In this diagram, BAC~EDF. If the area of BAC=15 in2, what is the area of EDF? 4, 5.

Mathematics

1 answer:

lesya692 [45]3 years ago

5 0

Answer:

9.6 square inches.

Step-by-step explanation:

We are given that ΔBAC is similar to ΔEDF, and that the area of ΔBAC is 15 inches. And we want to determine the area of ΔDEF.

First, find the scale factor k from ΔBAC to ΔDEF:

$\displaystyle 5 k = 4$

Solve for the scale factor k:

$\displaystyle k = \frac{4}{5}$

Recall that to scale areas, we square the scale factor.

In other words, since the scale factor for sides from ΔBAC to ΔDEF is 4/5, the scale factor for its area will be (4/5)² or 16/25.

Hence, the area of ΔEDF is:

$\displaystyle \Delta EDF = \frac{16}{25}\left(15\right) = 9.6\text{ in}^2$

In conclusion, the area of ΔEDF is 9.6 square inches.

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$cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}$

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