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## What is the perimeter of a triangle with corners at (7 ,5 ), (1 ,2 ), and (4 ,8 )?

Mr. Raj
Featured 6 months ago

$17.66 \text{ units}$

#### Explanation:

To find the perimeter, you need to find the distance of each side and add them up right?

So first you need to figure out the length of each side, and you can do this by plotting the corner points and assigning vectors to each side.
When you plot the points, it should look something like this:

To find a vector $\vec{P Q}$ between two points $P \left({x}_{1} , {y}_{1}\right) \text{ and } Q \left({x}_{2} , {y}_{2}\right)$ You can use the equation $\vec{P Q} = \left[\begin{matrix}{x}_{2} - {x}_{1} \\ {y}_{2} - {y}_{1}\end{matrix}\right]$

Hence to calculate the vectors between $A \left(7 , 5\right) , B \left(1 , 2\right) \text{ and } C \left(4 , 8\right)$, the below calculations can be conducted:
$\vec{A B} = \left[\begin{matrix}1 - 7 \\ 2 - 5\end{matrix}\right] = \left[\begin{matrix}- 6 \\ - 3\end{matrix}\right]$

$\vec{B C} = \left[\begin{matrix}4 - 1 \\ 8 - 2\end{matrix}\right] = \left[\begin{matrix}3 \\ 6\end{matrix}\right]$

$\vec{A C} = \left[\begin{matrix}4 - 7 \\ 8 - 5\end{matrix}\right] = \left[\begin{matrix}- 3 \\ 3\end{matrix}\right]$

Then the length of each vector needs to be calculated. To calculate the length of vector $\vec{M N} = \left[\begin{matrix}x \\ y\end{matrix}\right]$

You can use:
$\text{distance of } \vec{M N} = | \vec{M N} | = \sqrt{{x}^{2} + {y}^{2}}$

Hence when we are calculating the distances of the vectors of the triangle:

$| \vec{A B} | = \sqrt{{\left(- 6\right)}^{2} + {\left(- 3\right)}^{2}} = \sqrt{36 + 9} = \sqrt{45} \text{ units}$

$| \vec{B C} | = \sqrt{{\left(3\right)}^{2} + {\left(6\right)}^{2}} = \sqrt{9 + 36} = \sqrt{45} \text{ units}$

$| \vec{A C} | = \sqrt{{\left(- 3\right)}^{2} + {\left(3\right)}^{2}} = \sqrt{9 + 9} = \sqrt{18} \text{ units}$

Therefore the total perimeter is
$| \vec{A B} | + | \vec{B C} | + | \vec{A C} | = \sqrt{45} + \sqrt{45} + \sqrt{18} = 6 \sqrt{5} + 3 \sqrt{2} \approx 17.66 \text{ units}$

## An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (7 ,5 ) to (8 ,2 ) and the triangle's area is 27 , what are the possible coordinates of the triangle's third corner?

Featured 5 months ago

The coordinates are $\left(23.7 , 8.9\right)$ and $\left(- 8.7 , - 1.9\right)$

#### Explanation:

The length of side $A = \sqrt{{\left(7 - 8\right)}^{2} + {\left(5 - 2\right)}^{2}} = \sqrt{10}$

Let the height of the triangle be $= h$

The area of the triangle is

$\frac{1}{2} \cdot \sqrt{10} \cdot h = 27$

The altitude of the triangle is $h = \frac{27 \cdot 2}{\sqrt{10}} = \frac{54}{\sqrt{10}}$

The mid-point of $A$ is $\left(\frac{15}{2} , \frac{7}{2}\right)$

The gradient of $A$ is $= \frac{2 - 5}{8 - 7} = - 3$

The gradient of the altitude is $= \frac{1}{3}$

The equation of the altitude is

$y - \frac{7}{2} = \frac{1}{3} \left(x - \frac{15}{2}\right)$

$y = \frac{1}{3} x - \frac{5}{2} + \frac{7}{2} = \frac{1}{3} x + 1$

The circle with equation

${\left(x - \frac{15}{2}\right)}^{2} + {\left(y - \frac{7}{2}\right)}^{2} = {54}^{2} / 10 = 291.6$

The intersection of this circle with the altitude will give the third corner.

${\left(x - \frac{15}{2}\right)}^{2} + {\left(\frac{1}{3} x + 1 - \frac{7}{2}\right)}^{2} = 291.6$

${x}^{2} - 15 x + \frac{225}{4} + \frac{1}{9} {x}^{2} - \frac{5}{3} x + \frac{25}{4} = 291.6$

$1.11 {x}^{2} - 16.7 x - 229.1 = 0$

$x = \frac{16.7 \pm \sqrt{{16.7}^{2} + 4 \cdot 1.11 \cdot 229.1}}{2 \cdot 1.11}$

$= \frac{16.7 \pm 36}{2.22}$

${x}_{1} = 23.74$

${x}_{2} = - 8.69$

The points are $\left(23.7 , 9.65\right)$ and $\left(- 8.7 , - 1.9\right)$

graph{(y-1/3x-1)((x-7.5)^2+(y-3.5)^2-291.6)((x-7)^2+(y-5)^2-0.05)((x-8)^2+(y-2)^2-0.05)(y-5+3(x-7))=0 [-12, 28, -10, 10]}

## How can I find the length of a side of this regular star-sharped pentagon inscribed in a circle with a radius of 25 cm ? (see details for image)

CW
Featured 3 months ago

see explanation

#### Explanation:

1) to find length $B E$

Let $O$ be the center of the circle.
Let $\angle A O B = x$
$\implies \angle x = \frac{360}{5} = {72}^{\circ}$
Consider $\Delta O B E$,
As $O B = O E = r = 25$ cm
$\implies \Delta O B E$ is isosceles
$\implies \angle O B E = \angle O E B$,
$\implies M$ is the mid-point of $B E$
$\implies \Delta O M B \mathmr{and} \Delta O M E$ are two congruent right triangles.
$\implies B M = r \sin x$
$\implies B E = 2 \cdot B M = 2 r \sin x$
$= 2 \cdot 25 \cdot \sin 72 = 2 \cdot 25 \cdot \left(\frac{\sqrt{10 + 2 \sqrt{5}}}{4}\right)$
$= \frac{25 \sqrt{10 + 2 \sqrt{5}}}{2} \approx 47.55$ cm

2) to find length $B A$ or length $B F$

$\Delta B O E$ is isosceles
$x = \frac{360}{5} = {72}^{\circ}$
$\implies y = \frac{180 - 2 x}{2} = \frac{180 - 144}{2} = {18}^{\circ}$
$B O A$ is also isosceles.
$\implies \angle O B A = \frac{180 - x}{2} = \frac{180 - 72}{2} = {54}^{\circ}$
$\implies \angle F B A = z = 54 - 18 = {36}^{\circ}$
$\implies B A = 2 r \sin \left(\frac{x}{2}\right) = 2 \cdot 25 \cdot \sin 36$
$= 2 \cdot 25 \cdot \frac{\sqrt{10 - 2 \sqrt{5}}}{4} = \frac{25 \sqrt{10 - 2 \sqrt{5}}}{2} \approx 29.39$ cm
$\Delta B F A$ is also isosceles.
$\implies 2 \cdot B F \cdot \cos z = B A$
$\implies B F = \frac{B A}{2 \cdot \cos z} = \frac{B A}{2 \cdot \cos 36}$
$= \frac{1}{2} \cdot \frac{25 \sqrt{10 - 2 \sqrt{5}}}{2} \cdot \frac{4}{\sqrt{5} + 1}$
$= \frac{25 \sqrt{10 - 2 \sqrt{5}}}{\sqrt{5} + 1} \approx 18.16$ cm

## In triangle ABC AM is median prove that AB^2+AC^2=2×AM^2+1/2+BC^2 solve by apollonius theorem?

Shwetank Mauria
Featured 1 month ago

#### Explanation:

Let us consider the following figure, where we have the median $A M$ and $A P \bot B C$.

Now in $\Delta A B M$, we have

$A {B}^{2} = A {P}^{2} + P {B}^{2} = A {P}^{2} + {\left(B M - P M\right)}^{2}$

= $\textcolor{red}{A {P}^{2}} + B {M}^{2} + \textcolor{red}{P {M}^{2}} - 2 B M \times P M$

or $A {B}^{2} = \textcolor{red}{A {M}^{2}} + B {M}^{2} - 2 B M \times P M$ .....................(1)

and in $\Delta A C M$, we have

$A {C}^{2} = A {P}^{2} + P {C}^{2} = A {P}^{2} + {\left(M C + P M\right)}^{2}$

= $\textcolor{red}{A {P}^{2}} + M {C}^{2} + \textcolor{red}{P {M}^{2}} - 2 M C \times P M$

or $A {C}^{2} = \textcolor{red}{A {M}^{2}} + M {C}^{2} - 2 M C \times P M$

But as $B M = M C$, we can write it as

$A {C}^{2} = \textcolor{red}{A {M}^{2}} + B {M}^{2} - 2 B M \times P M$ .....................(2)

Adding (1) and (2), we get

$A {B}^{2} + A {C}^{2} = 2 A {M}^{2} + 2 B {M}^{2}$ and as $B M = \frac{B C}{2}$

$A {B}^{2} + A {C}^{2} = 2 A {M}^{2} + \frac{2 B {C}^{2}}{4}$

or $A {B}^{2} + A {C}^{2} = 2 A {M}^{2} + \frac{1}{2} B {C}^{2}$

This is Apollonius theorem.

## x^2+y^2 +6x =0 and x^2+y^2-2x=0 are two circles then the area of the triangle formed by their common tangent is?

CW
Featured 1 month ago

area $= 3 \sqrt{3} {\text{ units}}^{2}$

#### Explanation:

Circle 1 : ${x}^{2} + {y}^{2} + 6 x = 0 , \implies$ center ${C}_{1} = \left(- 3 , 0\right) , {r}_{1} = 3$,
Circle 2: ${x}^{2} + {y}^{2} - 2 x = 0 , \implies$ center ${C}_{2} = \left(1 , 0\right) , {r}_{2} = 1$
As the two circles touch each other externally, they have 3 common tangents.
Obviously, Y-axis $\left(x = 0\right)$ is one of the common tangents.
Let $y = m x + c$ be the equation of the other tangents.
Circle 1 : ${x}^{2} + {\left(m x + c\right)}^{2} + 6 x = 0$
$\implies {x}^{2} + {m}^{2} {x}^{2} + 2 m c x + {c}^{2} + 6 x = 0$
$\implies \left(1 + {m}^{2}\right) {x}^{2} + 2 \left(m c + 3\right) x + {c}^{2} = 0$
As the line $y = m x + c$ is tangent to the circle, it touches the circle at one point, so set the discriminant ${b}^{2} - 4 a c$ equal to $0$,
$\implies {\left(2 \left(m c + 3\right)\right)}^{2} - 4 \left(1 + {m}^{2}\right) {c}^{2} = 0$
$\implies 4 {\left(m c + 3\right)}^{2} - 4 \left(1 + {m}^{2}\right) {c}^{2} = 0$
$\implies {\left(m c + 3\right)}^{2} - \left(1 + {m}^{2}\right) {c}^{2} = 0$
$\implies {m}^{2} {c}^{2} + 6 m c + 9 - {c}^{2} - {m}^{2} {c}^{2} = 0$
$\implies 6 m c = {c}^{2} - 9$ ----------- $\textcolor{red}{E Q \left(1\right)}$
Similarly, Circle 2 : ${x}^{2} + {\left(m x + c\right)}^{2} - 2 x = 0$
$\implies \left(1 + {m}^{2}\right) {x}^{2} + 2 \left(m c - 1\right) x + {c}^{2} = 0$
similarly, set the discriminant ${b}^{2} - 4 a c = 0$,
$\implies 2 m c = - {c}^{2} + 1$ ------------$\textcolor{red}{E Q \left(2\right)}$
Solving $E Q \left(1\right) \mathmr{and} E Q \left(2\right)$ gives $c = \pm \sqrt{3}$
a) when $c = \sqrt{3} , \implies m = - \frac{1}{\sqrt{3}}$
b) when $c = - \sqrt{3} , \implies m = \frac{1}{\sqrt{3}}$
$\implies$ equations of the two tangents :
$y = - \frac{1}{\sqrt{3}} x + \sqrt{3} , \implies \sqrt{3} y = - x + 3$, and,
$y = \frac{1}{\sqrt{3}} x - \sqrt{3} , \implies \sqrt{3} y = x - 3$
Setting $x = 0 , \sqrt{3} y = - x + 3 , \mathmr{and} \sqrt{3} y = x - 3$ equal to one another, we get :
$P = \left(3 , 0\right) , Q = \left(0 , \sqrt{3}\right) \mathmr{and} R = \left(0 , - \sqrt{3}\right)$
As $P Q = P R = Q R = \sqrt{12} = 2 \sqrt{3}$,
$\implies \Delta P Q R$ is an equilateral triangle.
$\implies$ area of $\Delta P Q R = \frac{\sqrt{3}}{4} \cdot {\left(2 \sqrt{3}\right)}^{2} = 3 \sqrt{3} {\text{ units}}^{2}$

## If one side of a square is on line y=2*x-17, and other two points are on a parabola y=x^2, how is its minimum area detemined?

dk_ch
Featured 1 month ago

Given that

• the equation of parabola $y = {x}^{2}$
• the equation of a straight line $y = 2 x - 17$

Let the corner points $A \left({t}_{1} , {t}_{1}^{2}\right) \mathmr{and} D \left({t}_{2} , {t}_{2}^{2}\right)$ of the square $A B C D$ are on the parabola and other two corner points $B \mathmr{and} D$ are on the given line $y = 2 x - 17$

So slope of AD $= \frac{{t}_{1}^{2} - {t}_{2}^{2}}{{t}_{1} - {t}_{2}} = 2 \to \text{the slope of the given line}$ , since side AD is parallel to the given line.

Hence we have ${t}_{1} + {t}_{2} = 2$

The length of side $A D = \sqrt{{\left({t}_{1} - {t}_{2}\right)}^{2} + {\left({t}_{1}^{2} - {t}_{2}^{2}\right)}^{2}}$

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

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

Now length of the perpendicular from $A$ to the line

$A B = \frac{{t}_{1}^{2} - 2 {t}_{1} + 17}{\sqrt{{1}^{2} + {2}^{2}}} = \frac{{t}_{1}^{2} - 2 {t}_{1} + 17}{\sqrt{5}}$

$A B C D$ being a square

We have $A B = A D$

$\implies \frac{{t}_{1}^{2} - 2 {t}_{1} + 17}{\sqrt{5}} = 2 \sqrt{5} \left({t}_{1} - 1\right)$

$\implies {t}_{1}^{2} - 2 {t}_{1} + 17 - 10 \left({t}_{1} - 1\right) = 0$

$\implies {t}_{1}^{2} - 12 {t}_{1} + 27 = 0$

$\implies {t}_{1}^{2} - 9 {t}_{1} - 3 {t}_{1} + 27 = 0$

$\implies {t}_{1} \left({t}_{1} - 9\right) - 3 \left({t}_{1} - 9\right) = 0$

=>(t_1-9)((t_1-3)=0

Hence we get

${t}_{1} = 3 \mathmr{and} 9$

When ${t}_{1} = 3$, the area of the square will be minimum

So

$\text{AREA"_"min} = A {D}^{2}$ at ${t}_{1} = 3$

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

$= 20 \times {\left(3 - 1\right)}^{2}$

$= 80$squnit

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