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## Two corners of an isosceles triangle are at #(8 ,5 )# and #(6 ,7 )#. If the triangle's area is #15 #, what are the lengths of the triangle's sides?

Alan P.
Featured 3 months ago

Sides:$\left\{2.8284 , 10.7005 , 10.7005\right\}$

#### Explanation:

Side $\textcolor{red}{a}$ from $\left(8 , 5\right)$ to $\left(6 , 7\right)$
has a length of
$\textcolor{red}{\left\mid a \right\mid} = \sqrt{{\left(8 - 6\right)}^{2} + {\left(5 - 7\right)}^{2}} = 2 \sqrt{2} \approx 2.8284$

Not that $\textcolor{red}{a}$ can not be one of the equal length sides of the equilateral triangle since the maximum area such a triangle could have would be ${\left(\textcolor{red}{2 \sqrt{2}}\right)}^{2} / 2$ which is less than $15$

Using $\textcolor{red}{a}$ as the base and $\textcolor{b l u e}{h}$ as the height relative to that base, we have
$\textcolor{w h i t e}{\text{XXX}} \frac{\textcolor{red}{2 \sqrt{2}} \cdot \textcolor{b l u e}{h}}{2} = \textcolor{b r o w n}{15}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow \textcolor{b l u e}{h} = \frac{15}{\sqrt{2}}$

Using the Pythagorean Theorem:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{b} = \sqrt{{\left(\frac{15}{\sqrt{2}}\right)}^{2} + {\left(\frac{2 \sqrt{2}}{2}\right)}^{2}} \approx 10.70047$

and since the triangle is isosceles
$\textcolor{w h i t e}{\text{XXX}} c = b$

## 6 equal circular discs placed so that their centres lie on the circumference of a given circle with radius (r), and each disc touches its 2 neighbours. What is the radius of a 7th disc placed in the centre which will touch each of the each existing ones?

sente
Featured 3 months ago

Radius of ${7}^{\text{th}}$ disc = $\frac{r}{2}$

Perimeter of each outer disc = $\pi r$

#### Explanation:

If we sketch a picture of the situation, we can see that the centers of the six discs lie on the vertices of a regular hexagon inscribed in the circle.

As the triangles formed by connecting the vertices of the hexagon to its center are equilateral, we know that the sides of the hexagon are also of length $r$.

As the sides of the hexagons serve as distances between the centers of the discs, we know each disc must intersect at a midpoint of a side, meaning each disc has a radius of $\frac{r}{2}$.

Given that the discs each have radius $\frac{r}{2}$ and are centered at a distance of $r$ from the center, then the ${7}^{\text{th}}$ disc must also have a radius of $\frac{r}{2}$ to touch each of the surrounding discs.

Finally, the perimeter of each of the outside discs is given by the product of twice its radius and $\pi$. As they each have a radius of $\frac{r}{2}$, they each have a perimeter of $\pi r$

## Given the equilateral triangle inscribed in a square of side #s# find the ratio of #Delta BCR " to " DeltaPRD#?

dk_ch
Featured 2 months ago

$\frac{\Delta B C R}{\Delta P R D} = \frac{1}{2}$

#### Explanation:

As per given figure $\Delta B C R \cong \Delta A B P$
Since $\angle B C R = \angle A B P = {90}^{\circ} \to \text{angle of a square}$

$\text{hypotenuse "BR ="hypotenuse } B P$
( sides of equilateral $\Delta B R P$)

$A B = B C \to \text{sides of square ABCD}$

So $\angle C B R = \angle A B P$

Using trigonometry

In triangle BCR
$\angle C B R = \angle A B P = \frac{1}{2} \left(\angle A B C - \angle P B R\right)$

$= \frac{{90}^{\circ} - {60}^{\circ}}{2} = {15}^{\circ}$

and $\angle B R C = {90}^{\circ} - {15}^{\circ} = {75}^{\circ}$

If the length of each side of the equilareral triangle BPR be a,then
$B C = a \sin \angle C B R = a \cos {15}^{\circ}$

$C R = a \sin \angle C B R = a \sin {15}^{\circ}$

For $\Delta P R D$

$\angle P R D = \left(180 - \angle B R C - \angle B R P\right)$

$= {180}^{\circ} - {75}^{\circ} - {45}^{\circ} = {45}^{\circ}$

So $P D = D R = P R \cos {45}^{\circ} = a \cos {45}^{\circ} = a \sin {45}^{\circ}$

Now

$\frac{\Delta B C R}{\Delta P R D}$

$= \frac{\cancel{\frac{1}{2}} \times C R \times B C}{\cancel{\frac{1}{2}} \times P D \times D R}$

#=([email protected]/* */^@)/([email protected]/* */^@)#

#=([email protected]/* */^@)/([email protected]/* */^@)#

$= \sin {30}^{\circ} / \sin {90}^{\circ}$

$= \sin {30}^{\circ} = \frac{1}{2}$

Without using trigonometry

We have shown above $\Delta B C R \cong \Delta A B P$

So CR=AP

Now

$D R = C D - C R = A D - A P = D P$

Applying Pythagoras theorem for $\Delta P R D$ we get

$P {R}^{2} = D {P}^{2} + D {R}^{2} = 2 D {R}^{2}$

$\implies B {R}^{2} = 2 D {R}^{2} \to \text{since } P R = B R$

Niw applying Pythagoras theorem for $\Delta B R C$ we get

$B {R}^{2} = B {C}^{2} + C {R}^{2}$

So $2 D {R}^{2} = B {C}^{2} + C {R}^{2}$

$\implies 2 D {R}^{2} = B {C}^{2} + C {R}^{2}$

$\implies 2 D {R}^{2} = C {D}^{2} + C {R}^{2}$

$\implies 2 D {R}^{2} = {\left(C R + D R\right)}^{2} + C {R}^{2}$

$\implies 2 D {R}^{2} = C {R}^{2} + D {R}^{2} + 2 C R \cdot D R + C {R}^{2}$

$\implies 2 D {R}^{2} - D {R}^{2} = 2 C {R}^{2} + 2 C R \cdot D R$

$\implies D {R}^{2} = 2 C {R}^{2} + 2 C R \cdot D R$

$\implies \frac{1}{2} \cdot D {R}^{2} = C {R}^{2} + C R \cdot D R$

$\implies \frac{1}{2} \cdot D {R}^{2} = C R \left(C R + D R\right)$

$\implies \frac{1}{2} \cdot D R \cdot P D = 2 \cdot \frac{1}{2} \cdot C R \cdot C D$

$\implies \frac{1}{2} \cdot D R \cdot P D = 2 \cdot \frac{1}{2} \cdot C R \cdot B C$

$\implies \Delta P R D = 2 \cdot \Delta B C R$

$\implies \frac{\Delta B C R}{\Delta P R D} = \frac{1}{2}$

## Two circles have the following equations: #(x -6 )^2+(y -4 )^2= 64 # and #(x -5 )^2+(y -9 )^2= 49 #. Does one circle contain the other? If not, what is the greatest possible distance between a point on one circle and another point on the other?

Douglas K.
Featured 2 months ago

The farthest distance that between to points on the circle is the sum of the distance between the centers and the two radii:

${d}_{c} + {r}_{1} + {r}_{2} = \sqrt{26} + 8 + 7 \approx 20.1$

#### Explanation:

Here is a graph of the two circles:

The circles overlap but the larger circle does not contain the other. This could have been deduced by the following process.

We assign ${r}^{2}$ to the constant terms to the right of the equal sign in the given equations and then solve for r:

For circle 1:

${r}_{1}^{2} = 64$

${r}_{1} = 8$

For circle 2:

${r}_{2}^{2} = 49$

${r}_{2} = 7$

Compute the distance, ${d}_{c}$, between the two centers:

${d}_{c} = \sqrt{{\left(6 - 5\right)}^{2} + {\left(4 - 9\right)}^{2}}$

${d}_{c} = \sqrt{{\left(1\right)}^{2} + {\left(- 5\right)}^{2}}$

${d}_{c} = \sqrt{26} \approx 5.1$

If the larger circle contained the smaller, then the distance between the centers would be less than difference between the two radii, 1. This is not the case.

The farthest distance that between two points on the circle is the sum of the distance between the centers and the two radii:

${d}_{c} + {r}_{1} + {r}_{2} = \sqrt{26} + 8 + 7 \approx 20.1$

## An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(2 ,1 )# to #(3 ,7 )# and the triangle's area is #12 #, what are the possible coordinates of the triangle's third corner?

Tom G.
Featured 2 months ago

The coordinates of C are either ($- 1.3 \dot{91} \dot{8} , 4. \dot{64} \dot{8}$) or ($6.3 \dot{91} \dot{8} , 3. \dot{35} \dot{1}$)

This question took a lot longer to answer than I first anticipated!

#### Explanation:

There are two possible solutions for point C, both of which must lie on the line bisecting AB, as ABC is isosceles and all the points on this perpendicular line are equidistant from points A and B.

The mid-point of AB is straightforward enough:

$\left(\frac{2 + 3}{2} , \frac{1 + 7}{2}\right) = \left(\frac{5}{2} , \frac{8}{2}\right) = \left(2.5 , 4\right)$

AB has a slope of 6, because an increase of 1 on the $x$ axis results in an increase of 6 on the $y$ axis, and the slope of the perpendicular on which ${C}_{1}$ and ${C}_{2}$ lie has a slope of -1/6, because an increase of 6 along the $x$ axis results in a decrease of 1 on the $y$ axis.

Now triangle ABC has an area of 12

The area $A$ of a triangle is defined as $A = \frac{1}{2} b h$, where $b$ is the length of the base and $h$ is the height. In this case, we can consider side AB to be the base $b$, and the distance between the mid-point of AB and C to be the height $h$.

AB = $\sqrt{{\left(7 - 1\right)}^{2} + {\left(3 - 2\right)}^{2}} = \sqrt{{6}^{2} + {1}^{2}} = \sqrt{37} = b$

So with this value for $b$ we can solve for $h$ as follows:

$12 = \frac{1}{2} \times \sqrt{37} \times h$

$\therefore h = \frac{24}{\sqrt{37}}$

So point C must lie on one side or the other of AB, a distance of $\frac{24}{\sqrt{37}}$ along the line which bisects it.

In order to calculate the coordinates, we can imagine a right-angled triangle with a hypotenuse of length $\frac{24}{\sqrt{37}}$ with the opposite side length $n$ and the adjacent side length $6 n$

From Pythagorus, we know that ${h}^{2} = {a}^{2} + {o}^{2}$

$\therefore {\left(\frac{24}{\sqrt{37}}\right)}^{2} = 36 {n}^{2} + {n}^{2} = 37 {n}^{2}$

$\therefore {n}^{2} = {\left(\frac{24}{\sqrt{37}}\right)}^{2} / 37$

So $n = \frac{24}{37} = o$ and $6 n = \frac{144}{37} = a$

Point ${C}_{1}$ on the left of AB will have coordinates as follows:

$\left(2.5 - a , 4 + o\right)$

$2.5 - \frac{144}{37} = - 1.3 \dot{91} \dot{8}$

$4 + \frac{24}{37} = 4. \dot{64} \dot{8}$

So ${C}_{1} = \left(- 1.3 \dot{91} \dot{8} , 4. \dot{64} \dot{8}\right)$

Point ${C}_{2}$ on the right of AB will have coordinates as follows:

$\left(2.5 + a , 4 - o\right)$

$2.5 + \frac{144}{37} = 6.3 \dot{91} \dot{8}$

$4 - \frac{24}{37} = 3. \dot{35} \dot{1}$

So ${C}_{2} = \left(6.3 \dot{91} \dot{8} , 3. \dot{35} \dot{1}\right)$

## The common tangents to the circle x^2+y^2=2 and the parabola y^2=8x touch the circle at the points P,Q and the parabola at the points R,S. Then the area of quadrilateral PQRS is?

Cesareo R.
Featured 1 month ago

$15$ area units.

#### Explanation:

Given two conics

${C}_{1} \left({x}_{1} , {y}_{1}\right) = {x}_{1}^{2} + {y}_{1}^{2} - 2 = 0$ and
${C}_{2} \left({x}_{2} , {y}_{2}\right) = {y}_{2}^{2} - 8 {x}_{2} = 0$

The tangent space to those conics is

${m}_{1} = \frac{{\mathrm{dy}}_{1}}{{\mathrm{dx}}_{1}} = - {x}_{1} / {y}_{1}$
${m}_{2} = \frac{{\mathrm{dy}}_{2}}{{\mathrm{dx}}_{2}} = \frac{4}{y} _ 2$

$\left\{\begin{matrix}{y}_{2} - {y}_{1} = {m}_{1} \left({x}_{2} - {x}_{1}\right) \\ {y}_{1} - {y}_{2} = {m}_{2} \left({x}_{1} - {x}_{2}\right)\end{matrix}\right.$

Solving now

$\left\{\begin{matrix}{x}_{1}^{2} + {y}_{1}^{2} - 2 = 0 \\ {y}_{2}^{2} - 8 {x}_{2} = 0 \\ {y}_{2} - {y}_{1} = {m}_{1} \left({x}_{2} - {x}_{1}\right) \\ {y}_{1} - {y}_{2} = {m}_{2} \left({x}_{1} - {x}_{2}\right)\end{matrix}\right.$

for ${x}_{1} , {x}_{2} , {y}_{1} , {y}_{2}$ we obtain

$\left(\begin{matrix}Q & \left(- 1 - 1\right) \\ P & \left(- 1 1\right) \\ S & \left(2 - 4\right) \\ R & \left(2 4\right)\end{matrix}\right)$

Attached a plot focusing the main elements

The area is $\frac{3 \left(2 + 8\right)}{2} = 15$.

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