Shaded area = area of the hexagon – area of the pentagon + area of the square – area of the equilateral triangle. This can be obtained by finding each shaded area and then adding them.
<h3>Find the expression for the area of the shaded regions:</h3>
From the question we can say that the Hexagon has three shapes inside it,
Also it is given that,
An equilateral triangle is shown inside a square inside a regular pentagon inside a regular hexagon.
From this we know that equilateral triangle is the smallest, then square, then regular pentagon and then a regular hexagon.
A pentagon is shown inside a regular hexagon.
- Area of first shaded region = Area of the hexagon - Area of pentagon
An equilateral triangle is shown inside a square.
- Area of second shaded region = Area of the square - Area of equilateral triangle
The expression for total shaded region would be written as,
Shaded area = Area of first shaded region + Area of second shaded region
Hence,
⇒ Shaded area = area of the hexagon – area of the pentagon + area of the square – area of the equilateral triangle.
Learn more about area of a shape here:
brainly.com/question/16501078
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Answer:
Option D. 8.53 units
Step-by-step explanation:
we know that
The triangle OAB is congruent with the triangle OCD
because
OA=OB=OC=OD=radius circle O
AB=CD
therefore
The height of both triangles is equal to 8.53 units
The segment blue is equal to 8.53 units
Answer:
324
Step-by-step explanation:
Given:

Find:

First, find f(x):

Now,

You put it 48 divide 50 which equally 0.96 or 50 divide 48 which equal 1.04166666667
Rearrange x/4-(-5)=0
Simplify x/4
-5 = -5/1 = -5*4/4
x- (-5*4)/4 = x+20/4
x+ 20/4 = 0
x+20/4 * 4 = 0*4
The equation now takes the shape: x+20=0
subtract 20 from both sides of the equation: x=-20
Answer: x=-20