CBSE Class 10 - Maths

3 is a zero of the polynomial. So, it should satisfy the polynomial.

1) Assume that the given polynomial as p(x).

2) 3 is the zero of the polynomial, so it should satisfy the given polynomial.

3) Find the value of p(3) and equate it to zero.

4) Simplify to find the value of k.

Given that, 3 is a zero of the polynomial.

Let p(x) = 2$x^2$ + x + k

∴ p(3) = 0

⇒ 2(3${)}^2$ + 3 + k = 0

⇒ 18 + 3 + k = 0

⇒ k = −21

Hence, k = −21.

Consider the coefficients of the equations along with their signs for substituting in the condition for infinitely many solutions.

1) Compare the given equations with the pair of linear equations $a_1$x + $b_1$y + $c_1$ = 0 and $a_2$x + $b_2$y + $c_2$ = 0 and write the corresponding coefficients.

2) Write the conditions for the lines to have infinitely many solutions.

3) Substitute the values of the corresponding coefficients.

4) Consider any two terms of the equality and simplify to find the value of 'm'.

Given that the pair of linear equations has infinitely many solutions.

2x + 3y− 7 = 0

(m−1) x + (m + 1) y = (3m−1)⇒ (m−1) x + (m + 1) y − (3m−1) = 0

A pair of linear equations $a_1$x + $b_1$y + $c_1$ = 0 and $a_2$x + $b_2$y + $c_2$ = 0 have infinitely many solutions if, $\frac{a_1}{a_2}$ = $\frac{b_1}{b_2}$ = $\frac{c_1}{c_2}$.

Comparing the given equations with the above equations, $a_1$= 2, $b_1$= 3, $c_1$= −7.

$a_2$= m −1, $b_2$= m + 1, $c_2$= −(3m − 1).

∴ $\frac{2}{(m-1)}$ = $\frac{3}{(m+1)}$ = $\frac{(-7)}{-(3m-1)}$

$\frac{2}{(m-1)}=\frac{3}{(m+1)}$

⇒ 2(m + 1) = 3(m−1)

⇒ 2m+ 2 = 3m−3

⇒ 3m−2m = 2 + 3

⇒ m = 5

Hence, the value of m is 5 .

Use the formula to find the sum of n terms of an A.P. which includes the first term and the last term of the series.

1) Write all the given terms of the A.P. and the formula to find the sum of the n terms of an A.P.

2) Substitute the given values in the above formula and find the value of n which gives the number of terms of the A.P.

3) Write the formula to find the $n^{\mathrm{th}}$term of an A.P. and substitute the values of a, n, $t_{10}$to find the value of d.

Sum of n terms of an A.P. = $S_n$= $\frac{n(a+l)}{2}$, where a is the first term and l is the last term.

Given, a = 4 and l = 49.

$S_n$= 265.

∴ 265 = $\frac{n(4+49)}{2}$

⇒265 = $\frac{(n\times 53)}{2}$

⇒n = $\frac{(265\times 2)}{53}$= 10.

Hence, the number of terms in the series =10

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

a = 4, n = 10, $t_n$= 49

49 = 4 + (10 − 1) d

49 − 4 = 9d

45 = 9d

d = 5

Therefore, the common difference of the A.P. is 5.

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

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

2) Join OA, OB and OP.

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

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

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

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

Let us join OA, OB and OP.

OA = 5 cm (radius of outer circle)

OB = 3 cm (radius of inner circle)

AP = 12 cm

The radius is perpendicular to the tangent at the point of contact.

∴ OA ⊥ PA and OB⊥ PB

Hence, ∆OAP is a right angled triangle right angled at A and and ∆OBP is also a right angled triangle right-angled at B.

On applying Pythagoras theorem in ∆OAP,

${\mathrm{OP}}^2$ = ${\mathrm{AP}}^2$ + ${\mathrm{OA}}^2$

${\mathrm{OP}}^2$ = (12 ${)}^2$ + (5 ${)}^2$

${\mathrm{OP}}^2$= 144 + 25

${\mathrm{OP}}^2$ = 169 $$

∴ OP = $\sqrt{169}$√= 13 cm

On applying Pythagoras theorem in ∆OBP,

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

⇒ ${\mathrm{BP}}^2$ = ${\mathrm{OP}}^2$− ${\mathrm{OB}}^2$

⇒ ${\mathrm{BP}}^2$ = (13 ${)}^2$−(3 ${)}^2$

⇒ ${\mathrm{BP}}^2$ = 169 − 9

⇒ ${\mathrm{BP}}^2$ = 160 $$

⇒ BP = $\sqrt{160}$= 4 $\sqrt{10}$√cm

Therefore, the length of BP is 4 $\sqrt{10}$√cm.

Find the complementary angles of cosine and secant of the expression.

OR

Assume the length of each side of the equilateral triangle as 2a.

1) Apply complementary angles of cosine and secant in the expression.

2) Substitute the value of sin 90°.

3) Simplify and find the value of the expression.

OR

1) Assume the triangle ABC as an equilateral triangle.

2) Assume the length of each side as '2a'.

3) Draw a perpendicular AD from A on BC.

4) BD and CD are equal to a.

5) Each angle of an equilateral triangle measures 60°.

6) Find the secant of 60° and simplify the ratio of the sides to get the value.

Given $\frac{(\cos 70°)}{(3\sin 20°)}$ + $\frac{4(\mathrm{se}{c}^{2}59°-\mathrm{co}{t}^{2}31°)}{3}$ - $\frac{2}{3}$ sin 90°

⇒ $\frac{\cos (90°-20°)}{(3\sin 20°)}$+ $\frac{4(\mathrm{se}{c}^2(90°-31°)-\mathrm{co}{t}^{2}31°)}{3}$ - $\frac{2}{3}$× 1 (sin 90° = 1)

⇒ $\frac{(\sin 20°)}{(3\sin 20°)}$+ $\frac{4(\cos e{c}^{2}31°-\mathrm{co}{t}^{2}31°)}{3}$ - $(\frac{2}{3}\times 1)$ [cos ⁡(90°- θ) = sin θ, and sec ⁡(90° - θ ) = cosec θ]

= $(\frac{1}{3}\times 1)$ + $$ $\left(\frac{4\times 1)}{3}\right)$+ $$ $(-\frac{2}{3})$ [ ${\cos \mathrm{ec}}^2$θ - ${\cot }^2$θ = 1]

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

= $\frac{3}{3}$

= 1

Therefore, the value of given expression is 1.

OR

Consider the equilateral triangle ABC.

In an equilateral triangle, measure of each angle is 60°.

So, ∠A = ∠B = ∠C = 60°

Let AD⊥BC.

Assume the length of AB as 2a.

∴ BD = $\frac{1}{2}$ BC = a

cos 60° = $\frac{\mathrm{BD}}{\mathrm{AB}}$ = $\frac{a}{2a}$= $\frac{1}{2}$.

∴ sec 60° = $\frac{1}{\cos 60°}$= $\frac{1}{\left(\frac{1}{2}\right)}$= 2.

Hence, the value of sec 60° is 2.

If a number is a multiple of 3, its square is also a multiple of 3.

1) Assume that $\sqrt{3}$is a rational which is in the form of $\frac{p}{q}$, where q≠0 and p, q are co-primes.

2) Square the equation on both the sides and prove that p is a multiple of 3.

3) Assume p equal to 3r.

4) Square on both sides and substitute the value of $p^2$ as ${3q}^2$.

5) Simplify and prove that q is also a multiple of 3. So, p and q are not co-primes.

6) This contradicts that p and q are co-primes and $\sqrt{3}$is an irrational number.

Assume that $\sqrt{3}$ is a rational number.

If $\sqrt{3}$ is a rational number, there exists integers p and q for which $\sqrt{3}$= $\frac{p}{q}$, where q≠0 and p, q are co-primes.

Squaring on both sides we get, 3 = $\frac{p^2}{q^2}$.

⇒ $p^2$ = 3$q^2$----------- (i)

From (i), we can say that $p^2$ is a multiple of 3.

So, p is a multiple of 3.

Assume p = 3r, where r is any integer.

On squaring both sides:

$p^2$ = 9$r^2$

⇒3$q^2$= 9$r^2$ [from (i)]

So, q is a multiple of 3.

p and q are multiples of 3.

Hence, the numbers p and q are not co-primes, which is a contradiction.

This proves that our assumption that "$\sqrt{3}$ is a rational number" is wrong.

Therefore, $\sqrt{3}$ is an irrational number.

Write the given equations in the general form of the pair of linear equation. Consider the coefficients of the terms along with their signs.

OR

Ratio of the numerator and the denominator can be written as a fraction.

1) Write the given equations in the general form of the pair of linear equations.

2) Use the formula to solve the pair of linear equations in cross multiplication method.

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

OR

1) Assume the numerator and the denominator of the fraction as x and y.

2) Use the first condition and find the value of x in terms of y.

3) Use the second condition and obtain an equation in x and y.

4) Substitute the value of x in the equation and simplify to find the y.

5) Find the value of x using the value of y.

$\frac{b}{a}$x + $\frac{a}{b}$y = $a^2$+ $b^2$ … (i)

x + y = 2ab ------ (ii)

The equations can be written as

$\frac{b}{a}$x + $\frac{a}{b}$y - ($a^2$+ $b^2$) = 0 ------ (iii)

x + y - 2ab = 0 ------ (iv)

The general form of a pair of linear equations $a_1$x + $b_1$y + $c_1$ = 0 , and $a_2$x + $b_2$y + $c_2$= 0.

Using cross multiplication method, roots of the equation are x = $\frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1}$, y = $\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}$.

$\frac{x}{\{\frac{a}{b}\times (-2\mathrm{ab}\left)\right\}-\{-(a^2+b^2)\times 1\}}$ = $\frac{y}{\{-(a^2+b^2)\times 1\}-\{\frac{b}{a}\times (-2\mathrm{ab}\left)\right\}}$ = $\frac{1}{(\frac{b}{a}\times 1)-(\frac{a}{b}\times 1)}$

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

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

∴x = $\frac{\mathrm{ab}(b^2-a^2)}{(b^2-a^2)}$ = ab

y = $\frac{\mathrm{ab}(b^2-a^2)}{(b^2-a^2)}$ = ab

Therefore, we get x = ab and y = ab.

OR

Assume a fraction as $\frac{x}{y}$

Given that:

(Numerator of the fraction + Denominator of fraction) = (Twice the numerator + 4)

∴ x + y = 2x + 4

⇒ x = y − 4 …(i)

On adding 3 to the numerator and denominator, their ratio becomes 2 : 3.

∴$\frac{(x+3)}{(y+3)}$ = $\frac{2}{3}$

⇒$\frac{(y-4+3)}{(y+3)}$= $\frac{2}{3}$ [ using (i)]

⇒$\frac{(y-1)}{(y+3)}$= $\frac{2}{3}$

⇒3y - 3 = 2y + 6

⇒3y - 2y = 6 + 3

⇒y = 9

∴x = y - 4 = 9 - 4 = 5

Therefore, the fraction is $\frac{5}{9}$.

Sum of 'sum of the first ten terms' and the 'sum of the next ten terms' is equal to the sum of the first twenty terms.

1) Find the sum of first ten terms of the A.P. using the formula and obtain equation (i).

2) Find the sum of the first ten terms and the sum of the next ten terms of the A.P. which is equal to the sum of the first twenty terms of the A.P.

3) Find the sum of the first twenty terms of the A.P and equate it to the obtained value to get equation (ii).

4) Solve equations (i) and (ii) to get the values of a and d.

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

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

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

$S_{10}$= −80.

Given that the sum of next ten terms = −280

∴ $S_{20}$ = −80 + (− 280) = −360

$S_{10}$= $\frac{10}{2}$[2a + (10 - 1) d] = - 80

⇒5(2a + 9d ) = - 80.

⇒2a + 9d = -16 --------- (i)

$S_{20}$= $\frac{20}{2}$[2a + (20 - 1) d] = - 360

⇒10 (2a + 19d) = - 360

⇒2a + 19d = - 36 --------- (ii)

Solving equations (i) and (ii),

10d = - 36 + 16 = - 20

⇒d = - 2

Substituting the value of d in equation (i),

2a = - 16 - 9 (- 2) = - 16 + 18 = 2

∴ a = 1.

Hence the AP is , 1, -1, -3, -5, ….

Compare the corresponding angles of the triangles ADB and EFC.

1) ABC is an equilateral triangle so, ∠FCE and ∠ABD are equal.

2) Draw EF perpendicular to AC and AD perpendicular to BC,

3) Compare the angles of the triangles ABD and EFC and prove that they are similar.

4) Corresponding sides of the similar triangles are proportional.

5) Write the ratio of the corresponding sides of the triangles and simplify to get the required proof.

Given that AB = AC.

In a triangle, angles opposite to equal sides are equal.

∴∠ACB = ∠ABC

⇒∠FCE = ∠ABD ------- (i)

Consider ΔABD and ΔEFC:

∠ADB = ∠EFC = 90º

∠ABD = ∠ECF [from (i)]

ΔADB ~ ΔEFC (According to AA similarity criterion)

The corresponding sides of similar triangles are in proportion.

∴$\frac{\mathrm{AB}}{\mathrm{EC}}$= $\frac{\mathrm{AD}}{\mathrm{EF}}$

⇒AB × EF = AD × EC

Hence proved.

Add the fractions of trigonometric ratios by finding their LCM.

OR

Factorisation of expressions involving trigonometric ratios is performed as in algebra.

1) Consider the LHS of the equation and replace the ratios in terms of sine and cosine.

2) Add the terms of the expressions by finding their LCM.

3) Find the product of the expressions.

4) Simplify the product by applying the identities and get the result equal to RHS.

OR

1) Consider LHS of the equation and replace the ratios in terms of sine and cosine.

2) Simplify the terms in the bracket by finding their LCM.

3) Factorise the expressions as done is algebra.

4) Simplify using the identities to get the result equal to RHS.

Consider L.H.S of the equation:

(1+ cot A - cosec A)(1+ tan A + sec A)

= $[1+\frac{\cos A}{\sin A}$-$\frac{1}{\sin A}]$$[1+\frac{\sin A}{\cos A}$+ $\frac{1}{\cos A}]$

= $\left[\frac{(\sin A+\cos A-1)}{\sin A}\right]$$\left[\frac{(\cos A+\sin A+1)}{\cos A}\right]$

= $\frac{(\sin A+\cos A-1)(\sin A+\cos A+1)}{\sin A\cos A}$

= $\frac{\left\{\right(\sin A+\cos A{)}^2-{\left(1\right)}^2\}}{\sin A\cos A}$ [(x + y)( x - y ) = ($x^2$- $y^2$)]

= $\frac{({\sin }^{2}A+\mathrm{co}{s}^{2}A+2\sin A\cos A-1)}{\mathrm{Sin}A\mathrm{Cos}A}$

= $\frac{(1+2\sin A\cos A-1)}{\sin A\cos A}$ [${\sin }^2$ ⁡A + $\mathrm{co}{s}^2$A = 1 ]

= $\frac{2\sin A\cos A}{\sin A\cos A}$

= 2

= R.H.S

∴ (1 + cot A - cosec A)(1+ tan A + sec A) = 2.

Hence proved.

OR

Given LHS = sin A (1 + tan A) + cos A (1 + cot A)

= sin A (1 + $\frac{\sin A}{\cos A}$) + cos A (1 + $\frac{\cos A}{\sin A}$)

= sin A$\left[\frac{\cos A+\sin A}{\cos A}\right]$ + cos A $\left[\frac{\sin A+\cos A}{\sin A}\right]$

= (sin$A+\cos$A) $[\frac{\sin A}{\cos A}$+ $\frac{\cos A}{\sin A}]$

= (sin$A+\cos$A) $\left[\frac{\mathrm{si}{n}^{2}A+{\cos }^{2}A}{\sin A\cos A}\right]$ [ $\mathrm{si}{n}^{2}A+{\cos }^{2}A=1$]

= $\left(\frac{\sin A+\cos A}{\sin A\cos A}\right)$

= $\frac{1}{\cos A}$+ $\frac{1}{\sin A}$

= sec A + cosec A

= RHS.

Hence proved.

Locate five points on BX such that they are equidistant.

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

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

4) Draw a line $B_4$C' parallel to $B_5$C from $B_4$.

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

6) BC'A' is the required triangle.

Step 1:

Draw a line segment BC of length 10 cm. Then draw an arc taking B as centre and radius equal to 8 cm. Similarly, draw an arc, taking point C as centre and radius equal to 6 cm. Both arcs will intersect at point A. Join AB, AC and thus ABC is formed.

Step 2:

Draw a ray BX which makes an acute angle with line BC on the opposite side of vertex A.

Step 3:

On the ray BX, mark 5 points $B_1$, $B_2$, $B_3$, $B_4$, $B_5$such that

${\mathrm{BB}}_1$= $B_1{B}_2$= $B_2{B}_3$= $B_3{B}_4$= $B_4{B}_5$

Step 4:

Join ${\mathrm{CB}}_5$. Draw a line parallel to ${\mathrm{CB}}_5$to intersect BC at point C', through point $B_4$.

Step 5:

Draw a line parallel to line AC which intersects AB at A’ through point C'.

A' B C' is the required triangle.

Substitute the x and y coordinates of the point P for x and y in the equation x−y + 2 = 0 respectively.

1) Find the coordinates of the point P by substituting the coordinates of A, B and the given ratio in the section formula.

2) The point P lies on the line x−y + 2 = 0.

3) Substitute the x and y coordinates of the point P for x and y in the equation x−y + 2 = 0 respectively.

4) Simplify after substitution and find the value of k.

Given $\frac{\mathrm{AP}}{\mathrm{PB}}$= k : 1

⇒The Point P divides AB in the ratio of k:1

Coordinates of point P according to the section formula = $(\frac{9k+1(-1)}{k+1}$, $\frac{8k+1\left(3\right)}{k+1})$

P = $(\frac{9k-1}{k+1}$, $\frac{8k+3}{k+1})$

Given that point P lies on the line x−y + 2 = 0.

⇒ $\frac{9k-1}{k+1}$−$\frac{8k+3}{k+1}$+ 2 = 0

⇒ 9k −1 − 8k − 3 + 2k + 2 = 0

⇒ 3k − 2 = 0

∴ k = $\frac{2}{3}$.

Substitute the corresponding coordinates of points in the area of the triangle.

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

2) Write the formula for the area of the triangle.

3) Substitute the corresponding coordinates of points in the area of the triangle.

4) Simplify to get the required proof.

Three points with coordinates ($x_1$, $y_1$), ($x_2$, $y_2$) and ($x_3$,$y_3$) are collinear if:

$x_1$($y_2$ −$y_3$) + $x_2$ ($y_3$−$y_1$) + $x_3$($y_1$ − $y_2$ ) = 0.

The points given in the question are (p, q), (m, n) and (p − m, q − n)

∴ [p{n − (q − n)} + m(q − n − q) + (p − m)(q − n)] = 0

⇒ p(n − q + n) + m (− n) + (pq − mq −np + mn) = 0

⇒ p(2n − q ) + m (− n) + (pq − mq − np + mn) = 0

⇒ 2np − pq − mn + pq − mq − np + mn = 0

⇒ (2np − np ) + (− pq + pq ) − mq + (− mn + mn) = 0

⇒ np − mq = 0

⇒ np = mq

Hence proved.

Consider the total height of the cylindrical vessel to find the volume of the water in the vessel.

OR

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

1) Find the radius of the cylindrical vessel.

2) Assume the height of the water on the roof as 'h'.

3) Equate the volume of the water on the rectangular roof to the volume of the water filled in the cylindrical vessel.

4) Consider the total height of the vessel as the water is filled up to the brim in the vessel.

5) Simplify and find the value of h which gives the height of the rain water on the roof.

OR

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

2) Find the radius of the smaller circle.

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

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

Assume the height of the rain water on the roof as h and length of the roof as l, breadth as b and radius of the vessel as r.

l = 22 m

b = 20 m

Diameter of the cylindrical vessel = 2 m

∴r = $\frac{2}{2}$ m = 1m.

Height (H) of the cylindrical vessel = 3.5 m

Given that vessel is full of rain water up to the brim

Therefore,

Volume of rain water on the roof = volume of the cylindrical vessel

⇒ l × b × h = π $r^2$H

⇒ 22 × 20 × h = $\frac{22}{7}$× $1^2$× 3.5

⇒h = $\frac{22\times 1\times 3.5}{7\times 22\times 20}$

⇒ h = $\frac{1}{40}$ m = $\frac{1}{40}$× 100 = 2.5 cm

Therefore, the height of rain water on the roof = 2.5 cm.

OR

From the figure, radius of the circle =OA = OB = OC = OD = 7 cm.

∴ CD = OC + OD = 7 + 7 = 14 cm

Also, AB and CD are perpendicular diameters.

⇒OB is the altitude of △CDB corresponding to side CD.

Since, OA is the diameter of the small circle, radius of the small circle = $\frac{\mathrm{OA}}{2}$.

Area of shaded region = Area of the bigger circle − Area of smaller circle − Area of CDB

= π ${\left(\mathrm{OA}\right)}^2-$π( $\frac{\mathrm{OA}}{2}{)}^2$− $\frac{1}{2}$× CD × OB.

= $\frac{22}{7}$× $7^2$− $\frac{22}{7}\times \frac{49}{4}$− $\frac{1}{2}$× 14 × 7

= (22 × 7 − $\frac{77}{2}$−49)

= ( 154 − $\frac{77}{2}$−49)

= (105 − $\frac{77}{2}$)

= $\frac{(210-77)}{2}$

= $\frac{133}{7}$

= 66.5 $$

Hence, the area of the shaded region = 66.5 ${\mathrm{cm}}^2$.

Include the cards numbered 2 and 90 in finding the total number of cards.

1) Find the total number of cards in cards in the bag which is the number of possible outcomes.

2) Find the number of cards bearing two digit numbers from 2 to 90 which is the number of favourable outcomes.

3) The ratio of the above outcomes is the required probability.

4) Find the number of cards bearing a number which is a perfect square from 2 to 90 which is the number of favourable outcomes.

5) The ratio of the favourable outcomes and the total number of outcomes is the required probability.

Number of cards in the bag = 90 − (2 − 1) = 89 .

(i) The two - digit numbers from 2 to 90 = 10, 11, 12, 13,14 ............. 90.

Number of cards having a two-digit number = 81

∴Probability =$\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$ = $\frac{81}{89}$.

(ii) The perfect squares from 2 to 90 = 4, 9, 16, 25, 36, 49, 64 and 81.

Number of cards with a number which is a perfect square = 8

∴Probability =$\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}$ =$\frac{8}{89}$

Thus, the probability of randomly drawn card bears a number which is a perfect square is $\frac{8}{89}$.

Add 4 to the ages of the sisters to find their ages after four years.

1) Assume the age of girl as x

2) Age of her sister is half of her age.

3) Find the ages of the sisters after four years by adding 4 to both of their ages.

4) Find the product of the ages of the sisters and equate it to 160.

5) Solve the quadratic equation obtained by splitting the middle term method.

6) Find the value of x which gives the present age of the girl.

Let us assume that the age of the girl = x years

∴ Present age of the sister = $\frac{x}{2}$ years

After 4 years, age of girl = (x + 4) years

Age of girl’s sister = ($\frac{x}{2}$+ 4) years

It is given that after four years, the product of their ages will be 160

∴ (x + 4)($\frac{x}{2}$+ 4 ) = 160

$\frac{(x+4)(x+8)}{2}$= 160

⇒( x + 4)( x +8) = 320

⇒$x^2$+ 4x + 8x + 32 = 320

⇒ $x^2$+ 12x - 288 = 0

⇒ $x^2$+ 24x - 12x - 288 = 0

⇒ x( x +24) -12 ( x + 24) = 0

⇒ ( x +24)( x - 12) = 0

∴ x = -24 or 12.

Since the age can’t be negative, x = 12

Therefore, present age of the girl = 12 years

Present age of her sister = $\frac{12}{2}$ = 6 years.

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

1) Consider a triangle ABC where ${\mathrm{AC}}^2$= ${\mathrm{AB}}^2$ + ${\mathrm{BC}}^2$.

2) Construct another triangle PQR right angled at Q and PQ = AB, QR = BC.

3) Apply Pythagoras theorem in triangle PQR and find ${\mathrm{PR}}^2$.

4) From the given condition, prove that AC = PR.

5) So, the triangles ABC and PQR are congruent by SSS congruency criterion.

6) Angles Q and B are right angles according to CPCT.

7) Consider an isosceles triangle PQR, in which PQ = PR and ${\mathrm{PR}}^2=2{\mathrm{PQ}}^2$.

8) Use given condition and the theorem proved above to prove that ∠Q is a right angle.

Let us assume the triangle ABC where, ${\mathrm{AC}}^2$ = ${\mathrm{AB}}^2$ + ${\mathrm{BC}}^2$.

We need to prove that∠B = 90°.

Construct another triangle PQR with a right angle at Q and PQ = AB and QR = BC

In ∆PQR,

${\mathrm{PR}}^2$ = ${\mathrm{PQ}}^2$ + ${\mathrm{QR}}^2$ (Pythagoras theorem)

⇒ ${\mathrm{PR}}^2$ = ${\mathrm{AB}}^2$ + ${\mathrm{BC}}^2$ --------- (i) (Since PQ = AB and QR = BC)

However, ${\mathrm{AC}}^2$ = ${\mathrm{AB}}^2$ + ${\mathrm{BC}}^2$…(ii) (Given)

From (i) and (ii),

AC = PR … (iii)

From ∆ABC and ∆PQR,

AB = PQ (Construction)

BC = QR (Construction)

AC = PR [From (iii)]

ΔABC ≅ΔPQR (According to SSS congruency criterion)

∴∠B = ∠Q (CPCT)

However, ∠Q = 90° (Construction)

∴∠B = 90°

So, in a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Hence proved.

Consider an isosceles triangle PQR with PQ = PR. We need to prove that ∠Q = 90°.

PQ = QR --------- (i) (Given)

${\mathrm{PR}}^2$ = 2 ${\mathrm{PQ}}^2$--------- (ii) (Given)

⇒ ${\mathrm{PR}}^2$ = ${\mathrm{PQ}}^2$ + $P{Q}^2$

⇒ ${\mathrm{PR}}^2$ = ${\mathrm{PQ}}^2$ + (QR ${)}^2$ [From (i)]

⇒ ${\mathrm{PR}}^2$ = ${\mathrm{PQ}}^2$ + ${\mathrm{QR}}^2$.

From the above, in a triangle if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Hence, by applying the theorem proved above, ∠Q is a right angle.

1) Find the alternate angle of the angle of depression of the base of the cliff.

2) Find tangent of the angle of elevation and the angle of depression.

OR

1) Depth of the reflection of the cloud from the surface of the water of the lake is equal to the height of the cloud from the surface of the water in the lake.

2) Find tangent of the angle of elevation and the angle of depression.

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

2) Mark the alternate angle of the angle of depression of the base of the cliff.

3) Find the tangent of the angle and find the distance between the ship and the cliff.

4) Draw a horizontal line at the top of the ship and mark the angle of elevation.

5) Find the tangent of the angle of elevation and find the height of the cliff from the top of the deck of the ship.

OR

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

2) Mark the reflection of the cloud below the surface of the water at a depth equal to the height of the cloud from the surface of the water.

3) Find the tangent of the angle of depression of the reflection of the cloud and obtain the horizontal distance between the cloud and the point of observation in terms of height.

4) Find the tangent of the angle of elevation of the the cloud from the point of observation and simplify the ratio to find the height of the cloud from the surface of the lake.

The diagram below represents the data given in the question:

Here AB = the deck of the ship above water level. PC is the cliff and AC is the water level. Point B gives the position of the man.

According to the condition AB = 12 ,∠PBE = 60° .

Since ∠EBC = 30° then ∠BCA = 30°.

Let PC be 'h' m

∴PE = (h -12) m (since CE = AB = 12 m)

In a ∆ABC

Tan 30° = $\frac{\mathrm{AB}}{\mathrm{AC}}$

⇒ $\frac{1}{\sqrt{3}}$= $\frac{12}{\mathrm{AC}}$

⇒AC = 12 $\sqrt{3}m$

∴ AC = BE = 12 $\sqrt{3}$m .

In a ∆BE

Tan 60° = $\frac{\mathrm{PE}}{\mathrm{BE}}$

⇒ $\sqrt{3}$= $\frac{(h-12)}{12\sqrt{3}}$

⇒ ( h - 12) = 36

∴ h = 36 + 12 = 48 m

Hence height of cliff is 48 m.

Distance of the cliff from the ship = 12 $\sqrt{3}$m = (12 × 1.732) m = 20.784 m

OR

Let the surface of the lake be AB.

Point of observation = P where AP = 60 m

Position of cloud = C

Reflection of the cloud in water = C’

Therefore CB = C’B.

Draw a line PM which is perpendicular to CB and intersects CB at M.

If CM = h,

CB = h + 60 m

Consider ∆CPM,

Tan 30° = $\frac{\mathrm{CM}}{\mathrm{PM}}$

$\frac{1}{\sqrt{3}}$= $\frac{h}{\mathrm{PM}}$

PM = h $\sqrt{3}$ ----------- (i)

Consider ∆PMC',

Tan 60° = $\frac{C"M}{\mathrm{PM}}$

⇒Tan 60° = $\frac{(C"B+\mathrm{CB})}{\mathrm{PM}}$

$\sqrt{3}$= $\frac{(h+60+h)}{h\sqrt{3}}$ [from (i)]

⇒3h = ( 2h + 60)

∴ h = 60 m.

⇒ CB = h + 60 m = 60 m + 60 m = 120 m.

Therefore, height of the cloud from the surface of the lake is 120 m.

1) Volume of the sphere and the cone are equal.

OR

1) Consider that the hollow cylinder has no base.

2) Consider the length of the cylinder and the height.

1) Find the radius of the sphere from the surface area of the sphere.

2) Find the volume of the sphere using the radius.

3) Assume the height of the cone as 'H'.

4) Equate the volumes of the sphere and the cone.

5) Substitute the values and simplify to find the value of H.

OR

1) Equate the difference between the curved surface areas of the hollow cylinder to the given value and obtain equation (1).

2) Equate the volume of the hollow cylinder to the given value.

3) Substitute the value obtained in equation (1) and obtain equation (2).

4) Solve equations (1) and (2) to get the values of inner and outer radii of the hollow cylinder.

5) Find the inner and the outer diameter of the cylinder from the radii.

Given that, surface area of the sphere = 616 ${\mathrm{cm}}^2$.

Let the radius of the sphere be r .

Surface area of the sphere = 4π$r^2$ = 616 ${\mathrm{cm}}^2$.

$r^2=\frac{616}{4\pi }$ = $\frac{616\times 7}{4\times 22}$ ${\mathrm{cm}}^2$= 49 ${\mathrm{cm}}^2$.

⇒ r = 7 cm.

The sphere is melted and recast into a cone whose height is 28 cm.

∴Volume of the sphere = volume of the cone

$\frac{4}{3}$π$r^3$= $\frac{1}{3}$π$R^2$H where R = radius of cone and H = height of the cone.

$\frac{4}{3}$ × $\frac{22}{7}$ × (7${)}^3$= $\frac{1}{3}$ × $\frac{22}{7}$ × $R^2$ × 28

⇒$R^2=\frac{4\times 7\times 7\times 7}{28}$= $7^2$

⇒ R = 7 cm.

Since, the radius of cone R = 7 cm, diameter of the cone, 2R = 2 × 7 =14 cm.

OR

Let the height of the cylinder be h.

Given that, h = 14 cm.

Let the outer radius of the hollow cylinder be R and inner radius of hollow cylinder be r.

Curved surface area of cylinder = 2 × π × radius × height

Given that the difference between the outer and inner curved surface areas of a hollow right circular cylinder is 88 ${\mathrm{cm}}^2$

2πRh − 2πrh = 88 ${\mathrm{cm}}^2$

⇒ 2π (14) (R − r) = 88

⇒ 2×$\frac{22}{7}$× (14) (R − r) = 88

⇒ (R−r) = $\frac{88}{88}$ = 1 ----------------(i)

Given that the volume of the metal used in making the cylinder = 176 ${\mathrm{cm}}^3$

∴ π ($R^2$−$r^2$)h = 176

⇒ $\frac{22}{7}$(R−r)(R + r) ×14 = 176 from (i)

⇒ (R−r) = $\frac{176}{22\times 2}$= 4

( R + r) = 4 ------------------(ii)

Adding (i) and (ii) to get

⇒2R = 5

∴ R = $\frac{5}{2}$

From (ii) r = 4 - R

r = 4 - $\frac{5}{2}$

r = $\frac{3}{2}$cm

Outer radius of the cylinder, R =$\frac{5}{2}$ cm . Hence outer diameter, 2R = 5 cm.

Inner radius of the cylinder, r =$\frac{3}{2}$ cm . Hence inner diameter, 2r = 3 cm.

Use the upper class limits and less than cumulative frequencies for less than ogive and use the lower class limits and more than cumulative frequencies for more than ogive.

1) Draw less than type cumulative frequency table for drawing less than ogive.

2) Mark class limits on the horizontal axis and the cumulative frequencies on the vertical axis.

3) Plot the cumulative frequency for each upper class limit to obtain the required ogive of less than type.

4) Draw more than type cumulative frequency table for drawing more than ogive.

2) Mark class limits on the horizontal axis and the cumulative frequencies on the vertical axis.

3) Plot the cumulative frequency for each lower class limit to obtain the required ogive of more than type.

4) Draw a perpendicular on the horizontal axis from the point of intersection of the ogives to get the median of the data.

Construct the less than type cumulative frequency table and more than type cumulative frequency table .

Less than type cumulative frequency table:

Horizontal axis = Upper class limit.

Vertical axis = Cumulative frequency.

Plot the cumulative frequency for each upper class limit and obtain the required ogive of less than type.

More than type cumulative frequency table:

Horizontal axis = lower class limit

Vertical axis = cumulative frequency

Plot the cumulative frequency for each lower class limit and obtain the required ogive of more than type.

The given graph shows the drawings of both the ogives:-

From the above, observe that the less than and more than types of ogives of the given frequency distribution meets at point P (60, 50).

On drawing a perpendicular from point P to the x-axis, we get a value on the x-axis as 60.

Hence, median = 60.