Find the circumcentre of the triangle with the vertices A (-3,4) , B (3,4) and C (-4,-3). What is the circumradius and area of the circle?
Find the circumcentre of the triangle with the vertices A (-3,4) , B (3,4) and C (-4,-3). What is the circumradius and area of the circle?We want the equation of the circle which passes through the points [math](-3,4),[/math] [math](3,4),[/math] and [math](-4,3).[/math]The equation of a circle is [math]x^2+y^2+ax+by+c=0.[/math]From the coordinates of the three given points we get following three equations:[math]\qquad (-3)^2+4^2-3a+4b+c=0[/math][math]\qquad 3^2+4^2+3a+4b+c=0[/math][math]\qquad (-4)^2+3^2-4a+3b+c=0[/math]We can simplify the above equations, and put them in matrix form:[math]\qquad\begin{bmatrix} -3 & 4 & 1 \\ 3 & 4 & 1 \\ -4 & 3 & 1 \end{bmatrix}\begin{bmatrix} a \\ b \\ c \end{bmatrix}=\begin{bmatrix} -25 \\ -25 \\ -25 \end{bmatrix}[/math]Solving the above matrix equation using Cramer's method, we get [math]a=0,[/math] [math]b=0,[/math] [math]c =-25.[/math]Plugging in these values of [math]a,[/math] [math]b,[/math] and [math]c[/math] into the general form of the equation, we can now write the equation of the required circle as[math]\qquad x^2+y^2-25=0,[/math]Or, expressing it in center radius form,[math]\qquad \left(x+0\right)^2+\left(y+0\right)^2=25[/math]The center of the circle is [math]\left(0,0\right)[/math] and the radius of the circle is [math]5.[/math]The area of a circle with radius [math]5[/math] is [math]\pi r^2[/math] [math]=[/math] [math]25\pi.[/math]
A circle inscribed in a square.if the length of the diagonal is 14√2 cm. What is the area of the circle (in cm sq)?
Let us first take out the length of the square. As we know square has all its sides equal.So by Pythagoras theorem,(diagonal)^2=2*(side)^2(14√2)^2=2x^2392=2x^2196=x^2√196=x14=x=side.As circle is inscribed inside the square, it's diameter is equal to square's side.And radius =1/2 diameterRadius =1/2*14=7Area=πr^2=22/7*7*7=22*7=154
Is there a way to calculate the angle of a rectangle's diagonal?
Sure. If the rectangle has sides [math]a[/math] and [math]b[/math], then the diagonal has length [math]\sqrt{a^2 + b^2}[/math].So the diagonal divides the right angle into two pieces. One of these angles which I'll call [math]\theta_1[/math] has the property that [math]\sin(\theta_1) = \frac a {\sqrt{a^2 + b^2}}[/math].The other angle, [math]\theta_2[/math], has the property that [math]\sin(\theta_2) = \frac b {\sqrt{a^2 + b^2}}[/math].Since we know that both of these angles are acute, we immediately get that:[math]\theta_1 = \arcsin \left(\frac a {\sqrt{a^2 + b^2}}\right)[/math].and [math]\theta_2 = \arcsin \left(\frac b {\sqrt{a^2 + b^2}}\right)[/math].
I have the four sides of an irregular quadrilateral and none of its angles. How can I calculate its area?
Brahmagupta (c.598 - 665) was an Indian mathematician who gave his formula for finding the area of a quadrilateral, which does not require the angle measurements. Only the 4 lengths are sufficient to find the area. We should give him due credit!Brahmagupta's FormulaBrahmagupta's formula finds the area of a cyclic quadrilateral. The formula for the area of a cyclic quadrilateral with sides a, b, c, d is given bywhere s = [a+b+c+d]/2and a, b, c and d are the four sides.Example 1: Let ABCD be a kite. DA = AB = 3, DC = CB = 5. Let us find the area of the kite.s = (3+3+4+4)/2 = 14/2 = 7Area = [(7–3)(7–4)(7–4)(7–3)]^0.5 = (4x3x4x3)^0.5 = 12.As you can see DAC is a RAT as also ABC, with AC as the hypotenuse (5). Therefore area of DAC and ABC = 3*4/2 = 6 each, and so area of ABCD = area DAC + area ABC = 6+6 = 12, same as what we got from Brahmagupta’s formula.Example 2: Let PQRS be a kite. SP = PQ = 5, QR = RS = 12. Let us find the area of the kite.s = (5+5+12+12)/2 = 34/2 = 17Area = [(17–5)(17–12)(17–12)(17–5)]^0.5 = (12x5x5x12)^0.5 = 60.As you can see SPR is a RAT as also PQR, with PR as the hypotenuse (13). Therefore area of SPR and PQR = 5*12/2 = 30 each, and so area of PQRS =area SPR + area PQR = 30 + 30 = 2*30 = 60, same as what we got from Brahmagupta’s formula.
How do I calculate the radius of a circle with the length of a chord = 40 and the perpendicular distance from the midpoint of the chord to the perimeter of the circle = 14?
First we have the length the chord =40Now the radius is assumes to be equal to r So the perpendicular distance from the centre to the chord is = (r-14) After getting two distances we now have a right triangle with hypotenuse equal to r and one side equal to (r-14) The third side is given by the half of the chord =20Solve the hypotenuse equation given by r^2 = (r-14)^2 + 20^2 Solve for the value of r The radius value comes to ~ 21.287
ABC is a triangle whose AB=AC=20cm and angle BAC=30 degree. What is the area of the triangle?
Area of triangle1/2* x*y*sin thetaHere x and y are two sides and theta is the angle made by these two sides.Putting values1/2*20*20*sin 30= 100 cm^2
Four points are chosen uniformly at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points?
This tantalizing, wonderful question was Problem A6 at the 53rd Putnam competition in 1992. It is hard, as A6’s are, but there’s an elegant and short solution requiring almost no calculation.For a point X on a sphere, we will use X’ to denote the point antipodal to X. This is the point that lies the farthest from X, or more precisely it is the point where the sphere meets the line XO (O is the center of the sphere).The short version of the solution is this:Once three points A, B, C are chosen, the region in which you need to pick D in order for ABCD to contain O is the spherical triangle A’B’C’ opposite ABC.Therefore, the probability of success is simply the expected area of the (spherical) triangle A’B’C’, normalized so that the surface of the sphere has area 1. This is clearly the same as the expected area of ABC, and in fact it is also the expected area of A’BC, A’BC’ and so on since all of these triangles are spanned by three uniformly chosen points on the sphere.Now, there are 8 such triangles, and their total area is 1, so the expected area of each one is [math]1/8[/math].The answer, therefore, is simply [math]1/8[/math].The long version will need to wait… In the meantime, it turns out that 3Blue1Brown just published a video on that problem, so check it out as well.