istg my math tutor and I worked on these two problems:

1. Let $G$ be the center of equilateral triangle $XYZ.$ A dilation centered at $G$ with scale factor $-\frac{4}{3}$ is applied to triangle $XYZ,$ to obtain triangle $X'Y'Z'.$ Let $A$ be the area of the region that is contained in both triangles $XYZ$ and $X'Y'Z'.$ Find $\frac{A}{[XYZ]}.$

2. A conical frustum has bases with radii of $9$ and $12,$ and a height of $4.$ The total surface area of the frustum is $A$, in square units, and the volume of the frustum is $V$, in cubic units. Find $A + V.$

and we still couldn't figure them out. Now this is my last resort ;-;

tysm in advance and goodluck

(also im really sorry I posted more than one problem- ik im not really supposed to...)

Guest Jul 30, 2021

#1**+1 **

Since you know you are not supposed to put more than one question per post why did you do it?

When you do the wrong thing it cuts down on the number of good answers you may get.

You can always add a link to your second post. That would be an option.

Melody Jul 31, 2021

#2**+1 **

You can always post 2 seperate posts.

I'm going to try the second problem.

There are equations to use to find the volume/surface area of the frustum.

https://www.cuemath.com/measurement/volume-of-frustum/

https://www.varsitytutors.com/hotmath/hotmath_help/topics/surface-area-of-a-cone

Try to imagine the frustum as a complete cone, with the base being 12 radius and a line cutting it at 9 radius. The height would be from 12 radius to the tip (0 radius). For the height to increase by 4 the radius subtracted by 3. (12/3)*4 = 16, the height of the whole cone is 16.

To find the volume, we're looking for the big cone (radius of 12) subtracted by the little cone (radius of 9). The volume of the big cone is 12^2*pi*16/3 = 768pi.

The volume of the small cone is 9^2*pi*(16-4)/3 = 324pi.

768pi - 324pi = 444pi.

To find the surface area, we're going to start by focusing on the lateral surface.

The lateral surface of the big cone (radius of 12) is sqrt(16^2+12^2)*12*pi = 240pi.

The lateral surface of the little cone (raduis of 12) is sqrt(9^2+12^2)*9*pi = 135pi.

The lateral surface of the frustum is 240pi - 135pi = 105pi.

Next the bases. pi*9^2 + pi*12^2 = 225pi.

225pi + 105pi = 330pi.

330pi + 444pi = 774pi.

Sorry if the explanation is the bit confusing, please ask questions if you don't understand anything.

=^._.^=

catmg Jul 31, 2021

#4**+1 **

1/ Similar math problem::: https://web2.0calc.com/questions/help_54234

Scale factor = - 4/3

A / [XYZ] = 851 / 999

civonamzuk Jul 31, 2021