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## How do you use Heron's formula to determine the area of a triangle with sides of that are 15, 18, and 12 units in length?

Nimo N.
Featured 3 days ago · Trigonometry

 color(purple)( A ~~ 89.2941

#### Explanation:

Problem:
Use Heron's to determine the area of a triangle with sides 15, 18, and 12 units long.

Heron's formula:
 color(blue)( A = sqrt( s(s - a)(s - b)(s - c) ), and
 color(blue)( s = (a + b + c)/(2).

Calculate s and the stuff inside the radical.
$s = \frac{\left(15\right) + \left(18\right) + \left(12\right)}{2}$
$s = \frac{45}{2}$.
$s - a = \frac{45}{2} - \frac{30}{2} = \frac{15}{2}$

$s - b = \frac{45}{2} - \frac{36}{2} = \frac{9}{2}$

$s - c = \frac{45}{2} - \frac{24}{2} = \frac{21}{2}$

$s \left(s - a\right) \left(s - b\right) \left(s - c\right) = \frac{45}{2} \cdot \frac{15}{2} \cdot \frac{9}{2} \cdot \frac{21}{2}$

$s \left(s - a\right) \left(s - b\right) \left(s - c\right) = \frac{127575}{16}$

Now, the area.
$A = \sqrt{\frac{127575}{16}}$
 color(purple)( A ~~ 89.2941

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## If a=5 b=12 c=13 in a trangle ABC than what is the value of tan (B/4)=?

maganbhai P.
Featured 3 days ago · Trigonometry

$\tan \left(\frac{B}{4}\right) = \frac{\sqrt{13} - 3}{2}$

#### Explanation:

Here,

$a = 5 , b = 12 \mathmr{and} c = 13$

$\implies B C = 5 , C A = 12 , A B = 13$

$\implies \overline{A B}$ , is the hypotenuse of $\triangle A B C$

So, $m \angle C = {90}^{\circ} \implies \angle A \mathmr{and} \angle B$ are acute angles.

Now, color(red)(cosB=("side adjacent to B")/("hypotenus")=5/13

Using half angle formula :

$\cos \left(\frac{B}{2}\right) = \frac{1 + \cos B}{2} = \frac{1 + \frac{5}{13}}{2} = \frac{9}{13} = {\left(\frac{3}{\sqrt{13}}\right)}^{2}$

=>color(red)(cos(B/2)=+3/sqrt13...to[because mangleB < 90^circ]

Also ,we know the half angle formula:

color(blue)(tan^2(theta/2)=(1-costheta)/(1+costheta)

Let us take, $\theta = \left(\frac{B}{2}\right)$

:.color(blue)(tan^2((B/2)/2)=(1-cos(B/2))/(1+cos(B/2))

=>tan^2(B/4)=(1-3/sqrt13)/(1+3/sqrt13)=(sqrt13-3)/(sqrt13+3)=(sqrt13-3)/(sqrt13+3)xxcolor(green)((1)

=>tan^2(B/4)=(sqrt13-3)/(sqrt13+3)xxcolor(green)((sqrt13-3)/(sqrt13-3)

$\implies {\tan}^{2} \left(\frac{B}{4}\right) = {\left(\sqrt{13} - 3\right)}^{2} / \left({\left(\sqrt{13}\right)}^{2} - {\left(3\right)}^{2}\right)$

$\implies {\tan}^{2} \left(\frac{B}{4}\right) = {\left(\sqrt{13} - 3\right)}^{2} / 4$

$\implies {\left(\tan \left(\frac{B}{4}\right)\right)}^{2} = {\left(\frac{\sqrt{13} - 3}{2}\right)}^{2.} . . \to m \angle B < {90}^{\circ}$

$\implies \tan \left(\frac{B}{4}\right) = \frac{\sqrt{13} - 3}{2}$

1

## What does redundant mean? Can a person be redundant?

Featured 3 days ago · English Grammar

More than needed.

#### Explanation:

In the sentence "This is a single unpaired sock" there is redundancy in the form of twice saying the same (also known as tautology).

A person can be "made redundant", meaning he is no longer needed for a certain job at a certain office or factory (there are too many in that job at that place).

Calling a person redundant (overall) would not be a very nice thing to say: It would mean the person is good for nothing and might as well die.

1

## Who was Amerigo Vespucci and what did he discover?

John C. Thompson
Featured 3 days ago · World History

Amerigo Vespucci was a Florentine explorer and navigator who first determined that the east coast of North Amercia was part of a seperate landmass from Asia.

#### Explanation:

"America" largely results from the Latin version of Americao Vespucci's first name. Born in Florence in 1454, he was originally a merchant and entrepreneur but started to serve both the Portuguese and Spanish governments as a navigator.

In 1497-1504 Vespucci determined that the coast of South America extended much further than Cristopher Columbus had thought, and argued (in a widely circulated series of letters) that this meant South America might be a major landmass that was not connected to Asia.

Like many of his contemporaries (and other explorers), Vespucci was a little bit of a huckster, and some of his accomplishments on his four voyages may be questionable. Moreover, subsequent claims by others added to his lustre -- partly due to a scholarly debate over whether it was Vespucci or Columbus who truely discovered America.

Regardless, it remains that Columbus believed his discoveries were outlying parts of Asia, but Vespucci argued -- successfully that there was a major landmass, a continent in its own right, to the west of Europe and the east of Asia. Europe's leading cartographer Martin Waldseemüller named this land mass "America" in Vespucci's honor. Vespucci died in 1512.

2

## Why are video texts an example of multimedia?

Dakota
Featured 3 days ago · English Grammar

because is it 2 different kinds of media combined

#### Explanation:

whenever text, audio, still images, video, animation etc. are combined, the result is multimedia.
think of multimedia as multiple medias.

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## How do you use Heron's formula to determine the area of a triangle with sides of that are 15, 18, and 12 units in length?

Dean R.
Featured 3 days ago · Trigonometry

$16 {S}^{2} = \left(a + b + c\right) \left(- a + b + c\right) \left(a - b + c\right) \left(a + b - c\right) = \left(15 + 18 + 12\right) \left(- 15 + 18 + 12\right) \left(15 - 18 + 12\right) \left(15 + 18 - 12\right) = \left(45\right) \left(15\right) \left(9\right) \left(21\right) = 3 \left(15\right) \left(15\right) \left(9\right) \left(3\right) \left(7\right)$ or

$S = \frac{9 \left(15\right)}{4} \sqrt{7} = \frac{135 \sqrt{7}}{4}$

#### Explanation:

Seems a bit easier than the other answer.

2

## If tanx/(1-cotx) + cotx/(1-tanx) = -1, what is the value of tanx?

dk_ch
Featured 3 days ago · Trigonometry

$\tan \frac{x}{1 - \cot x} + \cot \frac{x}{1 - \tan x} = - 1$

$\implies - \tan \frac{x}{1 - \frac{1}{\tan} x} - \frac{\frac{1}{\tan} x}{1 - \tan x} = 1$

$\implies - {\tan}^{2} \frac{x}{\tan x - 1} - \frac{1}{\tan x \left(1 - \tan x\right)} = 1$

$\implies {\tan}^{2} \frac{x}{1 - \tan x} - \frac{1}{\tan x \left(1 - \tan x\right)} = 1$

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

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

$\tan x = 1$ will make LHS undefined so neglected.
Hence

$\implies - \frac{{\tan}^{2} x + \tan x + 1}{\tan} x = 1$

$\implies {\tan}^{2} x + 2 \tan x + 1 = 0$

$\implies {\left(\tan x + 1\right)}^{2} = 0$

$\implies \tan x = - 1$

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## Two circles having radii aand b touch each other externally. If c is the radius of another circle which touches these two circles as well as a common tangent to the two circles, how do we prove that 1/sqrtc=1/sqrta+1/sqrtb?

CW
Featured 3 days ago · Geometry

see explanation.

#### Explanation:

Let $A , B , C$ be the center of circle 1, circle 2 and circle 3, respectively, as shown in the figure.
$A B = a + b , A C = a + c , B C = b + c$
In $\Delta C D B , C {D}^{2} = B {C}^{2} - B {D}^{2}$
$\implies C {D}^{2} = {\left(b + c\right)}^{2} - {\left(b - c\right)}^{2}$
$\implies C {D}^{2} = \left(b + c + b - c\right) \cdot \left(b + c - b + c\right) = 2 b \cdot 2 c = 4 b c$
$\implies C D = \sqrt{4 b c} = 2 \sqrt{b c}$
Similarly, $C {F}^{2} = {\left(a + c\right)}^{2} - {\left(a - c\right)}^{2}$
$\implies C F = 2 \sqrt{a c}$
$\implies D F = C F + C D = 2 \sqrt{a c} + 2 \sqrt{b c} = 2 \sqrt{c} \left(\sqrt{a} + \sqrt{b}\right)$
In $\Delta A E B , B {E}^{2} = A {B}^{2} - A {E}^{2}$
$\implies B {E}^{2} = {\left(a + b\right)}^{2} - {\left(b - a\right)}^{2}$
$\implies B {E}^{2} = \left(a + b + b - a\right) \cdot \left(a + b - b + a\right) = 2 b \cdot 2 a = 4 a b$
$\implies B E = \sqrt{4 a b} = 2 \sqrt{a b}$
As $B E = D F$,
$\implies 2 \sqrt{a b} = 2 \sqrt{c} \left(\sqrt{a} + \sqrt{b}\right)$
$\implies \frac{1}{\sqrt{c}} = \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a b}}$
$\implies \frac{1}{\sqrt{c}} = \frac{\sqrt{a}}{\sqrt{a b}} + \frac{\sqrt{b}}{\sqrt{a b}}$
$\implies \frac{1}{\sqrt{c}} = \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{a}}$
$\implies \frac{1}{\sqrt{c}} = \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}}$

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## What is the orthocenter of a triangle with corners at (2 ,3 ), (6 ,1 ), and (6 ,3 )?

maganbhai P.
Featured 2 days ago · Geometry

Hence, the orthocentre of $\triangle A B C$ is $C \left(6 , 3\right)$

#### Explanation:

Let , $\triangle A B C$ ,be the triangle with corners at

$A \left(2 , 3\right) , B \left(6 , 1\right) \mathmr{and} C \left(6 , 3\right)$ .

We take, $A B = c , B C = a \mathmr{and} C A = b$

So,

${c}^{2} = {\left(2 - 6\right)}^{2} + {\left(3 - 1\right)}^{2} = 16 + 4 = 20$

${a}^{2} = {\left(6 - 6\right)}^{2} + {\left(1 - 3\right)}^{2} = 0 + 4 = 4$

${b}^{2} = {\left(2 - 6\right)}^{2} + {\left(3 - 3\right)}^{2} = 16 + 0 = 16$

It is clear that, ${a}^{2} + {b}^{2} = 4 + 16 = 20 = {c}^{2}$

 i.e. color(red)(c^2=a^2+b^2=>mangleC=pi/2#

Hence, $\overline{A B}$ is the hypotenuse.

$\therefore \triangle A B C$ is the right angled triangle.

$\therefore$The orthocenter coindes with $C$

Hence, the orthocentre of $\triangle A B C$ is $C \left(6 , 3\right)$

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## How will you compare 2 ratios: 5/4 and 11/10 in a visual diagram form?

Gió
Featured 2 days ago · Prealgebra