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.

HELP PLEASE Will give the Brainliest if shown work 1 In the figure below ABC PQR If the area of ABC is 40 cm2 what is the area of PQR Image down below 2 A scien class=

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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]

Answer: The answers are

(1) 90 cm²   and (2) 321 m³.

Step-by-step explanation:  The calculations are as follows:

(1) The triangles ABC and PQR are similar. And the area of ΔABC is 40 cm².

Also, AB = 4 cm  and PQ = 6 cm.

We are to find the area of triangle PQR.

Similarity ratio of two similar triangles is equal to the ratio of any two corresponding sides of the triangles.

So, the similarity ratio of ΔABC and ΔPQR is given by

[tex]\dfrac{AB}{PQ}=\dfrac{4}{6}=\dfrac{2}{3} =2:3\\\\\\\Rightarrow AB:PQ=2:3.[/tex]

Now, let the area of ΔPQR be denoted by [tex]A_{PQR}.[/tex]

We know that the ratios of the area of two similar triangles is equal to the ratios of the squares of any two corresponding sides of the triangles.

Therefore, we must have

[tex]\dfrac{\textup{area of triangle ABC}}{\textup{area of triangle PQR}}=\dfrac{AB^2}{PQ^2}\\\\\\\Rightarrow \dfrac{40}{A_{PQR}}=\left(\dfrac{AB}{PQ}\right)^2\\\\\\\Rightarrow \dfrac{40}{A_{PQR}}=\left(\dfrac{2}{3}\right)^2\\\\\\\Rightarrow \dfrac{40}{A_{PQR}}=\dfrac{4}{9}\\\\\\\Rightarrow 4\times A_{PQR}=40\times 9\\\\\\\Rightarrow A_{PQR}=90~\textup{cm}^2.[/tex]

Thus, the area of triangle PQR is 90 cm².

(2) Given that a science museum has a spherical model of the earth with a diameter of 8.5 m.

We are to find the volume of the model.

Since the model is spherical in shape, so will be using the following formula:

the volume of a sphere with radius 'r' units is given by

[tex]V=\dfrac{4}{3}\pi r^3.[/tex]

The diameter of the model is 8.5 cm, so the radius of the model will be

[tex]r=\dfrac{8.5}{2}=4.25~\textup{m}.[/tex]

Therefore, the volume of the model is given by

[tex]V\\\\\\=\dfrac{4}{3}\pi r^3\\\\\\=\dfrac{4}{3}\times3.14\times(4.25)^3\\\\\\=\dfrac{964.17625}{3}\\\\\\=321.39\sim 321~\textup{m}^3.[/tex]

Thus, the volume of the model is 321 m³.

Hence, the answers are

(1) 90 cm²   and (2) 321 m³.