Let h represent the height of the trapezoid, the perpendicular distance between AB and DC. Then the area of the trapezoid is
Area = (1/2)(AB + DC)·h
We are given a relationship between AB and DC, so we can write
Area = (1/2)(AB + AB/4)·h = (5/8)AB·h
The given dimensions let us determine the area of ∆BCE to be
Area ∆BCE = (1/2)(5 cm)(12 cm) = 30 cm²
The total area of the trapezoid is also the sum of the areas ...
Area = Area ∆BCE + Area ∆ABE + Area ∆DCE
Since AE = 1/3(AD), the perpendicular distance from E to AB will be h/3. The areas of the two smaller triangles can be computed as
Area ∆ABE = (1/2)(AB)·h/3 = (1/6)AB·h
Area ∆DCE = (1/2)(DC)·(2/3)h = (1/2)(AB/4)·(2/3)h = (1/12)AB·h
Putting all of the above into the equation for the total area of the trapezoid, we have
Area = (5/8)AB·h = 30 cm² + (1/6)AB·h + (1/12)AB·h
(5/8 -1/6 -1/12)AB·h = 30 cm²
AB·h = (30 cm²)/(3/8) = 80 cm²
Then the area of the trapezoid is
Area = (5/8)AB·h = (5/8)·80 cm² = 50 cm²
The diameter is 10:
To do this we must use the distance formula:
Distance =√(x2−x1)^2+(y2−y1)^2
So, if we substitute in our values for the origin and endpoint (origin is 1 values, endpoint is 2)
D=✓(-4-0)^2+(-3-0)^2
Simplified, this is
D=✓16+9
D=✓25
D=5
so, the distance from the center of the crcle to the endpoint is 5 (making the radius)
multiply by two, and the diameter of the circle is 10 :)
Answer:
X= 7 I'm sure
Step-by-step explanation:
If you want me to expand on the equation leave me a comment :)
