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.
a) the length of AD is _____________units b)the area of the entire figure to the nearest integer is ____________square units.
since DEA makes 90degrees, AD is just the hypotenuse from the pythagorean theorem
hypotenuse is \[sqrt(a^2+b^2)\]
so sqrt(3^2+4^2)
yes, so \[\sqrt(25)\]
\[\sqrt{25}=5\]
a is right but b wrong
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"
hold, the long line on the semicircle isn't 4 :/, is a bit more, since is not a straight line is slanted
so how to solve it
if you were to extend B by 3more units, EB would become 10, just like DC, a 10x4 rectangle, so, let me draw it
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 \[\sqrt(25)\]
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.
right
which is what you're already doing
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
it was wrong answer b 44
....hmm, I put in (3*4*1/2)+(4/2*(4+10))+(pi*5^2) and got something else
rechecking
is not 44 here :|
it is 44
Join our real-time social learning platform and learn together with your friends!