1m = 100 cm
so
40 m = 4,000 cm
25 m = 2,500 cm
10 m = 1,000 cm
perimeter of big rectangle = 2(4,000 + 2000) = 12,000 cm
perimeter of small rectangle on the left: 2(2,500 + 800) = 6,600 cm
perimeter of small rectangle on the right: 2(1,000 + 800) = 3,600 cm
perimeter of circle = πd = 3.141593 x 200 = 628.32 cm
total area of the pitch and seating area :
= 12,000 + 6,600 + 3,600 + 628.32 = 22,828.32<span> cm
answer
</span>22,828.32 cm<span>
</span>
The scale model of a building 3 inches tall if the building is 90 feet tall is 2:5
<h3>Scaling of measurement</h3>
Scaling is the way of enlarging or reducing the image of an object.
Given scale model of a building is 3 inches, if the building is 90feet tall, to write the scale;
Convert both measures to same units
Since 1 feet. = 12 in
90 feet = 90/12 in = 7.5 in
Take the ratio
3 : 7.5 = 30: 75
Reduce to lowest term
30: 75 = 2 : 5
Hence the scale model of a building 3 inches tall if the building is 90 feet tall is 2:5
Learn more on scale model here: brainly.com/question/261465
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Answer:
C
Step-by-step explanation:
Answer:
What?
Step-by-step explanation:
Answer:
X=18, Y=6
Step by Step Explanation:
This is a little long winded.. Let's solve for Y first.
First looking at the problem, you can note a few things. First of all, there is an Isosceles triangle, and there is an equilateral triangle. Where they connect, there is a 90° angle.
Now, all equilateral triangles have the same angle measurement. 60°. Now, if you look where the right angle is, it is showing a 60° angle, a total of 90°, and an unknown area. So simply subtract. 90-60=30. Divide that by 5, and you have your answer of 6.
Now let's solve X. X is very simple. On an isosceles triangle, the two top sides are the same length. And one of the top sides is the same size as the equilateral triangle. And on equilateral triangles, all sides are the same.
So the 11 transfers over to the X side. So let's make a small equation. 11=X-7. To make it even, let's add 7 to both sides. 11+7=X+7-7. Simplify to get 18=X, which is your answer.
