Number of cubes = \(\frac { { \left( 20 \right) }^{ 3 } }{ { \left( 5 \right) }^{ 3 } } =64\)

\(AC=CB=\frac { 1 }{ 2 } AB=\frac { 1 }{ 2 } \times 16=8\quad cm\) 
\(OA=\sqrt { OC^{ 2 }+AC^{ 2 } }\) 
\(=\sqrt { 6^{ 2 }+8^{ 2 } } =10\quad cm\)

2+3=5
Required angle =\(\frac { 2 }{ 5 } x180^{ 0 }\)

O(0,0) satisfies x+y=0

Steps of construction:
i) Draw a line segment BC = 4.7 cm. and at point B construct an angle of 45^o i.e ㄥXBC=45^o
ii) Cut the line segment BD = 2 cm (equal to AB-AC)on ray BX
iii) Join DC and draw the perpendicular bisector PQ of DC
iv) The perpendicular bisector intersects BXat point A. Jon AC ΔABC is the required triangle.

According to the question, E and F are the midpoints of sides AB and CD.
ஃ AE=\(\frac{1}{2}\)AB
CF=\(\frac{1}{2}\)CD
ஃ In the parallelogram opposite sides are equal, so AB=CD
ஃ AE=CF
Again, AB||CD
So, AE||FC
Hence AECF is a parallelogram.
In ΔABP,
E is the mid-point of AB.EQ || AP
ஃ Q is the mid-point of BP
Similarly, P is the mid-point of DQ
DP= PQ= QB
ஃ Line segments AF and EC trisect the diagonal BD.

Let, g(x)=x²-3x+2
=x²-2x-x+2
=x(x-2)-1(x-2)
=(x-2)(x-1)
Zero of x-2 is 2, as x-2=0\(\Rightarrow \)x=2
Zero of x-1 is 1, as x-1=0\(\Rightarrow \)x=1
Now, f(x)=2x⁴-6x³+3x²+3x-2
f(2)=2(2⁴)-6(2³)+3(2²)+3(2)-2
=32-48+12+6-2=0
f(1)=2(1)⁴-6(1)²+3(1)²+3(1)-2
=2-6+3+3-2=0
\(\Rightarrow \) (x-1) and (x-2) are the factors of f(x).
\(\therefore\) f(x) is exactly divisible by g(x).

Given equation
3x-5y-15=0
5y=3x-15
\(\Rightarrow y=\frac{3x-15}{5}\)
\(\Rightarrow y=\frac{3}{5}(x-5)\)
put x=0, then y=-3
put x=5, then y=0
put x=-5, then y=-6

To plot these points on graph paper, join these points. The graph of the line intersects x-axis at (5,0) and y-axis at (0,-3).

For one black colour sheet  a= 4 cm, b = 6 cm
\(\therefore \) Area \(=\frac { a }{ 4 } \sqrt { 4{ b }^{ 2 }-{ a }^{ 2 } } \)     | Sheet is an isosceles triangle
 \(=\frac { 4 }{ 4 } \sqrt { 4{ (6) }^{ 2 }-4^{ 2 } } \)  \(=\sqrt { 128 } =8\sqrt { 2 } \) cm²
\(\therefore \) Area of 2 black colour sheets = 2\(\times 8\sqrt { 2 } \) = \( 16\sqrt { 2 } \) cm²
Similarly, area of 2 white colour sheets =  \( 16\sqrt { 2 } \) cm²
\(\therefore \) Total area of black and white colour sheets =  \( 16\sqrt { 2 } \) + \( 16\sqrt { 2 } \) =  \( 32\sqrt { 2 } \) cm²

In right triangle PSQ,  PQ\(\frac { 1 }{ 2 } \)² = PS² + QS²     |By Pythagoras Theorem
= (12)² + (16)²
= 144 + 256 =400
\(\Rightarrow \)  PQ = \(\sqrt { 400 } \) = 20 cm
Now, for  \(\Delta \)PQR
a = 20cm, b = 48cm, c = 52cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 20+48+52 }{ 2 } =60\) cm
\(\therefore \)  Area of  \(\Delta \)PQR \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 60(60-20)(60-48)(60-52) }\)
\( \\ =\sqrt { (60)(40)(12)(8) } \)
 \(=\sqrt { \left( 6\times 10 \right) \left( 4\times 10 \right) \left( 6\times 2 \right) \left( 8 \right) } \)
\(=6\times 10\times 8=480\) cm²
Area of \(\Delta \) PSQ = \(\frac { 1 }{ 2 } \)\(\times \)Base\(\times \)Altitude
=\(\frac { 1 }{ 2 } \)\(\times \)16\(\times \)12=96 cm²
\(\therefore \) Area of the shaded portion =Area of \(\Delta \)PQR - Area of \(\Delta \)PSQ
= 480 - 96 = 384 cm²

Given: ABC is an isosceles triangle in which AB = AC. AD bisects L PAC and CD IIAB.
To Prove:
(i)  \(\angle \)DAC =\(\angle \)BCA
(ii) ABCD is a parallelogram.
Proof:
(i) In \(\Delta \) ABC,
\(\because\) AB = AC
\(\therefore\) \(\angle \)B =\(\angle \)C               .......(1)      I Angles opposite to equal sides of a triangle are equal
Also, Ext.   \(\angle \)PAC =\(\angle \)B +\(\angle \)C     I An exterior angle of a triangle is equal to the sum of its two interior opposite angles
⇒    \(\angle \)PAC = \(\angle \)C+\(\angle \)C   | From (1)
⇒ 2\(\angle \)CAD = 2\(\angle \)C   | \(\because\) AD bisects \(\angle \)PAC 
⇒ \(\angle \)CAD =\(\angle \)C
⇒  \(\angle \) DAC =\(\angle \)BCA
(ii) But these angles form a pair of equal alternate interior angles
\(\therefore\)  AD II BC
Also, CD II AB         I Given
\(\therefore\)   ABCD is a parallelogram   IA quadrilateral is a parallelogram if its both the pairs of opposite sides are parallel.

\(p(x)=2{ x }^{ 3 }-3{ x }^{ 2 }+ax-3a+9\)
By remainder theorem,
\(p(-1)=16\ x+1=0\quad \Rightarrow x=-1\)
\(\Rightarrow 2{ (-1) }^{ 3 }-3{ (-1) }^{ 2 }+a(-1)-3a+9=16\) 
\(\Rightarrow -2-3-a-3a+9=16\)
\(\Rightarrow 4a=-12\) 
\(\Rightarrow a=-3\) 
\(\therefore p(x)=2{ x }^{ 3 }-3{ x }^{ 2 }-3x-3 \times(-3)+9\) 
\(={ 2x }^{ 3 }-3{ x }^{ 2 }-3x+18\)
\(\therefore \) Remainder when p(x) is divided by x+2 =p(-2)
By remainder theorem: \(x+2=0\Rightarrow x=-2\)
\(={ 2(-2) }^{ 3 }-3{ (-2) }^{ 2 }-3(-2)+18\) 
\(=-16-12+6+18=-4\) 

16.5

\(\frac { 77 }{ 12 } { m }^{ 3 }\)

7 : 3

\(36°\),\(144°\),\(36°\), \(144°\)

\(2(2x-y)(4x^2+2xy+y^2)\)

0

(i) There are 4 bags in which seeds are more than 40.
P(more than 40 seeds in a bag)\(=\frac{4}{5}\)
(ii) There is no bag of 49 seeds.
P(49 seeds in a bag) = 0
(iii) There is only one bag out of 5 in which seeds are less than 40.
P(less than 40 seeds in a bag)\(=\frac{1}{5}\)

In \(\triangle\)PQS,
PQ + QS > PS ...(1)
(Sum of any two sides is greater than the third side)
In \(\triangle\)PSR,
PR + SR > PS ...(2)
Adding (1) & (2), 
PQ + QS + PR + SR >2 PS   (\(\because\) QR=QS+SR)
PQ + QR + RP > 2PS.

Construction: Draw PM \(\bot \)QS and RN \(\bot \) QS

\(ar(\Delta PSA)\times ar(\Delta QAR)\)
\(=\left( \frac { 1 }{ 2 } \times AS\times PM \right) \times \left( \frac { 1 }{ 2 } \times AQ\times RN \right) \)
\(=\left( \frac { 1 }{ 2 } \times RN\times AS \right) \times \left( \frac { 1 }{ 2 } \times PM\times AQ \right) \)
\(=ar\left( \Delta SAR \right) \times ar\left( \Delta PAQ \right) \)
Hence Proved.

Draw, EH||AB

Now, EH||AB and AB||CD,
Therefore, EH||CD
Now, \(\angle BGE+\angle GEH=180^o\) (Co-interior \(\angle S)\)
(AB||EH, Co-interior angles)
\(135^o+\angle GEH=180^o\)
\(\angle GEH=180^o-135^o=45^o...(1)\)
Again, \(\angle DFE+\angle FEH=180^o\)
(CD||EH, Co-interior angles)
\(\Rightarrow 125^o+\angle FEH=180^o\)
\(\Rightarrow \angle FEH=180^o-125^o\)
=55^o
Adding (1) and (2), we get
\(\angle GEH+\angle FEH=45^o+55^o\)
x=100^o

\(\frac { 30 }{ 4\sqrt { 3 } +3\sqrt { 2 } } =4\sqrt { 3 } -a\sqrt { 2 } \)
\(\Rightarrow \frac { 30 }{ 4\sqrt { 3 } +3\sqrt { 2 } } \times \frac { 4\sqrt { 3 } -3\sqrt { 2 } }{ 4\sqrt { 3 } -3\sqrt { 2 } } =4\sqrt { 3 } -a\sqrt { 2 } \)
\(\Rightarrow \frac { 30\left( 4\sqrt { 3 } -3\sqrt { 2 } \right) }{ 30 } =4\sqrt { 3 } -a\sqrt { 2 } \)
\(4\sqrt { 3 } -3\sqrt { 2 } =4\sqrt { 3 } -a\sqrt { 2 } \)
On comparing a = 3

Base diameter = 40 cm
\(\therefore\) Base radius (r) = \(\frac { 40 }{ 2 } cm=20cm\)
\(=\frac { 20 }{ 100 } m=0.2m\)
Height (h) = 1 m
 \(\therefore l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(\\ =\sqrt { { \left( 0.2 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 } } =\sqrt { 0.04+1 } \)
\(\\ =\sqrt { 1.04 } =1.02m\)
\(\therefore\)  Curved surface area = \(\pi rl\)
= 3.14 \(\times\) 0.2 \(\times\) 1.02
= 0.64056 m²
\(\therefore\)  Curved surface area of 50 cones
= 0.64056 \(\times\) 50 m²
= 32.028 m²
\(\therefore\)  Cost of painting all these cones
= 32.028 \(\times\) 12
= 384.336 = Rs 384.34 (approximately).

110°

Given ABCD is a quadrilateral in which AD=BC and \(\angle DAB=\angle CBA\) 
To prove: (i) \(\triangle ABD\cong \triangle BAC\)
(ii) BD = AC
(iii) \(\angle ABD=\angle BAC\)
Proof: (i) In \(\triangle ABD\) and \(\triangle BAC\)
AD = BC
AB = BA
\(\angle DAB=\angle CBA\)
\(\triangle ABD\cong \triangle BAC\)  |SAS Rule
(ii) \(\triangle ADB\cong \triangle BAC\)  |Proved in (i)
BD = AC  | C.P.C.T
(iii) \(\triangle ABD\cong \triangle BAC\) |Proved in (i)
\(\angle ABD=\angle BAC\)  | C.P.C.T

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