SOLUTION: please illustrate the figure..the longer base of an isosceles trapezoid is 40, one leg is 20 and one of the base angles is 63 degrees. (a) find the altitude and shorter base of th

Algebra ->  Surface-area -> SOLUTION: please illustrate the figure..the longer base of an isosceles trapezoid is 40, one leg is 20 and one of the base angles is 63 degrees. (a) find the altitude and shorter base of th      Log On


   

Question 29178: please illustrate the figure..the longer base of an isosceles trapezoid is 40, one leg is 20 and one of the base angles is 63 degrees.
(a) find the altitude and shorter base of the trapezoid
(b) using the results obtained in (a), find the area of the trapezoid
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
SEE THE TRAPEZOID ....ABCD below
AREA OF TRAPEZOID IS GIVEN BY
AREA=(1/2)*ALTITUDE*(SUM OF PARALLEL SIDES)=(1/2)*DP*(AB+CD)
WHERE ALTITUDE IS THE PERPENDICULAR DISTANCE BETWEEN PARALLEL SIDES =DP
PARALLEL SIDES ARE AB AND CD.
WE HAVE AB=40
BC=AD=20
ANGLE DAP=63
DAP IS A RIGHT ANGLED TRIANGLE...DP/DA=SIN(63)
DP=DA*SIN(63)=20*0.89=17.8
SO ALTITUDE=17.8
AP/DA=COS(63)
AP=DA*COS(63)=20*0.4545=9.1
SO CD =AB-2*AP SINCE TRAPEZOID IS ISOCELLES..
CD=40-2*9.1=21.8
AREA=(1/2)17.8(40+21.8)=550 SQUARE UNITS.