What the base b1 of an Isosceles trapezoid if b2=57ft,h=17ft6inches and both legs are 23ft? or the formula for the solution?thank you.
At each side of the trapezoid is a righ triangle with height 17.5 and hypotenuse 23. the base leg a is given by
a^2 + 17.5^2 = 23^2
so, a = 14.92
There are two of these triangles, so their combined base legs make 29.84
Subtract that from 57 and we see that the inner rectangle has height 17.5 and width 27.16 top and bottom. That's the upper base of the trapezoid.
thanks
To find the base b1 of an isosceles trapezoid, you can use the formula:
b1 = b2 - (2h / Φ)
Where:
- b2 is the length of the top base of the trapezoid (given as 57 feet).
- h is the height or altitude of the trapezoid (given as 17 feet 6 inches).
- Φ is the diagonal or slant height of the trapezoid, which can be calculated using the Pythagorean theorem.
Since both legs of the trapezoid are given as 23 feet, we can use this as the value for Φ.
First, let's convert the height from feet and inches to feet. Since there are 12 inches in a foot, 6 inches is equal to 6/12 = 0.5 feet. So, the height h is 17.5 feet.
Now, let's calculate the value of Φ:
Φ = √(h^2 + ((b2 - b1) / 2)^2)
Since the trapezoid is isosceles, we can assume that b1 and b2 are equal, so the equation simplifies to:
Φ = √(h^2 + ((b2 - b2) / 2)^2)
= √(h^2 + 0^2)
= h
Now substitute the known values into the formula for b1:
b1 = b2 - (2h / Φ)
= 57 - (2 * 17.5 / 23)
Using a calculator to evaluate the expression:
= 57 - (35 / 23)
= 57 - 1.52
= 55.48
Therefore, the base b1 of the isosceles trapezoid is approximately 55.48 feet.