MATH SOLVE

5 months ago

Q:
# The area of the smaller triangle is about 270 ft2. Which is the best approximation for the area of the larger triangle?

Accepted Solution

A:

the assumption here is that, both triangles are similar.Β Now, if that's the case,

[tex]\bf \qquad \qquad \textit{ratio relations} \\\\ \begin{array}{ccccllll} &\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\ &-----&-----&-----\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}\\\\ -----------------------------[/tex]

[tex]\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\ -------------------------------\\\\ \cfrac{small}{large}\qquad \cfrac{25}{35}\implies \stackrel{simplified}{\cfrac{5}{7}}\qquad \qquad \cfrac{5}{7}=\cfrac{\sqrt{270}}{\sqrt{x}}\implies \cfrac{5}{7}=\sqrt{\cfrac{270}{x}} \\\\\\ \left( \cfrac{5}{7} \right)^2=\cfrac{270}{x}\implies \cfrac{5^2}{7^2}=\cfrac{270}{x}\implies x=\cfrac{7^2\cdot 270}{5^2}[/tex]

[tex]\bf \qquad \qquad \textit{ratio relations} \\\\ \begin{array}{ccccllll} &\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\ &-----&-----&-----\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}\\\\ -----------------------------[/tex]

[tex]\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\ -------------------------------\\\\ \cfrac{small}{large}\qquad \cfrac{25}{35}\implies \stackrel{simplified}{\cfrac{5}{7}}\qquad \qquad \cfrac{5}{7}=\cfrac{\sqrt{270}}{\sqrt{x}}\implies \cfrac{5}{7}=\sqrt{\cfrac{270}{x}} \\\\\\ \left( \cfrac{5}{7} \right)^2=\cfrac{270}{x}\implies \cfrac{5^2}{7^2}=\cfrac{270}{x}\implies x=\cfrac{7^2\cdot 270}{5^2}[/tex]