I need help...
Did I end up with the correct answer?
A trapezoid has equal nonparallel sides, The upper base is 6, the lower base is 16 ( 11,5) and the diagonal is 12. What is the altitude of the trapezoid? Round your answer to the one decimal place.
I was told this is how to get the
answer.
h(squared)+ 11(sqared)=12(squared)
___I worked it out___
12(squared)-11(sqared)= 23
sqare root of 23 is 4.8
=4.8
Is this right?
Yes, it is correct. Drop a verical line from an upper correr perpendicualr to the base. The diagonal, h and 8+3 = 11 form a right triangle.
In order to find the altitude (height) of the trapezoid, you need to use the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
In this case, the diagonal of the trapezoid acts as the hypotenuse of a right triangle. Let's label the altitude as "h" and the other two sides of the triangle as 11 and 12.
To find the altitude, you can set up the equation:
h² + 11² = 12²
Simplifying this equation, you get:
h² + 121 = 144
Now, subtract 121 from both sides of the equation:
h² = 144 - 121
h² = 23
Next, take the square root of both sides of the equation to solve for h:
√(h²) = √23
h = √23
After taking the square root of 23, you get approximately 4.8. Therefore, the altitude of the trapezoid is approximately 4.8.
So, you are correct! Your answer of 4.8 is the correct altitude of the trapezoid.