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Three vertices of parallelogram WXYZ are W(3,1), X(2,7),and Z(4,0). How do I find the coordinates of vertex Y?

CW
Featured 4 months ago

$\left(3 , 6\right) \mathmr{and} \left(1 , 8\right)$

Explanation:

Some of the properties of a parallelogram:
a) Opposite sides are parallel.
b) Opposite sides are congruent.
c) Opposite angles are congruent.
d) The diagonals bisect each other, $\implies$ the two diagonals have the same midpoint.

The Midpoint Formula: The midpoint of two points, $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , y 2\right)$ is the point M found by the following formula :

$M = \left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$

First, let $X Z \mathmr{and} W Y$ be the two diagonals,
$\implies X Z \mathmr{and} W Y$ have the same midpoint,

$\implies \frac{2 + 4}{2} = \frac{3 + x}{2} , \implies x = 3$
$\implies \frac{7 + 0}{2} = \frac{1 + y}{2} , \implies y = 6$

Hence, the first coordinates of vertex $Y$ are $\left(3 , 6\right)$

Then, let $X W \mathmr{and} Y Z$ be the two diagonals,
$\implies X W \mathmr{and} Y Z$ have the same midpoint.

$\implies \frac{2 + 3}{2} = \frac{x + 4}{2} , \implies x = 1$
$\implies \frac{7 + 1}{2} = \frac{y + 0}{2} , \implies y = 8$

So the second coordinates for vertex Y are $\left(1 , 8\right)$

Hence, the two possible coordinates for vertex Y are $\left(3 , 6\right) \mathmr{and} \left(1 , 8\right)$

A dog is tied to the corner of a house with a regular hexagonal base that measures 6 ft. on each side. If the rope is 12 ft. in length, what is the area in square feet of the region outside the house that the dog can reach?

Shwetank Mauria
Featured 2 months ago

Area of the region outside the house that the dog can reach is $339.29$ square feet

Explanation:

As the dog is tied to a corner of a hexagon, with a rope of length of $12$ feet, his movement is indicated in the following figure.

As may be seen, he has freedom of movement in a circle of $12$ feet radius but within a sector of ${240}^{\circ}$ only. Then he reaches next corner, on either side and has a movement restricted to $6$ feet in an arc of ${60}^{\circ}$ only.

Hence the area of the region outside the house that the dog can reach is given by

$\frac{240}{360} \times \pi \times {12}^{2} + 2 \times \frac{60}{360} \times \pi \times {6}^{2}$

= $\frac{2}{3} \times 144 \times \pi + 2 \times \frac{1}{6} \times {6}^{2} \times \pi$

= $96 \pi + 12 \pi$

= $108 \pi$

= $108 \times 3.1416$

= $339.29$ square feet

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

Douglas K.
Featured 2 months ago

The equation of the circle is

${\left(x - - \frac{12}{10}\right)}^{2} + {\left(y - \frac{3}{2}\right)}^{2} = {\left(\frac{\sqrt{369}}{10}\right)}^{2}$

Explanation:

The standard Cartesian form for the equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \text{ [1]}$

where $\left(x , y\right)$ is any point on the circle, $\left(h , k\right)$ is the center point and r is the radius.

Given: The tangent line $4 x - 5 y = 0$ is tangent to the circle at the point $\left(0 , 0\right)$

Because the radius is the circle is perpendicular to the tangent, we can use the tangent line to find an equation that pass through the center. Write the tangent line in slope-intercept form:

$- 5 y = - 4 x$

$y = \frac{4}{5} x$

This makes the center line:

$y = - \frac{5}{4} x$

Evaluate this at the point $\left(h , k\right)$:

$k = - \frac{5}{4} h \text{ [2]}$

Evaluate equation [1] and the points $\left(0 , 0\right)$ and $\left(0 , 3\right)$

${\left(0 - h\right)}^{2} + {\left(0 - k\right)}^{2} = {r}^{2} \text{ [3]}$
${\left(0 - h\right)}^{2} + {\left(3 - k\right)}^{2} = {r}^{2} \text{ [4]}$

Expand the squares:

${h}^{2} + {k}^{2} = {r}^{2} \text{ [5]}$
${h}^{2} + 9 - 6 k + {k}^{2} = {r}^{2} \text{ [6]}$

Subtract equation [5] from equation [6]:

$9 - 6 k = 0$

$k = \frac{3}{2}$

Substitute into equation [2]:

$\frac{3}{2} = - \frac{5}{4} h$

$h = - \frac{12}{10}$

Substitute into equation [5]:

${\left(\frac{3}{2}\right)}^{2} + {\left(- \frac{12}{10}\right)}^{2} = {r}^{2}$

${\left(\frac{15}{10}\right)}^{2} + {\left(- \frac{12}{10}\right)}^{2} = {r}^{2}$

$\frac{225}{100} + \frac{144}{100} = {r}^{2}$

$r = \frac{\sqrt{369}}{10}$

The equation of the circle is

${\left(x - - \frac{12}{10}\right)}^{2} + {\left(y - \frac{3}{2}\right)}^{2} = {\left(\frac{\sqrt{369}}{10}\right)}^{2}$

Here is a graph of, the circle (red), the center point (blue), the tangent line (green) and the center line (yellow):

A triangle is divided into seven triangles. The areas of four of them are 420 #cm^2#, 80 #cm^2#, 60 #cm^2# and 30 #cm^2# as shown in the diagram on the right. Find the area of triangle AEF, in #cm^2#?

CW
Featured 2 months ago

Area $A E F = 1512 c {m}^{2}$

Explanation:

When a line from one of the vertex of a triangle, meets the opposite side, the ratio between the areas of the two triangles is same as the ratio of the two parts in which it divides the opposite side.

By corollary, in a quadrilateral whose diagonals divide it in four triangles, areas of two adjacent triangles are in the same ratio as the ratio of other two triangles.

Let $\Delta X Y Z$ denotes area of triangle $X Y Z$.
Given $\Delta B D G : \Delta G D E = 30 : 60 = 1 : 2$
$\implies \Delta B D C : \Delta D C E = \Delta B D C : 80 = 1 : 2$
$\implies \Delta B D C = 40$

As $\Delta G C B : B C F = 70 : 420 = 1 : 6$
$\implies G B : B F = 1 : 6$
$\implies \Delta G A B : \Delta B A F = 1 : 6$
let $\Delta G A B = 1 a , \mathmr{and} \Delta B A F = 6 a$

As $\Delta E B C : \Delta C B F = 120 : 420 = 2 : 7$
$\implies E C : C F = 2 : 7$
$\implies \Delta E A C : \Delta C A F = 2 : 7$
$\implies \left(1 a + 210\right) : \left(6 a + 420\right) = 2 : 7$
$\implies \left(1 a + 210\right) \times 7 = \left(6 a + 420\right) \times 2$
$\implies 1470 - 840 = 12 a - 7 a$
$\implies 5 a = 630$
$\implies a = 126$

Area of triangle $A E F = 1 a + 6 a + 210 + 420 = 7 a + 210 + 420$
$= 7 \times 126 + 210 + 420 = 1512 c {m}^{2}$

Can someone help me with this problem set? My professor haven't taught us that much and she already gave us this. I want to be able to solve this in the near future. Thank you so much!

CW
Featured 1 month ago

see explanation.

Explanation:

1.
Area of a cricle $A = \pi {r}^{2}$, where $r$ is the radius of the circle.
$\implies 16 \pi = \pi {r}^{2} , \implies r = \sqrt{16} = 4$ m
circumference of a circle $c = 2 \cdot \pi \cdot r$
$\implies c = 2 \cdot \pi \cdot 4 = 8 \pi$ m

2.

Let ${r}_{1}$ be the radius of the inscribed circle and $s$ be the side of the square.
As shown in the figure, the inscribed circle has a diameter $d = s$
$\implies {r}_{1} = \frac{d}{2} = \frac{s}{2} = \frac{4}{2} = 2$ cm
Let ${r}_{2}$ be the radius of the circumsribed circle,
${r}_{2}^{2} = {r}_{1}^{2} + {r}_{1}^{2} = 2 {r}_{1}^{2}$
$\implies {r}_{2} = \sqrt{2} {r}_{1} = 2 \sqrt{2}$ cm

3.

The largest circle that can fit insides a square of edge of $4$ in. has a diameter $d = 4$ in.
Area of circle ${A}_{c} = \pi {r}^{2} = \pi {\left(\frac{d}{2}\right)}^{2} = \frac{\pi {d}^{2}}{4} = \frac{16 \pi}{4} = 4 \pi$ sq. in.
Let ${A}_{s}$ be the area of the square, $\implies {A}_{s} = {4}^{2} = 16$ sq. in.
Area of the materail wasted = shaded area = ${A}_{s} - {A}_{c}$
$= 16 - 4 \pi = 4 \left(4 - \pi\right) \approx 3.434$ sq. in.

4.

Area of washer (one face) = shaded area $= \pi {r}_{1}^{2} - \pi {r}_{2}^{2}$
given ${r}_{1} = \frac{1}{2} , \mathmr{and} {r}_{2} = \frac{1}{4}$
$= \pi \left({r}_{1}^{2} - {r}_{2}^{2}\right) = \pi \left({\left(\frac{1}{2}\right)}^{2} - {\left(\frac{1}{4}\right)}^{2}\right) = \frac{3}{16} \pi$ sq. in.

5.

An equilateral triangle is a triangle in which all 3 sides are equal and all 3 angles are each ${60}^{\circ}$
As two tangents to a circle from an external point are equal,
$\implies B D = B E$
As $D \mathmr{and} E$ are points of tangency and $B O$ is a common side,
$\implies \Delta O D B \mathmr{and} \Delta O E B$ are congruent.
$\implies B O$ bisects $\angle D B E$, which is ${60}^{\circ}$
$\implies \angle O B D = \angle O B E = {30}^{\circ}$
$B E = \frac{1}{2} B C = \frac{6}{2} = 3$
$B O \cos 30 = B E = 3$
$B O = \frac{3}{\cos} 30 = \frac{3}{\frac{\sqrt{3}}{2}} = \frac{6}{\sqrt{3}} = \frac{6 \sqrt{3}}{3} = 2 \sqrt{3}$
$O E = B O \sin 30 = 2 \sqrt{3} \times \frac{1}{2} = \sqrt{3}$
$B O$ is the radius of the circumsribed circle and $O E$ the radius of the inscribed circle.
annulus area = shaded area $= \pi {\left(B O\right)}^{2} - \pi {\left(O E\right)}^{2}$
$= \pi {\left(2 \sqrt{3}\right)}^{2} - \pi {\left(\sqrt{3}\right)}^{2}$
$= 12 \pi - 3 \pi = 9 \pi {\text{ cm}}^{2}$

Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A ? plzz give explanation with appropriate image

dk_ch
Featured 2 weeks ago

Let the equation of the circle be

${x}^{2} + {y}^{2} = {a}^{2.} \ldots \ldots \ldots \ldots \left[1\right]$,

where $a$ is the radius and $O \left(0 , 0\right)$is the center of the circle.

Let $A \left(a \cos \theta , a \sin \theta\right)$ be any point on the circle parametrically and $\theta$ is the parameter.

So the equation of the tangent on the circle at $A$ will be given by

$x \times a \cos \theta + y \times a \sin \theta = {a}^{2}$

$\mathmr{and} , x \cos \theta + y \sin \theta = a$

So the equations of set of chords parallel to that tangent can be written as

$x \cos \theta + y \sin \theta = c \ldots \ldots \ldots . \left[2\right]$,

where c is another parameter which changes from chord to chord.

Multiplying {1} by ${\cos}^{2} \theta$ we get

${x}^{2} {\cos}^{2} \theta + {y}^{2} {\cos}^{2} \theta = {a}^{2} {\cos}^{2} \theta \ldots \ldots \ldots \ldots . \left[3\right]$

From [2] and [3] we can write

${\left(c - y \sin \theta\right)}^{2} + {y}^{2} {\cos}^{2} \theta = {a}^{2} {\cos}^{2} \theta$

$\implies {y}^{2} - 2 c \sin \theta y + \left({c}^{2} - {a}^{2} {\cos}^{2} \theta\right) = 0$

If ${y}_{1} \mathmr{and} {y}_{2}$ are two roots of this quadratic equation then

${y}_{1} + {y}_{2} = 2 c \sin \theta$

If $\left(h , k\right)$ represents the coordinates of the mid point of the chord then $k = \frac{{y}_{1} + {y}_{2}}{2} = c \sin \theta$

Now $\left(h , k\right)$ is on [2} so we get

$h \cos \theta + k \sin \theta = c \ldots \ldots \ldots . \left[4\right]$

$\implies h \cos \theta + c \sin \theta \times \sin \theta = c$

$\implies h \cos \theta = c \left(1 - {\sin}^{2} \theta\right)$

$\implies h \cos \theta = c {\cos}^{2} \theta$

$\implies h = c \cos \theta$
So

$\frac{k}{h} = \tan \theta$

Hence $y = x \tan \theta$ will represent the locus of the mid point of parallel chords of a particular tangent for particular value of $\theta$.

Now equation of the diameter of the circle passing through $O \left(0 , 0\right) \mathmr{and} P \left(a \cos \theta , a \sin \theta\right)$ is given by

$\frac{y - 0}{x - 0} = \frac{a \sin \theta - 0}{a \cos \theta - 0}$

$\implies y = x \tan \theta$, which is also the locus of the mid point of the chords parallel to the tangent.

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