8
Active contributors today

## Point A is at #(1 ,-1 )# and point B is at #(-2 ,1 )#. Point A is rotated #(3pi)/2 # clockwise about the origin. What are the new coordinates of point A and by how much has the distance between points A and B changed?

sankarankalyanam
Featured 1 week ago

New coordinates of #color(red)(A ((1),(1))#

Change (reduction) in distance from $\overline{A B} , \overline{A ' B}$ is

$\implies \sqrt{13} - 3 \approx \textcolor{g r e e n}{0.61}$

#### Explanation:

A (1, -1), B (-2,1) and rotated by $\frac{3 \pi}{2}$ clockwise about the origin.

$A \left(\begin{matrix}1 \\ - 1\end{matrix}\right) \to A ' \left(\begin{matrix}1 \\ 1\end{matrix}\right)$

Distance #vec(AB) = sqrt((1+2)^2 + (-1-1)^2) = color(blue)(sqrt13#

Distance $\vec{A ' B} = \sqrt{{\left(1 + 2\right)}^{2} + {\left(1 - 1\right)}^{2}} = \textcolor{b l u e}{3}$

Change (reduction) in distance from $\overline{A B} , \overline{A ' B}$ is

$\implies \sqrt{13} - 3 \approx \textcolor{g r e e n}{0.61}$

## What is the impact on area of a trapezoid if only one base is halved?

sankarankalyanam
Featured 1 week ago

${A}_{2} = \left({A}_{1} / 2\right) \left(1 + \left(\frac{b}{a + b}\right)\right)$

#### Explanation:

Area of a trapizoin $A = \left(\frac{a + b}{2}\right) \cdot h$ where a, b are the parallel sides and h the distance between them.

${A}_{1} = \left(\frac{a + b}{2}\right) \cdot h$ Eqn (1)

Now base a has been halved or the new base is a/2
${A}_{2} = \left(\frac{\left(\frac{a}{2}\right) + b}{2}\right) \cdot h$

${A}_{2} = \left(\frac{a + 2 b}{4}\right) \cdot h$ Eqn (2)

Dividing Eqn(2) by (1),

${A}_{2} / {A}_{1} = \frac{\left(\frac{a + 2 b}{\cancel{4}} ^ 2\right) \cdot \cancel{h}}{\left(\frac{a + b}{\cancel{2}}\right) \cdot \cancel{h}}$

${A}_{2} = \left(\frac{a + 2 b}{2 \cdot \left(a + b\right)}\right) \cdot {A}_{1}$

${A}_{2} = \left({A}_{1} / 2\right) \left(1 + \left(\frac{b}{a + b}\right)\right)$

## What is the value of x? Enter your answer in the box.

sankarankalyanam
Featured 6 days ago

Case 1 : ED is the altitude (perpendicular to GH) #x = color(brown)(78.92# ft

Case 2 : ED is the median #x = color(blue)(60# ft

Case 3 :ED is the angular bisector of $G \hat{E} H$ #x = color(green)(68# ft

#### Explanation:

Assuming

Since it has not been given, ED is perpendicular to GH or D is the mid point of GH or ED is the angular bisector of $G \hat{E} H$

Let's consider all these three cases.

Case 1 : ED is the altitude (perpendicular to GH)

In right triangle DEG,

$D {E}^{2} = G E 62 - D {G}^{2} = {99.2}^{2} - {62}^{2}$ Eqn (1)

In right triangle DEH,

$D {E}^{2} = E {H}^{2} - D {H}^{2} = {112}^{2} - {\left(x + 2\right)}^{2}$ Eqn (2)

Comparing Eqns (1), (2),

${112}^{2} - {\left(x + 2\right)}^{2} = {99.2}^{2} - {62}^{2}$

${\left(x + 2\right)}^{2} = {112}^{2} - {99.2}^{2} + {62}^{2} = 6547.36$

$\left(x + 2\right) \approx 80.92$

#x = 80.92 - 2 = color(brown)(78.92# ft

Case 2 : ED is the median

i.e DH = DG #

$x + 2 = 62$ or #x = 62 - 2 = color(blue)(60# ft

Case 3 : ED is the angular bisector of #GhatEH

As per angular bisector theorem,

$\frac{E G}{E H} = \frac{D G}{G H}$

$\frac{99.2}{112} = \frac{62}{x + 2}$

$\left(x + 2\right) = \frac{62 \cdot 112}{99.2} = 70$

$x = 70 - 2 = \textcolor{g r e e n}{68}$ ft

## Find the equation of the circle whose centre is (3,-1) and which cuts off an intercept of length 6 from the line 2x-5y+18=0?

Shwetank Mauria
Featured 1 week ago

${x}^{2} + {y}^{2} - 6 x + 2 y - 28 = 0$

#### Explanation:

The length of perpendicular from $\left(3 , - 1\right)$ to $2 x - 5 y + 18 = 0$ is

$| \frac{2 \times 3 - 5 \times \left(- 1\right) + 18}{\sqrt{{2}^{2} + {5}^{2}}} | = \frac{29}{\sqrt{29}} = \sqrt{29}$

Now as circle cuts off an intercept of length $6$, it intercepts at a distance of $3$ on each side from the foot of the perpendicular.

Hence radius of circle is $\sqrt{{\left(\sqrt{29}\right)}^{2} + {3}^{2}} = \sqrt{38}$

and equation of circle is

${\left(x - 3\right)}^{2} + {\left(y + 1\right)}^{2} = 38$

or ${x}^{2} + {y}^{2} - 6 x + 2 y - 28 = 0$

graph{(x^2+y^2-6x+2y-28)(2x-5y+18)((x-3)^2+(y+1)^2-0.03)=0 [-9.75, 10.25, -4.04, 6.38]}

## A circle has a radius of 3 feet and a central angle DEF that measures 60°. What is the length of the intercepted arc DF?

Jim G.
Featured yesterday

$\text{arc DF "=pi~~3.14" feet to 2 dec. places}$

#### Explanation:

$\text{the length of the arc is calculated using}$

#• " length of arc "="circumference "xx"fraction of circle"#

$\textcolor{w h i t e}{\times \times \times \times \times \times} = 2 \pi r \times \frac{60}{360}$

$\textcolor{w h i t e}{\times \times \times \times \times \times} = 6 \pi \times {\cancel{60}}^{1} / {\cancel{360}}^{6}$

$\textcolor{w h i t e}{\times \times \times \times \times \times} = \frac{\cancel{6} \pi}{\cancel{6}} = \pi \approx 3.14$

## It has a triangle equal to 180 degrees and I don’t understand this, can you help me?

LucaGasparro
Featured yesterday

See below.

#### Explanation:

Here we are formulating an equation to solve for $x$.

We know that the interior angles of any triangle adds up $180$ degrees.

We have three angles given:

$60$
$x$
$3 x$

This means that :

$60 + 3 x + x = 180$

Now we collect like terms to simplify.

$60 + 4 x = 180$

Now we solve like any linear equation by isolating the variable on one side of the equation with the constant on the other.

Here we must subtract $60$ from both sides to isolate the $x$.

$\therefore 60 + 4 x - 60 = 180 - 60$

$\implies 4 x = 120$
We want one $x$, therefore we divide by the coefficient of $x$ on both sides.

Here we divide by $4$

$4 x = 120$
$\implies x = 30$

We can check if we are right by putting our value of $x$ back into our formulated equation above.

$60 + \left(4 \cdot 30\right) = 60 + 120 = 180$

##### Questions
• · An hour ago
• · An hour ago
• · 3 hours ago
• · 3 hours ago
• · 3 hours ago
• · 5 hours ago
• · 8 hours ago
• · 8 hours ago
• · 12 hours ago
• · 17 hours ago
• · 17 hours ago
• · 18 hours ago
• · 20 hours ago
• · 20 hours ago