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sp2606 [1]
3 years ago
6

What is the area of the triangle in the coordinate plane?

Mathematics
2 answers:
xeze [42]3 years ago
8 0

Answer:

The area of the triangle is 16\ units^{2}

Step-by-step explanation:

we know that

The area of a triangle is equal to

A=\frac{1}{2}bh

where

b is the base of triangle

h is the height of triangle

In this problem

Observing the graph

b=(12-8)=4\ units

h=(10-2)=8\ units

substitute the values

A=\frac{1}{2}(4)(8)=16\ units^{2}

katovenus [111]3 years ago
6 0
To solve for the area of a triangle you use the formula A=b*h/2. The base for this triangle is 4 and the height is 8. Take 4*8 which is 32 and divide that answer in half to get 16 which is your answer.
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How do I do 8b(ii) ? Please help me thank you!
Mila [183]
Step One
======
Find the length of FO (see below)

All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)

Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ

Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.

FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)]                                           \
OJ = ??

[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
7 + 2sqrt(10) = 1/4 (7 + 2sqrt(10)) + OJ^2      Multiply through by 4
28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2    Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2   Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
21/4 + 3/2 * sqrt(10) = OJ^2 Take the square root of both sides.
sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )

Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.


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