MATH SOLVE

4 months ago

Q:
# HELP PLEASE! Will give the Brainliest if shown work!!1.) In the figure below, ABC ~ PQR.If the area of ABC is 40 cm*2, what is the area of PQR? (Image down below)2.) A science museum has a spherical model of the earth with a diameter of 8.5 m. What is the volume of the model? Use 3.14 for and round your answer to the nearest whole number. Show your work.

Accepted Solution

A:

Answer:[tex]\large\boxed{Q1.\ A_{\triangle PQR}=90\ cm^2}\\\\\boxed{Q2.\ V\approx321\ m^3}[/tex]Step-by-step explanation:[tex]Q1.\\\\\text{If}\ \triangle ABC\sim\triangle PQR\ \text{then the quotient of the areas is equal}\\\text{the square of the similarity scale}\ k.\\\\\text{The sides}\ AB\ \text{and}\ QP\ \text{are corresponding}.\ \text{Calculate the scale:}\\\\k=\dfrac{4}{6}=\dfrac{4:2}{6:2}=\dfrac{2}{3}\\\\\text{The area of }\ \triangle ABC=40\ cm^2.\\\\\text{Let the area of}\ \triangle PQR=x,\ \text{then}\\\\\dfrac{40}{x}=\left(\dfrac{2}{3}\right)^2\\\\\dfrac{40}{x}=\dfrac{4}{9}\qquad\text{cross multiply}\\\\4x=(9)(40)\qquad\text{divide both sides by 4}\\\\x=(9)(10)\\\\x=90\ cm^2[/tex][tex]Q2.\\\\\text{The formula of a volume of a sphere:}\\\\V=\dfrac{4}{3}\pi R^3\\\\R-radius\\\\\text{We have the diameter}\ 2R=8.5\ m\to R=\dfrac{8.5}{2}\ m=4.25\ m.\\\\\text{Substitute:}\\\\V=\dfrac{4}{3}\pi(4.25)^3=\dfrac{4}{3}\pi(76.765625)=(4)(25.588541)\pi=102.354164\pi\\\\\pi\approx3.14\\\\V=(102.354164)(3.14)\approx321\ m^3[/tex]