CAT Geometry Questions

Upon drawing a rough sketch of the coordinates given, we realise that this is a right-angled triangle.

The three side lengths are 6, 8 and 10 units

The inradius of a circle can be calculated using the formula: $$rs=Area$$, where s is the semi-perimeter of the triangle 

The Area would be $$\frac{1}{2}\times\ 6\times\ 8=24$$
and the semi-perimeter would be $$\frac{10+6+8}{2}=12$$

Giving the inradius to be $$\frac{24}{12}=2$$ units

Therefore, 2 is the correct answer. 

This is the situation that is described in the question above, Angle AOB is 120 degrees and the chord AB is of length 10cm. 

Using Cosine rule we can find the length of AO and AB
$$\cos\left(AOB\right)=\frac{\left(r^2+r^2-300\right)}{2r^2}$$
$$-\frac{1}{2}=\frac{\left(2r^2-300\right)}{2r^2}$$
$$3r^2=300$$
$$r=10$$

Area of the Sector AOB is $$\frac{120}{360}\times\ \pi\ \times\ r^2$$
$$\frac{1}{3}\times\ \pi\ \times\ \frac{100}{1}$$ which is $$\frac{100\pi}{3}\ $$

Area of triangle AOB is $$\frac{1}{2}\times\ r^2\times\ \sin\left(120\right)$$

$$\frac{1}{2}\times\ \frac{100}{1}\times\ \frac{\sqrt{3}}{2}$$

Area of triangle AOB is $$\frac{75}{\sqrt{3}}$$

The smaller region will be, $$\frac{100\pi\ }{3}-\frac{75}{\sqrt{3}}$$
Taking 25 common we will get, 
$$25\left(\cfrac{4 \pi}{3} - \sqrt{3}\right)$$

Let's take the radius of the original circles to be R and that of the circle in between the three circles to be r. 

Joining the centres of the three circles, we will get an equilateral triangle or length 2R. 
The distance between the circle's centre and the original circle's centre would be R+r.

Using this right angle triangle, we can get the relation: $$\frac{R}{R+r}=\frac{\sqrt{\ 3}}{2}$$
We can take $$R=\sqrt{\ 3}a$$ and R+r as 2a, this would give us r as $$R=\left(2-\sqrt{\ 3}\right)a$$

The outer circle will have a radius of 2R+r
We need to find the ratio of $$\frac{2R+r}{r}$$

This will be equal to $$\frac{\left(2\sqrt{\ 3}+2-\sqrt{\ 3}\right)a}{\left(2-\sqrt{\ 3}\right)a}=\frac{2+\sqrt{\ 3}}{2-\sqrt{\ 3}}=4+3+4\sqrt{\ 3}=7+4\sqrt{\ 3}:1$$

Therefore, Option A is the correct answer. 

The simplest way to visualise this would be with symmetry. 

We can take AD to be x and CB to be 3-x; since the perimeter of ABCD is 6, and AB and CD are given as 2 and 1 cm, respectively, that leaves AD+BC as three only. 

We can see that triangles AEB and DEC are similar, with the lengths of AB being double of CD, essentially making D the mid-point of AE and C the mid-point of EB. 

Through this, we get the length of DE and CE to be x and 3-x. 
Give us the AE and BE lengths as 2x and 6-2x, respectively. 

 Giving the perimeter of AEB as 2x+6-2x+2 = 8 cm

Therefore, Option A is the correct answer. 

Drawing the figure described in the question, we get the following representation, 

Since E is the midpoint of CD, the length of each half will be 28cm. 

We need to find the incircle of the triangle ADE, which is a right angled triangle, we can do that with the formula, 
$$\dfrac{\left(a+b-h\right)}{2}$$

$$h^2=28^2+45^2$$
$$h^2=784+2025=2809$$
$$h=53$$

$$r=\dfrac{\left(28+45-53\right)}{2}=10$$

This is the figure in the question, 

We are given that each side is 6cm long, 
To find the side AC, we can use cosine rule, since we know each interior angle of the octagon(which is 135 degrees)

$$\cos\left(\angle ABC\right)=\dfrac{\left(AB^2+BC^2-AC^2\right)}{2\left(AB\right)\left(AC\right)}$$

$$\cos\left(135\right)=\dfrac{\left(36+36-AC^2\right)}{2\left(36\right)}$$

$$-\dfrac{1}{\sqrt{2}}=\dfrac{\left(72-AC^2\right)}{72}$$

$$AC^2=72+36\sqrt{2}$$

$$AC^2=36\left(2+\sqrt{2}\right)$$

Since AC is the side of the square, and the area of a square is square of the side. 

Answer is $$36\left(2+\sqrt{2}\right)$$

The question describes a figure of the following nature, 

We are told that the area of the triangle ABC is 1440. 
By proportionality theorem, the area of MPN should be one fourth of that. 
Since we know that, $$\frac{\left(Area\ \triangle_1\right)}{Area\ \triangle_2}=\frac{\left(side\ of\ \triangle_1\right)^2}{\left(side\ of\ \triangle_2\right)^2}$$

And since it is given that MPN are midpoints of the respective sides, area of triangle MPN is $$\triangle\ \frac{ABC}{4}=\frac{1440}{4}=360$$

Now, the medians from each side intersect, MP MN and NP are X Y and Z respectively, since these sides are proportional to the main triangle, the point of intersection of the medians with these sides will divide the side MP, MN and NP in half. 

Then, Area of Triangle XYZ will be one fourth the area MPN, 
$$\triangle\ \frac{MPN}{4}=\frac{360}{4}=90$$

The answer is 90. 

Given that, The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm
So, $$2(lb+bh+hl)=846$$. 

And $$4(l+b+h)=144$$
$$(l+b+h)=36$$
$$\left(l+b+h\right)^2=l^2+b^2+h^2+2\left(lb+bh+hl\right)$$
$$1296=\left(l^2+b^2+h^2\right)+846$$
$$450=l^2+b^2+h^2$$

We are told that this cuboid is inscribed in a sphere, the body diagonal of the cuboid equals the diameter of the sphere, this can be visualised as:

This is nothing but, $$\sqrt{l^2+b^2+h^2}=2R$$
$$l^2+b^2+h^2=4R^2$$
$$450=4R^2$$
$$R^2=\frac{225}{2}$$
$$R=\frac{15}{\sqrt{2}}$$

Volume of sphere will be $$\dfrac{4}{3}\times\ \pi\ \times\ \left(\dfrac{15}{\sqrt{2}}\right)^3$$

$$\dfrac{4}{3}\pi\ \left(\dfrac{3375}{2\sqrt{\ 2}}\right)$$

$$\pi\ \times\ 1125\sqrt{\ 2}$$

Since BC is the diameter of the circle, which implies angle BAC is 90 degrees. Let AB = a cm, which implies AC = b cm. Hence, $$BC\ =\ \sqrt{\ a^2+b^2}$$, which is diameter of the circle (2r).

Hence, $$2r\ =\sqrt{\ a^2+b^2}$$

=> $$4r^2=a^2+b^2$$

The area of the triangle is $$\frac{1}{2}\times\ a\times\ b$$, which can be written as 

=> $$\ \frac{\ a\cdot b}{2\left(a^2+b^2\right)}\times\ \left(a^2+b^2\right)$$

=> $$\frac{\ a\cdot b}{2\left(a^2+b^2\right)}\times\ 4r^2$$

=> $$\frac{\ a\cdot b}{a^2+b^2}\times\ 2r^2$$

The correct option is B

Given equation of circle = $$x^{2} + y^{2} + 4x - 6y - 3 = 0$$

Center of the circle is (-2,3) and radius of the circle = $$\sqrt{\ g^2+f^2-c}=\sqrt{\ 4+9+3}=4$$

Let us assume the point of the intersection of the tangents is ( h,k)

The angle made by the line joining (h,k) to the centre makes an angle of 30 degrees with the tangent, and sin(30) will be the ratio of the radius and the distance between the center and (h,k)

=> $$\sin\left(30\right)=\dfrac{4}{\sqrt{\left(\ h+2\right)^2+\left(k-3\right)^2}}$$

Squaring on both sides:

$$\dfrac{1}{4}=\dfrac{16}{\left(h+2\right)^2+\left(k-3\right)^2}$$

=> $$\left(h+2\right)^2+\left(k-3\right)^2=64$$

When x = 6 => h = 6 => $$64+\left(k-3\right)^2=64$$ => k = 3.

=> required point is (6,3)

Given that AB = AC => Angle C = Angle B (1)

AD and BE are altitudes => they make 90 degrees with the sides

Angle AOB = 105 => Angle EOD = 105 (Vertically Opposite Angles)

In quadrilateral DOEC

Angle C = 360 - 105 - 90 - 90 = 75 => Angle B = 75 (from 1)

We know that from the area of the triangle AD * BC = BE * AC

=> $$\dfrac{AD}{BE}=\dfrac{AC}{BC}=\dfrac{2R\sin B}{2R\sin A}=\dfrac{\sin\left(75\right)}{\sin\left(30\right)}=2\sin\left(75\right)$$ = 2cos(15)

[sin(x) = cos(90-x)]

Given ABCD is a cyclic quadrilateral.

Angle ADB = Angle ACB (Angle subtended by chord on the same side of arc)

Angle DAC = Angle DBC (Angle subtended by chord on the same side of arc)

=> Triangles AED and BEC are similar triangles

Similarly triangles AEB and DEC are also similar using AA similarity property.

Now, given that AB : CD = 2 : 1 and BC : AD = 5 : 4

AE/BE = AD/BC = 4/5 (Similar Triangles AED and BEC)

BE/CE = AB/CD = 2/1 (Similar Triangles AEB and DEC)

Multiplying both, we get AE/CE = 8/5.

Let us assume the length of the rectangle is 'l' and breadth of the rectangle is 'b'.

The radius, l/2 and b in the above diagram form a right-angled triangle.

=> $$\left(\frac{l}{2}\right)^2+b^2=2^2$$

We know that the area of the rectangle is l*b, which can be obtained by considering 2 times the geometric mean of $$\left(\frac{l}{2}\right)^2$$ and $$b^2$$.

Therefore, for the maximum area, the equality condition of AM-GM inequality should be satisfied

=> $$\left(\frac{l}{2}\right)^2=b^2$$ => l = 2b.
=> l/b = 2/1.

It is given that AB = 9 cm, BC = 6 cm.

It is also known that the areas of the figures ABP, APQ, and AQCD are in geometric progression.

Hence, the area of the ABP, APQ, and AQCD are k, 2k, and 4k respectively.

The ratio of BP, PQ, QC will be the ratio of the respective triangles. Hence, we can draw a line from point A to point C.

Let the area of triangle AQC be x, which implies the area of triangle ADC = ADQC - AQC = 4k -x, which is equal to the sum of the area of triangle APB, AQP, and ACQ, respectively.

Therefore, 4k-x = 3k+x 

=> x = k/2

Hence the ratio of BP: PQ: CQ = k:2k: k/2 =2:4:1

The sum of the interior angles of a polygon of 'n' sides is given by $$\left(2n-4\right)\times\ 90$$, and the sum of the exterior angles of a polygon is 360 degrees.

So, the difference between them will be 120 * n

=> $$\left(2n-4\right)90-360=120n$$

=> 60n = 720 => n = 12.

We know that the number of diagonals of a regular polygon is nC2 - n = 12C2 - 12 = 66 - 12 = 54.

Given that ABC is a right-angled triangle with AB = 5 and BC = 12 => Area of the triangle = 0.5 * 5 * 12 = 30.

Let us assume BP = p, BQ = q

=> Area of ABP = 0.5 * 5 * p = 2.5p

=> Area of ABQ = 0.5 * 5 * q = 2.5q

Given the area of ABC is 1.5 times that of ABP

=> 30 = 1.5 * 2.5p => 20 = 2.5p => p = 8.

Given Areas of ABP, ABQ and ABC are in A.P. => 2 * 2.5q = 2.5 * 8 + 30 => 5q = 50 => q = 10.

PQ = BQ - BP = q - p = 10 - 8 = 2.

It is given, Angle BAE = 45 degrees

This implies AE = BE

Let AE = BE = x 

In right-angled triangle ABD, it is given $$\angle ABC=\theta\ $$

$$\sin\theta=\dfrac{AD}{AB}\ $$

$$\sin\theta=\dfrac{AD}{x\sqrt{\ 2}}\ $$

$$\sqrt{\ 2}\sin\theta=\dfrac{AD}{BE}\ $$

The answer is option D.

Area of ABD : Area of BDC = 1:1

Therefore, area of ABD = 54

Area of ADE : Area of EDB = 1:1

Therefore, area of ADE = 27

O is the centroid and it divides the medians in the ratio of 2:1

Area of BEO : Area of EOD = 2:1

Area of EOD = 9

CD = $$\sqrt{\ y^2+25}$$

$$11+y+\sqrt{y^2+25}=36$$

$$\sqrt{y^2+25}=25-y$$

$$y^2+25=25^2+y^2-50y$$

2y = 24

y = 12

Area of trapezium = $$3y+\frac{5y}{2}=\frac{11y}{2}=\frac{11}{2}\left(12\right)=66$$

Let the number of sides of polygons A and B be n and 2n, respectively.

$$\ \dfrac{\ \ \dfrac{\ \left(n-2\right)180}{n}}{\ \dfrac{\ \left(2n-2\right)180}{2n}}=\dfrac{3}{4}$$

$$\ \dfrac{\ \ \ n-2}{\ \ n-1}=\dfrac{3}{4}$$

$$\ \ \ \ 4n-8=3n-3$$

$$n=5$$

The number of sides of polygon B is 2*5, i.e. 10.

Sum of the three sides of a quadrilateral is greater than the fourth side.

Therefore, let the fourth side be

1+2+4>d or d<7

1+2+d>4 or d>1

Possible values of d are 2, 3, 4, 5 and 6.

$$A=lb$$

$$l^2+b^2=4r^2$$

$$P\ =2\left(l+b\right)$$

$$\frac{P}{2}=l+b$$

Squaring on both the sides, we get

$$\frac{P^2}{4}=l^2+b^2+2lb$$

$$\frac{P^2}{4}=4r^2+2lb$$

$$\frac{P^2}{8}-2r^2=lb$$

The answer is option B.

BC is the diameter of circle C2 so we can say that $$\angle BAC=90^{\circ\ }$$ as angle in the semi circle is $$90^{\circ\ }$$

Therefore overlapping area = $$\frac{1}{2}$$(Area of circle C2) + Area of the minor sector made be BC in C1

AB= AC = 8 cm and as $$\angle BAC=90^{\circ\ }$$, so we can conclude that BC= $$8\sqrt{2}$$ cm

Radius of C2 = Half of length of BC = $$4\sqrt{2}$$ cm

Area of C2 = $$\pi\left(4\sqrt{2}\right)^2=32\pi$$ $$cm^2$$

A is the centre of C1 and C1 passes through B, so AB is the radius of C1 and is equal to 8 cm

Area of the minor sector made be BC in C1 = $$\frac{1}{4}$$(Area of circle C1) - Area of triangle ABC = $$\frac{1}{4}\pi\left(8\right)^2-\left(\frac{1}{2}\times8\times8\right)=16\pi-32$$ $$cm^2$$

Therefore,

Overlapping area between the two circles= $$\frac{1}{2}$$(Area of circle C2) + Area of the minor sector made be BC in C1

= $$\frac{1}{2}\left(32\pi\right)\ +\left(16\pi-32\right)=32\left(\pi-1\right)$$ $$cm^2$$

In a parallelogram, two diagonals of parallelogram bisects each other, which concludes that mid-point of both diagonals are the same.

Midpoint of AC = $$\left(\ \frac{\ 1-2}{2},\ \frac{\ 1+8}{2}\right)$$

Let the coordinates of vertex D be (x,y)

$$\left(\ \frac{\ x+3}{2},\ \frac{\ y+4}{2}\right)=\left(\ \frac{\ 1-2}{2},\ \frac{\ 1+8}{2}\right)$$

x = -4 and y = 5

The answer is option D.

Area of triangle ABD is twice the area of triangle ACD

$$\angle\ ADB=\theta\ $$

$$\dfrac{1}{2}\left(AD\right)\left(BD\right)\sin\theta\ =2\left(\dfrac{1}{2}\left(AD\left(CD\right)\sin\left(180-\theta\ \right)\right)\right)$$

$$BD\ =2CD$$

Therefore, BD = 2 and CD = 1

$$\angle\ ABC=\angle\ ACB=60^{\circ\ }$$

Applying cosine rule in triangle ADC, we get

$$\cos\angle\ ACD=\ \dfrac{\ AC^2+CD^2-AD^2}{2\left(AC\right)\left(CD\right)}$$

$$\dfrac{1}{2}=\ \dfrac{\ 9+1-AD^2}{6}$$

$$AD^2=7$$

$$AD=\sqrt{\ 7}$$

The answer is option C.

Area of a regular hexagon = $$\frac{3\sqrt{3}}{2}x^2$$

Area of an equilateral triangle = $$\frac{\sqrt{3}}{4}\left(a\right)^2$$ ; where a = side of the triangle

Since the area of the two figures are equal, we can equate them as folllows: $$\frac{3\sqrt{3}}{2}x^2=\frac{\sqrt{3}}{4}\left(12\right)^2$$

On simplifying: $$x^2=24\ $$

$$\therefore\ x=2\sqrt{6}$$

Since a regular hexagon can be considered to be made up of 6 equilateral triangles, a line joining the farthest vertices of a hexagon can be considered to be made up using the sides of two opposite equilateral triangle forming the hexagon. Hence, its length should be twice the side of the hexagon, in this case, 4 cm.

Now, AD divided the hexagon into two symmetrical halves. Hence, AD bisects angle D, and hence, angle ADC is $$60^{\circ\ }$$.

We can find out the value of AT using cosine formula:

$$AT^2=\ 4^2+1^2-2\times\ 1\times\ 4\cos60$$

$$AT^2=\ 17-4=13$$

$$AT=\sqrt{\ 13}$$

All the sides of the rhombus are equal.

The area of a rhombus is $$12\ cm^2$$

Considering d1 to be the length of the longer diagonal, d2 to be the length of the shorter diagonal.

The area of a rhombus is $$\left(\frac{1}{2}\right)\left(d1\right)\cdot\left(d2\right)\ =\ 12$$

d1*d2 = 24.

The length of the side of a rhombus is given by $$\frac{\sqrt{\ d1^2+d2^2}}{2}$$. This is because the two diagonals and a side from a right-angled triangle with sides d1/2, d2/2 and the side length.

$$\frac{\sqrt{\ d1^2+d2^2}}{2}=\ 5$$

Hence $$\sqrt{\ d1^2+d2^2}\ =\ 10$$

$$d1^2+d2^2\ =\ 100$$

Using d1*d2 = 24, 2*d1*d2 = 48.

$$d1^2+d2^2\ +2\cdot d1\cdot d2=\ 100+48\ =\ 148$$

$$d1^2+d2^2\ -2\cdot d1\cdot d2=\ 100-48\ =\ 52$$

$$d1+d2\ =\ \sqrt{\ 148}$$  (1)

d1-d2 = $$\sqrt{52}$$   (2)

(1) + (2)= 2*(d1) = 2*($$\sqrt{\ 37}+\sqrt{\ 13}$$)

d1 = $$\sqrt{\ 37}+\sqrt{\ 13}$$

or 

In a rhombus the area of a Rhombus is given by :

The diagonals perpendicularly bisect each other. Considering the length of the diagonal to be 2a, 2b.

The area of a Rhombus is : $$\left(\frac{1}{2}\right)\cdot\left(2a\right)\cdot\left(2b\right)\ =\ 12$$

ab =6.

The length of each side is : $$\sqrt{\ a^2+b^2}$$ = 5, $$a^2+b^{2\ }=\ 25,\ $$

$$\left(a+b\right)^2=\ 37,\ \left(a+b\right)\ =\ \sqrt{\ 37}$$

($$\left(a-b\right)^2=\ 13,\ a-b\ =\ \sqrt{\ 13}$$

$$2a\ =\ \left(\sqrt{\ 37}+\sqrt{\ 13}\right)$$,  $$2b\ =\ \left(\sqrt{\ 37}-\sqrt{\ 13}\right)$$.

2a is longer diagonal which is equal to $$\ \left(\sqrt{\ 37}+\sqrt{\ 13}\right)$$

Let us draw the rectangle.

Now, definitely, three sides should be fenced at Rs 100/ft, and one side should be fenced at Rs 200/ft.

In this question, we are going to assume that the L is greater than B.

Hence, the one side painted at Rs 200/ft should be B to minimise costs.

Hence, the total cost = 200B + 100B + 100L + 100L = 300B + 200L

Now, L x B = 60000

B = 60000/L

Hence, total cost = 300B + 200L = 18000000/L + 200L

To minimise this cost, we can use AM>=GM,

$$\frac{\frac{18000000}{L}+200L}{2}\ge\sqrt{\ \frac{18000000}{L}\times\ 200L}$$

$$\frac{18000000}{L}+200L\ge2\sqrt{\ 18000000\times\ 200}$$

$$\frac{18000000}{L}+200L\ge2\times\ 60000$$

Hence, minimum cost = Rs 120000.

We can say 40m is the perimeter of the park 
so side of rhombus = 10 
Now $$\frac{1}{2}\times\ d_1\times\ d_2\ =\ 96$$
so we get $$\ d_1\times\ d_2\ =\ 192$$   (1)
And as we know diagonals of a rhombus are perpendicular bisectors of each other :
so $$\ \frac{d_1^2}{4}+\ \frac{d_2^2}{4}=\ 100$$
so we get $$\ d_1^2+\ d_2^2=\ 400$$    (2)
Solving (1) and (2)
We get $$d_1=12$$ and$$d_2=16$$
Now the cost, in INR, of laying electric wires along its two diagonals, at the rate of ₹125 per m, is= (12+16)(125) =3500

We are given that :

Let DP =x 
So AB =x 
Now DP=CP 
So CD = 2x 
Now let height of trapezium be h
we can say A(Parallelogram ABPD ) = xh
And A (BPC) =$$\frac{1}{2}xh$$
Now by condition$$xh-\frac{1}{2}xh=10$$
$$\frac{xh}{2}=10$$
so xh =20
Now therefore area of trapezium ABCD = $$\frac{1}{2}\left(x+2x\right)h\ =\ \frac{3}{2}xh\ =\ 30$$

We know that Inradius=$$\frac{\left(Perpendicular+Base-Hypotenuse\right)}{2}$$

$$4=\frac{\left(p+10-h\right)}{2}$$

h-p = 2 or h= p+2.

Now, $$p^2+100\ =\ h^2$$

$$p^2+100\ =\ \left(p+2\right)^2$$

$$p^2+100\ =\ p^2+4p+4$$

4p = 96

p=24.

Hence, Area = $$\frac{1}{2}\times\ 10\times\ 24\ =\ 120$$.

We need to find out p.

Angle ADE = 180 - 2x

Angle BDF = 180 - 2y

Now, 180 - 2y + p + 180 - 2x = 180 [Straight line = 180 deg]

p = 2x + 2y - 180

Also, x + y + 50 = 180 [Sum of the angles of triangle = 180]

x + y = 130

p = 260 - 180 = 80 degrees.

We have :

Now area of ADE = $$\frac{1}{2}\times\ AD\times\ AE\times\ \sin A$$
=$$\frac{1}{2}\times\ 2x\times\ 2y\times\ \sin A=8$$
we get xy sinA =4
Now Area of triangle ABC = $$\frac{1}{2}AB\times\ AC\times\ \sin A$$
we get $$\frac{1}{2}\times3x\ \times5y\ \sin A=\frac{15}{2}xy\ \sin A=\ \frac{15}{2}\times\ 4$$
we get Area of ABC = 30

Applying cosine rule in triangle ACD, 

$$100+X^2-2\times\ 10\times\ X\cos30=400$$

$$X^2-10X\sqrt{\ 3}-300=0$$

Solving, we get X = $$\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)$$

Hence, area = 10Xsin 30 = $$\frac{\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)10}{2}$$

= $$25(\sqrt{3}+\sqrt{15})$$

We can form the following figure based on the given information:

Since OA = 4 m and OB=3 m; AB = 5 m. OR bisects the chord into PC and QC. 

Since AB = 5 m, we have $$a+b = 5      ...(i)$$  Also, $$4^2\ -k^2=a^2...\left(ii\right)$$ and $$3^2\ -k^2=b^2...\left(iii\right)$$

Subtracting (iii) from (ii), we get: $$a^2\ -b^2=7...\left(iv\right)$$

Substituting (i) in (iv), we get $$a - b = 1.4      ...(v)$$; $$\left[\left(a+b\right)\left(a\ -b\right)=7;\ \therefore\ \left(a-b\right)=\frac{7}{5}\right]$$ 

Solving (i) and (v), we obtain the value of $$a=3.2$$ and $$b=1.8$$

Hence, $$k^2\ =\ 5.76$$

Moving on to the larger triangle $$\triangle\ POC$$, we have $$5^2-k^2=\left(x+a\right)^2$$; 

Substituting the previous values, we get: $$(25-5.76)=\left(x+3.2\right)^2$$ 

$$\sqrt{19.24}=\left(x+3.2\right)$$ or $$x = 1.19 m$$

Similarly, solving for y using $$\triangle\ QOC$$, we get $$y=2.59 m$$

Therefore, $$PQ = 5+2.59+1.19 = 8.78 \approx\ 8.8 m$$

Hence, Option A is the correct answer.

Let the base radius be 3r.

Height of upper cone is 9 so, by symmetry radius of upper cone will be r.

Volume of frustum=$$\frac{\pi}{3}\left(9r^2\cdot27-r^2.9\right)$$

Volume of upper cone = $$\frac{\pi}{3}.r^2.9$$

Difference= $$\frac{\pi}{3}\cdot9\cdot r^2\cdot25=225$$ => $$\frac{\pi}{3}\cdot r^2=1$$

Volume of larger cone = $$\frac{\pi}{3}\cdot9r^2\cdot27=243$$

Let ABCD be the rectangle with length 2l and breadth 2b respectively.

Area of the circle i.e. area of painted region = $$\pi\ b^2$$.

Given, 4lb-$$\pi\ b^2$$=(2/3)$$\pi\ b^2$$.

=> 4lb=(5/3)$$\pi\ b^2$$.

=>l=$$\frac{5\pi}{12}b$$.

Given, 4lb=135 => 4*$$\frac{5\pi}{12}b^2$$=135 => b= $$\frac{9}{\sqrt{\ \pi\ }}$$

=> l=$$\frac{15}{4}\sqrt{\ \pi\ }$$
Perimeter of rectangle =4(l+b)=4($$\frac{15}{4}\sqrt{\ \pi\ }$$+$$\frac{9}{\sqrt{\ \pi\ }}$$ )=$$3\sqrt{\pi}(5+\frac{12}{\pi})$$.

Hence option A is correct. 

Let the sides of the rectangle be "a" and "3a" m. Hence the perimeter of the rectangle is 8a.

Let the side of the equilateral triangle be "m" cm. Hence the perimeter of the equilateral triangle is "3m" cm. Now we know that 8a+3m=90......(1)

Moreover area of the equilateral triangle is $$\frac{\sqrt{\ 3}}{4}m^2$$ and area of the rectangle is $$3a^2$$

According to the relation given $$\left(\frac{\sqrt{\ 3}}{4}m^2\right)^{^2}=\ 3a^2$$

$$\frac{3}{16}m^4=\ 3a^2\ or\ a^2=\frac{m^4}{16}$$

$$a=\frac{m^2}{4}$$  

Substituting this in (1) we get $$2m^2+3m-90\ =0$$ solving this we get m=6 (ignoring the negative value since side can't be negative)

Hence a=9 and the longer side of the rectangle will be 3a=27cm

Let the length of radius be 'r'.

From the above diagram,

$$x^2+r^2=6^2\ $$....(i)

$$\left(10-x\right)^2+r^2=8^2\ $$----(ii)

Subtracting (i) from (ii), we get: 

x=3.6 => $$r^2=36-\left(3.6\right)^2$$ ==> $$r^2=36-\left(3.6\right)^2\ =23.04$$.

Area of circle = $$\pi\ r^2=23.04\pi\ $$

Area of rhombus= 1/2*d1*d2=1/2*12*16=96.

.'. Ratio of areas = 23.04$$\pi\ $$/96=$$\frac{6\pi}{25}$$

Now we know that the perpendicular from the centre to a chord bisects the chord. Hence at the point of intersection of tangent, the chord will be divided into two parts of 3 cm each. As you can clearly see in the diagram, a right angled triangle is formed there.

Hence   $$\left(r+1\right)^2=r^2+9$$ or $$r^{2\ }+1+2r\ =\ r^2+9\ or\ 2r=8\ or\ r=4cm$$

Hence the radius of the larger circle is 5cm and diameter is 10cm.

The midpoints of two diagonals of a parallelogram are the same

Hence the midpoint of (2,1) and (-3,-4) lie on $$x+9y+c=0$$

midpoint of (2,1) and (-3,-4) = ($$\frac{2-3}{2},\frac{1-4}{2}$$) = (-1/2 , -3/2)

Keeping this cordinates in the above line equation, we get c = 14

Based on the question: AD, CE and BF are the three altitudes of the triangle. It has been stated that {GD+GE+GF = s}

Now since the triangle is equilateral, let the length of each side be "a". So area of triangle will be

$$\frac{1}{2}\times\ GD\times\ a+\ \frac{1}{2}\times\ GE\times\ a\ +\frac{1}{2}\times\ GF\times\ a=\frac{\sqrt{\ 3}}{4}a^2$$

Now $$GD+GE+GF=\frac{\sqrt{\ 3}a}{2}$$ or $$s=\frac{\sqrt{\ 3}a}{2}\ or\ a=\frac{2s}{\sqrt{\ 3}}$$

Given the area of the equilateral triangle = $$\ \frac{\sqrt{3}}{4}a^2\ $$ ; substituting the value of 'a' from above, we get the area {in terms 's'}= $$\frac{s^2}{\sqrt{3}}$$

Given, BC = DE = 4

CD = BE = 5

In triangle ADE, EAD=45^{0}$$

$$\tan\ 45\ =\ \frac{DE}{AE}$$ => AE = 4

Area of trapezium = Area of rectangle BCDE + Area of triangle AED

= 20 + 8 = 28

Equation of circle $$x^2+y^2+2gx+2fy+c=0$$

It passes through (0,0), (4,0) and (3,9). Substitute each point in the above equation:

=> On substituting the value (0,0) in the above equation, we obtain: $$c=0$$

=> On substituting the value (4,0) in the above equation, we obtain:  $$16+0+8g+0 = 0$$ ; $$g=-2$$

=> On substituting the value (3,9) in the above equation, we obtain: $$9+81-12+18f = 0$$ ; $$f= -13/3$$

Radius of the circle r = $$\sqrt{\ g^2+f^2-c}$$ => $$r^2=\frac{205}{9}$$

Therefore, Area = $$\pi\ r^2=\frac{205\pi\ }{9}$$

In any right triangle, the circumradius is half of the hypotenuse. Here,L=$$\ \frac{\ 1}{2}\ $$* the length of the hypotenuse = $$\ \frac{\ 1}{2}$$($$\sqrt{\ 15^2+9^2}$$) = $$\ \frac{\ 1}{2}\ $$*$$\sqrt{\ 306}$$ = $$\ \frac{\ 1}{\ 2}\times\ $$17.49 = 8.74

Hence, the integer close to L = 9

Since AB is a diameter,  AQB and APB will right angles.

In right triangle APB, AP = $$\sqrt{10^2-6^2}=8$$

Now, 2AQ=AP  => AQ= 8/2=4

In right triangle AQB, AP = $$\sqrt{10^2-4^2}=9.165$$ =9.1 (Approx)

Let 'h' be the height of the triangle ABC, semiperimeter(S) $$= \frac{4+4+r+4+4+r}{2} = 8+r$$, 
$$a=4+r, b=4+r, c=8$$

Area of triangle ABC $$=\ \ \sqrt{\ s\cdot\left(s-a\right)\left(s-b\right)\left(s-c\right)}=$$
$$= \sqrt{\left(\ 8+r\right)\times\ 4\times\ 4\times\ r}$$ = $$\ \frac{\ 1}{2}\times\ \left(4+4\right)\times\ height$$

Height (h) = $$\sqrt{\ \left(8+r\right)r}$$

Now, $$ h + r = 4 \longrightarrow  \sqrt{\ \left(8+r\right)r} + r = 4$$ (Considering the height of the triangle)

$$\sqrt{\ \left(8+r\right)r}$$=4-r

16r=16

r=1


Alternatively,

$$\text{AE}^@+\text{EC}^2=\text{AC}^2 \longrightarrow 4^2+\left(4-r\right)^2 = \left(4+r\right)^2 \longrightarrow\ \longrightarrow\ \longrightarrow\ r=1$$

It is given that the base of the pyramid is square and each of the four sides are equilateral triangles.

Length of each side of the equilateral triangle = 20cm

Since the side of the triangle will be common to the square as well, the side of the square = 20cm

Let h be the vertical height of the pyramid ie OA

OB = 10 since it is half the side of the square 

AB is the height of the equilateral triangle i.e 10$$\sqrt{\ 3}$$

AOB is a right angle, so applying the Pythagorean formula, we get

$$OA^2+ OB^2= AB^2$$

$$h^2$$ + 100= 300

h=10$$\sqrt{\ 2}$$

Let p be the length of AP.

It is given that $$\angle\ BAC\ =90\ $$ and $$\angle\ APC\ =90\ $$

Let $$\angle\ ABC\ =\theta\ $$, then $$\angle\ BAP\ =90-\theta\ $$ and $$\angle\ BCA\ =90-\theta\ $$

So $$\angle\ PAC\ =\theta\ $$

Triangles BPA and APC are similar

$$p^2=x\left(20-x\right)$$

We have to maximize the value of p, which will be maximum when x=20-x

x=10

It is given that the volume of all the cylinders is the same, so the volume of each cylinder = HCF of (405, 783, 351)

=27

The number of iron cylinders = $$\ \frac{\ 405}{27}$$ = 15

The number of aluminium cylinders = $$\ \frac{\ 783}{27}$$ = 29

The number of copper cylinders = $$\ \frac{351\ }{27}$$ = 13

15*$$\pi\ r^2h=$$ 405  

15*$$\pi\ 9*h=$$ 405

$$\pi\ h=3$$

Now we have to calculate the total surface area of all the cylinders 

Total number of cylinders = 15+29+13 = 57

Total surface area of the cylinder = 57*($$2\pi\ rh+2\pi\ r^2$$)

=57(2*3*3 + 2*9*$$\pi\ $$)

=$$1026(1 + \pi)$$

In figure AE*BE=CE*DE    (The   intersecting chords theorem)

=> 7*15= x(20.5-x)                  (Assuming AE=x)

=> 210=x(41-2x)

=> $$2x^2$$-41x+210=0

=> x=10 or x=10.5   => AE=10 or AE=10.5      Hence BE = 20.5-10=10.5  or BE = 20.5-10.5=10

Required difference= 10.5-10=0.5

It is given that AD and BE are medians which are perpendicular to each other.

The lengths of AD and BE are 12cm and 9cm respectively.

It is known that the centroid G divides the median in the ratio of 2:1

Area of $$\triangle$$ ABC = 2* Area of the triangle ABD

Area of $$\triangle\ $$ABD = Area of $$\triangle\ $$ AGB + Area of $$\triangle\ $$ BGD

Since $$\angle\ AGB\ =\ \angle\ BGD\ =\ 90$$ (Given)

Area of $$\triangle\ $$ AGB = $$\ \frac{\ 1}{2}\times\ 8\times\ 6$$ = 24

Area of $$\triangle\ $$ BGD = $$\ \frac{\ 1}{2}\times\ 6\times\ 4$$ = 12

 Area of $$\triangle\ $$ABD = 24+12=36

Area of $$\triangle\ ABC\ =\ 2\times\ 36=72$$

The given figure can be divided into 9 regions or equilateral triangles of equal areas as shown below,

Now the hexagon consists of 6 regions and the triangle consists of 9 regions.

Hence the ratio of areas = 6/9 =2:3

Each interior angle in an n-sided polygon = $$\frac{\left(n-2\right)180\ }{n}$$

It is given that each interior angle of B is $$\frac{3}{2}$$ times each interior angle of A and b = 2a

$$\frac{\left(b-2\right)180\ }{b}$$ = $$\ \frac{\ 3}{2}\times\ $$ $$\frac{\left(a-2\right)180\ }{a}$$

$$2\times\ \left(b-2\right)\times\ a\ =\ 3\times\ \left(a-2\right)\times\ b$$

2(ab-2a) = 3(ab-2b)

ab-6b+4a=0

a*2a-12a+4a=0

$$2a^2-8a=0$$

a(2a-8) = 0

a cannot be zero so 2a=8

a=4, b = 4*2=8

a+b = 12

Each interior angle of a regular polygon with 12 sides = $$\ \frac{\ \left(12-2\right)\times\ 180}{12}$$

=150

Assuming the dimensions of the brick are a, b and c and the diagonals are 3, 2 $$\surd3$$ and $$\surd{15}$$

Hence, $$a^{2\ }+\ b^2$$ = $$3^2$$   ......(1)

$$b^{2\ }+\ c^2$$ = $$(2\sqrt{3})^2$$ ......(2)

$$c^{2\ }+\ a^2$$ = $$(\sqrt{15})^2$$ ......(3)

Adding the three equations, 2($$a^2+b^2+c^2$$) = 9+12+15=36

=>$$a^2+b^2+c^2$$ = 18......(4)

Subtracting (1) from (4), we get $$c^2$$ = 9    =>c=3

Subtracting (2) from (4), we get $$a^2$$ = 6    =>a=$$\sqrt{6}$$

Subtracting (3) from (4), we get $$b^2$$ = 3    =>b=$$\sqrt{3}$$

The ratio of the length of the shortest edge of the brick to that of its longest edge is = $$\ \frac{\ \sqrt{3}}{3}$$ = $$1 : \sqrt{3}$$

Given that two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm.

In the diagram we can see that AB = 6 cm, CD = 4 cm and MN = 1 cm.

We can see that M and N are the mid points of AB and CD respectively. 

AM = 3 cm and CD = 2 cm. Let 'OM' be x cm. 

In right angle triangle AMO,

$$AO^2 = AM^2 + OM^2$$ 

$$\Rightarrow$$ $$AO^2 = 3^2 + x^2$$  ... (1)

In right angle triangle CNO,

$$CO^2 = CN^2 + ON^2$$ 

$$\Rightarrow$$ $$CO^2 = 2^2 + (OM+MN)^2$$ 

$$\Rightarrow$$ $$CO^2 = 2^2 + (x+1)^2$$  ... (2)

We know that both AO and CO are the radius of the circle. Hence $$AO = CO$$

Therefore, we can equate equation (1) and (2)

$$3^2+x^2$$=$$2^2+(x+1)^2$$

$$\Rightarrow$$ x = 2 cm

Therefore, the radius of the circle 

$$AO = \sqrt{AM^2 + OM^2}$$ 

$$\Rightarrow$$ $$AO=\sqrt{3^2+2^2}=\sqrt{13}$$. Hence, option A is the correct answer. 

Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?

We know that AC is the diameter and $$\angle$$ ABC = 90°. AC = 2*13 = 26 cm

In right angle triangle ABC,

$$AC^2 = AB^2 + BC^2$$

$$\Rightarrow$$ $$AB^2+BC^2=26^2$$

$$\Rightarrow$$ $$AB^2+BC^2=676$$

Let us check with the options.

Option (A): $$24^2+10^2 = 676$$. Hence, this is a possible answer. 

Option (B): $$25^2+9^2 = 706 \neq 676$$.  Hence, this is an incorrect pair.

Option (C): $$25^2+10^2 = 725 \neq 676$$.  Hence, this is an incorrect pair.

Option (D): $$24^2+12^2 = 720 \neq 676$$. Hence, this is an incorrect pair.

Therefore, we can say that option A is the correct answer.

It is given that EFGH is also a square whose area is 62.5% of that of ABCD. Let us assume that E divides AB in x : 1. Because of symmetry we can se that points F, G and H divide BC, CD and DA in x : 1.

Let us assume that 'x+1' is the length of side of square ABCD. 

Area of square ABCD = $$(x+1)^2$$ sq. units.

Therefore, area of square EFGH = $$\dfrac{62.5}{100}*(x+1)^2$$ = $$\dfrac{5(x+1)^2}{8}$$ ... (1)

In right angle triangle EBF,

$$EF^2 = EB^2 + BF^2$$

$$\Rightarrow$$ $$EF = \sqrt{1^2+x^2}$$

Therefore, the area of square EFGH = $$EF^2$$ = $$x^2+1$$ ... (2)

By equating (1) and (2),

$$x^2+1 = \dfrac{5(x+1)^2}{8}$$

$$\Rightarrow$$ $$8x^2+8 = 5x^2+10x+5$$

$$\Rightarrow$$ $$3x^2-10x+3 = 0$$

$$\Rightarrow$$ $$(x - 3)(3x - 1) = 0$$

$$\Rightarrow$$ $$x = 3$$ or $$1/3$$

The ratio of length of EB to that of CG = 1 : x 

EB : CG = 1 : 3 or 3 : 1. Hence, option D is the correct answer. 

We can see that area of parallelogram ABCD = 2*Area of triangle ACD 

48 = 2*Area of triangle ACD 

Area of triangle ACD = 24

$$(1/2)*CD*DA*sinADC=24$$

$$AD*sinADC=6$$

We know that $$sin\theta$$ $$\leq$$ 1, Hence, we can say that AD $$\geq$$ 6

$$\Rightarrow$$ s $$\geq$$ 6

We are given that AB = 5 cm and $$\angle$$ AOB = 60°

Let us draw OM such that OM $$\perp$$ AB. 

In right angle triangle AMO,

$$sin 30° = \dfrac{AM}{AO}$$

$$\Rightarrow$$ AO = 2*AM = 2*2.5 = 5 cm. Therefore, we can say that the radius of the circle = 5 cm. 

In right angle triangle PNO,

$$sin 60° = \dfrac{PN}{PO}$$

$$\Rightarrow$$ PN = $$\dfrac{\sqrt{3}}{2}$$*PO = $$\dfrac{5\sqrt{3}}{2}$$

Therefore, PQ = 2*PN = $$5\sqrt{3}$$ cm

We can see that T$$_{2}$$ is formed by using the mid points of T$$_{1}$$. Hence, we can say that area of triangle of T$$_{2}$$ will be (1/4)th of the area of triangle T$$_{1}$$.

Area of triangle T$$_{1}$$ = $$\dfrac{\sqrt{3}}{4}*(24)^2$$ = $$144\sqrt{3}$$ sq. cm

Area of triangle T$$_{2}$$ = $$\dfrac{144\sqrt{3}}{4}$$ = $$36\sqrt{3}$$ sq. cm 

Sum of the area of all triangles =  T$$_{1}$$ + T$$_{2}$$ + T$$_{3}$$ + ... 

$$\Rightarrow$$ T$$_{1}$$ + T$$_{1}$$/$$4$$ + T$$_{1}$$/$$4^2$$ + ... 

$$\Rightarrow$$ $$\dfrac{T_{1}}{1 - 0.25}$$

$$\Rightarrow$$ $$\dfrac{4}{3}*T_{1}$$

$$\Rightarrow$$ $$\dfrac{4}{3}*144\sqrt{3}$$

$$\Rightarrow$$ $$192\sqrt{3}$$

Hence, option D is the correct answer. 

We are given that diameter of base = 8 ft. Therefore, the radius of circular base = 8/2 = 4 ft 

In triangle OAB and OCD

$$\dfrac{OA}{AB} = \dfrac{OC}{CD}$$

$$\Rightarrow$$ AB = $$\dfrac{3*4}{12}$$ = 1 ft.

Therefore, the volume of remaining part = Volume of entire cone - Volume of smaller cone

$$\Rightarrow$$ $$\dfrac{1}{3}*\pi*4^2*12-\dfrac{1}{3}*\pi*1^2*3$$

$$\Rightarrow$$ $$\dfrac{1}{3}*\pi*189$$

$$\Rightarrow$$ $$\dfrac{22}{7*3}*189$$

$$\Rightarrow$$ $$198$$ cubic ft

Let 'a' and 'b' be the length of sides of the rectangle. (a > b)

Area of the rectangle = a*b

Perimeter of the rectangle = 2*(a+b)

$$\Rightarrow$$ $$\dfrac{a*b}{(2*(a+b))^2}=\dfrac{1}{25}$$

$$\Rightarrow$$ $$25ab=4(a+b)^2$$

$$\Rightarrow$$ $$4a^2-17ab+4b^2=0$$

$$\Rightarrow$$ $$(4a-b)(a-4b)=0$$

$$\Rightarrow$$ $$a = 4b$$ or $$\dfrac{b}{4}$$

We initially assumed that a > b, therefore a $$\neq$$ $$\dfrac{b}{4}$$.

Hence, a = 4b

$$\Rightarrow$$ b : a = 1 : 4

Let us draw the diagram according to the available information.

We can see that triangle BPC and BQC are inscribed inside a semicircle. Hence, we can say that

$$\angle$$ BPC = $$\angle$$ BQC = 90°

Therefore, we can say that BQ  $$\perp$$ AC and CP $$\perp$$ AB.

In triangle ABC,

Area of triangle = (1/2)*Base*Height = (1/2)*AB*CP = (1/2)*AC*BQ

$$\Rightarrow$$ BQ = $$\dfrac{AB*CP}{AC}$$ = $$\dfrac{30*20}{25}$$ = 24 cm.

We know that area of the triangle = 32 sq. units, BC = 8 units

Therefore, the height of the perpendicular drawn from point A to BC = 2*32/8 = 8 units.

Let us draw a possible diagram of the given triangle. 

We can see that if A coincide with (-4, 0) then the distance between A and (0, 0) = 4 units.

If we move the triangle up or down keeping the base BC on x = 4, then point A will move away from origin as vertical distance will come into factor whereas horizontal distance will remain as 4 units.

Hence, we can say that minimum distance between A and origin (0, 0) = 4 units. 

Area of the semicircle with AB as a diameter = $$\dfrac{1}{2}*\pi*(\dfrac{AB^2}{4})$$

$$\Rightarrow$$ $$\dfrac{1}{2}*\pi*(\dfrac{AB^2}{4})$$ = $$72*\pi$$

$$\Rightarrow$$ $$AB = 24 cm$$

Given that area of the rectangle ABCD = 768 sq.cm

$$\Rightarrow$$ AB*BC = 768

$$\Rightarrow$$ BC = 32 cm

We can see that the perimeter of the remaining shape = AD + DC + BC + Arc(AB)

$$\Rightarrow$$ 32+24+32+$$\pi*24/2$$

$$\Rightarrow$$ $$88+12\pi$$

In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is

Given that area of parallelogram = 72 sq cm

Area of triangle ABC = $$\dfrac{1}{2}$$*area of parallelogram

(1/2)*AB*BC*sinABC = $$\dfrac{1}{2}$$*72

sinABC = $$\dfrac{1}{2}$$

$$\angle$$ ABC = 30°

Let us draw a perpendicular CQ from C to AB. 

$$\angle\ B\ =\ \angle\ D\ =\ 30^{\circ\ }$$ ( opposite angles in a parallelogram)

$$\angle\ QCB\ =\ \angle\ DAP\ =\ 60^{\circ\ }$$  ( Sum of angles in a triangle = 180)

In triangles APD and CQB, AP= CQ , AD=CB ( opposite sides of a parallelogram) , $$\angle\ QCB\ =\ \angle\ DAP\ =\ 60^{\circ\ }$$

Hence APD and CQB are congruent triangles using SAS property.

Therefore, we can say that area of triangle APD = area of triangle CQB

In right angle triangle CQB,

QB = CBcos30° = $$16*\dfrac{\sqrt{3}}{2}$$ = $$8\sqrt{3}$$ cm

CQ = CBsin30° = $$16*\dfrac{1}{2}$$ = $$8$$ cm

Therefore, area of triangle CQB = 1/2*CQ*QB = $$1/2*8*8\sqrt{3}$$ = $$32\sqrt{3}$$

Hence, we can say that area of triangle APD = $$32\sqrt{3}$$.

It is given that radius of the circle = 1 cm

Chord AB subtends an angle of 60° on the centre of the given circle. R be the region bounded by the radii OA, OB and the arc AB.

Therefore, R = $$\dfrac{60°}{360°}$$*Area of the circle = $$\dfrac{1}{6}$$*$$\pi*(1)^2$$ = $$\dfrac{\pi}{6}$$ sq. cm

It is given that OC = OD and area of triangle OCD is half that of R. Let OC = OD = x.

Area of triangle COD = $$\dfrac{1}{2}*OC*OD*sin60°$$

$$\dfrac{\pi}{6*2}$$ = $$\dfrac{1}{2}*x*x*\dfrac{\sqrt{3}}{2}$$

$$\Rightarrow$$ $$x^2 = \dfrac{\pi}{3\sqrt{3}}$$

$$\Rightarrow$$ $$x$$ = $$(\frac{\pi}{3\sqrt{3}})^{\frac{1}{2}}$$ cm. Hence, option C is the correct answer. 

The length of the diagonals of a regular hexagon with side s are $$\sqrt{3}s$$.
Here length of AC = $$\sqrt{3}s$$ = $$\sqrt{3}$$ cms
Hence area of the square = $$\sqrt{3}^2$$ = 3 sq cm

The following equation will form a square of side $$ 2\sqrt{2} $$.

The area of the square = $$(2\sqrt{2})^2$$ = 8 units.

The lengths are given as 40, 25 and 35.

The perimeter = 100

Semi-perimeter, s = 50

Area = $$ \sqrt{50 * 10 * 25 * 15}$$ = $$250\sqrt{3}$$

The triangle formed by the centroid and two vertices is removed.

Since the cenroid divides the median in the ratio 2 : 1

The remaining area will be two-thirds the area of the original triangle.

Remaining area = $$\frac{2}{3} * 250\sqrt{3}$$ = $$\frac{500}{\sqrt{3}}$$

See the diagram below

Length of side AD = $$\sqrt{12^2 + 5^2 } = 13$$
Area of the trapezium = 12 * (10 + 20)/2 = 180
Perimeter of the trapezium = 10 + 20 + 13 + 13 = 56
Area of the sides of the pillar = 56 * height  = 56 * 20 = 1120.
Total are of the pillar = 1120 + area of base + area of the top = 1120 + 180 + 180 = 1480

The image of the figure is as shown. 

AB = AC = 6cm. Thus, BC = $$\sqrt{6^2 + 6^2}$$ = 6√2 cm

The required area = Area of semi-circle BQC - Area of quadrant BPC + Area of triangle ABC

Area of semicircle BQC

Diameter BC = 6√2cm

Radius = 6√2/2 = 3√2 cm

Area = $$\pi r^2 $$/2 = $$ \pi$$ * $$(3 \sqrt{2})^2 $$/2 = 9$$\pi$$

Area of quadrant BPC

Area = $$\pi r^2$$/4 = $$\pi*(6)^2$$/4 = 9$$\pi$$

Area of triangle ABC

Area = 1/2 * 6 * 6 = 18

The required area = Area of semi-circle BQC - Area of quadrant BPC + Area of triangle ABC

= 9$$\pi$$ - 9$$\pi$$ + 18 = 18

The midpoint of one diagonal lies on the other diagonal.

Midpoint is ((2+6)/2, (5+3)/2) = (4,4)
Hence 4 = 3 * 4 + c => c = -8

Let the volumes of the five cubes be x, x, 8x, 27x and 27x.

Let the sides be a, a, 2a, 3a and 3a. ($$x = a^3$$)

Let the side of the original cube be A.

$$A^3 = x + x + 8x + 27x + 27x$$

$$A = 4a$$

Original surface area = $$96a^2$$

New surface area = $$6(a^2 + a^2 + 4a^2 + 9a^2 + 9a^2)$$ = $$144a^2$$

Percentage increase = $$\frac{144 - 96}{96}*100 =$$ 50%

$$\angle COD = 120$$ => $$\angle CAD = 120/2 = 60$$ (The angle subtended by the chord DC at the major arc is half the angle subtended at the centre of the circle.)
$$\angle BAC = 30$$
$$\angle BAD = \angle BAC + \angle CAD$$ = 30 + 60 = 90.
$$\angle BCD = 180 - \angle BAD$$ = 180 - 90 = 90

The volume of a cylinder is $$\pi * r^2 * 3 = 9 \pi$$

r = $$\sqrt{3}$$ cm.

Radius of the ball is 2 cm. Hence, the ball will lie on top of the cylinder.

Lets draw the figure.

Based on the Pythagoras theorem, the other leg will be 1 cm.

Thus, the height will be 3 + 1 + 2 = 6 cm

Let the length and breadth of the park be l,b, l > b
Case 1: 2l + b = 400
Area = lb. Area is maximum when 2l * b is maximum, which is maximum when 2l = b (using AM $$ \geq $$ GM inequality) => l = 100, b = 200. Which can't happen since l > b

Case 2: l + 2b = 400
Area = lb. Area is maximum when l *2 b is
maximum, which is maximum when l = 2b (using AM $$ \geq $$ GM
inequality) => l = 200, b = 100.

Hence length of the longer side is 200 ft

The length of the altitude from A to the hypotenuse will be the shortest distance.

This is a right triangle with sides 3 : 4 : 5. 

Hence, the hypotenuse = $$\sqrt{20^2+15^2}$$ = 25 Km.

Length of the altitude = $$\frac{15 * 20}{25}$$ = 12 Km

(This is derived from equating area of triangle, $$\frac{15\cdot20}{2}\ =\ \frac{25\cdot\text{altitude}}{2}$$)

The time taken = $$\frac{12}{30} * 60$$ = 24 minutes

Let the length of non-hypotenuse sides of the right angled triangle be $$a$$. Then the hypotenuse h = $$\sqrt{2}a$$
P is equidistant from all the side of the triangle. Hence P is the incenter and the perpendicular distance is the inradius.
In a right angled triangle, inradius = $$\frac{a + b - h}{2}$$
=> $$\frac{a + a - \sqrt{2}a}{2} = 4(\sqrt{2}-1)$$
=> $$ \sqrt{2}a( \sqrt{2} - 1) = 8(\sqrt{2} -1)$$
=> $$ a = 4\sqrt{2}$$
Area of the triangle = $$\frac{1}{2}a^2$$ = 16 sq cm

The graph of the given function is as shown below.

We can see that the shortest distance of the point (1/2, 1) will be 1 unit.

Let x be the value of third side of the triangle. Now we know that Area = 17.5*9*x/(4*R), where R is circumradius.

Also Area = 0.5*x*3 .

Equating both, we have 3 = 17.5*9 / (2*R)

=> R = 26.25.

For obtuse-angles triangle, $$c^2 > a^2 + b^2$$ and c < a+b
If 15 is the greatest side, 8+x > 15 => x > 7 and $$225 > 64 + x^2$$ => $$x^2$$ < 161 => x <= 12
So, x = 8, 9, 10, 11, 12
If x is the greatest side, then 8 + 15 > x => x < 23
$$x^2 > 225 + 64 = 289$$ => x > 17
So, x = 18, 19, 20, 21, 22
So, the number of possibilities is 10

Consider side of square as 10 units. So HD=5 and HP=$$\frac{5}{\sqrt3}$$ . So now area of triangle HPD=$$\frac{12.5}{\sqrt3}$$. Also Area APD=Area BQC=2*AreaHPD=$$\frac{25}{\sqrt3}$$. So numerator of required answer is $$100-\frac{50}{\sqrt3}$$ and denominator as $$\frac{50}{\sqrt3}$$. Solving we get answer as $$2\sqrt{3}-1$$.

The circumferences of the two circle pass through each other's centers. Hence, O1A = O1B=O1O2 = 1cm

By symmetry, the line joining the two centres would be bisect AB and would be bisected by AB. As the line joining the center to the midpoint of a chord is perpendicular to the chord, O1O2 and AB are perpendicular bisectors of each other. Suppose they intersect at point P.

O1P = Half of O1O2 = 1/2 cm
So, the angle AO1P = 60 degrees as cos 60 = 1/2
By symmetry, BO1P = 60 degrees.
So, angle AO1B= 120 degrees

In the above, the required area is 2 times A(segment ABO2)(blue region). And A(segment ABO2)(blue region) = A(sector O2AO1B)(blue + red) - A(triangleO1AB )(red)

Area of sector = 120°/360° * $$\pi * 1^2 $$ = $$\pi/3$$

Area of triangle = 1/2 * b * h = 1/2 * (2* 1 cos 30°) * (1/2) = √3/4

Hence, required area = $$\frac{\pi}{3}-\frac{\sqrt 3}{4}$$ . Hence so the required area is 2 times the above value which is $$\frac{2\pi}{3}-\frac{\sqrt 3}{2}$$

Trangle AGF and triangle FHC are similar.
So, $$\frac{10-h}{x}=\frac{h}{4-x}$$
=> $$\frac{10-h}{h}=\frac{x}{4-x}$$
=> $$\frac{10}{h}-1=\frac{x}{4-x}$$
=> $$\frac{10}{h}=\frac{4}{4-x}$$
=> $$h=\frac{10}{4}(4-x)$$

Total surface area= $$2 \pi x^2 + 2 \pi x h= 2 \pi (x^2 + x \frac{10}{4}(4-x)=2 \pi (10x - \frac{3}{2} (x^2))$$

=$$2\pi\frac{-3}{2}[x^{2}-\frac{20}{3}x] $$. We can add and subtract 100/9 to get a perfect square.

=$$2\pi\frac{-3}{2}[x^{2}-\frac{20}{3}x+\frac{100}{9}-\frac{100}{9}]$$

=$$2 \pi (-\frac{3}{2}(x-\frac{10}{3})^{2}+ \frac{50}{3})$$

This value will be maximum when the negative factor is minimum i.e. 0 because it is multiplied by a square. Thus, when x=10/3 the surface area is max.

So, the surface area is $$\frac{100}{3} \pi$$

To know the range, we have to take the limiting case.

The limiting case in this case is when the circles pass through each other's centers. 

In this case, PQ = AP = AQ => They form an equilateral triangle => angle AQP = 60 degrees.

So, the maximum possible angle is 60 degrees.

Another limiting case is when the circles touch each other externally.

In this case, angle AQP = 0 degrees.

Hence, the range is 0 to 60.

In triangle CDO, which is a right anged triangle, we can use pythagoras theorem.

$$6^2$$ + $$(r-2)^2$$ = $$r^2$$

=> 36 + $$r^2 - 4r + 4$$ = $$r^2$$

=> 4r = 40

=> r = 10

=> Area = $$\pi *10*10/2$$ = $$50\pi $$

Triangle AOD is isosceles. So, angle DAO = angle DOA = 75. Similarly, angle BOC = 75. So, angle AOB = 150

The area of the four walls is length*height*2 + breadth*height*2
Initial area = 3*1*2 + 2*1*2 = 10
Final area = 6*1/2*2 + 1*1/2*2 = 7
So, the area decreased by 30%

The area of triangle ABC is 1/2 * $$\sqrt2$$ * $$\sqrt2$$ = 1

Area of semi-circle ABC = $$\pi/2$$

So, area of circle outside the square = $$\pi/2$$ - 1 = ($$\pi$$ -2)/2

So, area of circle inside the sheet = $$\pi$$ - ($$\pi/2$$ - 1) = 1 + $$\pi/2$$

Area of original square = 2*2 = 4

So, area of the sheet after punching = 4 - 1 - $$\pi/2$$ = 3 - $$\pi/2$$

So, proportion of sheet that remains after punching = (3 - $$\pi/2$$)/4 = (6 - $$\pi$$)/8

The area of triangle ABC is 1/2 * $$\sqrt2$$ * $$\sqrt2$$ = 1

Area of semi-circle ABC = $$\pi/2$$

So, area of circle outside the square = $$\pi/2$$ - 1 = ($$\pi$$ -2)/2

The number of points on x = 1 is 39. The number of points on x = 2 is 38 and so on till x = 39, which has one point.

So, the total is 1+2+3+...+39 = $$\frac{39*40}{2}$$ = 780.

Since ant cant go more near than 1 m.

So it'll have to travel in circular path from A till it is south of pt.

B and hence travels a quarter circle with radius = 1 and center as B, so travel $$\frac{\pi }{2}$$ m distance.

Then from B to C it travels 1m in straight line till C.

Then again from C to D in circular path way with distance $$\frac{\pi }{2}$$ m.

Hence total distance traveled = $$\frac{\pi }{2}$$ + $$\frac{\pi }{2}$$ +1 = $$\pi $$ +1.

Alternate solution : 

Let the given figure represent the situation. A,B,C,D are the points with distance between them 1 m. 

The repellents are at B and C respectively. The circles drawn with centre at B and C and radius equal to AB = 1 m. 

Let E and F be the points of intersection of the circles and G and H be the points on the circle perpendicular to B and C. 

Therefore, the ant can only travel at the circumference of the circles.

The shortest path for the to take will take is : A-G-H-D.

This is because taking the curve G-E and E-H will be a longer path than travelling straight from G to H. 

While travelling from G to H, the distance of the ant will be more than 1 m from the repellant. 

Thus, the distance travelled by the ant on the circumference of the circle ie 

arc AG = $$\dfrac{90}{360} \times 2 \times \pi 1 = \dfrac{\pi}{2} $$ 

Length of arc HD = length of arc AG = $$\dfrac{\pi}{2} $$ 

Total length = $$(2 \times \dfrac{\pi}{2})+ BC = \pi + 1$$  

As seen from the fig. If following configuration is used max 6 number of tiles that can be accommodated on the floor. 

Let EO = x, So, AE = 1.5 - x
AE : EB = 1:2 => x = 1/2

(1.5-x):(1.5+x) = 1:2.

x=0.5.

So, EO = 0.5
Similarly, OL = 0.5
Now, EOLH is a parallelogram and EO = OL = 0.5
In triangle DOL, DO = radius = 1.5 and OL = 0.5
So, DL = $$\sqrt2$$
=> DH = $$(2\sqrt2 - 1)/2$$

Consider triangles ABC and BDC
Angle B is common in both triangles. Also $$\angle A = \angle C$$
Since 2 angles are equal, the third angles $$\angle{ACB} = \angle{BDC}$$
Hence, triangle BCA $$\sim$$ BDC (BCA and BDC are similar)
AC/DC = BC/BD = AB/BC 
BC = 12 cm, DB = 9 cm, CD = 6 cm
=> AC = 12/9 * 6 = 8 cm
AB = 8/6 * 12 = 16 cm
So, AD = 16 - 9 = 7 cm
Perimeter of ADC = 7+8+6 = 21 cm
Perimeter of BCD = 27 cm
Ratio = 21/27 = 7/9

Let PQR be an equilateral triangle with side equal to x and let the intersection point of PS and QR be M.

Clearly, the circle is the circumcircle of the triangle PQR.

QR = x => QM = $$\frac{x}{2}$$ because a perpendicular from the centre to any chord bisects the chord.

Angle OQM = 30 degrees and QM is equal to $$\frac{x}{2}$$ => OQ = $$\frac{\frac{x}{2}}{cos(30)}$$ = $$\frac{x}{\sqrt{3}}$$

Hence the radius of the circumcircle of an equilateral triangle is equal to $$\frac{x}{\sqrt{3}}$$.

Angle PQS = 90 degrees as it is an angle in a semicircle. PS bisects angle QPR => angle QPS is 30 degrees. Hence QS subtends an angle of 30 degrees in the major arc => QS subtends an angle of 60 degrees at the centre because angle subtended by a chord at the centre is twice the angle subtended by the chord in the major arc.

Angle QOS = 60 degrees => Triangle QOS is equilateral and hence QS is equal to radius of the circle => QS = $$\frac{x}{\sqrt{3}}$$

Given that radius is r => r = $$\frac{x}{\sqrt{3}}$$ => x = $$r\sqrt{3}$$

=> Perimeter of PQRS = PQ+QS+SR+RP= $$ r\sqrt{3} + r + r + r\sqrt{3}$$ = $$2r(1+\sqrt{3})$$

The distances of the chords from the center are 12 cm and 16 cm respectively.

If the chords lie on the same side of the center, the distance between the chords is 4 cm, if they lie on opposite sides of the center, the distance between them is 28 cm.

We know that quad ambn is a square of side 1.

Area of the sector a-mqn is $$\frac{90}{360}* \pi *1*1$$ = $$\frac{\pi }{4}$$.

Area of square = 1*1 = 1

Area of common portion = 2 * Area of sector - Area of square

= 2 * $$\frac{\pi }{4}$$ - 1 =  $$\frac{\pi }{2}$$ - 1

Let the radius of 2 circles be r . Speed of A would be 12r/t and Speed of B would be 4*Pi*r/t . To find percentage faster B have to run than A we have : (4*Pi - 12)*100/ (12) = 4.7% approx. Hence, option D.

The three triangles are similar.

Let the distance of the tip of the shadow from the child be y. Let the child be standing at distance x from the father.

So, 6/(2.1+x+y) = 1.8/(x+y) = 0.9/y

=> 2y = x+y => x = y

=> 6/(2.1 + 2x) = 0.9/x

=> 6x = 0.9*(2.1+2x)

=> 6x = 1.89 + 1.8x

=> 4.2x = 1.89

=> x = 1.89/4.2 = 0.45

Radius of the circles $$C_1, C_2, C_3,...$$ would be in GP with (R/4),(R/8),(R/16) and so on. Radii of circles are in the ratio 1:4. 

Ratio of unshaded region to the ratio of original circle = 1-$$\frac{Ratio\ of\ shaded\ region}{Ratio\ of\ original\ circle}$$

= 1-$$\frac{\pi r^2/16 + \pi r^2/64+.....}{\pi r^2}$$ = 1-$$\frac{1/16}{(1-1/4)}$$ = 1- 1/12 = $$\frac{11}{12}$$ = 11:12

Consider side of the cube as x.

So diagonal will be of length $$\sqrt{3}$$ * x.

Now if diagonals are side of equilateral triangle we get area = 3*$$\sqrt{3}*x^2$$ /4 .

Also in a triangle

4 * Area * R = Product of sides

4* 3*$$\sqrt{3}*x^2$$ /4 * R = .3*$$\sqrt{3}*x^3$$

R = x

In the triangle BEO, BE = $$\sqrt{4^2-2^2}$$ = $$2\sqrt3$$

==> BC = $$4\sqrt3$$

Using pythagoras we can find diagonal of small square shown in fig. = 2*$$\sqrt2$$ . Now this is equal to 2 + r +  $$\sqrt2$$ *r. Equating we get r = $$6- 4 \sqrt{2}$$

If EBC = 65 then EOC = 130 then OEC = OCE = 25. NOw since OC and ED are parallel we have OCE = OED = 25. Hence option D. 

Let the longer side of the original rectangle be x. Therefore, breadth of the modified rectangle = x/2.
=> x:2 = 2:(x/2)
=> $$x^2$$ = 8
Area of smaller rectangle = (x/2)*2 = x = $$2 \sqrt{2}$$

Triangles PRO and QSO are similar.

PR/QS = 4/3.

Therefore, PO/QO = 4/3 =>

Let the radius of 2 circles be 4R and 3R respectively.

28/(28-7R) = 4/3

4-R = 3

R = 1

Required ratio of PQ/QO = (4R+3R)/(28-7R) = 7R/(28-7R) = 7/21 = 1:3

Let the radius of circles I and II be 4R and 3R respectively.

Triangles PRO and QSO are similar.

PR/QS = 4/3.

Therefore, PO/QO = PR/QS

=> PO/QO = 4/3

=> 28/(28-7R) = 4/3

=> 4-R = 3

=> R = 1

Radius of smaller circle = 3R = 3

Triangles PRO and QSO are similar.

PR/QS = 4/3.

Therefore, PO/QO = 4/3 =>

Let the radius of 2 circles be 4R and 3R respectively.

28/(28-7R) = 4/3

4-R = 3

R = 1

Therefore, SO = $$\sqrt{21^2 - 3^2}$$ = $$\sqrt{432}$$ = $$12 \sqrt{3}$$ cm

 

Let the total number of sides be x.

Sum of internal angles in a polygon = (x-2)*180 where x is the number of sides.

It is given that the polygon has 25 convex sides, then the number of concave sides = x-25

(25*90)+(x-25)*270 = (x-2)180

x = 46

Number concave corners = x-25 = 46-25 = 21

Let R ,r be radius of big and small circles respectively. We know that R=2r.

And since area = 12 ;

$$R^2 = \frac{12}{\pi}$$.
$$r^2$$ = $$\frac{3}{\pi}$$

By pythagoras theorem in the small triangle with side 'x' we have x = $$\sqrt{\ 3}r$$.

Angle OAB = $$\tan\ \theta\ \ =\ \ \frac{\ r}{\sqrt{\ 3}r}\ =\ 30$$
Hence Angle CAB = 60.

So area = $$\ \frac{\ 1}{2}\ \times\ 2\sqrt{\ 3}r\times\ 2\sqrt{\ 3}r\times\ \sin\ 60$$.
On substituting the value of sin 60 and $$r^2$$, we get

Area = $$9\sqrt3/\pi$$.

Let R be radius of big circle and r be radius of circle with centre S. Radius of 2 semicircles is R/2.

 From Right angled triangle OPS, using pythagoras theorem we get

$$(r+0.5R)^2 = (0.5R)^2 + (R-r)^2$$ . We get R=3r .

Now the area of big semicircle that cannot be grazed is Area of big S.C - area of 2 semicircle - area of small circle = $$\pi*R^2$$/2 - 2*$$\pi*(0.5R)^2$$/2-$$\pi*r^2$$ = $$\pi*R^2$$/2 - 2*$$\pi*(0.5R)^2$$/2-$$\pi*(R/3)^2$$= $$\pi*R^2$$/2 - $$\pi*(R)^2$$/4-$$\pi*(R)^2$$/9 = 5*$$\pi*R^2$$/36.  this is about 28 % of the area $$\pi*R^2$$/2 . Hence option B.

When the hexagon is divided into number of similar triangle AOF we get 12 such triangles . Hence required ratio of area is 1/12.
Alternate Approach :

Let us take side of the hexagon as a
Now We get AF = a
Now As OF || ED
Angle OFE +Angle E = 180
We know Angle E =120
So angle OFE=60 degrees.
Now therefore Angle AFO=120-60=60 degrees
Now in triangle AFO
Cos60=$$\frac{OF}{AF}$$
$$\frac{1}{2}=\frac{OF}{AF}$$
We get OF = $$\frac{a}{2}$$
Now therefore area of triangle AOF =$$\frac{1}{2}\times\ AF\times\ OF\times\ \sin60$$
So we get $$\frac{1}{2}\times\ a\times\ \frac{a}{2}\times\ \frac{\sqrt{\ 3}}{2}$$
=$$\frac{\sqrt{\ 3}}{8}a^2$$     (1)
Now area of hexagon = $$6\times\ \frac{\sqrt{\ 3}}{4}a^2$$      (2)
Dividing (1) and (2) we get ratio as $$\frac{1}{12}$$
Hence option A is the correct answer.

Consider the triangle APB.  $$\angle$$P = 60 and AP = BP => APB is an equilateral triangle.

Hence AP = $$b$$    ...(1)

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

$$\text{AC}^2 = b^2 + b^2$$  =>  $$\text{AC} = \sqrt{2}\times b$$  =>  $$\text{AO} = \dfrac{\text{AC}}{2} = \dfrac{b}{\sqrt{2}}$$

$$\text{AP}^2 = \text{AO}^2 + \text{OP}^2$$

$$b^2 = \dfrac{b^2}{2} + h^2$$    ...From (1)

$$2h^2 = b^2$$

Hence, option B is the correct answer.

Let BD = x .Semi-perimeter of triangle ABC = 12. Now by herons formula area of ABC is 24. Also Area = 0.5*x*10 . We get x = 24/5 .  AP = 6/5  and CQ = 16/5 . Hence the required ratio is 3:8. 

Let BQ = z , QD = y , PQ = x.

From similar triangles PQD and ABD we have

(y/x) = (z+y)/3 .

Also from  similar triangles PQB and CBD we have

(z/x) = z+y .

Solving we get z = 3*y.

Now required ratio is (z+y)/z.

We get eual to 4/3 which is equal to 1:0.75. 

Since Angle BOC = Angle BCO = y.

Angle OBC = 180-2y .

Hence Angle ABO =  z = 2y = Angle OAB.

Now since x is exterior angle of triangle AOC .

We have x = z + y = 3y.

Hence option A. 

As seen in the fig. we have a right angled triangle with sides  r ,r-10 , r-20.

Using pythagoras we have $$r^2 = (r-10)^2 + (r-20)^2$$.

Solving the equation, we get r = 10 or 50.

But 10 is not possible , so r = 50.

Hence radius is 50.

The radii of both the circles and the line joining the centers of the two circles form a right angled triangle. So, the length of the common chord is twice the length of the altitude dropped from the vertex to the hypotenuse.

Let the altitude be h and let it divide the hypotenuse in two parts of length x and 25-x
So, $$h^2 + x^2 = 15^2$$ and $$h^2 + (25-x)^2 = 20^2$$
=> $$225 - x^2 = 400 - x^2 + 50x - 625$$
=> 50x = 450 => x = 9 and h = 12
So, the length of the common chord is 24 cm.

Length of the rope tied to each horse = 7 m.
Total area of the portion that the horses can graze = 4* $$\pi 7^2$$/4 = 49$$\pi$$
Area of the circular pond = 20 $$m^2$$
So, area left ungrazed = $$14^2 - 20 - 49\pi m^2$$ = 22 $$m^2$$ (approx)

The weights are at distances of 0 m, 2 m, 4 m, 6 m and 8 m from X.

Let us first keep all the weights at a distance of 8 m from X. This would be 8 + 6*2 + 4*2 + 2*2 = 32 m.

Now from the point where all the weights are kept is at a distance of 92 m from Y. So total distance required = 184+184+184+184+92 = 828 m.

So in all 860 m.

Let AB = BE = x

Area of triangle ABE = $$\dfrac{x^2}{2}$$ = 7; we get x = $$\sqrt{14}$$

So we have side BC = 4*$$\sqrt{14}$$

Now area is AB*BC = 14 *4 = 56 $$cm^2$$

The triangle we have is :

The length of three sides is $$\sqrt 2, \sqrt 2$$ and $$2$$.
This is a right-angled triangle.
Hence, it's area equals $$1/2 * \sqrt 2 * \sqrt 2 = 1$$
So, the correct answer is b)
Alternate Approach :
Area of triangle = $$\frac{1}{2}\times\ base\times\ height$$
So we get $$\frac{1}{2}\times2\times\ 1$$=1 square units

Let x be the shorter side and y be the longer one. The shortcut route would be of length $$\sqrt{x^2+y^2}$$. According to given condition we know that (x+y)- $$\sqrt{x^2+y^2}$$ = y/2 . Solving, we get 1=(x/y)+(1/4) 

=> x : y = 3 : 4. Hence option D is the correct answer.

To mow half of lawn is to mow 400 squares of 1m width each. Neeraj mows 40+18+40+18 = 116 squares of 1m width each in 1st round, in 2nd round he mows 38+16+38+16 = 108 squares and in 3rd round, he mows 36+14+36+14 = 100 squares. So, in total, he mows 324 squares; but he needs 400-324 = 76 more.

If he covers the 4th round completely, he mows 92 squares, to cover only 76, he needs to cover 76/92 = 0.826 round.

So in total 3.826 $$\approx$$ 3.83 rounds are required.

By Pythagoras theorem we get BC = 25 . Let BD = x

Triangle ABD is similar to triangle CBA => AD/x = 20/15 

Also triangle ADC is similar to triangle ABC => AD/(25-x) = 15/20

From the 2 equations, we get x = 9, DC = 16 and AD = 12

We know that AREA = (semi perimeter ) * inradius

For triangle ABD, Area = 1/2 x BD X AD = 1/2 x 9 x 12 = 54 and semi perimeter = (15 + 9 + 12)/2 = 18. On using the above equation we get, inradius, r = 3.

Similarly for triangle ADC we get inradius R = 4 .

PQ = R + r = 7 cm

By using cosine rule we can find BC = $$\sqrt{13}$$ . By angle bisector theorem we have BA / BD = AC / DC . Also BD + DC = $$\sqrt{13}$$. So by substitution we get we get BD = 4*$$\sqrt{13}$$/7 . Now using cosine rule in triangle ABD taking AD = x, we get

$$x^2 - 4*\sqrt3*x + 16*(36/49) = 0$$. Solving the equation we get x = $$\frac{12\sqrt 3}{7}$$ .

The length of FI is twice the length of IG.

So, the sides of the triangle FIG are in the ratio 2:1:$$\sqrt5$$.

So, angle FGO = angle FGI, which is definitely not equal to 30 or 45 or 60.

Hence, option D is the answer.

Area of ABCD = Area of triangle + Area of rectangle = Let AB = x. So, area = $$3/2 * x^2 / 2 = 3/4 * x^2$$
Area of ODE = $$\frac{1}{2}* \frac{x}{2}*\frac{x}{2}$$=$$x^2/8$$
Area of OEFI = $$ \frac{x}{2}*\frac{x}{2}$$=$$x^2/4$$
Area of FGI = $$x^2/16$$
Total area of DEFG = $$7x^2/16$$
Required ratio = 12:7

B is the south gate. A is the house and BC = 9 km.