Altitude AD which is perpendicular to BC, then D is the mid-point of BC.
BD = CD = \(\frac{1}{2}\) BC = \(\frac{8}{2}\) = 4 cm.
Apply, Pythagoras theorem to find AD.
AD = \(4\sqrt{3}\) cm.

Given, PQRS is a trapezium in which \(PQ\parallel RS\) and PQ = 3RS.

\(\Rightarrow \frac { PQ }{ RS } =\frac { 3 }{ 1 } \) ...(i) 
In \(\triangle POQ\) and \(\triangle ROS\),
\(​​\angle SOR=\angle QOP \)  [vertically opposite angles]
\(​​\angle SRP=\angle RPQ\) [alternate angles]
\(\therefore \triangle POQ\sim \triangle ROS\)     [by AA similarity criterion]
By property of area of similar triangle,
\(\frac { ar\left( \triangle POQ \right) }{ ar\left( \triangle ROS \right) } =\frac { { \left( PQ \right) }^{ 2 } }{ { \left( RS \right) }^{ 2 } } =\left( \frac { 3 }{ 1 } \right) ^{ 2 }\)  [from Eq.(i)]
\(\Rightarrow \frac { ar\left( \triangle POQ \right) }{ ar\left( \triangle SOR \right) } =\frac { 9 }{ 1 } \)
Hence, the required ratio is 9 : 1.

Given, \(DE\parallel BC\) , DE = 6 cm and BC = 12 cm
In \(\triangle ABC\) and \(\triangle ADE\),
\(\angle ABC=\angle ADE\)  [corresponding angles]
\(\angle ACB=\angle AED\) [corresponding angles]
and \(\angle A=\angle A\)  [common angle]
\(\therefore \triangle ABC\sim \triangle ADE\)  [by AAA similarity criterion]
We know that, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
\(\therefore \frac { ar\left( \triangle ADE \right) }{ ar\left( \triangle ABC \right) } =\frac { { \left( DE \right) }^{ 2 } }{ { \left( BC \right) }^{ 2 } } =\frac { { \left( 6 \right) }^{ 2 } }{ { \left( 12 \right) }^{ 2 } } =\left( \frac { 1 }{ 2 } \right) ^{ 2 }\)
\(\Rightarrow \frac { ar\left( \triangle ADE \right) }{ ar\left( \triangle ABC \right) } =\left( \frac { 1 }{ 2 } \right) ^{ 2 }=\frac { 1 }{ 4 } \)
Let \(ar\left( \triangle ADE \right) =k\) , then \(ar\left( \triangle ABC \right) =4k\)
Now, \(ar\left( \triangle DECB \right) =ar\left( \triangle ADC \right) - ar\left( \triangle ADE \right) \)
= 4k - k = 3k
\(\therefore \)  Required ratio = \(ar\left( \triangle ADE \right) : ar\left( \triangle DECB \right) \)
= k : 3k = 1 : 3

Prove \(\triangle ABC\) and \(\triangle AMP\)  are similar.
then take ratio \(\frac{AC}{AP}=\frac{BC}{MP}\)

Given, \(\triangle ABC\sim \triangle EDF\)
Also, AB = 5 cm, AC = 7 cm, DF = 15 cm
and DE = 12 cm                         .... (i)
Since, \(\triangle ABC\sim \triangle EDF\)
\(\therefore \frac { AB }{ ED } =\frac { AC }{ EF } =\frac { BC }{ DF } \)
[ ∵ Corresponding sides of similar triangles are proportions]
\(\Rightarrow \frac { 5 }{ 12 } =\frac { 7 }{ EF } =\frac { BC }{ 15 } \)  [from Eq. (i)]

On taking first and second terms, we get
\(\frac { 5 }{ 12 } =\frac { 7 }{ EF } \Rightarrow EF=\frac { 7\times 12 }{ 5 } \) = 16.8 cm
On taking first and third terms, we get
\(\frac { 5 }{ 12 } =\frac { BC }{ 15 } \Rightarrow BC=\frac { 5\times 15 }{ 12 } \) = 6.25 cm
Hence, lengths of the remaining sides of the triangles are EF = 16.8 cm and BC = 6.25 cm.

Draw the figure according to the given conditions and use Pythagoras theorem to find the value of x, then required saved distance will be equal to the difference of (AC + BC) and 26.
8 km.

Diagonals of a rhombus bisect each other at right angles.

So, OA = OC = 8 cm
and OB = OD = 6 cm
Now use pythagoras theorem in  ΔAOB
10 cm.

Given, \(\sqrt { 2 } \) is one of the zeroes of the cubic polynomial. Thus, \((x-\sqrt { 2 } )\) is one of the factors of the given polynomial p(x)=\(6x^{ 3 }+\sqrt { 2x^{ 2 } } -10x-4\sqrt { 2 } \).
Then, division process is
Here, quotient \(=6x^{ 2 }+7\sqrt { 2x } +4\) and remainder =0
Now, factorise the quotient by splitting the middle term i.e. write
\(6x^{ 2 }+7\sqrt { 2x } +4=6x^{ 2 }+4\sqrt { 2x } +4\\ =2x(3x+2\sqrt { 2 } )+\sqrt { 2 } (3x+2\sqrt { 2 } )\\ =(2x+\sqrt { 2 } )(3x+2\sqrt { 2 } )\)
For other zeroes of p(x), put \(6x^{ 2 }+7\sqrt { 2x } +4=0\)
\((2x+\sqrt { 2 } )(3x+2\sqrt { 2 } )=0\)
⇒ \(x=-\frac { \sqrt { 2 } }{ 2 } ,\frac { -2\sqrt { 2 } }{ 3 } \)
Hence, other two zeroes of p(x) are \(\frac { -1 }{ \sqrt { 2 } } \) and \(\frac { -2\sqrt { 2 } }{ 3 } \) .

Let O be the centre and P be the mid-point of BC.Then, OP丄BC.

Since, ∆ABC is an isosceles triangle and P is the mid-point of BC. Therefore, AP .L BC as median from the vertex in an isosceles triangle is perpendicular to the base.
Let AP = x
and PB = CP = y
In ΔAPB and ΔOPB,
AB² = BP² + AP²
36 = y² + x²
and OB²=OP²+BP²
81= (9- x)² + y²
On subtracting Eqq.(i) from Eq.(ii) we get
81- 36 = [(9 - x)² + y²]- (y² + x2)
45 =81-18x+x²+y² -y² -x²
45=81-18x
18x=81-45
18x = 36
\(x={36\over 18}\)
x=2cm
On putting x = 2 in Eq.(i), we get
36=y²+4
y²=32
\(y=4\sqrt{2}cm\)
BC=2BP
=2y=\(8\sqrt{2}cm\)
Now, area of ∆ABC=\({1\over2}\times BC\times AP\)
\(={1\over2}\times8\sqrt{2}\times2\)
\(=8\sqrt{2}cm^2\)
Hence, the area of ∆ABC is \(=8\sqrt{2}cm^2\)

Consider 10, 20, 30, 40, ... 490. According to question
490 = 10 + (n - 1) x 10
[\(\because\) a = 10, an = 490]
and \(S_{n}=\frac{n}{2}[2 \times 10+(n-1) \times 10]\)
On solving, we get n = 49.
= 12,250

 
Given, scale factor = \(\frac { 2 }{ 3 } <1\)
Steps of Construction:
1. Draw a line segment Be = 12 cm.
2. From B, draw a line AB = 5 cm, which makes right angle at B.
3. Join AC. Thus, \(\Delta ABC\) is the required right angled triangle.
4. From B, draw an acute \(\angle CBY\)  downwards.
5. On ray BY, mark three points B₁ , B₂ and B₃ such that BB₁ = B₁B₂ = B₂B₃
6. Join B₃C,
7. From point B₂, draw B₂N || B₃C intersecting BC at N.
8. From point N, draw NM II CA intersecting BA at M.Thus \(\Delta MBN\) is the required triangle.
Hence, \(\Delta MBN\) is also a right angled triangle, right-angled at B.

Rs.230.12

30 hec

To draw a rhombus AB' C' D' similar to rhombus ABCD, we use the following steps of construction
1. First, draw a rhombus ABCD, in which AB = 4 cm and \(\angle ABC=60°\)  and join its diagonal ACwhich divides it into two triangles ABC and ADC
2. Now, construct MB' C' similar to MBC with scale factor 2/3
(i) First, draw a ray AX making an acute angle with AB downwards.
(ii) Mark 3 points A₁ A₂  and A₃ on AX such that
 AA₁ = A₁A₂ = A₂A₃
(iii) Join A₃B and then draw A ₂B' II A₃B which intersects AB at B'.
(iv) Draw B^' C^' parallel to BC which intersects AC at C^' . Thus,\(\Delta A{ B }^{ ' }{ C }^{ ' }\) similar, to \(\Delta ABC\) is constructed.
3. Draw the line segment D^'C^' II DC. Thus, AB' C' D' is the required rhombus similar to given rhombus ABCD.
   

Given, S₅ + S₇ = 167 and S₁₀ = 235
\(\Rightarrow 12 a+31 d=167 \text { and } 2 a+9 d=47\)
Now, find a and d, and then S₂₀
S₂₀ = 970

Let  P(9a-2, -b) divides AB internally in the ratio 3: 1.
\(\therefore \) By section formula, we get 9a-2 = \(\frac { 3(8a)+1(3a+1) }{ 3+1 } \) and -b = \(\frac { 3(5)+1(-3) }{ 3+1 } \) [\(\because \) for internally ratio, coordinates are \(\left( \frac { { m }_{ 1 }{ x }_{ 2 }+{ m }_{ 2 }{ x }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } ,\frac { { m }_{ 1 }{ y }_{ 2 }+{ m }_{ 2 }{ y }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) \)]
\(\Rightarrow 9a-2=\frac { 24a+3a+1 }{ 4 } \) and \(-b=\frac { 15-3 }{ 4 } \)
\(\Rightarrow 9a-2=\frac { 27a+1 }{ 4 }\)  and \(-b=\frac { 12 }{ 4 } \)
\(\Rightarrow \) 36a-8 = 27a+1 and b = -3
\(\Rightarrow \)  9a-9 =0  and b=-3
\(\Rightarrow \)  a=1   and  b=-3
Now,\({ a }^{ 2 }+{ b }^{ 2 }={ (1) }^{ 2 }+{ (-3) }^{ 2 }\) = 1+9 = 10 units.

k=0 and 8

Volume of the wood in the entire stand =Volume of cuboid - 4 x Volume of each cone - Volume of a cube
170.8 cm³

Let her pocket money be Rs x. Now, she takes Rs 1 on day 1, Rs 2 on day 2, Rs 3 on day 3 and so on till the end of the month, from this money.
i.e.  1 + 2 + 3 + 4 + ..... + 31
which form an AP, in which number of terms is 31 and first term 
\(\therefore\) Sum of first terms = S ₃₁
Sum of n terms,  \({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
\(\therefore\)  \({ S }_{ 31 }=\frac { 31 }{ 2 } \left[ 2\times 1+\left( 31-1 \right) \times 1 \right] \)
\(=\frac { 31 }{ 2 } \left( 2+30 \right) \)
\(=\frac { 31\times 32 }{ 2 } \)
\(=31\times 16=496\)
So, Kanika takes Rs 496 till the end of the month from this money.
Also, she spent Rs 204 of her pocket money and found that at the end of the month, she still has Rs 100 with her.
Now, according to the condition
\(\left( x-496 \right) -204=100\)
\(\Rightarrow\) x - 700 = 100
\(\Rightarrow\)   x = Rs 800
Hence, Rs 800 was her pocket money for the month.

Total number of bulbs=24
Number of defective buls=6
Number of good ones=24-6=18
P (not defective)=\({18\over24}={3\over4}\)
P (2nd bulb is defective) =\(5\over23\)
[first bulb is defective and not replaced]

Let the first three terms of an A.P. be a - d, a, a + d
\(\therefore\)  a - d + a + a + d = 33
\(\Rightarrow\)     3a = 33
a = 11
Now, according to the given condition|
( a - d ) ( a + d ) = a + 29
( 11 - d ) ( 11 + d ) = 11 + 29
\(\Rightarrow\)     121 - d1 = 40
\(\Rightarrow\)    d² = 81
\(\Rightarrow\)  d = \(\pm\) 9
\(\therefore\)  The required A.P is 2, 11, 20, ... or 20, 11, 2, ...

Possible outcomes 
(0,0),(0,1),(0,1),(0,1),(0,6),(0,6)
(1,0),(1,1),(1,1),(1,1),(1,6),(1,6)
(1,0),(1,1),(1,1),(1,1),(1,6),(1,6)
(1,0),(1,1),(1,1),(1,1),(1,6),(1,6)​​​​​​​
(6,0),(6,1),(6,1),(6,1),(6,6),(6,6)
(6,0),(6,1),(6,1),(6,1),(6,6),(6,6)​​​​​​​
Different total scores are 0, 1, or 12
Let A = getting a total of 7
No. of favourable outcomes are = 12
\(\therefore\) P(A) = \(\frac{12}{36}=\frac{1}{3}\)​​​​​​​

Total number of slips = 25 + 50 = 75
Number of slips marked with Rs 1 = 19 + 45 = 64
\(\therefore\) Number of slips marked other than 1 = 75 - 64 = 11
\(\therefore\) Required probability  = \(\frac{11}{75}\)

PA= 10 cm. 
PA = PB [If P is external point]   .....(i) 
[ ∵ From an external point tangents drawn to a circle are equal in length]
If C is external point, then CA = CE 
If D is external point, then 
DB = DE       ......(ii)
Perimeter of triangle \(\triangle\)PCD
= PC + CD + PD 
= PC + CE + ED + PD 
= pc + CA + DB + PD 
= PA + PB 
-PA + PA = 2 PA 
= 2 x 10 = 20 cm.[From (ii)]

\(\angle \)PCA = 110°
\(\angle \)CBA = ?

\(\angle \)ACB = 90°e   [Angle in a semicircle is right agle]
\(\angle \)PCB = 110° - 90° = 20°
\(\angle \)PCB = \(\angle \)CAB = 20° [Alternate segment theorm]
In \(\triangle\)ABC
\(\angle \)ABC + \(\angle \)CAB + \(\angle \)BCA = 180°   [∵ Sum of angles of a is 180°]
⇒ \(\angle \)ABC = 180° - 110° = 70°