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telo118 [61]
2 years ago
15

Find the area of a regular octagon with a side length of 4 cm.​

Mathematics
1 answer:
goblinko [34]2 years ago
6 0

Answer:

77.25 cm^2

Step-by-step explanation:

Area of a regular octagon :

2a^2(1 + (sqrt2))

where a is the side

substitute 4 in for 'a'

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Solve for x<br> 3x + 6 over 3 = 6
Svetradugi [14.3K]

Answer:

X= 4

Step-by-step explanation:

3x+6/3=6

multiply each side by 3 to get rid of the fraction bar

3x+6=6*3

3x +6=18

-6 from both sides

3x= 18-6

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The area of ABED is 49 square units. Given AGequals9 units and ACequals10 ​units, what fraction of the area of ACIG is represent
stiks02 [169]

Answer:

The fraction of the area of ACIG represented by the shaped region is 7/18

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

In the square ABED find the length side of the square

we know that

AB=BE=ED=AD

The area of s square is

A=b^{2}

where b is the length side of the square

we have

A=49\ units^2

substitute

49=b^{2}

b=7\ units

therefore

AB=BE=ED=AD=7\ units

step 2

Find the area of ACIG

The area of rectangle ACIG is equal to

A=(AC)(AG)

substitute the given values

A=(9)(10)=90\ units^2

step 3

Find the area of shaded rectangle DEHG

The area of rectangle DEHG is equal to

A=(DE)(DG)

we have

DE=7\ units

DG=AG-AD=9-7=2\ units

substitute

A=(7)(2)=14\ units^2

step 4

Find the area of shaded rectangle BCFE

The area of rectangle BCFE is equal to

A=(EF)(CF)

we have

EF=AC-AB=10-7=3\ units

CF=BE=7\ units

substitute

A=(3)(7)=21\ units^2

step 5

sum the shaded areas

14+21=35\ units^2

step 6

Divide the area of  of the shaded region by the area of ACIG

\frac{35}{90}

Simplify

Divide by 5 both numerator and denominator

\frac{7}{18}

therefore

The fraction of the area of ACIG represented by the shaped region is 7/18

5 0
2 years ago
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Lisa [10]

Answer:

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