find the area of triangle whose vertices are (5,0),(8,0),(8,4)
Answers
AnswEr:
Let the two points be A(5,0) ,B(8,0) &C(8,4)
By using Formula to find area of triangle = 1/2 [x1 ( y2 - y3 ) + x2 ( y3 - y1 ) +x3 ( y1 - y2)]
Here , x1 = 5 ,x2 = 8 & x3 = 8
y1 = 0 ,y2 = 0 & y3 = 4
Putting Values :
Area of ∆ABC = 1/2 [ 5(0 -4) +8(4 - 0)+8( 0 -0)]
= 1/2 [5 × (-4) + 8 × 4 + 0]
= 1/2 [-20 + 32 + 0]
= 1/2 [ 12 ]
= 6 square units
Area of triangle is 6 square units.
The area is "6 square units". Further explanation is given below.
Step-by-step explanation:
Given vertices are:
(5,0), (8,0), (8,4)
(x₁,y₁) = (5,0)
(x₂,y₂) = (8,0)
(x₃-y₃) = (8,4)
As we know,
Area of a Triangle = 1/2 [x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
On putting the values in the above formula, we get
Area = 1/2 [5(0-4)+8(4-0)+8(0-0)]
= 1/2 [5(-4)+8(4)+8(0)]
= 1/2 (-20+36+0)
= 1/2 (12)
= 6 square units
Learn more:
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