CBSE Class 10 - Maths

1) Find the reaminder by the long division method of polynomials.

2) Avoid calculation mistakes.

3) Equate both the terms of the remainder to ax + b.

1) Divide the given polynomial by 3$x^2$-1 using long division method.

2) Equate the remainder to ax + b .

3)find the values of 'a' and 'b'.

Given polynomial 6 $x^4$+ 8 $x^3$+ 17 $x^2$+ 21x + 7 can be divided by 3 $x^2$ + 4x +1 as

The remainder obtained is x + 2.

Since it is given that the remainder is of the form ax + b.

Now, equating both remainders to get a = 1 and b = 2.

1) Write the pair of linear equations are in correct order by comparing the standard form.

2) Find the ratio of corresponding coefficients of the equations.

3) Relate the exact condition for finding the relation between the pair of lines.

1) Write the coefficients of the equations by comparing the standard form of the linear equations.

2) Find the ratios of the corresponding coefficients if the equations have no solution.

Write the given equations in the form of $a_1$x + $b_1$y + $c_1$ = 0 and $a_2$x + $b_2$y + $c_2$ = 0

These equations have no solution if $\frac{a_1}{a_2}$ = $\frac{b_1}{b_2}$ ≠$\frac{c_1}{c_2}$.

Let us analyze the given equations and for that first rewrite it as,

kx + 3y − (k − 2) = 0, and 12x + ky − k = 0

On comparing the given equations, we get

$a_1$ = k, $a_2$ = 12, $b_1$ = 3, $b_2$= k, $c_1$ = −(k − 2), $c_2$ = −k

Therefore $\frac{k}{12}$= $\frac{3}{k}$≠$\frac{-(k-2)}{-k}$

On comparing first part of the equation $\frac{k}{12}$= $\frac{3}{k}$, we get $k^2$ = 36, which gives k = ± 6.

So that , When k = 6 and −6, the given pair of equations will have no solutions.

1) Write the correct formula to find its $n^{\mathrm{th}}$ term, If the sum of first n terms of an A.P. is given.

1) Write the correct formula of $n^{\mathrm{th}}$ term, If the sum of first n terms of an A.P is given.

2) Substitute values and simply will get n th term of the given A.P.

Let the $n^{\mathrm{th}}$ term of the A.P. be $a_n$.

Since it is given that $S_n$ = 3$n^2$−4n.

But $S_n$-$S_{n-1}$= $a_n$.

Therefore

⇒3$n^2$- 4n - { 3(${n-2)}^2$- 4( n - 1)} = $a_n$

⇒3$n^2$- 4n - { 3($n^2$+ 1 - 2n) - 4(n- 1)} = $a_n$$a_n$

⇒3$n^2$- 4n - { 3(n2+ 1 - 2n) - 4(n- 1)} = $a_n$

⇒3$n^2$- 4n - 3$n^2$-7 + 10n = $a_n$

⇒ 6n - 7 = $a_n$

Therefore , the n th term of the given A.P. is (6n − 7).

1) Draw a line OB.

2) Observe OAPB is a quadrilateral.

or

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

2) Group the tangents according to the sides of the parallelogram.

1) Given that two tangents PA and PB are drawn to a circle with centre O and P is an external point.

2) join O and B i.e. OB.

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

4) Observe OAPB is a quadrilateral. Therefore sum of the angles in a quadrilateral is 360 degrees.

5) Find ∠AOB .

6) Observe OAB is a triangle and OA and OB are radii then ∠OAB = ∠OBA.

7) The sum of angles in a triangle is equal to 180 °.

8) Find ∠APB = 2∠OAB by using ∠AOB.

OR

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

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

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

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

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

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

To complete the quadrilateral by joining OB.

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

∴∠OAP = ∠OBP = 90º

Now,

∠OAP + ∠APB + ∠OBP + ∠AOB = 360º [Angle sum property of quadrilaterals]

⇒ 90º + ∠APB + 90º + ∠AOB = 360º

⇒∠AOB = 360º − 180º − ∠APB = 180º − ∠APB ---------------------- (1)

Now, in ΔOAB,

OA is equal to OB as both are radii.

⇒∠OAB = ∠OBA [In an isosceles triangle, angles opposite to equal sides are equal]

The sum of angles in a triangle is equal to 180 °.

In ΔAOB,

∠OAB +∠OBA + ∠AOB = 180º

⇒ 2∠OAB + ∠AOB = 180º

⇒ 2∠OAB + (180º − ∠APB) = 180º [From (1)]

⇒ 2∠OAB = ∠APB

⇒∠APB = 2∠OAB

Hence proved.

OR

Let the parallelogram be ABCD and let the circle touch the sides AB, BC, CD and AD at P, Q, R and S respectively.

Since, ABCD is a parallelogram.

AB = CD … (1)

BC = AD … (2)

Lengths of tangents drawn from an external point to the circle are equal. If A, B, C, D are the external points then,

DR = DS ----- (3)

CR = CQ ----- (4)

BP = BQ … (5)

AP = AS … (6)

On adding equations (3), (4), (5) and (6), we obtain

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

Grouping these tangent lengths according to the sides of the parallelogram.

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

∴ CD + AB = AD + BC

From equations (1) and (2), we obtain

2AB = 2BC

∴ AB = BC

∴ AB = BC = CD = DA

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

Hence, the parallelogram ABCD is also a rhombus.

1) Write the correct formula and apply the identity.

1) Apply the $a^3$+ $b^3$ formula.

2) Cancel the similar terms and simplify using identity to get required solution.

Given $\frac{{\sin }^3\theta +{\cos }^3\theta }{\sin \theta +\cos \theta }$+ sinθ cosθ

Since$a^3+b^3$ = ( a + b)($a^2+b^2$ - 2ab)

= $\frac{(\sin \theta +\cos \theta )({\sin }^2\theta +{\cos }^2\theta -\sin \theta \cos \theta )}{(\sin \theta +\cos \theta )}$ + sin θ cos θ

= ${(\sin }^2$ θ +${\cos }^2$ θ - sinθ cos θ) + sin θ cos θ

= 1 - sinθ cosθ + sinθ cos θ = 1

Hence, the value of the given expression is 1.

1) Substituting appropriate values properly.

1)Assume that √5 is rational number.

2) To prove that we assume there are positive co-primes a and b such that √5 = $\frac{a}{b}$.

3) This shows that 5 is a factor of $a^2$, which comes down to the fact that 5 is factor of a.

4) Let a = 5c. Similarly apply we get $b^2$ is a multiple of 5, or b is a multiple of 5.

5) Since both a and b are multiples of 5 or 5 is a factor of both a and b.

7) Therefore, √5 is an irrational number.

For proving that √5 is an irrational number try to prove the contrary.

Let us assume on the contrary that √5 is rational. To prove that we assume there are positive co-primes a and b such that

√5 = $\frac{a}{b}$

5 = $\frac{a^2}{b^2}$

$\Rightarrow b^2$x 5 = $a^2$

This shows that 5 is a factor of $a^2$, which comes down to the fact that 5 is factor of a. So that a = 5c for a positive integer c.

${\Rightarrow b}^2$x 5 = $a^2$

⇒$b^2$x 5 =$5^2$ x $c^2$

$\Rightarrow b^2$= 5 x $c^2$

So that $b^2$ is a multiple of 5, or b is a multiple of 5.

Since both a and b are multiples of 5 or 5 is a factor of both a and b.

This contradicts shows that a and b as co-primes, so our assumption incorrect .

Hence, √5 is an irrational number.

1) Assuming the given equations$\frac{1}{(x-1)}=u$ and $\frac{1}{(y-2)}$= v .

2) Avoid calculation mistakes.

3) Substitute the u and v values to get x and y values.

1) Let the equations be (1) and (2) .

2) Assuming the given equations$\frac{1}{(x-1)}=u$ and $\frac{1}{(y-2)}$= v .

3) We get (3) and (4) equations.

4) Solving (3) and (4) equations to get u and v values.

5) Substitute u and v values to get x and y values.

Let

$\frac{5}{(x-1)}+\frac{1}{(y-2)}=2$----------------(1)

$\frac{6}{(x-1)}$- $\frac{3}{(y-2)}$ = 1 ---------------- (2)

Assume that $\frac{1}{(x-1)}=u$ and $\frac{1}{(y-2)}$= v

So now the equations will become

5u + v = 2 …......... (3)

6u − 3v = 1 …......... (4)

Solving equations (3) and (4) as linear equations we get that

u = $\frac{1}{3}$ and v = $\frac{1}{3}$

That means

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

From which we get that x = 4 and y = 5 which is the solution for the given pair of equation.

1) Write the correct formula of $n^{\mathrm{th}}$ term of an A.P.

2) By solving equations avoid calculation mistakes.

3) Write the correct A.P. series by using difference value.

1) Let us assume that a is the first term and d is the common difference of the A.P

2) Write the $n^{\mathrm{th}}$term of an A.P.

3) Find $4^{\mathrm{th}}$, $6^{\mathrm{th}}$, $8^{\mathrm{th}}$and ${10}^{\mathrm{th}}$ terms.

4) Adding 4th and 8th term is 24 and 6th term and 10th term is 44, to get two equations .

5) Solving these equations to get a and d values.

6) Finally write the correct A.P. series by using difference value.

Let us assume that a is the first term of the A.P. and d is the common difference of the A.P.

It is known that the $n^{\mathrm{th}}$term of an A.P. is given by $a_n$ = a + (n -1)d.

The fourth term of the A.P. $a_4$ = a + (4 -1)d = a + 3d.

8th term, $a_8$ = a + (8 - 1)d = a + 7d

6th term, $a_6$ = a + (6 -1)d = a + 5d

10th term, $a_{10}$ = a + (10 - 1)d = a + 9d

It is given that $a_4$ + $a_8$= 24

a + 3d + a + 7d = 24

2a + 10d = 24

a + 5d= 12 ------------ (1)

And also

$a_6$ + $a_{10}$ = 44

a + 5d + a + 9d = 44

2a + 14d = 44

a + 7d = 22 ----------- (2)

On solving linear equations (1) and (2) we get,

d = 5 and a = -13

The A.P. is a, a + d, a + 2d …

⇒ −13, −13 + 5, −13 + 10…

Thus, the A.P. is −13, −8, −3…

1) Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

2) Locate 4 points$B_1$, $B_2$, $B_3$, and $B_4$ on line segment BX.

3) Join B, C and draw a line through $B_3$ such that it is parallel to $B_4$C and intersecting BC at C '.

4) Draw a line through C ' parallel to AC and intersecting AB at A'.

1) Draw a line segment BC of length 6.5 cm.

2) Draw an arc of any radius while taking B as the centre. Taking O as the centre, draw another arc to cut the previous arc at point O'.

3) Taking B as the centre, draw an arc of radius 4.5 cm intersecting the extended at point A. Join AC.

4) Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

5) Locate 4 points on line segment BX.

6) Join two lines which is parallel to AC and $B_4$C then to get a required similar triangle.

7) Hence the required construction will satisfy the condition.

In a triangle whose sides are $\frac{3}{4}$of the corresponding sides of ΔABC and it can be drawn by following the given steps.

Step 1: Draw a line segment BC of length 6.5 cm.

Step 2: Draw an arc of any radius while taking B as the centre. Let it intersect line BC at point O. Taking O as the centre, draw another arc to cut the previous arc at point O'. Join BO', which is the ray making 60° with line BC.

Step 3: Taking B as the centre, draw an arc of radius 4.5 cm intersecting the extended line segment BO' at point A. Join AC. In ΔABC, AB = 4.5 cm, BC = 6.5 cm, and ∠ABC = 60°.

Step 4: Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 5: Locate 4 points (as 4 is greater in 3 and 4) $B_1$, $B_2$, $B_3$, and $B_4$ on line segment BX.

Step 6: Join $B_4$C and draw a line through $B_3$ such that it is parallel to $B_4$C and intersecting BC at C '.

Step 7: Draw a line through C ' parallel to AC and intersecting AB at A'. ΔA'BC' is the required triangle.

This is possible because

A' B = $\frac{3}{4}$AB, BC ' = $\frac{3}{4}$BC, A' C ' = $\frac{3}{4}$AC

In ΔA′BC′ and ΔABC,

∠A′C′B = ∠ACB ( they are corresponding angles)

∠A′BC′ = ∠ABC ( common angle)

∴ ΔA′BC′ ∼ ΔABC (by AA similarity criterion)

Therefore $\frac{(A\text{'}B)}{\mathrm{AB}}$= $\frac{\mathrm{BC}\text{'}}{\mathrm{BC}}$= $\frac{(A\text{'}C\text{'})}{\mathrm{AC}}$ ----------- (1)

In ΔB $B_3$C′ and ΔB $B_4$C,

∠ $B_3$BC′ = ∠ $B_4$BC (Common angle)

∠B $B_3$C′ = ∠B $B_4$C (Alternate interior angles)

∴ ΔB $B_3$C′ ∼ ΔB $B_4$C (byAA similarity criterion)

$\frac{\mathrm{BC}\text{'}}{\mathrm{BC}}$= $\frac{B{B}_3}{B{B}_4}$

$\frac{\mathrm{BC}\text{'}}{\mathrm{BC}}$ = $\frac{3}{4}$ ----------(2)

Therefore from equations (1) and (2), we obtain

$\frac{A\text{'}B}{\mathrm{AB}}$ = $\frac{\mathrm{BC}\text{'}}{\mathrm{BC}}$ = $\frac{A\text{'}C\text{'}}{\mathrm{AC}}$= $\frac{3}{4}$

Therefore A'B = $\frac{3}{4}$AB, BC ' = $\frac{3}{4}$ BC, A' C ' = $\frac{3}{4}$AC

Hence the required construction will satisfy the condition.

1) Comparing the similarity of traingles

2) Check the ratio corresponding sides of similar triangles are proportional.

3) Pythagoras theorem applied to ΔABC.

or

1) Let ∠DBG = α. In each triangle find the remaining angles using sum of angles in a triangle = 180°.

2) Comparing the AA similarity criterion.

3) Check the ratio corresponding sides of similar triangles are proportional.

1) Given that ΔABC is right angled at C and DE ⊥ AB.

2) Comparing the similarity of traingles.

4) Substitute thevalues and using Pythagoras theorem to find the side AB.

5) Using side AB to find DE and AE.

or

1) In a ΔABC is right angled at A and DEFG is a square.

2) Find angles ∠BDG and ∠CEF .

3) Let ∠DBG = α then find ∠DGB .

4) ∴In a ΔABC to find ∠ECF and In a ΔFEC to find ∠CFE.

⇒By AA similarity criterion, ΔBDG ∼ ΔFEC.

6) Corresponding sides of similar triangles are proportional then we get ${\mathrm{DE}}^2$= BD × EC.

Consider ΔABC and ΔADE

∠ACB = ∠AED (both 90°)

∠BAC = ∠DAE (as common angle)

∴ ΔABC ∼ ΔADE (By AA similarity criterion)

Since the corresponding sides of similar triangles are proportional,

∴ $\frac{\mathrm{AB}}{\mathrm{AD}}$ = $\frac{\mathrm{BC}}{\mathrm{DE}}$ = $\frac{\mathrm{AC}}{\mathrm{AE}}$

∴ $\frac{\mathrm{AB}}{\mathrm{AD}}$ = $\frac{\mathrm{BC}}{\mathrm{DE}}$= $\frac{(\mathrm{AD}+\mathrm{DC})}{\mathrm{AE}}$

⇒ $\frac{\mathrm{AB}}{3}$ = $\frac{12}{\mathrm{DE}}$= $\frac{(3+2)}{\mathrm{AE}}$

⇒ $\frac{\mathrm{AB}}{3}$ = $\frac{12}{\mathrm{DE}}$= $\frac{5}{\mathrm{AE}}$ ------------(1)

Pythagoras theorem applied to ΔABC, we get

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

⇒ ${\mathrm{AB}}^2$ = 52 + 122 = 25 + 144 = 169

⇒ AB = 13 cm

Substituting the value of AB in equation (1), we obtain

$\frac{13}{3}$= $\frac{12}{\mathrm{DE}}$ = $\frac{5}{\mathrm{AE}}$

$\frac{13}{3}$ = $\frac{12}{\mathrm{DE}}$

DE = 2.77 cm

Similarly AE = 1.15 cm

Hence DE and AE are approximately 2.77cm and 1.55cm.

OR

In a figure given that DEFG is a square.

∴∠EDG = ∠FED = 90°

Now, ∠BDG + ∠EDG = 180° (straight line)

⇒∠BDG + 90° = 180°

⇒∠BDG = 90°

Similarly, ∠CEF = 90°

Let ∠DBG = α

Sum of angles in a triangle = 180°

In a ΔBDG:

∠DGB + ∠DBG + ∠BDG = 180°

⇒∠DGB = 90° − α .

∴ a ΔABC:

∠ECF + ∠ABC + ∠BAC = 180° [∠ECF = ∠BCA]

⇒∠ECF = 90° − α.

In a ΔFEC:

∠CEF + ∠ECF + ∠CFE = 180°

⇒∠CFE = α

Now, in ΔBDG and ΔFEC:

∠BDG = ∠FEC ( = 90°)

∠DBG = ∠EFC ( = α)

By AA similarity criterion, ΔBDG ∼ ΔFEC

As corresponding sides of similar triangles are proportional.

$\frac{\mathrm{BD}}{\mathrm{FE}}$ = $\frac{\mathrm{DG}}{\mathrm{EC}}$

$\frac{\mathrm{BD}}{\mathrm{DE}}$ = $\frac{\mathrm{DE}}{\mathrm{EC}}$ [In square DEFG, DE = DG = EF = GF]

${\mathrm{DE}}^2$ = BD × EC

Hence proved.

1) Draw a line AD which is perpendicular to side BC

2) Comparing two triangles.

3) Let the length of AB be 2a.

or

1) Using complimentary angles to find trigonometric ratios.

1) Consider an equilateral triangle ABC, each angle in an equilateral triangle is 60°.

2) Draw a line AD which is perpendicular to side BC.

3) Now Comparing two triangles.

4) By RHS congruency criterion to get ΔABD ≅ ΔACD.

5)∴BD = DC ( by CPCT)∠BAD = ∠CAD = 30º.

6) Let the length of AB be 2a. therefore AB = BC = CD = 2a. There fore, BD = half of BC = a.

7) To find Sin 30° using right angled triangle.

or

1) Using complimentary angles find the trigonometric ratios.

2) Substitute and simplify to get solution.

Consider an equilateral triangle ABC, each angle in an equilateral triangle is 60°.

Therefore, ∠A = ∠B = ∠C = 60°

Draw a line AD which is perpendicular to side BC.

Now, in ΔABD and ΔACD

∠ADB = ∠ADC (both are 90°)

AB = AC (ABC is an equilateral triangle)

AD = AD (Common line)

ΔABD ≅ ΔACD (By RHS congruency criterion)

∴ BD = DC ( by CPCT)

∠BAD = ∠CAD = 30º

Let the length of AB be 2a. therefore AB = BC = CD = 2a

BD = half of BC = a.

Therefore, Sin 30° = $\frac{\mathrm{BD}}{\mathrm{AB}}$ = $\frac{a}{2a}$= $\frac{1}{2}$.

OR

$\frac{\cos 58°}{\sin 32°}$+ $\frac{\sin 22°}{\cos 68°}$ - $\frac{\cos 38°\cos \mathrm{ec}52°}{\tan 18°\tan 35°\tan 60°\tan 72°\tan 55°}$

$\frac{\cos (90°-32°)}{\sin 32°}$+ $\frac{\sin (90°-68°)}{\cos 68°}$- $\frac{\cos 38°\cos \mathrm{ec}(90°-38°)}{\tan 18°\tan 35°\tan 60°\tan (90°-18°)\tan (90°-35°)}$

We know that sin (90° - θ) = cos θ, cos(90° - θ) =sin θ, cosec (90° - θ) = sec θ, tan (90° - θ) = cot θ

$\frac{\sin 32°}{\sin 32°}$+ $\frac{\cos 68°}{\cos 68°}$ - $\frac{\cos 38°\sec 38°}{\tan 18°\tan 35°\surd 3\cot 18°\cot 35°}$

⇒ (1+1) - $\frac{1}{\surd 3}$

⇒ $\frac{(2\surd 3-1)}{\surd 3}$x $\frac{\surd 3}{\surd 3}$

⇒ $\frac{(6-\surd 3)}{3}$.

1) Take correct coordinates of y - axis.

2) Write the correct formula for distance between two points.

OR

1) Write the correct section formula.

1) Assume that the coordinates of y - axis are (0, y).

2) Write the distance between two points.

3) Substitute in given condition.

4) Squaring on both sides to get required point.

OR

1) Let P, Q are two points of trisection of line segment AB.

2) The point P divides AB internally in the ratio 1:2.

3) Using section formula find the point P.

4) Substitute P in given line to get k value.

The required point is at equal distance from points (5, −2) and (−3, 2).

The point lies on the y-axis and hence its x-coordinate will be 0.

Assume that the coordinates are (0, y).

Distance between two points ( $x_1$, $y_1$) and ( $x_2$, $y_2$) is given by√ ${{(x}_2-x_1)}^2$ + ${{(y}_2-y_1)}^2$ )

Distance between (0,y) and (5, -2) = √( ${(5-0)}^2+({-2-y)}^2$).

And the distance between (0, y) and (−3, 2) =√( ${(-3-0)}^2+({2-0)}^2$)

Given that √( ${(5-0)}^2$+ ${(-2-y)}^2$) = √( ${(-3-0)}^2$+ ${(2-y)}^2$) squaring on both sides

25 + 4 + $y^2$+ 4y = 9 + 4 + $y^2$ - 4y

8y = -16

y = - 2

Hence the point is (0,-2)

OR

Let P ( $x_1$, $y_1$) and Q ( $x_2$, $y_2$) be the points of trisection of line segment AB, i.e., AP = PQ = QB

Hence the point P divides AB internally in the ratio 1:2.

By using section formula

( $x_1,y_1)$= [ $\frac{(1\times 5+2\times 2)}{(1+2)}$, $\frac{(1\times -8+2\times 1)}{(1+2)}$]

$x_1$= 3 and $y_1$ = -2

Therefore the coordinates are P(3,-2)

It is given that P(3,-2) on 2x − y + k = 0

⇒2(3) - (-2) + k = 0

k = -8

Hence k = - 8

1) Write the correct formula.

2) Avoid calculation mistakes.

1) Given that three points A (a, 0), P (x, y), and C (0, b) lie on the same line, hence the area of the triangle is zero.

2) Substitute three points convert into line intercept form.

It is given that the three points A (a, 0), P (x, y), and C (0, b) lie on the same line, hence the area of the triangle formed by these three points is zero.

⇒$\frac{1}{2}$[a (y - b) + x ( b - 0) + 0 (0 - y) ] = 0

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

⇒ay - ab + xb = 0

⇒$\frac{\mathrm{ay}}{\mathrm{ab}}$ + $\frac{\mathrm{xb}}{\mathrm{ab}}$= 1

$\frac{x}{a}$ + $\frac{y}{b}$ = 1

Hence Proved.

1) ΔPQR represents a right-angled triangle. Find QR.

2) Diameter convert into radius.

2) Area of the shaded region = Area of semicircle − Area of ΔPQR

1) Given that PQ = 24 cm, PR = 7 cm and O is the centre of the circle then QOR is diameter of the circle.

2) The angle in a semicircle is a right angle.Therefore, ΔPQR represents a right-angled triangle.

3) Apply pythagoras theorem to find QR.

4) Find Area of semicircle and Area of triangle.

5) Finally to find area of the shaded region.

Given that QOR is the diameter of the circle.

The angle in a semicircle is a right angle.

∴∠RPQ = 90°

Therefore, ΔPQR represents a right-angled triangle where PR = 7 cm and PQ = 24 cm.

When Pythagoras theorem is applied to right ΔPQR, we obtain

${\mathrm{QR}}^2$ = ${\mathrm{PQ}}^2$ + ${\mathrm{PR}}^2$ = ${24}^2$ + $7^2$ = (576 + 49) = 625 ${\mathrm{cm}}^2$

QR = √625 = 25 cm

Area of semicircle = $\frac{1}{2}$ π$r^2$,where r = $\frac{25}{2}$ cm = 12.5 cm

Area of semicircle = $\frac{1}{2}$×3.14×12.5×12.5 ${\mathrm{cm}}^2$= 245.3125 ${\mathrm{cm}}^2$

Area of ΔPQR = $\frac{1}{2}$× Base × Height = $\frac{1}{2}$×7×24 ${\mathrm{cm}}^2$= 7 × 12 ${\mathrm{cm}}^2$ = 84 ${\mathrm{cm}}^2$.

Hence , area of the shaded region = Area of semicircle − Area of ΔPQR

= 245.3125 − 84 = 161.3125 ${\mathrm{cm}}^2$.

1) Out of 52 cards only king, queen and jack of clubs are removed.

2) Total$\mathrm{number}\mathrm{of}\mathrm{remaining}\mathrm{card}$s are 49.

1) Remove king, queen and jack of clubs in a pack of 52 cards.Remaining 49 cards left and in that 13 are hearts, 13 are spades, 13 are diamonds and 10 are clubs.

2) Applying the formula for i) Probability of getting a heart ii) Probability of getting a queen iii) Probability of getting a club card,

When we remove the King, Queen and Jack of Clubs we have about 49 cards left and in that 13 are hearts, 13 are spades, 13 are diamonds and 10 are clubs. In this

Probability of getting a heart = $\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{hearts}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{remaining}\mathrm{card}s}$= $\frac{13}{49}$

Probability of getting a queen = $\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{Queens}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{remaining}\mathrm{card}s}$= $\frac{3}{49}$

Probability of getting a club card = $\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{remaining}\mathrm{club}\mathrm{card}s}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{remaining}\mathrm{card}s}$ = $\frac{10}{49}$.

1) Write the two consecutive odd numbers perfectly.

2) Avoid calculation mistakes.

or

1) Consider two conditions when both cars travel in the same direction and when both cars travel in opposite directions.

1) Let the speed of car A be u and the car B be v.

2) Consider two conditions when both cars travel in the same direction and when both cars travel in opposite directions.

3) Using distance = time × speed formula to get two equations.

4) Solving two equations to get speed of two cars.

We can represent the two odd numbers as (2n − 1) and (2n + 1), where n is a natural number.

It is given that (2n − 1 ${)}^2$ + (2n + 1 ${)}^2$ = 394

⇒ 4 $n^2$ + 1 − 4n + 4 $n^2$+ 1 + 4n = 394

⇒ 8 $n^2$ + 2 = 394

⇒ 8 $n^2$ = 392

⇒ $n^2$ = 49

⇒ n = 7

⇒ 2n − 1 = 2 (7) − 1 = 13

⇒ 2n + 1 = 2 (7) + 1 = 15

Therefore, the odd numbers are 13 and 15.

OR

Consider the speed of car A be u and the car B be v.

Condition 1 : when both cars travel in the same direction.

Let the cars meet at point C in 5 hours.

Car A travels distance AC, whereas car B travels distance BC.

Distance = time × speed

⇒ AC = 5 × u

⇒BC = 5 × v

⇒AC − BC = 100

⇒ 5u − 5v = 100

⇒ u − v = 20 … (1)

Condition 2 : When both cars travel in opposite directions

Let both cars meet at point D.

Car A will travel AD, while car B will travel BD when they meet.

Distance = time × speed

⇒ AD = 1 × u

⇒ BD = 1 × v

AD + BD = 100

⇒ u + v = 100 ----------((2)

On adding equations (1) and (2), we obtain

2u = 120

⇒ u = 60

From equation (2), we obtain

60 + v = 100

⇒ v = 40

Hence, the speed of car A is 60 km/h and the car B is 40 km/h.

1) Draw a perpendicular lines from T and S to PS and PT i.e. TU and SV respectively and also join QT and RS.

2) Observe that height of triangles will take out side and find area of a triangle.

3) Compare the ratio of similar triangles.

In a triangle PQR as shown below a line ST drawn parallel to the base QR. ST is intersecting the sides PQ and PR at points S and T.

A line ST divides PQ into two parts PS and SQ, similarly ST divides PR into two parts PT and TR.Prove that $\frac{\mathrm{PS}}{\mathrm{SQ}}$= $\frac{\mathrm{PT}}{\mathrm{TR}}$.

Draw a perpendicular lines from T and S to PS and PT i.e. TU and SV respectively and also join QT and RS.

Area or a triangle = $\frac{1}{2}$×base ×height

Area of ∆ PST = $\frac{1}{2}$×PS ×TU

If PT as base and SV as height, so that

Area of ∆ PST= $\frac{1}{2}$×PT ×SV

In the similar way

Area of ∆ QST = $\frac{1}{2}$×QS ×TU

Area of ∆ RST = $\frac{1}{2}$×SV ×TR

$\frac{\mathrm{Area}\mathrm{of}△\mathrm{PST}}{\mathrm{Area}\mathrm{of}△\mathrm{QST}}$ = $\frac{\frac{1}{2}\times \mathrm{PS}\times \mathrm{TU}}{\frac{1}{2}\times \mathrm{QS}\times \mathrm{TU}}$

Therefore

$\frac{\mathrm{Area}\mathrm{of}△\mathrm{PST}}{\mathrm{Area}\mathrm{of}△\mathrm{QST}}$= $\frac{\mathrm{PS}}{\mathrm{QS}}$ ------------------------- (1)

$\frac{\mathrm{Area}\mathrm{of}△\mathrm{PST}}{\mathrm{Area}\mathrm{of}△\mathrm{RST}}$= $\frac{\frac{1}{2}\times \mathrm{PT}\times \mathrm{SV}}{\frac{1}{2}\times \mathrm{TR}\times \mathrm{SV}}$

$\frac{\mathrm{Area}\mathrm{of}△\mathrm{PST}}{\mathrm{Area}\mathrm{of}△\mathrm{RST}}$ = $\frac{\mathrm{PT}}{\mathrm{TR}}$ ----------------------- (2)

In the diagram given above ΔQST and ΔRST are drawn on the same base i.e., ST and between the same parallel lines i.e., ST and QR.

∴area of (ΔQST) = area of (ΔRST)------------------------- (3)

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

$\frac{\mathrm{PS}}{\mathrm{QS}}$= $\frac{\mathrm{PT}}{\mathrm{TR}}$.

Hence proved.

In ΔABC a line DE parallel to its base (i.e., BC) is drawn such that it intersects sides AB and BC at points D and E respectively.

From the above result

$\frac{\mathrm{AD}}{\mathrm{BD}}$= $\frac{\mathrm{AE}}{\mathrm{CE}}$

It is given that BD = CE ------------------------------- (4)

Therefore AD = AE ------------------------------- (5)

Lets add equations (4) and (5), we obtain

BD + AD = CE + AE

⇒ AB = AC

Hence, ΔABC is an isosceles triangle.

1) Write the angles in correct order.

2) Since the car is moving in uniform speed. Take the ratio will be constant.

1) Using angles find the distance between foot of the tower and car starting point.

2) Since the car is moving in uniform speed then Speed =$$$\frac{\mathrm{dis}\tan \mathrm{ce}}{\mathrm{speed}}$

3) Since ratio will be constant because of uniform speed.

4) Hence to get time taken by the car to reach the tower.

Consider that AB is the tower.

The car was originally at point C.

It travelled for 6 sec and reached point D and hence CD is the distance travelled.

The angle of depression changes from ∠XAC to ∠XAD.

∠ACB = ∠XAC = 30° (Alternate interior angles)

And, ∠ADB = ∠XAD = 60° (Alternate interior angles)

In ΔABD

Tan 60° = $\frac{\mathrm{AB}}{\mathrm{BD}}$

$\frac{\mathrm{AB}}{\mathrm{BD}}$ = √3 --------------------------- (1)

In ΔABC,

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

$\frac{\mathrm{AB}}{\mathrm{BC}}$= $\frac{1}{\sqrt{3}}$------------------------ (2)

Divide (1) by (2)

$\frac{\mathrm{BC}}{\mathrm{BD}}$= 3

$\frac{(\mathrm{BD}+\mathrm{DC})}{\mathrm{BD}}$= 3

1+ $\frac{\mathrm{DC}}{\mathrm{BD}}$= 3

$\frac{\mathrm{DC}}{\mathrm{BD}}$= 2

BD = $\frac{\mathrm{DC}}{2}$---------------------------(3)

Since the car is moving in uniform speed

Speed = $\frac{\mathrm{dis}\tan \mathrm{ce}}{\mathrm{time}}$

This ratio will be constant because of uniform speed.

$\frac{\mathrm{DC}}{6}$= $\frac{\mathrm{DC}}{2t}$

T = 3 sec

Hence the time taken by the car to reach the tower = 3 seconds.

1) Write the correct formulas.

2) Avoid calculation mistakes.

or

1) The tangents drawn from an external point to the circle are equal.

1) Obtain the volume of the remaining solid from subtracting the volume of conical cavity from the volume of cylinder.

2) Find slant height .

3) By adding curved surface area of cylinder , area of top face and curved surface area of cone to get the total surface area of remaining solid.

or

1) By applying Pythagoras theorem to find remaining side.

2) The circle touches all sides then the radius of the centre of the circle is perpendicular to the tangent through the point of contact.

3) Find the radius if the tangents drawn from an external point to the circle are equal.

4) Area of shaded region from subtracting the area of circle from the area of triangle.

Since the conical cavity is hollowed out from the cylinder, we can obtain the volume of the remaining solid from subtracting the volume of conical cavity from the volume of cylinder

i.e;Volume of the remaining solid = Volume of cylinder − Volume of conical cavity

Volume of the remaining solid = π $r^2$h - $\frac{1}{3}$π $r^2$h

= $\frac{2}{3}$π $r^2$h

= $\frac{2}{3}$×3.1416×(6 ${)}^2$× 8

Volume of the remaining solid = 603.1872 ${\mathrm{cm}}^3$.

The total surface area of remaining solid = Curved surface area of cylinder + area of top face + curved surface area of cone

= 2πrh + πr2 + πrl

= πr (2h + r + l)

We know that the slant height (l) of the cone is given by√( $r^2$+ $h^2)$.

∴ Slant height of conical cavity =√( $6^2$+ $8^2$) = 10 cm

Hence, total surface area of the solid = 3.1416 × (6 cm) (2 × 8 cm + 6 cm + 10 cm)

= 3.1416 × (6 cm) (16 cm + 6 cm + 10 cm)

= 3.1416 × 6 cm × 32 cm

= 603.1872 ${\mathrm{cm}}^2$

Therefore the total surface area of the remaining solid is approximately 603.18 ${\mathrm{cm}}^2$.

OR

In ΔABC, applying Pythagoras theorem we obtain

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

⇒ ${\mathrm{AC}}^2$ = ${\mathrm{BC}}^2$− ${\mathrm{AB}}^2$= ( ${10\mathrm{cm})}^2$ − ${(6\mathrm{cm})}^2$

⇒ ${\mathrm{AC}}^2$ = 64 ${\mathrm{cm}}^2$

⇒ AC = 8 cm

Let us assume the radius of the incircle to be ‘r’.

The circle touches side AB of the triangle at P, side AC at Q, and side BC at R.

The radius from the centre of the circle is perpendicular to the tangent through the point of contact.

∴ OP⊥AB, OQ⊥AC, and OR⊥BC

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

∴ AP = AQ (Tangents from point A)

BP = BR (Tangents from point B)

CR = CQ (Tangents from point C)

Now, in quadrilateral OPAQ, AQ = AP and ∠AQO = ∠APO = ∠PAQ = 90°

Therefore, OPAQ is a square.

∴ OP = AQ = AP = OQ = r

∴ AB = 6 cm − r

⇒ BR = 6 cm − r

⇒CQ = 8 cm − r

⇒ CR = 8 − r

Now, BC = BR + CR

⇒ 10 cm = 6 cm − r + 8 cm − r

⇒ 10 cm = 14 cm − 2r

⇒ r = 2 cm

Area of shaded region = Area of ΔABC − Area of circle

= $\frac{1}{2}$×AB×AC - π $r^2$

= $\frac{1}{2}$×8×6 - π $2^2$= 24 - 12.56 = 11.44 ${\mathrm{cm}}^2$.

Hence, the area of the shaded region is 11.44 cm2.

1) Take correct assumed mean.

2) In the mode class interval depends on corresponds to the maximum frequency .

3) In the median class interval depends on number of observations.

1) Find the Cumulative frequency and deviations.

2) Substitute all values in appropriate formulas of Mean, Mode and Median to get the solution.

Let us represent the assumed mean by ‘a’ and assume it as 150.

The table for the given data can be drawn as

Mean = a + $\frac{\Sigma f_i{d}_i}{f_i}$

Therefore Mean = 150 + $\frac{(-240)}{50}$= 150 + (- 4.8) = 145.2

The class that corresponds to the maximum frequency of 14 is 120 -140.

Modal class = 120 - 140

Lower limit (l) = 120

Class size (h) = 140 − 120 = 20

Frequency of modal class ( $f_1$) = 14

Frequency of class preceding the modal class ( $f_0$) = 12

Frequency of class succeeding the modal class ( $f_2$) = 8

Mode = l + $\frac{(f_1-f_0)}{(2f_1-f_0-f_2)}$× h

Therefore, Mode = 120 + $\frac{(14-12)}{(2\times 14-12-8)}$× 20

Mode = 120 + $\frac{2}{8}$×20

= 120 + 5

= 125.

Hence, mode of the given data is 125.

Number of observations (n) = 50

$\frac{n}{2}$= 25.

This value lies in class interval 120−140.

Therefore, the median class is 120−140.

Lower limit of median class (l) = 120

Cumulative frequency of class preceding the median class (c.f) = 12

Frequency of median class (f) = 14

Median of the data = l + $\frac{\frac{n}{2}-c.f}{f}$× h

Therefore median = 120 + $\frac{(125-12)}{14}$× 20

= 120 + $\frac{130}{7}$ = 120 + 18.57 = 138.57

Hence, median = 138.57