Notes: find exact area of the triangle ABC?Points: here's how: -...

Friday, January 31, 2014

find exact area of the triangle ABC?Points: here's how: -...

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class="AM"> $\displaystyle{c}={\left|{A}{B}\right|}=\sqrt{{{{\left(-{3}-{2}\right)}}^{{2}}+{{\left({4}-{7}\right)}}^{{2}}}}=\sqrt{{{25}+{9}}}=\sqrt{{{34}}}$

$\displaystyle{a}={\left|{B}{C}\right|}=\sqrt{{{{\left({2}-{5}\right)}}^{{2}}+{{\left({7}-{1}\right)}}^{{2}}}}=\sqrt{{{9}+{36}}}=$
sqrt(45)=3sqrt(5)
$\displaystyle{b}={\left|{A}{C}\right|}=\sqrt{{{\left(-{3}-\right.}}}$
5)^2+(4-1)^2)) = sqrt(64+9) = sqrt(73)

We can
use Heron's formula for area

$\displaystyle{A}=$
sqrt(s(s-a)(s-b)(s-c)) where $\displaystyle{s}=$
(a+b+c)/2

$\displaystyle{s}=$
(sqrt(34)+3sqrt(5)+sqrt(73))/2
class="AM"> $\displaystyle{s}-{a}=\frac{{\sqrt{{{34}}}+{3}\sqrt{{{5}}}+\sqrt{{{73}}}}}{{2}}-$
3sqrt(5)=(sqrt(34)+sqrt(73)-3sqrt(5))/2

class="AM"> $\displaystyle{s}-{b}=\frac{{\sqrt{{{34}}}+{3}\sqrt{{{5}}}-\sqrt{{{73}}}}}{{2}}$

class="AM"> $\displaystyle{s}-{c}=\frac{{{3}\sqrt{{{5}}}+\sqrt{{{73}}}-\sqrt{{{34}}}}}{{2}}$

class="AM"> $\displaystyle{A}=\sqrt{{{\left(\frac{{\sqrt{{{34}}}+{3}\sqrt{{{5}}}+\sqrt{{{73}}}}}{{2}}\right)}{\left(\frac{{\sqrt{{{34}}}+\sqrt{{{73}}}-{3}\sqrt{{{5}}}}}{{2}}\right)}{\left(\frac{{\sqrt{{{34}}}+{3}\sqrt{{{5}}}-\sqrt{{{73}}}}}{{2}}\right)}{\left(\frac{{{3}\sqrt{{{5}}}+\sqrt{{{73}}}-\sqrt{{{34}}}}}{{2}}\right)}}}$ class="AM"> $\displaystyle{\quad\text{or}\quad}$

class="AM"> $\displaystyle{A}=\frac{{1}}{{4}}\sqrt{{{\left(\sqrt{{{34}}}+{3}\sqrt{{{5}}}+\sqrt{{{73}}}\right)}{\left(\sqrt{{{34}}}+\sqrt{{{73}}}-{3}\sqrt{{{5}}}\right)}{\left(\sqrt{{{34}}}+{3}\sqrt{{{5}}}-\sqrt{{{73}}}\right)}{\left({3}\sqrt{{{5}}}+\sqrt{{{73}}}-\sqrt{{{34}}}\right)}}}$

Notes

Friday, January 31, 2014

find exact area of the triangle ABC?Points: here's how: -...

$\displaystyle{s}=$
(sqrt(34)+3sqrt(5)+sqrt(73))/2
class="AM"> $\displaystyle{s}-{a}=\frac{{\sqrt{{{34}}}+{3}\sqrt{{{5}}}+\sqrt{{{73}}}}}{{2}}-$
3sqrt(5)=(sqrt(34)+sqrt(73)-3sqrt(5))/2

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What is the meaning of the 4th stanza of Eliot's Preludes, especially the lines "I am moved by fancies...Infinitely suffering thing".

Friday, January 31, 2014

find exact area of the triangle ABC?Points: here's how: -...

(sqrt(34)+3sqrt(5)+sqrt(73))/2 class="AM"> 3sqrt(5)=(sqrt(34)+sqrt(73)-3sqrt(5))/2

No comments:

Post a Comment

What is the meaning of the 4th stanza of Eliot&#39;s Preludes, especially the lines &quot;I am moved by fancies...Infinitely suffering thing&quot;.

$\displaystyle{s}=$
(sqrt(34)+3sqrt(5)+sqrt(73))/2
class="AM"> $\displaystyle{s}-{a}=\frac{{\sqrt{{{34}}}+{3}\sqrt{{{5}}}+\sqrt{{{73}}}}}{{2}}-$
3sqrt(5)=(sqrt(34)+sqrt(73)-3sqrt(5))/2

What is the meaning of the 4th stanza of Eliot's Preludes, especially the lines "I am moved by fancies...Infinitely suffering thing".