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exis [7]
3 years ago
9

What is Pi approximately equal to?

Mathematics
2 answers:
Harman [31]3 years ago
8 0
Pi is equal to 180 degrees.
anygoal [31]3 years ago
5 0

here is what it is approximately equal to

3.14159265358979323846264338327950288419716939937510  

58209749445923078164062862089986280348253421170679  

82148086513282306647093844609550582231725359408128  

48111745028410270193852110555964462294895493038196  

44288109756659334461284756482337867831652712019091  

45648566923460348610454326648213393607260249141273  

72458700660631558817488152092096282925409171536436  

78925903600113305305488204665213841469519415116094  

33057270365759591953092186117381932611793105118548  

07446237996274956735188575272489122793818301194912  

98336733624406566430860213949463952247371907021798  

60943702770539217176293176752384674818467669405132  

00056812714526356082778577134275778960917363717872  

14684409012249534301465495853710507922796892589235  

42019956112129021960864034418159813629774771309960  

51870721134999999837297804995105973173281609631859  

50244594553469083026425223082533446850352619311881  

71010003137838752886587533208381420617177669147303  

59825349042875546873115956286388235378759375195778  

18577805321712268066130019278766111959092164201989

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The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is units is? What is the area?
11111nata11111 [884]

we have

A(-2, 2),B(6, 2),C(0, 8)

see the attached figure to better understand the problem

we know that

The perimeter of the triangle is equal to

P=AB+BC+AC

and

the area of the triangle is equal to

A=\frac{1}{2}*base *heigth

in this problem

base=AB\\heigth=DC

we know that

The distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Step 1

<u>Find the distance AB</u>

A(-2, 2),B(6, 2)

Substitute the values in the formula

d=\sqrt{(2-2)^{2}+(6+2)^{2}}

d=\sqrt{(0)^{2}+(8)^{2}}

dAB=8\ units

Step 2

<u>Find the distance BC</u>

B(6, 2),C(0, 8)

Substitute the values in the formula

d=\sqrt{(8-2)^{2}+(0-6)^{2}}

d=\sqrt{(6)^{2}+(-6)^{2}}

dBC=6\sqrt{2}\ units

Step 3

<u>Find the distance AC</u>

A(-2, 2),C(0, 8)

Substitute the values in the formula

d=\sqrt{(8-2)^{2}+(0+2)^{2}}

d=\sqrt{(6)^{2}+(2)^{2}}

dAC=2\sqrt{10}\ units

Step 4

<u>Find the distance DC</u>

D(0, 2),C(0, 8)

Substitute the values in the formula

d=\sqrt{(8-2)^{2}+(0-0)^{2}}

d=\sqrt{(6)^{2}+(0)^{2}}

dDC=6\ units

Step 5

<u>Find the perimeter of the triangle</u>

P=AB+BC+AC

substitute the values

P=8\ units+6\sqrt{2}\ units+2\sqrt{10}\ units

P=22.81\ units

therefore

The perimeter of the triangle is equal to 22.81\ units

Step 6

<u>Find the area of the triangle</u>

A=\frac{1}{2}*base *heigth

in this problem

base=AB=8\ units\\heigth=DC=6\ units

substitute the values

A=\frac{1}{2}*8*6

A=24\ units^{2}

therefore

the area of the triangle is 24\ units^{2}

4 0
3 years ago
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