What is the ratio for the surface areas of the cones shown below, given that

Mathematics

1 answer:

andreyandreev [35.5K]3 years ago

5 0

Answer:

The ratio for the surface areas is equal to 4

Step-by-step explanation:

step 1

Find the scale factor

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and this ratio is called the scale factor

Let

z -----> the scale factor

In this problem

The scale factor is equal to

$z=\frac{2}{1}$ ----> the ratio of its corresponding radii or its corresponding altitudes

step 2

Find the ratio for the surface areas of the cones

we know that

If two figures are similar, then the ratio of its surface areas is equal to the scale factor squared

Let

z -----> the scale factor

x ----> the surface area of the larger cone

y ----> the surface area of the smaller cone

$z^{2} =\frac{x}{y}$

we have

$z=\frac{2}{1}$

substitute

$(\frac{2}{1})^{2}=\frac{x}{y}$

$4=\frac{x}{y}$

The ratio for the surface areas is equal to 4

That means-----> The surface area of the larger cone is 4 times the surface area of the smaller cone

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$\cot^4 A - \cot^2 A = 1\\\\\cot^4 A = 1 + \cot^2 A\\\\\frac{\cos^4 A}{\sin^4 A} = 1 + \frac{\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{\sin^2 A}{\sin^2 A}+\frac{\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{\sin^2 A+\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{1}{\sin^2 A}\\\\\cos^4 A = \frac{\sin^4 A}{\sin^2 A}\\\\\cos^4 A = \sin^2 A\\\\$