Let radius and height of the cone be 3x and 4x respectively.

Therefore,

(1/3)πr²h = 301.44

(1/3) × 3.14 × 3x × 3x × 4x = 301.44

x³ = 8

x = 2

Then, Radius = 3x = 3 × 2 = 6 cm and Height = 4x = 4 × 2 = 8 cm

Now, slant height (l) = (r² + h²)^1/2

             = (6² + 8²)^1/2 = 10 cm

CSA of cone = πrl = 3.14 × 6 × 10 = 188.40 cm²

Let the radius be 3x and height of the cone be 4x 

Use pi=3.14

and using the formula (1/3)(pi)r²h = Volume 

Find x , It comes x=2

Therefore radius is 6 and height is 8. Then FInd its CSA

Ans. Let radius of cone is r and height of cone is h and slant height of cone is l.

Then $\frac{r}{h} = \frac{3}{4}$ or $h = \frac{4r}{3} ..........\left(1\right)$ 

Volume of Cone = 301.44 cm³

=> We Know Volume of cone = $\frac{1}{3}\pi r^2 h$

=> $301.44 = \frac{1}{3}\times 3.14\times r^2\times \frac{4r}{3}$

=> $r^3 = \frac{301.44 \times 9 }{3.14\times 4}$

=> $r^3 = 216$

=> $r = 6$ cm

Put value of r in (1), We get 

$h = \frac{4\times 6}{3} = 8cm$

Slant Height of cone (l) = $\sqrt{{\left(r\right)}^2+{\left(h\right)}^2 } = \sqrt{{\left(6\right)}^2+{\left(8\right)}^2}=\sqrt{36+64} = \sqrt{100} = 10cm$

Curved Surface Area = $\pi \times r\times l = 3.14 \times 6 \times 10 = 188.4 c{m}^2$

The perpendicular from the centre to the chord bisects the chord

In the diagram below, AB is the chord of a circle with centre O.

OM is the perpendicular from the centre to the chord.

Look at triangles OAM and OBM.

The hypotenuses (OA and OB) are the same, as they are both the radius of the circle.

OM is common to both triangles.

OMA and OMB are both right angles.

Triangles OAM and OBM are congruent (RHS), so it follows that AM = MB.

Therefore, M is the midpoint of AB, and the chord has been bisected.

Draw AB and CD, the parallel chords of a circle such that M and N are the midpoints of the chords respectively.
The perpendicular bisector of the chord passes through the centre.
So, ON ⊥ CD.
and OM ⊥ AB.
But AB || CD
Therefore, NOM is a straight line.
Hence the line joining the midpoints of two chords of a circle parallel to each other passes through the centre of the circle.

Ans. Let Probability of Success = x

Probability of Failure = y

According To Ques

x - y = 3/17       ......(1)

Also We Know 

x + y = 1             ......(2)

adding (1) and (2), We Get

2x = 20/17

x = 10/17

y = 7/17

We can not draw an angle of 65° with the compass as we can only draw angles which are a multiple of 15 with the help of the compass.

Class mid point = Class mark = 20

Class width = Class size = 8

Lower limit of class interval = Class mark -$\frac{Class size}{2}$

= 20 - (8/2) = 20-4 = 16

Upper limit of class interval = Class mark +$\frac{Class size}{2}$

= 20 + (8/2) = 20+4 = 24

And. Class mid-point (a) = 20

width (h) = 8

we know,

Lower Limit = a - h/2

put given values, we get

Lower Limit = 20 - 8/2

=> Lower Limit = 20-4= 16

Ans. Normal range = x

Overweight range = y

As per question

x = 8y

=> x - 8y = 0

required linear equation

Ans. We know 

x³+y³= (x+y)(x²+y²+xy)  ......(1)

and

x²+y² = (x+y)²  - 2xy .....(2)

put value x²+y² from (2) in (1), we get

x³+y³ =(x+y)[(x+y)² -xy]   ...(3)

we need to find 

a³+8b³ can be written as

=> a³+ (2b)³ = (a+2b)[(a+2b)² -2a*b]

put given values, we get

=> a³+8b³= (7)[(7)² - 2*5]

=> (7)(39)

=> 273 

Ans.

present Age if father = x

present age of sheena = y

According to first condition

y = x/3

=> x =3y ........(1)

After 12 year 

Age of father = x + 12

age of sheena = y +12

According to 2nd condition

y + 12 = (x+12)/2

=> 2y + 24 = x + 12

=> x - 2y = 12 ....(2)

put value of x from (1) we get

=> 3y - 2y = 12

=> y =12

x = 36

present Age of father = 36 years

present Age of Sheena = 12 years 

Solution: Given : Arc AB of a circle with center O, Subtending $\angle AOB$ at the centre and $\angle APB$ at a point P on the remaining part of circle.

To Prove : $\angle AOB = 2\angle APB$

Proof :

We can have three cases;

i) AB is a minor arc

ii) AB is a diameter 

iii) AB is a major arc

we are taking first case only at this time.

We know that, an exterior angle of a triangle is equal to the sum of the interior opposite angles.

$In ∆OPB, \angle QOB = \angle OPB + \angle OBP ..... \left(1\right)$

OB = OP (Radius of the circle)

=> $\angle OPB = \angle OBP$ (In a triangle, equal sides have equal angle opposite to them)

=> $\angle QOB = \angle OPB + \angle OPB$

=> $\angle QOB = 2\angle OPB ...... \left(2\right)$

$in ∆OPA, \angle QOA = \angle OPA + \angle OAP ........\left(3\right)$

OA = OP (Radius of the circle)

=> $\angle OPA=\angle OAP$   (In a triangle, equal sides have equal angle opposite to them)

=> $\angle QOA = \angle OPA + \angle OPA$

=> $\angle QOA = 2\angle OPA ......\left(4\right)$

Adding (2) and (4), We get 

$\angle QOB + \angle QOA = 2\angle OPB + 2\angle OPA$

=> $\angle AOB=2\angle APB$

Hence Proved

A leap year has 366 days. This is equivalent to 52 weeks and 2 extra days. These 2 days can be combination of (sunday, Monday),

(Monday, Tuesday),

(Tuesday, Wednesday),

(Wednesday, Thursday),

(Thursday, Friday),

(Friday, Saturday) or

(saturday, sunday)

So,probability of getting a 53rd sunday = 2/7

How To Find if Triangles are Congruent

There are five ways to find if two triangles are congruent: SSS, SAS, ASA, AAS and HL.

1. SSS   <i>(side, side, side)</i>

SSS stands for "side, side, side" and means that we have two triangles with all three sides equal.

For example:

is congruent to:

<i>(See <a href="http://www.mathsisfun.com/algebra/trig-solving-sss-triangles.html">Solving SSS Triangles</a> to find out more)</i>

If three sides of one triangle are equal to three sides of another triangle, the triangles are congruent.

2. SAS   <i>(side, angle, side)</i>

SAS stands for "side, angle, side" and means that we have two triangles where we know two sides and the included angle are equal.

For example:

is congruent to:

<i>(See <a href="http://www.mathsisfun.com/algebra/trig-solving-sas-triangles.html">Solving SAS Triangles</a> to find out more)</i>

If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent.

3. ASA   <i>(angle, side, angle)</i>

ASA stands for "angle, side, angle" and means that we have two triangles where we know two angles and the included side are equal.

For example:

is congruent to:

<i>(See <a href="http://www.mathsisfun.com/algebra/trig-solving-asa-triangles.html">Solving ASA Triangles</a> to find out more)</i>

If two angles and the included side of one triangle are equal to the corresponding angles and side of another triangle, the triangles are congruent.

4. AAS   <i>(angle, angle, side)</i>

AAS stands for "angle, angle, side" and means that we have two triangles where we know two angles and the non-included side are equal.

For example:

is congruent to:

<i>(See <a href="http://www.mathsisfun.com/algebra/trig-solving-aas-triangles.html">Solving AAS Triangles</a> to find out more)</i>

If two angles and the non-included side of one triangle are equal to the corresponding angles and side of another triangle, the triangles are congruent.

5. HL   <i>(hypotenuse, leg)</i>

This one applies only to <a href="http://www.mathsisfun.com/right_angle_triangle.html">right angled-triangles</a>!

or

HL stands for "Hypotenuse, Leg" (the longest side of a right-angled triangle is called the "hypotenuse", the other two sides are called "legs")

It means we have two right-angled triangles with

• the same length of hypotenuse and
• the same length for one of the other two legs.

It doesn't matter which leg since the triangles could be rotated.

For example:

is congruent to:

<i>(See <a href="http://www.mathsisfun.com/pythagoras.html">Pythagoras' Theorem</a> to find out more)</i>

If the hypotenuse and one leg of one right-angled triangle are equal to the corresponding hypotenuse and leg of another right-angled triangle, the two triangles are congruent.

Caution ! Don't Use "AAA" !

AAA means we are given all three angles of a triangle, but no sides.

This is not enough information to decide if two triangles are congruent!

Because the triangles can have the same angles but be different sizes:

is not congruent to:

Without knowing at least one side, we can't be sure if two triangles are congruent.

Let Radius of Cone is r and slant height is l.

height of Cone (h) = 24 cm

CSA of cone = 550 cm²

We know, 

$l = \sqrt{r^2+ h^2}$

=> $l = \sqrt{r^2+ 576} ............\left(1\right)$

Also,

$CSA = \pi \times r\times l$

=> $550 = \frac{22}{7}\times r \times \sqrt{r^2 +\hspace{0.17em}576}$

Squaring both side,

=> $302500 = \frac{484 r^2 (r^2 + 576)}{49}$

=> $30625 = r^4 + 576r^2$

=> $r^4 + 576r^2 - 30625 = 0$

Put $r^{2 }= t$

=> $t^2 + 576t - 30625 = 0$

=> $t^2 + 625t - 49t - 30625 = 0$

=> $t(t + 625) -49(t+ 625) = 0$

=> $(t + 625) (t-49) = 0$

=> $t + 625 = 0 or t-49 = 0$

=> t = - 625 0r t = 49 

=> r²  = -625 is not possible so r² = 49 

=> r = 7 

put in (1), we get 

$l = \sqrt{49+576}$

=> $l = \sqrt{625}$

=> l = 25

=>$Volume of Cone = \frac{1}{3}\pi r^{2}h$

=> $Volume of Cone = \frac{1\times 22\times 49\times 24}{3\times 7}$

=> Volume = 1232cm³

Ans. Let A denote the probability of guessing correct answer and B denote the probability of not guessing correct answer.

Then, $p\left(A\right) = \frac{x}{2} and p\left(B\right) = \frac{2}{3}$

 $p\left(A\right) +\hspace{0.17em}p\left(B\right) = 1$

=> $\frac{x}{2}+ \frac{2}{3} = 1$

=> $\frac{x}{2} = 1 - \frac{2}{3}$

=> $\frac{x}{2} = \frac{1}{3}$

=> $x = \frac{2}{3}$

Area of field=13:5×2:5=33:75

Upper area of tank=5×4:5=22:5

Area of field on which soil is spread=33:75_22:5

=11:25m   Volume of tank=5×4:5×2:1=47:25m^3

Raised in level=volume of tank/area of field on which

soil spread=47:25/11:25=4:2

So,raised  level=4:2m

3√60=2

Diagonal of cube =√3a units

When the rectangular piece of paper is rolled along its length, we find that the length of paper forms the circumference of its base and breadth of paper becomes the height of the cylinder

Let r cm be the radius of the base and h cm be the height then h=10cm

Now $circumference of the base=length of the paper$

$\Rightarrow 2\mathrm{\pi r}=22$

$\Rightarrow 2\times \frac{22}{7}\times r=22$

$\Rightarrow r=\frac{7}{2}cm$

$Volume of cylinder={\mathrm{\pi r}}^{2}h$

$=\frac{22}{7}\times {\left(\frac{7}{2}\right)}^2\times 10$

$=385 c{m}^3$

Given: a rectangular piece of paper of length=22cm breadth=10cm.

As we will roll the rectangular piece along its length, then length of rectangle becomes height of cylinder and width of rectangular piece would become circular faces of diameter 10cm. So radius will be =5cm. 

So, volume of cylinder=πr²h 

22/7*5*5*22 =12100/7

=1728.57cm² .

Volume of cylinder =1728.57cm^{2 .}

$Given: A ∆ABC in which AD is median$

$To Prove: area\left(∆ABD\right)=area\left(∆ADC\right)$

$Construction: draw AE\perp BC$

$Proof:$

$Since AD is Median of ∆ABC$

$\Rightarrow BD=DC$

$\Rightarrow BD\times AE=DC\times AE \left(Multiplying both sides by AE\right)$

$\Rightarrow \frac{1}{2}\left(BD\times AE\right)=\frac{1}{2}\left(DC\times AE\right)$

$\Rightarrow Area(∆ABD)=area(∆ADC)$

$Hence Proved$

Given : A quadrilateral ABCD in which the diagonals AC and BD intersect each other at O. It is given that the diagonals of ABCD are equal and bisect each other at right angles AC = BD, OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90º.

To prove ABCD is a square, we have to prove that ABCD is a parallelogram, AB = BC = CD = AD, and one of its interior angles is 90º.

In ΔAOB and ΔCOD,

AO = CO (Diagonals bisect each other)

OB = OD (Diagonals bisect each other)

∠AOB = ∠COD (Vertically opposite angles)

∴ ΔAOB ≅ ΔCOD (SAS congruence rule)

∴ AB = CD (By CPCT) ... (1)

And, ∠OAB = ∠OCD (By CPCT)

However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel.

∴ AB || CD ... (2)

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

ABCD is a parallelogram.

In ΔAOD and ΔCOD,

AO = CO (Diagonals bisect each other)

∠AOD = ∠COD (Given that each is 90º)

OD = OD (Common)

∴ ΔAOD ≅ ΔCOD (SAS congruence rule)

∴ AD = DC ... (3)

However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD)

∴ AB = BC = CD = DA

Therefore, all the sides of quadrilateral ABCD are equal to each other.

In ΔADC and ΔBCD,

AD = BC (Already proved)

AC = BD (Given)

DC = CD (Common)

∴ ΔADC ≅ ΔBCD (SSS Congruence rule)

∴ ∠ADC = ∠BCD (By CPCT)

However, ∠ADC + ∠BCD = 180° (Co-interior angles)

⇒ ∠ADC + ∠ADC = 180°

⇒ 2∠ADC = 180°

⇒ ∠ADC = 90°

One of the interior angles of quadrilateral ABCD is a right angle.

Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angles is 90º. Therefore, ABCD is a square

Given that A(3, 5) and B(1, 4) lie on the graph of the line ax + by = 7 → (1)
Hence both the points satisfy the given equation
Put x = 3 and y = 5 in (1)

⇒ 3a + 5b = 7 ........ (2)

Similarly, B(1, 4) satisfy (1)
Put x = 1 and y = 4 in (1), we get
a + 4b = 7 
Multiply above equation with '3', we get
3a + 12b = 21 ....... (3)

Subtract (2) and (3)

3a + 12b = 21
3a + 5b = 7
---------------
  7b = 14
∴ b = 2
Put b = 2 in (2), we get

3a + 5(2) = 7
⇒ 3a = – 3
∴  a = – 1

Given: point (1,-2) lies on on the graph of linear equation = x-2y+k=0

Put x=1 ,y=-2 in the given equation. 

So,  1-2(-2)+k=0

        1+4+k=0

       5+k=0

K=-5

-------------------

Value of k= -5.

p(x)=x³+8x²+17x+ax

As the remainder same in both case :

then p(-2) = p(-1)

(-2)³+ 8(-2)² + 17(-2) + a(-2)  = (-1)³ + 8(-1)² + 17(-1) + a(-1)

=> -8 + 32 - 34 - 2a = -1 + 8 - 17 - a

=> -10  -2a = - 10 - a

=> -2a + a = -10 + 10

=> -a = 0

=> a = 0