semicircle is equal to the breadth of the rectangle.

1) Find the area of a quadrant of the circle using the given radius of the quadrant.

2) Find the area of the square and the area of the circle using the given dimensions.

3) Subtract the areas of four quadrants and the area of the circle from the area of the square to find the area of the shaded region.

OR

1) Find the area of the rectangle using the given measurements.

2) Breadth of the rectangle is the diameter of the semicircular region cut off.

3) Find the radius of the semicircle and the area of the semicircle.

4) Find the area of the remaining portion of the rectangle by subtracting the area of the semicircle from the area of the rectangle.

Given ABCD is a square. Quadrant of a circle of radius 1 cm is drawn at each vertex of a square.

The quadrant of a circle is a sector with angle 90°.

∴ Area of each quadrant = $\frac{90°}{360°}$× π$r^2$

= ( 3.14) × $\frac{1}{4}$ ×${\left(1\right)}^2$ = 0.785 ${\mathrm{cm}}^2$

Area of the square = ${\left(\mathrm{side}\right)}^2$= ${\left(4\right)}^2$= 16 ${\mathrm{cm}}^2$.

Diameter of the circle = 2 cm ⇒Radius of the circle = 1 cm.

Area of the circle = π$r^2$= 3.14 ×${\left(1\right)}^2$= 3.14 ${\mathrm{cm}}^2$

Area of the shaded region

= Area of square − Area of circle − (4 × Area of the quadrant)

= [16 − 3.14 − (4 × 0.785 ) ]$$

= [16 − 3.14 − 3.14]$$

= [16 − 6.28]

= 9.72 ${\mathrm{cm}}^2$.

Hence, the area of the shaded region is 9.72 ${\mathrm{cm}}^2$.

OR

Area of a rectangle = length x breadth

∴ Area of rectangular sheet of paper = AB × AD = 40 cm × 28 cm = 1120${\mathrm{cm}}^2$.

BC = AD = 28 (Opposite sides of a rectangle are equal)

∴ Diameter of the semicircle = 28 cm ( BC forms the diameter of the semicircle)

Radius of the semicircle (r ) = $\frac{28}{2}$ = 14 cm

Area of the semicircle = $\frac{\pi r^2}{2}$= $\frac{22}{7\times 2}$×${\left(14\right)}^2$

= $\frac{22}{14}$ × 196$$= 22 ×14 = 308 ${\mathrm{cm}}^2$.

Area of the remaining paper = Area of the rectangular sheet − Area of the semicircle

= 1120 − 308

= 812 $$

∴ Area of remaining sheet of paper = 812 ${\mathrm{cm}}^2$.

Simplify the fractions involving fractions correctly.

1) Find the volume of the sphere using the formula to find the volume of a sphere.

2) Find the volume of each smaller cone using the formula to find the volume of a cone.

3) Assume the number of smaller cones formed as 'n'.

4) Equate the volume of 'n' number of smaller cones to the volume of the sphere.

5) Simplify and find the value of 'n'.

Let n be the number of cones formed from the sphere

Let r be the radius of the sphere, h be the height of the cone and R be the radius of the cone.

r = 10.5 cm (Given)

Assume the volume of the sphere as $V_1$

$V_1$= $\frac{4}{3}\pi r^3$= $\frac{4}{3}\pi$${\left(10.5\right)}^3{\mathrm{cm}}^3$

Given that h = 3 cm and R = 3.5 cm

Volume of each smaller cone, $V_2$= $\frac{1}{3}\pi R^{2}h$

$V_2$= $\frac{1}{3}\times \pi \times {\left(3.5\right)}^2\times 3{\mathrm{cm}}^3$

As the solid sphere is melted and made into smaller cones, the volume of ‘n’ smaller cones is equal to the volume of the sphere.

n × Volume of each smaller cone = Volume of the sphere

n ×$\frac{1}{3}$× π ×${\left(3.5\right)}^2$× 3 = $\frac{4}{3}$× π ×${\left(10.5\right)}^3$

n = $\frac{4\times {\left(10.5\right)}^3}{{\left(3.5\right)}^2\times 3}$= 126.

Hence, the number of smaller cones formed = 126.

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

1) Point P is equidistant from A and B. So, equate the distances AP and BP.

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

3) Simplify and find the value of k.

The point P (2, 4) is equidistant from the points A(5, k) and B (k, 7).

∴ AP = BP (squaring on both sides)

${\mathrm{AP}}^2$= ${\mathrm{BP}}^2$

When we apply distance formula,

⇒ ${\left(2-5\right)}^2$+ ${\left(4-k\right)}^2$= ${\left(2-k\right)}^2$+ ${\left(4-7\right)}^2$

⇒ ${\left(-3\right)}^2$+ 16 − 8k +$k^2$= 4 − 4k +$k^2$ + ${\left(-3\right)}^2$

⇒ 9 + 16 − 8k +$k^2$ = 4 − 4k +$k^2$+ 9

⇒$k^2$−8k + 25 =$k^2$−4k + 13

⇒ $k^2$−$k^2$ − 8k + 4k = 13 − 25

⇒ − 4k = − 12

⇒ k = $\frac{-12}{-4}$= 3

Therefore, the value of k = 3.

Consider the queen and jack of both the colours and all the four flowers.

1) Total number of cards in a pack of cards is 52. So, the number of possible outcomes is 52.

2) Number of red kings in the pack is 2.

3) Assume the number of favourable outcomes of getting a red king as A.

4) Ratio of the number of favourable outcomes and the number of possible outcomes gives P(A).

3) There are 4 queens and 4 jacks in the pack of cards.

4) Assume the number of favourable outcomes of getting a queen or a jack as B.

5) Ratio of the number of favourable outcomes and the number of possible outcomes gives P(B).

Total number of cards = 52

When one card is drawn at random, number of possible outcomes = 52

(i) In a pack of 52 cards, there are two red kings, one is king of heart and another is king of diamond.

∴ Number of favourable outcomes = 2

If A is the event that the card drawn at random is a red king, then probability of E

P(A) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$ = $\frac{2}{52}$ = $\frac{1}{26}$

Hence, the probability of drawing a red king = $\frac{1}{26}$.

(ii) Number of queens in the pack of cards = 4 (1 queen each of heart, club, spade and diamond)

Number of jacks in the pack of cards = 4 (1 jack each of heart,club, spade and diamond)

∴ Number of favourable outcomes = 8

Let B be the event that the card drawn is a queen or a jack.

Therefore, the probability of drawing a queen or a jack is

P(B) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$= $\frac{8}{52}$= $\frac{2}{13}$

Therefore, the probability of drawing a queen or a jack is $\frac{2}{13}$.

Remaining part of the quadratic equation other than $x^2$term and x term is the constant term.

OR

Obtain a single pair of numbers from the numbers obtained.

1) Compare the given quadratic equation with the standard form of the quadratic equation and write the values of a, b and c.

2) Write the formula to find the roots of the quadratic equation and substitute the values of a, b and c.

3) Simplify and find the roots of the equation.

OR

1) Assume the required numbers as a and b.

2) Write the equations for the sum of the numbers and the product of the numbers.

3) Find the value of a in terms of b from the product of the roots.

4) Substitute the value of b in the equation of the sum of the roots and simplify to form a quadratic equation.

5) Solve the quadratic equation by splitting the middle term and find the required numbers.

Given quadratic equation$x^2$−4ax − $b^2$+${4a}^2$ = 0

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

$x^2$+ (−4a)x + (${4a}^2$−$b^2$) = 0 [a = 1, b = −4a, c =${4a}^2$−$b^2$]

x = $\frac{\_-b\pm \sqrt{b^2-4\mathrm{ac}}}{2a}$

x = $\frac{4a\pm \sqrt{{16a}^2-{16a}^2+{4b}^2}}{2}$

x = $\frac{4a\pm \sqrt{{4b}^2}}{2}$

x = $\frac{4a\pm 2b}{2}$

x = 2a + b or x = 2a - b

Hence, solution of the given quadratic equation x = 2a + b or x = 2a − b.

OR

Let us assume that two natural numbers are a and b.

Given that sum of two natural numbers = 8

∴ a + b = 8

⇒ a = 8 − b ---------- (1)

Given that product of two natural numbers = 15

∴ ab = 15

⇒ (8 − b)b = 15 [From (1)]

⇒ 8b −$b^2$= 15

⇒$b^2$−8b + 15 = 0

⇒$b^2$−5b − 3b + 15 = 0

⇒ b(b − 5) − 3(b − 5) = 0

⇒ (b − 3) (b − 5) = 0

⇒ b = 3 or b = 5

When b = 3, then a = 8 − b = 8 − 3 = 5

When b = 5, then a = 8 − b = 8 − 5 = 3

Therefore, the two numbers are 3 and 5.

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

1) List the numbers which are divisible by 7 and lying between 500 and 900.

2) Equate the last term of the A.P. to the formula of finding the $n^{\mathrm{th}}$ term of the A.P. and simplify to find the value of n.

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

4) Find the sum of all the terms of the A.P. using the formula and the values obtained.

The numbers which are multiples of 7 between 500 and 900 are 504, 511, 518, 525… , 896.

This forms an A.P with first term (a ) = 504 and common difference (d ) = 7.

If there are n terms in the A.P.

Last term =$a_n$= 896

⇒a + ( n -1)d = 896

⇒504 + ( n - 1)7 = 896

⇒( n - 1)7 = 896 - 504

⇒( n - 1)7 = 392

⇒( n - 1) = $\frac{392}{7}$

⇒( n - 1) = 56

⇒ n = 57

Sum of n terms =$S_n$= $\frac{n}{2}(a+l)$

Sum of all the numbers which are multiples of 7 between 500 and 900:

$S_n$= $\frac{n}{2}$( 504 + 896)

$S_n$= $\frac{57}{2}$(1400)

$S_n$= 57 × 700

$S_n$= 39900

Therefore, the sum of all the multiples of 7 lying between 500 and 900 = 39900.

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 $X_4$C' parallel to $X_5$C from $X_4$.

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

6) A'BC' is the required triangle.

1. Draw a line BC of length 7 cm.

2. Draw a ray CN which makes an angle of 60° at C.

3. Draw a ray BM which makes an angle of 45° at B.

4. Mark the point of intersection of rays CN and BM as A.

5. The resulting triangle ABC is the triangle whose similar triangle has to be drawn.

6. Draw any ray BX which makes an acute angle with the line BC on the side opposite to the vertex A.

7. Locate 5 (Greater of 3 and 5 in $\frac{3}{5}$) points $X_1$, $X_2$, $X_3$, $X_4$ and $X_5$ on BX so that B $X_1$= $X_1{X}_2$= $X_2{X}_3$= $X_3{X}_4$= $X_4{X}_5$.

8. Join $X_5$to the point C to get the line $X_5$C and draw a line through $X_3$(Since 3 is the smaller of 3 and 5 in $\frac{3}{5}$) parallel to $X_5$C such that it intersects BC at C '.

9.Draw a line through C’ parallel to the line AC such that it intersects the line BA.

11. Mark the point of intersection as A’.

12. A’BC’ is the required triangle which is similar to the ΔABC and whose sides are $\frac{3}{5}$times the corresponding sides of ΔABC.

Length of PL i.e. z can be found by subtracting the sum of x and y from the sum of x, y and z.

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

2) Assume the lengths of the equal tangents from three sides as x, y and z.

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

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

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

6) Find the length of x, y and z by subtracting the sum of two of them from the sum of all the three.

7) Lengths of x, y and z gives the required lengths.

Given that in a ΔPQR, PQ = 10, QR = 8 cm and PR = 12 cm.

Since the lengths of the tangents drawn from an external point to a circle are equal.

Assume the tangents from the external point Q, QM and QL as x cm.

Assume the tangents from the external point R, RM and RN as y cm.

Assume the tangents from the external point P, PL and PN as z cm.

QR = QM + MR = x + y = 8 cm …(1)

PQ = PL + LQ = z + x = 10 cm …(2)

PR = PN + NR = z + y = 12 cm …(3)

Add (1), (2) and (3) we get,

(x + y) + (z + x) + (z + y) = (8 + 10 + 12) = 30 cm

∴ 2(x + y + z) = 30 cm

⇒ x + y + z = 15 cm …............(4)

From (2) and (4), we have

⇒10 + y = 15

∴ y = 5

From (3) and 4, we have

⇒12 + x = 15

∴ x = 3

From (1) and (4), we have

⇒ z + 8 = 15

⇒ z = 7.

Hence we get, QM = x = 3 cm, RN = y = 5 cm and PL = z = 7 cm.

Hypotenuse of the triangle is the diameter of the circle.

OR

Side of the square is the diameter of the semicircle.

1) Angle in a semicircle is a right angle. Therefore, triangle ABC is a right angled triangle right angled at A.

2) Find the length of the hypotenuse BC using the Pythagoras theorem.

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

4) Hypotenuse is the diameter of the circle. Find the radius of the circle.

5) Find the area of the quadrant of the circle using the radius obtained.

6) Subtract the sum of the areas of the right angled triangle and the area of the quadrant of the circle from the area of the circle to find the area of the shaded region of the circle.

OR

1) Side of the square is the diameter of the semicircle.

2) Find the area of the semicircle.

3) Subtract twice the area of the semicircle from the area of the square to find the area of the shaded region.

In the figure, AC = 24 cm, AB = 7 cm and ∠BOD = 90°

Angle in a semicircle is 90°.

Hence ∠BAC = 90°

∴ ΔABC is a right-angled triangle.

Area of ΔABC, $A_1$= $\frac{1}{2}$ × Base × Height = $\frac{1}{2}$ × 7 × 24 = 84 ${\mathrm{cm}}^2$

Therefore in ΔABC, we have

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

∴${24}^2$ +$7^2$ = ${\mathrm{BC}}^2$

⇒${\mathrm{BC}}^2$ = (576 + 49)

⇒${\mathrm{BC}}^2$ = 625

⇒ BC = 25 cm

BC is the diameter of the circle and hence to get the radius OC.

OC = Radius of the circle = $\frac{25}{2}$cm.

$A_2$= Area of the sector COD with angle 90° at ∠BAC

= $\frac{\theta }{360°}$ × π$r^2$ = $\frac{90°}{360°}$ × $\frac{22}{7}$ × $\frac{25}{2}$ × $\frac{25}{2}$

= $\frac{1}{4}$ × $\frac{22}{7}$ × $\frac{25}{2}$ × $\frac{25}{2}$

= $\frac{11\times 25\times 25}{56}$

= 122.77 ${\mathrm{cm}}^2$

$A_3$ = Area of circle = π$r^2$

= $\frac{22}{7}$ × $\frac{25}{2}$ × $\frac{25}{2}$

= $\frac{11\times 25\times 25}{14}$

= 491.07 ${\mathrm{cm}}^2$

Area of the shaded region

= Area of the circle−(Area of ΔABC + Area of sector COD)

=$A_3$−($A_1$ +$A_2$)

= 491.07−(84 + 122.77)

= 491.07− 206.77

= 284.30

Therefore, area of the shaded region = 284.30 ${\mathrm{cm}}^2$.

or

Area of the shaded region

= Area of the square -(Area of$1^{\mathrm{st}}$ semicircle + Area of $2^{\mathrm{nd}}$ semicircle)

= Area of the square − (Area of semicircle APD + Area of semicircle BPC)

Given that side of the square = 14 cm

Area of square = ${\left(\mathrm{side}\right)}^2$= ${\left(14\right)}^2$ = 196${\mathrm{cm}}^2$

Area of semicircle APD = $\frac{1}{2}$π$r^2$

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

Diameter of semicircles APD and BPC = 14 cm

Radius of semicircles APD and BPC = 7 cm

∴ Area of semicircle BPC = 77 ${\mathrm{cm}}^2$

Area of the shaded region

= Area of square − (Area of semicircle APD + Area of semicircle BPC)

= 196$$−(77 + 77)

= 196$$−154

= 42

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

Find the volumes of the hemisphere and the cylinder in terms of π.

1) Find the volume of the hemisphere using the given radius in terms of π.

2) Assume the height of the water in the cylinder as 'h'.

3) Find the volume of the cylinder in terms of π and h.

4) Equate the volumes of the hemisphere and the volume of the cylinder.

5) Simplify and find the value of h.

Assume that h is the height of the water in the cylindrical vessel and r is the radius of the bowl.

Given that r = 9 cm

Let$V_1$ be the volume of the water in the bowl which is hemispherical in shape.

Hence$V_1$ = $\frac{2}{3}$ π$r^3$

$V_1$= $\frac{2}{3}$ × π ×${\left(9\right)}^3$

Let R be the radius of the cylinder.

Given that R = 6 cm.

Volume of water in the cylindrical vessel,$V_2$ = π$r^2$h

⇒$V_2$ = π${\left(6\right)}^2$h

Volume of water in cylindrical vessel = Volume of the water in hemispherical bowl

(As the water in the bowl has been emptied into the cylindrical vessel)

⇒π${\left(6\right)}^2$h = $\frac{2}{3}$ × π × $9^3$

⇒ h = $\frac{2\times {\left(9\right)}^3}{3\times {\left(6\right)}^2}$= $\frac{27}{2}$

⇒ h = 13.5

So, the height of the water in the cylindrical vessel is 13.5 cm.

Find tangent of the alternate angle of the angles of depression of the top and bottom of the tower from the cliff.

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

2) Draw a horizonal line at the top of the tower and mark the alternate angles of the angles of depression of the tower.

3) Assume the height of the tower as h. Height of the cliff is $60\sqrt{3}$m and the height of the cliff excluding the height of the tower is (60$\sqrt{3}$-h) m.

4) Find the tangent of the alternate angle of the angle of depression of the top of the tower.

5) Substitute the values and simplify to find the distance of the foot of the tower from the foot of the cliff in terms of h.

6) Find the tangent of the alternate angle of the angle of depression of the bottom of the tower.

7) Substitute the values and simplify to find the distance of the foot of the tower from the foot of the cliff.

8) Equate the values of distance of the foot of the tower from the foot of the cliff and simplify to find the height of the tower.

Let us assume that AD is the tower with height ‘h’ and BC is the cliff.

Let AB = x , denote the distance of the foot of the tower from the foot of the cliff.

Angles of depression of the top D of the tower from top C of the cliff is 45° and bottom A of the tower from top C of the cliff is 60°.

∴ ∠CDE = 45° and ∠CAB = 60°

BC = 60 $\sqrt{3}$(BC is the height of the cliff)

∴ CE = BC − BE

CE = (60 $\sqrt{3}$ − h) m (since BE = AD = h)

In a ΔDEC,

tan 45° = $\frac{\mathrm{CE}}{\mathrm{DE}}$

⇒1 = $\frac{60\sqrt{3}-h}{x}$ ( DE = AB = x)

⇒x = 60 $\sqrt{3}$−h ---------(1)

In a ΔCBA,

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

⇒ $\sqrt{3}$ = $\frac{60\sqrt{3}}{x}$ ( DE = AB = x)

⇒x = 60 ---------(2)

From (1) and (2), we get

60 = 60 $\sqrt{3}$−h

⇒ h = 60 $\sqrt{3}$−60

⇒h = 60( $\sqrt{3}$−1)

∴ Height of the tower = 60 ( $\sqrt{3}$−1) m.

Substitute the coordinates of the points in the section formula along with their signs.

OR

Substitute the coordinates of the points in the formula to find the area of the triangle along with their signs.

1) Find the ratio in which the point P divides the line segment AB using the given condition.

2) Use the section formula to find the coordinate of the point P.

3) Substitute the values of the coordinates and simplify to find the coordinates.

OR

1) Find the area of the triangle ABC using the formula to find the area of the triangle.

2) Find the area of the triangle ADC using the formula to find the area of the triangle.

3) Sum of the areas of the triangles ABC and ADC gives the area of the quadrilateral ABCD.

The points A, P and B are points on the line segment AB.

Given that AP = $\frac{3}{7}$AB.

⇒ $\frac{\mathrm{AB}}{\mathrm{AP}}$= $\frac{7}{3}$

⇒ $\frac{\mathrm{AB}}{\mathrm{AP}}$ - 1 = $\frac{7}{3}$ - 1

⇒ $\frac{\mathrm{AB}-\mathrm{AP}}{\mathrm{AP}}$= $\frac{7-3}{3}$

⇒ $\frac{\mathrm{PB}}{\mathrm{AP}}$ = $\frac{4}{3}$.

Hence, AP : PB = 3 : 4

We can find the coordinates of P using section formula on the points A(−2, −2) and B(2, −4).

Therefore, coordinates of the point P = $(\frac{(3\times 2)+(4\times -2)}{3+4},\frac{(3\times -4)+(4\times -2)}{3+4})$

= $(\frac{6-8}{7},\frac{-12-8}{7})$ = $(\frac{-2}{7},\frac{-20}{7})$

Therefore, coordinates of the point P are $(\frac{-2}{7},\frac{-20}{7})$

OR

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD

Area of a triangle with vertices ( $x_1$, $y_1$), ( $x_2$, $y_2$), ( $x_3$, $y_3$)

= $\frac{1}{2}{\mid x}_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\mid$

Area of triangle ABC = $\frac{1}{2}$| (−3)( −4 −1) + (−2)(−1 − (−1)) + 4(−1 − (−4))|

= $\frac{1}{2}$|(−3)(−3) + (−2)(0) + (4)(3)|

= $\frac{1}{2}$| 9 + 12|

= $\frac{1}{2}$| 21| sq unit

Area of triangle ACD = $\frac{1}{2}$ | (−3)(−1−4) + (4)(4 − (−1)) + 3( −1 − (−1))|

= $\frac{1}{2}$ |(−3)(−5) + (4)(5) + (3)(0)|

= $\frac{1}{2}$|15 + 20|

= $\frac{1}{2}$ |35| sq unit $$

So, area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD

= $(\frac{21}{2}+\frac{35}{2})$

= $\frac{56}{2}$

= 28 sq unit $$

Therefore, area of the quadrilateral ABCD = 28 sq unit.

Consider the coefficients of the coordinates of the points with their signs.

1) Given that the points A(x, y), B(3, 6) and C(-3, 4) are collinear. So the area of the triangle formed by the points is zero.

2) Substitute the values of the coordinates in the formula for the area of the triangle.

3) Simplify and obtain the required equation.

If the given points A (x, y), B (3, 6) and C (−3, 4) are collinear, then area of the triangle ABC = 0

Area of a triangle of vertices ($x_1$, $y_1$), ($x_2$, $y_2$) and ($x_3$, $y_3$) = $\frac{1}{2}\mid (y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\mid$ = 0

⇒ x(6 − 4) + 3 (4 − y) + (−3)(y − 6) = 0

⇒2x + 12 − 3y − 3y + 18 = 0

⇒2x − 6y + 30 = 0

⇒ x − 3y + 15 = 0

Hence proved.

Consider the kings, queens and aces of both the colour to be removed.

1) Find the number of cards of kings, queens and aces and subtract from the total number of cards in the pack. This gives the number of possible outcomes of the event.

2) Find the number of black face cards in the remaining cards.

3) Ratio of the number of black face cards and the number of remaining cards gives the required probability.

4) Find the red cards in the remaining cards.

5) Ratio of the number of red cards in the remaining cards and the number of remaining cards gives the required probability.

The pack has a total of 52 cards.

Number of kings in the pack = 4

Number of queens in the pack = 4

Number of aces in the pack = 4

Number of cards removed = 4 + 4 + 4 = 12

Number of remaining cards in the pack = 52−12 = 40

(i) In the remaining 40 cards, the number of black face cards are 2 jacks of spade and 2 jacks of club.

P(a black face card) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{black}\mathrm{face}\mathrm{card}s}{\mathrm{Number}\mathrm{of}\mathrm{card}s\mathrm{remaining}\mathrm{in}\mathrm{the}\mathrm{pack}}$= $\frac{2}{40}$= $\frac{1}{20}$.

Therefore, the probability of getting a black face card = $\frac{1}{20}$

(ii) Total Number of red cards in the 52 card pack = 26

After removing 12 cards, the remaining number of red cards

= 26 − (2 + 2 + 2) = 26−6 = 20.

P(a black face card) = $\frac{\mathrm{Remaining}\mathrm{number}\mathrm{of}\mathrm{red}\mathrm{card}s}{\mathrm{Number}\mathrm{of}\mathrm{card}s\mathrm{remaining}\mathrm{in}\mathrm{the}\mathrm{pack}}$ = $\frac{20}{40}$ = $\frac{1}{2}$.

Hence, the probability of getting a red card = $\frac{1}{2}$.

Find two fractions using the two values of the denominator.

OR

Consider the positive value as the speed of the aircraft.

1) Assume the denominator of the fraction as x.

2) Find the fraction in terms of x.

3) Find the new fraction according to the given condition.

4) Apply the given relation between the original fraction and the new fraction.

5) Solve the quadratic equation obtained.

6) find the fractions from the roots of the equation.

OR

1) Assume the original speed of the plane as x.

2) Find the decreased speed of the weather.

3) Find the time taken by the plane in terms of x.

4) FInd the time taken by the decreased speed of the plane.

5) equate the difference between the speeds to the given time.

6) SOlve the quadratic equation obtained.

7) Consider the positive value as the original speed of the plane.

Let x be the denominator of the fraction

Hence numerator = x − 3

Original fraction = $\frac{x-3}{x}$

To get the new fraction, add 1 to the denominator

New fraction = $\frac{x-3}{x+1}$

Given that the new fraction is decreased by $\frac{1}{15}$

According to the condition,

⇒$\frac{x-3}{x+1}$= $\frac{x-3}{x}$ − $\frac{1}{15}$

⇒$\frac{x-3}{x}$ − $\frac{x-3}{x+1}$= $\frac{1}{15}$

⇒$\frac{(x-3)(x+1)-x(x-3)}{x(x+1)}$ = $\frac{1}{15}$

⇒15[$x^2$−2x − 3 − $x^2$+ 3x] = $x^2$+ x

⇒ 15( x − 3) = $x^2$+ x

⇒$x^2$+ x − 15x + 45 = 0

⇒$x^2$−14x + 45 = 0

⇒$x^2$−9x − 5x + 45 = 0

⇒x( x − 9) − 5( x - 9) = 0

⇒x − 5 = 0 or x − 9 = 0

⇒ x = 5 or x = 9

Hence denominator = 5 or 9

Numerator = 5 − 3 = 2 or 9 − 3 = 6

Thus,the fraction = $\frac{2}{5}$ or $\frac{6}{9}$ = $\frac{2}{3}$.

OR

Assume that x km/hr is the original speed of the plane.

When speed is 100 km/hr reduced due to bad weather,

Speed of plane = (x −100) km/hr

Let$t_1$be the time taken by the plane to reach destination when it is flying at original speed $t_1$= $\frac{2800}{x}$ hr

Let$t_2$be the time taken by the plane to reach destination when it is flying at reduced speed.$t_2$= $\frac{2800}{x-100}$ hr

Given that$t_2$ - $t_1$ = 30 min = $\frac{1}{2}$hr

∴ $\frac{2800}{x-100}-$ $\frac{2800}{x}$= $\frac{1}{2}$

⇒$\frac{2800x-2800(x-100)}{x(x-100)}$= $\frac{1}{2}$

⇒2800×100×2 = $x^2$- 100x

⇒$x^2$- 100x - 560000 = 0

⇒$x^2$- 800x + 700x - 560000 = 0

⇒x( x - 800) + 700( x - 800) = 0

⇒( x - 800) ( x + 700) = 0

⇒ x - 800 = 0 or x + 700 = 0

⇒ x = 800 or x = -700

∴ x = 800 ( Since speed cannot be negative)

Hence the original speed of the plane = 800 km/hr

∴Original time taken to reach destination, $t_1$= $\frac{2800}{800}$= $\frac{7}{2}$= 3$\frac{1}{2}$ hrs

Therefore, original duration of flight = 3 hours 30 minutes.

Sum of next four terms after the fourth term is obtained by subtracting the sum of first four terms from the sum of first eight terms.

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

2) Find the sum of the first four terms using the formula to find the sum of n terms of the A.P.

3) Find the sum of the next four terms by subtracting the sum of first four terms from the sum of first eight terms.

4) Substitute the values of $S_8$and $S_4$in the given condition and simplify to find the value of d.

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

Given that, first term (a) = 5.

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

∴ Sum of the first four terms ($S_4)$= $\frac{4}{2}$[ 2×5 + ( 4 −1)d ]

= 2[10 + 3d]

= 20 + 6d

Sum of the next four terms = $S_8$−$S_4$= $\frac{8}{2}$[2 × 5 + (8 − 1)d] − ( 20 + 6d)

= 40 + 28d − 20 − 6d

= 20 + 22d

Given that sum of its first four terms is half the sum of the next four terms i.e $S_4$= $\frac{1}{2}$[$S_8$- $S_4$]

⇒ 20 + 6d = $\frac{1}{2}$[ 20 + 22d]

⇒20 + 6d = 10 + 11d

⇒11d - 6d = 20 - 10

⇒5d = 10

⇒d = 2

Therefore, the common difference of the A.P. = 2

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

Product of the rate of flow of water and the time taken gives the volume of the water.

OR

Units of the capacity of the glass are ${\mathrm{cm}}^3$.

1) Find the radius of the hemispherical tank from the given diameter.

2) Find the volume of the water in the tank.

3) Assume the time taken to empty half of the tank as t sec.

4) Equate the product of the rate of flow of water and t with the half of the volume of the tank.

5) Simplify the equation and find the value of t.

OR

1) Find the radii of the glass from the given diameters.

2) Substitute the values in the formula for the volume of a frustum of a cone.

3) Simplify to find the capacity of the glass.

Diameter of the base of the hemispherical tank = 3 m

Radius of of the hemispherical tank = $\frac{3}{2}$ m

Since it is hemispherical shaped tank, the volume of water in the tank = $\frac{2}{3}$π $r^3$

= $\frac{2}{3}$× $\frac{22}{7}$× ${\left(\frac{3}{2}\right)}^3$ = $\frac{99}{14}{m}^3$

It is given that water flows at a rate of $\frac{25}{7}$ litres per sec.

Let t sec be the time taken to empty half of the tank.

The rate of flow of water × t = $\frac{1}{2}$× Volume of water in the hemispherical tank.

⇒ $\frac{25}{7}$ × t = $\frac{1}{2}$× $\frac{99}{14}$

⇒ $\frac{25}{7}$ × $\frac{1}{1000}$ × t $$= $\frac{1}{2}$ × $\frac{99}{14}$

⇒t = 990 sec

Hence, the time taken to empty half the tank = 960 + 30 = 990 sec = 16 min 30 sec

OR

Radius ( $r_1$) of upper base of glass = $\frac{4}{2}$= 2 cm

Radius ( $r_2$) of lower base of glass = $\frac{2}{2}$= 1 cm

To calculate the capacity of glass we need to find out the volume of frustum of cone.

V = $\frac{1}{3}$πh( ${r_1}^2$+ ${r_2}^2$+ $r_1{r}_2$)

= $\frac{1}{3}$πh( $2^2$+ $1^2$+ 2)

= $\frac{1}{3}$× $\frac{22}{7}$×14[ 4 + 1 + 2]

= $\frac{308}{3}$= 102 $\frac{2}{3}{\mathrm{cm}}^3$

Hence, the capacity of the glass is 102 $\frac{2}{3}{\mathrm{cm}}^3$.

Difference between the total height of the tent and the height of the cylinder gives the height of the cone.

1) Find the radius of the cylinder.

2) Find the height of the cone.

3) Find the slant height of the cone using the Pythagoras property.

4) Find the curved surface area of the cone and the cylinder.

5) Sum of both the surface areas gives the area of the canves required.

6) Assume the length of the canvas required as 'l'.

7) Equate the area of the rectangular canvas to the area of the canvas required and find the value of l.

Since diameter of the cylinder = 30 m.

Let $r_1$be the radius of cylinder = $\frac{30}{2}$= 15 m

Let $h_1$denote the height of the cylinder, H the height of the tent, $h_2$ the height of the cone, $r_2$ the radius of the cone and l the slant height of the cone (in m)

$h_1$= 5.5 m, H = 8.25 m, $h_2$= 8.25 m − 5.50 m = 2.75 m and $r_2$= 15 m

To find l = $\sqrt{{\left(15\right)}^2+{\left(2.75\right)}^2}$= $\sqrt{225+7.5625}$= $\sqrt{232.5625}$m

l = 15.25 m

Curved surface area of the tent = Curved surface area of cylinder + Curved surface area of cone

= 2π $r_1{h}_1$+ π $r_2$l

= $\left(2\times \frac{22}{7}\times 15\times 5.5\right)$ + $(\frac{22}{7}\times 15\times 15.25)$

= ( 518.57 + 718.93)

= 1237.5 $m^2$

Now, the curved surface area of tent = area of rectangular piece of canvas

Given breadth of canvas = 1.5 m

Area of the rectangular piece = 1237.5 $m^2$

If l is the length of canvas,

∴ l × 1.5 = 1237.5

⇒l = $\frac{1237.5}{1.5}$ = 825 m

Therefore, the length of canvas used in making the tent = 825 m.

Find tangent of the angle of depression and the elevation of the light house.

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

2) Assume the height of light-house as h.

3) Draw the horizontal line at the top of the building.

4) Find the tangent of the angle of elevation of the light-house and find the distance between the light-house and the building in terms of h.

5) Find the tangent of the angle of depression of the light-house and find the distance between the light-house and the building.

6) Equate both the above values and find the value of h.

7) Also find the difference in height between the light house and building and the distance between the light house and building.

AB and CD are the building and light house respectively.

If the height of the light house be h m.

Given that AB = 60 m, ∠EAD = 30° and ∠CAE = 60°.

CE = AB = 60 m

Therefore DE = CD - CE = (h - 60) m

Therefore in ΔEAD,

tan 30° = $\frac{\mathrm{DE}}{\mathrm{AE}}$

$\frac{1}{\sqrt{3}}$= $\frac{h-60}{\mathrm{AE}}$

AE = $\sqrt{3}$(h - 60) ----------(1)

Consider ΔACE,

tan 60° = $\frac{\mathrm{CE}}{\mathrm{AE}}$

$\sqrt{3}$ $$= $\frac{60}{\mathrm{AE}}$

AE = $\frac{60}{\sqrt{3}}$ ----------(2)

Solving (1) and (2), we get

$\sqrt{3}$( h - 60) = $\frac{60}{\sqrt{3}}$

⇒3( h - 60) = 60

⇒h - 60 = 20

⇒h = 80

Hence the difference in height between the light house and building = CD - AB = 80 m - 60 m = 20 m

Distance between the light house and building = BC = AE = $\frac{60}{\sqrt{3}}$= 20 $\sqrt{3}$m