find the area of the triangle whose vertices are (-10, 6), (-6,6), and ( -6, 15)?

Question

geometry

Answer 1

This is very similar to the last question you asked. Plot the points, connect them to create a triangle, and figure out the lengths of the base and height. You can use that to figure out the area of the triangle using the formula for an area.

Answer 2

First you need to know the lengths of the three sides of the triangle...

A (-10,6)
B (-6,6)
C (-6,15)

AB is 4 units long
BC is 9 units long
AB and BC form a right angle (see how the y's in AB and the x's in BC are equal?)

To find the area of a triangle, it is 1/2(Base*Height) or

(1/2)(4)(9)=18

Answer 3

This is a righttrianle with legs = 9 and 4.
Therefore area= 9*4/2 = 18 square units.

Answer 4

A=bh/2
A=4*9/2
A=18

Answer 5

For a triangle with vertices (x?, y?), (x?, y?), and (x?, y?), define: A = | x? y? one million | | x? y? one million | | x? y? one million |. Then, the component of the triangle is: section = one million/2 * |A|. hence, we've: A = | -8 10 one million | | 6 17 one million | | 2 -4 one million | = [(-8)(17)(one million) + (10)(one million)(2) + (one million)(6)(-4)] - [(-8)(one million)(-4) + (10)(6)(one million) + (one million)(17)(2)] = -266. this gives the section to be: section = one million/2 * |A| = one million/2 * |-266| = 133 sq. instruments (option C). i wish this facilitates!

Answer 6

Use the matrices on your calculator and you end up with 18 units square.

Answer 7

you Crammer's rule

............l-10 . 6 . 1l -10 6
+/-1/2. l-6.....6...1l -6...6
............l-6....15..1l -6...15

A = +/-1/2 (-60+(-36)+(-90)) - (-36+(-150)+(-36)
A= +/-1/2 (-186-(-222))
A= +/-1/2 (36)

since 36 is already positive, use +1/2

A= 1/2(36)
A= 18 unit^2

Answer 8

ok first find the distance between each of the points so you could the the image of the triangle, then youd know each side. take two sides then divide it by 2.

Answer 9

(1/2)*4*9
=18 sq.units