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Geometry 22 Online
OpenStudy (anonymous):

In the diagram, ABCD is an isosceles trapezoid. Its bases are AB and CD. BA is extended to E, and DE and EB are perpendicular. Side BC is a diameter of semicircle O, AB=4, AE= 3,DE=4, and DC=10.

OpenStudy (anonymous):

a) the length of AD is _____________units b)the area of the entire figure to the nearest integer is ____________square units.

OpenStudy (jdoe0001):

since DEA makes 90degrees, AD is just the hypotenuse from the pythagorean theorem

OpenStudy (jdoe0001):

hypotenuse is \[sqrt(a^2+b^2)\]

OpenStudy (jdoe0001):

so sqrt(3^2+4^2)

OpenStudy (jdoe0001):

yes, so \[\sqrt(25)\]

OpenStudy (anonymous):

\[\sqrt{25}=5\]

OpenStudy (anonymous):

a is right but b wrong

OpenStudy (jdoe0001):

yes, as far as the full area of the whole composite, just add all adjacent figures you have there a triangle, a trapezoid, and a circle/2[semicircle], so \[1/2(3)(4)\] for the triangle PLUS \[4/2(4/10)\] for the trapezoid PLUS \[2\pi*r^2\] now keep in mind that the semicircle has a diameter of 4, the long line to the left, so half that is the radius, or "r"

OpenStudy (jdoe0001):

hold, the long line on the semicircle isn't 4 :/, is a bit more, since is not a straight line is slanted

OpenStudy (anonymous):

so how to solve it

OpenStudy (jdoe0001):

if you were to extend B by 3more units, EB would become 10, just like DC, a 10x4 rectangle, so, let me draw it

OpenStudy (jdoe0001):

OpenStudy (jdoe0001):

from the graph, notice that the hypotenuse for THAT triangle, will be the same as the one on the other side, and thus that diameter will be \[\sqrt(25)\]

OpenStudy (anonymous):

Hint ad is a hypotenuse of aright triangle the general strategy for computing the area of a combined figure is to divide the region up into individual components whose areas are easier to find, and then add up the areas.In this case, the region is a combination f a right triangle a trapezoid and a semicircle.

OpenStudy (jdoe0001):

right

OpenStudy (jdoe0001):

which is what you're already doing

OpenStudy (jdoe0001):

so the diameter is 5 for the semicircle, the radius is half that and thus \[2pi*r^2\] added to the other 2 areas will be it

OpenStudy (anonymous):

it was wrong answer b 44

OpenStudy (jdoe0001):

....hmm, I put in (3*4*1/2)+(4/2*(4+10))+(pi*5^2) and got something else

OpenStudy (jdoe0001):

rechecking

OpenStudy (jdoe0001):

is not 44 here :|

OpenStudy (anonymous):

it is 44

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