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1 year ago

11

An artist makes a scale drawing of a parallelogram-shaped sculpture. The scale is 10 cm on the drawing for every 8 meters on the

sculpture. What is the area of the scale drawing? Show your work.

1 answer:

8_murik_8 [283]1 year ago

3 0

Answer:

Area of drawing= 0.315 m²

Step-by-step explanation:

Base = 6m

Height = 4.2 m

Area = height* base

Area= 4.2*6

Area= 25.2 m²

The scale is 10 cm on the drawing for every 8 meters on the sculpture.

Which is 10 cm to 800 cm

Or

0.1m to 8m

Arra of sculpture= 25.2 m²

area of drawing= (25.2*0.1)/8

Area of drawing= 0.315 m²

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a. To solve the first part, we are going to use the formula for the surface area of a sphere: $A=4 \pi r^2$
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We know from our problem that $r=5ft$ ; so lets replace that value in our formula:
$A=4 \pi (5ft)^2$
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To solve the second part, we are going to use the formula for the volume of a sphere: $V= \frac{4}{3} \pi r^3$
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$V=523.6ft^3$

We can conclude that the surface area of the sphere is 314.16 square feet and its volume is 523.6 cubic feet.

b. To solve the first part, we are going to use the formula for the surface area of a square pyramid: $A=a^2+2a \sqrt{ \frac{a^2}{4} +h^2}$
where
$A$ is the surface area
$a$ is the measure of the base
$h$ is the height of the pyramid
We know form our problem that $a=8ft$ and $h=12ft$ , so lets replace those value sin our formula:
$A=(8ft)^2+2(8ft) \sqrt{ \frac{(8ft)^2}{4} +(12ft)^2}$
$A=266.39ft^2$

To solve the second part, we are going to use the formula for the volume of a square pyramid: $V=a^2 \frac{h}{3}$
where
$V$ is the volume
$a$ is the measure of the base
$h$ is the height of the pyramid
We know form our problem that $a=8ft$ and $h=12ft$ , so lets replace those value sin our formula:
$V=(8ft)^2 \frac{(12ft)}{3}$
$V=256ft^3$

We can conclude that the surface area of our pyramid is 266.39 square feet and its volume is 256 cubic feet.

c. To solve the first part, we are going to use the formula for the surface area of a circular cone: $A= \pi r(r+ \sqrt{h^2+r^2}$
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$A$ is the surface area
$r$ is the radius of the circular base
$h$ is the height of the cone
We know form our problem that $r=5ft$ and $h=8ft$ , so lets replace those values in our formula:
$A= \pi (5ft)[(5ft)+ \sqrt{(8ft)^2+(5ft)^2}]$
$A=226.73ft^2$

To solve the second part, we are going to use the formula for the volume os a circular cone: $V= \pi r^2 \frac{h}{3}$
where
$V$ is the volume
$r$ is the radius of the circular base
$h$ is the height of the cone
We know form our problem that $r=5ft$ and $h=8ft$ , so lets replace those values in our formula:
$V= \pi (5ft)^2 \frac{(8ft)}{3}$
$V=209.44ft^3$

We can conclude that the surface area of our cone is 226.73 square feet and its surface area is 209.44 cubic feet.

d. To solve the first part, we are going to use the formula for the surface area of a rectangular prism: $A=2(wl+hl+hw)$
where
$A$ is the surface area
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$l$ is the length
$h$ is the height
We know from our problem that $w=6ft$ , $l=10ft$ , and $h=16ft$ , so lets replace those values in our formula:
$A=2[(6ft)(10ft)+(16ft)(10ft)+(16ft)(6ft)]$
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To solve the second part, we are going to use the formula for the volume of a rectangular prism: $V=whl$
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- A cone has 1 face.
- A rectangular prism has 6 faces.
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<span>We can conclude that there are 12 faces in on the four geometric shapes on the holes.
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f. Remember that an edge is a line segment on the boundary of the polygon.
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- A rectangular pyramid has 8 edges.
- A rectangular prism has 12 edges.
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g. Remember that the vertices are the corner points of a polygon.
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We can conclude that there are 0 vertices for the sphere and the cone; there are 5 vertices for the pyramid, and there are are 8 vertices for the solid (rectangular prism). We can also conclude that your boss will need 13 brackets for the vertices of the four figures.

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