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## A circle has a chord that goes from #( pi)/6 # to #(5 pi) / 6 # radians on the circle. If the area of the circle is #135 pi #, what is the length of the chord?

CW
Featured 5 months ago

Length of chord $= 9 \sqrt{5} \approx 20.125$

#### Explanation:

As area of a circle is given by $\pi {r}^{2}$, and it is $135 \pi$, we have $r = \sqrt{135} = 3 \sqrt{15}$

As shown in the figure, the angle $\Theta$ subtended by the chord at the centre is :
$\Theta = \frac{5 \pi}{6} - \frac{\pi}{6} = \frac{2 \pi}{3}$
$\implies \frac{\Theta}{2} = \frac{\pi}{3}$
$\implies A M = r \sin \left(\frac{\Theta}{2}\right)$
$\implies$ Length of chord $A B = 2 \cdot A M = 2 \cdot r \cdot \sin \left(\frac{\Theta}{2}\right)$
$= 2 \cdot 3 \sqrt{15} \cdot \sin \left(\frac{\pi}{3}\right) = 2 \cdot 3 \sqrt{15} \cdot \frac{\sqrt{3}}{2} = 9 \sqrt{5}$

## Find eq. of circle passes through (0,0),(0,3) and the line 4x-5y=0 is tangent to it at (3,0)?

Shwetank Mauria
Featured 4 months ago

Equation of circle is ${x}^{2} + {y}^{2} - 3 x - 3 y = 0$, but tangent at $\left(3 , 0\right)$ is $y = x - 3$. In fact $4 x - 5 y = 0$ is a secant of the circle.

#### Explanation:

Let the equation of circle be ${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$.

As it passes through $\left(0 , 0\right)$, $\left(0 , 3\right)$ and $\left(3 , 0\right)$. Note we are having a tangent $4 x - 5 y = 0$ at point $\left(3 , 0\right)$, hence this too is on circle. Putting these in the equation of circle, we get

Putting $x - 0$ and $y = 0$, we get $c = 0$.

Putting $x = 0$ and $y = 3$, we get $9 + 6 f + c = 0$. But as $c = 0$, $f = - \frac{3}{2}$.

Now putting $x = 3$ and $y = 0$, we get $9 + 6 g + c = 0$. But as $c = 0$, $g = - \frac{3}{2}$.

Hence, equation of circle is ${x}^{2} + {y}^{2} - 3 x - 3 y = 0$, whose center is $\left(\frac{3}{2} , \frac{3}{2}\right)$.

Slope of the radius joining $\left(\frac{3}{2} , \frac{3}{2}\right)$ and $\left(3 , 0\right)$ is

$\frac{0 - \frac{3}{2}}{3 - \frac{3}{2}} = - 1$ and hence slope of tangent at $\left(3 , 0\right)$ is $1$

and equation of tangent at $\left(3 , 0\right)$ is $y = x - 3$ and not $4 x - 5 y = 0$ ( note that it passes through $\left(0 , 0\right)$ and hence, this cannot be a tangent, but is a secant ) as mentioned.

As is seen from figure below, the circle passes through the three points, but $4 x - 5 y = 0$ is not a tangent. Tangent at $\left(3 , 0\right)$ is $y = x - 3$.

graph{(y-x+3)(x^2+y^2-3x-3y)(4x-5y)=0 [-3.63, 6.37, -0.96, 4.04]}

graph{(y+x)(x^2+y^2-3x-3y)(4x-5y)=0 [-3.63, 6.37, -0.96, 4.04]}

## What is the orthocenter of a triangle with corners at #(1, 6)#, #(9, 1)#, and #(5, 3)#?

Dave G.
Featured 3 months ago

The point of intersection is $\left(- 22.5 , - 41\right)$

#### Explanation:

The orthocenter is where the altitudes intersect in a triangle. Technically, it's where all three altitudes meet. But we only need to find two of them - and it doesn't matter which two.

I've included two pictures - the first one is of the triangle (which is barely a triangle) and the other is a picture of the triangle, zoomed out, so we can see where the three altitudes intersect (so far outside the triangle!

So.... Let's find two of the altitudes. We'll need to know two things to find one altitude - one of the vertices, and the slope of the line that the altitude belongs to (we can just use the "rise over run" method for that.

Arbitrarily, let's start with the point $\left(5 , 3\right)$ and the line segment between $\left(1 , 6\right)$ and $\left(9 , 1\right)$. Some quick math, and the slope of that segment is $- \frac{5}{8}$ and the negative reciprocal is $\frac{8}{5}$. So let's use those two pieces of information to get the equation of that altitude!

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 3 = \frac{8}{5} \left(x - 5\right)$

$y - 3 = \frac{8}{5} x - 8$

$y = \frac{8}{5} x - 5$

Sweet. Now let's do the same for a different altitude. Again, it doesn't matter which one. Let's pick the point $\left(9 , 1\right)$ and the segment with endpoints $\left(5 , 3\right)$ and $\left(1 , 6\right)$. Some counting gives us a slope of $- \frac{3}{4}$ so the slope of the altitude is $\frac{4}{3}$. Let's do this!

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 1 = \frac{4}{3} \left(x - 9\right)$

$y - 1 = \frac{4}{3} x - 12$

$y = \frac{4}{3} x - 11$

Sweet.

Now we just need to find the point of intersection of those two altitudes. We can do that by setting them equal to each other.

$\frac{4}{3} x - 11 = \frac{8}{5} x - 5$

$\frac{20}{15} x - \frac{165}{15} = \frac{24}{15} x - \frac{75}{15}$

$15 \left[\frac{20}{15} x - \frac{165}{15}\right] = 15 \left[\frac{24}{15} x - \frac{75}{15}\right]$

$20 x - 165 = 24 x - 75$

$- 90 = 4 x$

$x = - 22.5$

So the x-coordinate is $- 22.5$. Let's plug that in to one of those two equations and find the y-coordinate:

$y = \frac{8}{5} x - 5$

$y = \frac{8}{5} \left(- 22.5\right) - 5$

$y = - 36 - 5$

$y = - 41$

So it looks like the point of intersection is $\left(- 22.5 , - 41\right)$

## Find the base and the altitude of the isoscele triangle of minimum area which circumscribes the ellipse #b^2x^2+a^2y^2=a^2b^2#, and whose base is parallel to the x-axis. How do I solve it?

dk_ch
Featured 1 month ago

The given equation of ellipse

${b}^{2} {x}^{2} + {a}^{2} {y}^{2} + {a}^{2} {b}^{2}$

Its standard form is

$\textcolor{b l u e}{{x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1}$

which has its center at origin $\left(0 , 0\right)$ and major axis and minor axis of lengths $2 a \mathmr{and} 2 b$ are along x-axis and y-axis respectively.

Parametrically any point on this ellipse is represented by $\left(a \cos \theta , b \sin \theta\right)$, where $\theta$ is the parameter

The equation of any tagent to this ellipse can be written as

$\textcolor{g r e e n}{\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1}$

Again by the given condition base of the triangle circumscribing the ellipse is parallel to X-axis.
The tangent for $\theta = {270}^{\circ}$ represents the base of the isoscele triangle and its equation becomes

$\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$

$\implies \frac{x \cos 270}{a} + \frac{y \sin 270}{b} = 1$

$\implies 0 - \frac{y}{b} = 1 \implies y = - b$

Now if we put $y = - b$ in the equation of the tangent then we can get the abscissa of the corner points of the base $\left(B C\right)$ of the triangle and this abscissa in the direction of postitive x-axis $\left(D C\right)$ represents the half of the length of the base of the triangle.

$\implies \frac{x \cos \theta}{a} + \frac{- b \sin \theta}{b} = 1$
$\implies x = \frac{a \left(1 + \sin \theta\right)}{\cos} \theta$

Hence $\textcolor{red}{\frac{1}{2} \text{Base} = \frac{a \left(1 + \sin \theta\right)}{\cos} \theta}$

It is obvious from the symmetry of the figure that ordinate of the point of intersection of the tangent with y-axis i.e. $x = 0$ is the distance of the top vertex from the center $\left(0 , 0\right)$

and then this ordinate

$O A = \frac{b}{\sin \theta}$

So height of the triangle

$\textcolor{m a \ge n t a}{\text{Height} = b + \frac{b}{\sin} \theta}$

$= \frac{b \left(1 + \sin \theta\right)}{\sin} \theta$

Hence area

$\textcolor{red}{S = \frac{1}{2} \text{Base"xx} H e i g h t}$

$\implies S = \frac{a \left(1 + \sin \theta\right)}{\cos} \theta \times \frac{b \left(1 + \sin \theta\right)}{\sin} \theta$

$\implies S = \frac{a \left(1 + \sin \theta\right)}{\cos} \theta \times \frac{b \left(1 + \sin \theta\right)}{\sin} \theta$

$\implies S = a b \left(\sec \theta \csc \theta + 2 \sec \theta + \tan \theta\right)$

Now imposing the condition of minimization i.e.$\frac{\mathrm{dS}}{d \theta} = 0$ we get

So

$\frac{d}{d \theta} \left(\left(a b \left(\sec \theta \csc \theta + 2 \sec \theta + \tan \theta\right)\right)\right) = 0$

$\implies \frac{d}{d \theta} \left(\sec \theta \csc \theta + 2 \sec \theta + \tan \theta\right) = 0$

$\implies \csc \theta \sec \theta \tan \theta - \sec \theta \csc \theta \cot \theta + 2 \sec \theta \tan \theta + {\sec}^{2} \theta = 0$

$\implies {\sec}^{2} \theta - \frac{1}{\sin} ^ 2 \theta + 2 \sin \frac{\theta}{\cos} ^ 2 \theta + {\sec}^{2} \theta = 0$

$\implies 2 {\sec}^{2} \theta - \frac{1}{\sin} ^ 2 \theta + 2 \sin \frac{\theta}{\cos} ^ 2 \theta = 0$

multiplying bothsides by ${\sin}^{2} \theta {\cos}^{2} \theta$

$\implies 2 {\sin}^{2} \theta - {\cos}^{2} \theta + 2 {\sin}^{3} \theta = 0$

$\implies 2 {\sin}^{2} \theta - 1 + {\sin}^{2} \theta + 2 {\sin}^{3} \theta = 0$

$\implies 2 {\sin}^{3} \theta + 3 {\sin}^{2} \theta - 1 = 0$

$\implies 2 {\sin}^{3} \theta - {\sin}^{2} \theta + 4 {\sin}^{2} \theta - 1 = 0$

$\implies {\sin}^{2} \theta \left(2 \sin \theta - 1\right) + \left(2 \sin \theta - 1\right) \left(2 \sin \theta + 1\right) = 0$

$\implies \left(2 \sin \theta - 1\right) \left({\sin}^{2} \theta + 2 \sin \theta + 1\right) = 0$

$\implies \left(2 \sin \theta - 1\right) {\left(\sin \theta + 1\right)}^{2} = 0$

Hence $\sin \theta = \frac{1}{2}$

$\sin \theta = - 1 \to \text{not possible}$ as the area becomes zero then.

Inserting the value $\sin \theta = \frac{1}{2}$ in the equation of Height and equation of Base we get

$\text{Height} = b + \frac{b}{\sin} \theta = b + \frac{b}{\frac{1}{2}} = 3 b$

$\frac{1}{2} \text{Base} = \frac{a \left(1 + \sin \theta\right)}{\cos} \theta$

$\implies \text{Base} = \frac{2 a \left(1 + \sin \theta\right)}{\sqrt{1 - {\sin}^{2} \theta}}$

$\implies \text{Base} = \frac{2 a \left(1 + \frac{1}{2}\right)}{\sqrt{1 - \frac{1}{4}}} = \frac{3 a}{\frac{\sqrt{3}}{2}} = 2 a \sqrt{3}$

## The equation x-y=4 and x^2+4xy+y^2=0 represents the side of which triangle?

dk_ch
Featured 2 weeks ago

Given

$x - y = 4. \ldots \left[1\right]$ as one side of the triangle.

and

${x}^{2} + 4 x y + {y}^{2}$ ,the equation of a pair of straight line representing other two sides of the triangle.

Now

${x}^{2} + 4 x y + {y}^{2} = 0$

$\implies {x}^{2} + 2 \cdot x \cdot 2 y + {\left(2 y\right)}^{2} - {\left(\sqrt{3} y\right)}^{2} = 0$

$\implies {\left(x + 2 y\right)}^{2} - {\left(\sqrt{3} y\right)}^{2} = 0$

$\implies \left(x + 2 y + \sqrt{3} y\right) \left(x + 2 y - \sqrt{3} y\right) = 0$

So

$x + \left(2 + \sqrt{3}\right) y = 0. \ldots . . \left[2\right]$

and

$x + \left(2 - \sqrt{3}\right) y = 0. \ldots . . \left[3\right]$ are the equations of other two sides of the triangle.

Now

slope of straight line [1]

${m}_{1} = 1$

slope of straight line [2]

${m}_{2} = - \frac{1}{2 + \sqrt{3}} = - \frac{2 - \sqrt{3}}{\left(2 + \sqrt{3}\right) \left(2 - \sqrt{3}\right)}$

$= - \frac{2 - \sqrt{3}}{4 - 3} = - \left(2 - \sqrt{3}\right)$

slope of straight line [3]

${m}_{3} = - \frac{1}{2 - \sqrt{3}} = - \frac{2 + \sqrt{3}}{\left(2 + \sqrt{3}\right) \left(2 - \sqrt{3}\right)}$

$= - \frac{2 + \sqrt{3}}{4 - 3} = - \left(2 + \sqrt{3}\right)$

Angle between line [1] and [2]

$\alpha = {\tan}^{-} 1 \left(\frac{{m}_{1} - {m}_{2}}{1 + {m}_{1} \times {m}_{2}}\right)$

$= {\tan}^{-} 1 \left(\frac{1 + 2 - \sqrt{3}}{1 - \left(2 - \sqrt{3}\right)}\right)$

$= {\tan}^{-} 1 \left(\frac{3 - \sqrt{3}}{\sqrt{3} - 1}\right)$

$= {\tan}^{-} 1 \left(\sqrt{3}\right) = \frac{\pi}{3}$

Angle between line [1] and [3]

$\beta = {\tan}^{-} 1 \left(\frac{{m}_{3} - {m}_{1}}{1 + {m}_{1} \times {m}_{3}}\right)$

$= {\tan}^{-} 1 \left(\frac{- \left(2 + \sqrt{3}\right) - 1}{1 - \left(2 + \sqrt{3}\right)}\right)$

$= {\tan}^{-} 1 \left(\frac{- \left(3 + \sqrt{3}\right)}{- \left(\sqrt{3} + 1\right)}\right)$

$= {\tan}^{-} 1 \left(\sqrt{3}\right) = \frac{\pi}{3}$

So the two angles of the triangle are same and each equals to $\frac{\pi}{3}$.

Hence the triangle is an equilateral triangle.

## What will be the area of shaded region (grey colored) if the given figure is square of side 6cm?

CW
Featured 3 weeks ago

shaded area $= 6 \cdot \left(3 \sqrt{3} - \pi\right) \approx 12.33 {\text{ cm}}^{2}$

#### Explanation:

See the figure above.

Green area $=$area of sector $D A F$ - yellow area

As $C F \mathmr{and} D F$ are the radius of the quadrants,
$\implies C F = D F = B C = C D = 6$
$\implies \Delta D F C$ is equilateral.
$\implies \angle C D F = {60}^{\circ}$
$\implies \angle A D F = {30}^{\circ}$
$\implies E F = 6 \sin 60 = 6 \cdot \frac{\sqrt{3}}{2} = 3 \sqrt{3}$

Yellow area = area of sector $C D F -$area $\Delta C D F$
$= \pi \cdot {6}^{2} \cdot \frac{60}{360} - \frac{1}{2} \cdot 3 \sqrt{3} \cdot 6$
$= 6 \pi - 9 \sqrt{3}$

Green area = $=$area of sector $D A F$ - yellow area
$= \pi \cdot {6}^{2} \cdot \frac{30}{360} - \left(6 \pi - 9 \sqrt{3}\right)$
$= 3 \pi - \left(6 \pi - 9 \sqrt{3}\right)$
$= 9 \sqrt{3} - 3 \pi$

Hence, the shaded area ${A}_{s}$ in your figure $= 2 \times$ green area
$\implies {A}_{s} = 2 \cdot \left(9 \sqrt{3} - 3 \pi\right)$
$= 18 \sqrt{3} - 6 \pi = 6 \left(3 \sqrt{3} - \pi\right) \approx 12.33 {\text{ cm}}^{2}$

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