Consider the following figure-

The graph cuts x-axis at (4,0) and (-4,0). It cuts the y-axis at (0,4) and (0,-4).
The area enclosed is a square with vertices at the above mentioned points.
The number of points lying in Q1 is 6. The number of points on positive x-axis is 4.
Therefore, by symmetry, the number of points lying in all the four quadrants = 6*4 = 24.
The number of points lying on all four axes is 4*4 = 16. Plus the point (0,0).
Thus the total number of points enclosed by the graph = 41.

EO=OF=O"D=O"C

Draw perpendiculars from the end-points of the chord to the diameter. Let the distance from the center of the circle to the foot of one of the perpendiculars be x. The remaining part of the diameter will be 4-x.

The chord parallel to diameter is CD. Let CE be the perpendicular drawn from C to AB.

CE^2 = AC^2 - AE^2 = 2^2 - (4-x)^2

CE^2 = OC^2 - OE^2 = 4^2-x^2

These both can be equated.

$$4^2 - x^2 =2^2 - (4-x)^2$$.

Solving the equation gives x=3.5. The length of the chord is 2x = 7cm.

The centre of the circle $$x^2+y^2=25$$ is (0,0) 

The centre of the circle $$(x-6)^2+(y-8)^2=25$$ is (6,8) 

The distance between centers =$$\sqrt{(6-0)^2+(8-0)^2}$$=10

Distance between the centres = 10
radius of first circle = 5
radius of second circle = 5
distance between the centres = $$r_1+r_2$$
=> The circles touch each other => 3 tangents possible

Let a, b and c be the three sides of a triangle.

When $$a^2+b^2 > c^2$$, the triangle is acute angled.
When $$a^2+b^2 = c^2$$, the triangle is right angled.
When $$a^2+b^2 < c^2$$, the triangle is obtuse angled.

In this case,
$$(7a+24b)^2$$ + $$(24a+7b)^2$$
= $$49a^2+576b^2+336ab+576a^2+49b^2+336ab$$
= $$625a^2+625b^2+672ab$$
< $$625a^2+625b^2+1250ab$$ =
$$(25a+25b)^2$$
Therefore, the triangle is obtuse angled.

Consider triangle ABF. Here, DH is parallel to BF. Thus, by proportionality theorem, AD:DB = AH:HF 

We have been given that AD:DB = 3:2. So AH:HF = 3:2

Consider triangles AHC and FHC. Their bases AH and HF lie along the same line and the vertex C is in common. So altitude from C to AH or HF would be equal in length. Therefore, height of both the triangles is same.

Area = 1/2 * base * height

As height is same for both triangles, Area AHC / Area FHC = Base AH / Base FH = AH = HF = 3:2

Hence, the required ratio is 3:2.

The area of a cyclic quadrilateral is given by $$\sqrt{(s-a)(s-b)(s-c)(s-d)}$$, where s is the semi-perimeter.
In this case, s = (14+17+18+23)/2 = 36.
Area = $$\sqrt{(36-14)(36-17)(36-18)(36-23)}$$ = 6$$\sqrt{2717}$$.

Let’s draw CH parallel to AE intersecting DF at G.
As CE || AH and AE || CH, AECH is also a parallelogram.
Hence, CE = AH and H is the midpoint of AD.
In triangle ADF, GH || AF, therefore, G is the midpoint of FD.
In triangle CFD, since EF || CG and ∠CGF = 90 degrees, CG is the median as well as the altitude.
Hence, ΔCDF is an isosceles triangle and CF = CD = 4 cm.

This problem is relatively simple.

Firstly we will calculate the coordinates of the centroid of the triangle.

Let the co-ordinates of centroid be (x,y)

x = (3-1+1)/3 = 1
y= (4-2+4)/3 = 2

Centroid = (1,2)

Now we will find the equation of the line BC

y -4 =$$\frac{4-(-2)}{1-(-1)}(x-1)$$
3x - y+1 = 0

Now distance of (1,2) from BC(3x - y+1 = 0) = $$\frac{[3(1)-2+1]}{\sqrt{10}}$$
= $$\frac{2}{\sqrt{10}}$$

The volume of the football is 8 times the volume of the cricket ball. So, the radius of the football is twice the radius of the cricket ball. Hence the surface area of the football is 4 times (300% more) that of the cricket ball.

Consider this figure. Let angle BYX be a. angle BXY=90, so this means that angle YBX=90-a which means angle angle ABD=a.

So triangles BYA and ABD are similar by AAA similarity.

This means, BY/AB = AB/AD So, BY is 12*12/16 = 9 cm and CY = 16-9 = 7 cm

Let the total number of angles of the polygon be n.
Hence, the sum of the interior angles is 180*(n-2)

We know that the number of angles which are 90 degrees is 31.
Hence, the number of angles which are 270 is n-31

Hence, the sum of the interior angles is 90*31 + 270*(n-31) which equals 180(n-2)

Therefore, 90*31 + 270*n - 270*31 = 180*n - 360

Or, 90*n = 180*31 - 360

So,n = 2*31 - 4 = 58

We know that 31 angles are 90 degrees. Hence, the number of angles which are 270 degrees is 58 - 31 = 27

We see that a1/a2 = b1/b2 =/= c1/c2. Therefore, the two lines are parallel.
Express the two equations as: 14x - 33y +15 = 0 and 14x - 33y +26 = 0.
Therefore, the distance between the lines is $$\frac{|c1-c2|}{\sqrt{a^2+b^2}}= \frac{11}{\sqrt{1285}}$$.
Therefore, the radius of the circle is $$\frac{11}{2*\sqrt{1285}}$$.

The angle between the tangent and the radius at the point of contact is equal to 90°. So ABC is a right angled triangle.
Thus,
$$ AC^{2} = AB^{2} + BC^{2}$$
By tangent secant theorem, we know that
$$AB^{2} = AC*AD$$
$$BC^{2} = 8^{2} = 64$$
=> $$ AC^{2} = AC*AD + 64$$
=> AC(AC - AD) = 64
=> AC*DC = 64 cm

Let the length be l and breadth be b. Hence l-4=b+2 => l=b+6. Hence b(b+6)=432. Solving the eqn we get b=18. Hence side of sq = b+2 =20

Using the concept of relative speed, let B be at rest and A be moving towards B at 15m/s. He has to cover a distance of 60m. Therefore, time taken = 60/15 = 4 s.
Since A, B and C are equidistant from each other at the start, they remain equidistant throughout the course of their travel. They meet at the centroid of the triangle. Distance between the centroid and vertex of the triangle is (2/3)*altitude = (2/3)*60*sin 60 = 60/$$\sqrt{3}$$. Distance traveled by A = speed*time = 15*4 =60. Therefore, difference = 60-60/$$\sqrt{3}$$.

$$(n+1)^{2}$$ - $$n^{2} = 49$$. So, n = 24 metres and area of swimming pool is 576 sq metres

There are two parts to the quadrilateral; the first one is a right triangle with 6, 8 and 10 (diameter). Area = 24. The second part is an isosceles triangle with the diameter as the base, and radius as the height on the base. Area = ½*10*5 = 25. Therefore, largest area possible = 49.

Let the radius of football be R and cricket ball be r

It is given,

Volume of football is 700% more than the volume of cricket ball.

This implies,

Volume of football = 8*(Volume of cricket ball)

$$\frac{4}{3}\pi R^3=8\left(\frac{4}{3}\pi\ r^3\right)$$

$$R^3=8r^3$$

$$R^3=\left(2r\right)^3$$

$$R = 2r$$

Surface area of football = $$4\pi R^2=4\pi\left(2r\right)^2=16\pi r^2$$

Surface area of cricket ball = $$4\pi r^2$$

Required percentage = $$\ \frac{\ 16\pi r^2-4\pi r^2}{4\pi r^2}\times100=300\%$$

The answer is option D.

The side of the equilateral triangle is r $$\sqrt{3}$$ and the side of the square is r $$\sqrt{2}$$ where r is the radius of the circle.
So, area of the triangle is (3 $$\frac{\sqrt{3}}{ 4}$$) $$r^{2}$$ and the area of the square is $$2r^{2}$$.

Hence, the required ratio is (3√3/4):2 = 0.65 : 1

$$(h+2)^{2}$$ = $$h^{2} + 100$$ or $$h = 24$$ feet

Mid-point of the line joining (0,0) and (16,6) is (8,3). The slope of the line joining (0,0) and (16,6) is 6/16 = 3/8. Therefore, slope of its perpendicular bisector is -8/3. Equation of its perpendicular bisector is y-3 = (-8/3)(x-8). Similarly, the equation of the perpendicular bisector of the line joining (0,0) and (10,-4) is y+2 = (5/2)*(x-5). The point of intersection of these two lines is (233/31,133/31).