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3 years ago

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A triangle has a side length of 4 inches and an area of 18 square inches and a larger similar triangle has a corresponding side

length of 8 inches. Find the area of the larger tringle ?
...?

1 answer:

cricket20 [7]3 years ago

6 0

For this problem, we can use ratio since these triangle are said to be similar triangles. We calculate as follows:

L1/A1 = L2/A2
4/18 = 8/A2
2/9 = 8/A2
A2 = 36 square inches

The area of the larger triangle would be 36 square inches. Hope this answers the question. Have a nice day.

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$Method\ 1:$

$\dfrac{1}{3}=\dfrac{1\cdot5}{3\cdot5}=\dfrac{5}{15}\\\\\dfrac{3}{5}=\dfrac{3\cdot3}{5\cdot3}=\dfrac{9}{15}\\\\\dfrac{3}{15} < \dfrac{9}{15}\ therefore\ \dfrac{3}{5}\ is\ bigger$

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$\dfrac{1}{3} < \dfrac{1}{2}=\dfrac{1.5}{3}\\\\\dfrac{3}{5} > \dfrac{1}{2}=\dfrac{2.5}{5}\\\\therefore\ \dfrac{3}{5}\ is\ bigger$

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4/5 divided by 1/3 plus 1/5 minus 3/5

r-ruslan [8.4K]

Answer:

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}=-12$

Step-by-step explanation:

Considering the expression

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}$

Solution Steps:

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}$

as

$\mathrm{Combine\:the\:fractions\:}\frac{1}{5}-\frac{3}{5}:\quad -\frac{2}{5}$

so

$=\frac{\frac{4}{5}}{\frac{1}{3}-\frac{2}{5}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}$

$=\frac{4}{5\left(\frac{1}{3}-\frac{2}{5}\right)}$

join $\frac{1}{3}-\frac{2}{5}:\quad -\frac{1}{15}$

so

$=\frac{4}{5\left(-\frac{1}{15}\right)}$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$=\frac{4}{-5\cdot \frac{1}{15}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{-b}=-\frac{a}{b}$

$=-\frac{4}{5\cdot \frac{1}{15}}$

$\mathrm{Multiply\:}5\cdot \frac{1}{15}\::\quad \frac{1}{3}$

so

$=-\frac{4}{\frac{1}{3}}$

$\mathrm{Simplify}\:\frac{4}{\frac{1}{3}}:\quad \frac{12}{1}$

so

$=-\frac{12}{1}$

$\mathrm{Apply\:rule}\:\frac{a}{1}=a$

$=-12$

Therefore

$\frac{\frac{4}{5}}{\frac{1}{3}+\frac{1}{5}-\frac{3}{5}}=-12$

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