Answer: 5.6 cm
Step by Step: I used the same formula as the question before and I got the answer for this question. When solving questions like this, you can work backwards from the problem and then solve
Answer:
30q or 30
Step-by-step explanation:
Answer:
6.1
Step-by-step explanation:
Draw a picture of an equilateral triangle. Cut the triangle in half, so that you get two 30-60-90 triangles. The area of these smaller triangles is 8 square inches.
The short leg of these triangles (the base) is half the side length: ½ s.
According to properties of 30-60-90 triangles, the long leg (the height) is √3 times the short leg: ½ s√3.
Area of a triangle is half the base times the height:
A = ½bh
8 = ½ (½ s) (½ s√3)
8 = ⅛ s²√3
64 = s²√3
s² = 64/√3
s = √(64/√3)
s ≈ 6.1
From calculations, we can say that the given tiles will not fit together perfectly.
<h3>How to find the sum of interior angles of a Polygon?</h3>
If the tiles join perfectly at a point, sum of all angles around the joining point should be 360°.
Expression for the measure of the interior angle of a polygon,
Interior angle of a polygon = [(n - 2) * 180]/n
Interior angle of a pentagon = [(5 - 2) * 180]/5 = 108°
Interior angle of a hexagon = [(6 - 2) * 180]/6 = 120°
Interior angle of an octagon = [(8 - 2) * 180]/8 = 135°
To prove that the given tiles fit together perfectly → Sum of all the angles around the common point should be 360°
Sum of all interior angles = 108° + 120° + 135° = 363°
Therefore, given tiles will not fit together perfectly.
Read more about Interior angles of a Polygon at; brainly.com/question/224658
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This is the concept of areas of solid materials; the surface area of the cylinder whose radius is 2.5 cm and lateral area is 20 pi cm^2 will be: Surface area of cylinder is given by:
SA=(area of cyclic sides)+(lateral area)
SA=2πr^2+πrl
Area of the cyclic sides will be:
Area=2πr^2
=2*π*2.5^2
=12.5π cm^2
The lateral area is given by:
Area=20π cm^2
Therefore the surface area of cylinder will be:
SA=(12.5π+20π) cm^2
SA=32.5π cm^2
The answer is 32.5π cm^2
