SOLUTION: 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 r

Algebra ->  Surface-area -> SOLUTION: 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 r      Log On


   

Question 29127: 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):
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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
LET ABCD BE THE TRAPEZOID
AB=40
BC=20
ANGLE ABC=63
LET CP BE ALTITUDE FROM C ON TO AB
CP=?
CD=?
AREA =?
CPB IS A RIGHT ANGLED TRIANGLE.
CP/CB=SIN (ANGLE CBP)=SIN(ANGLE CBA)=SIN(63)
CP=CB*SIN(63)=20*SIN63=APPROXIMATELY 17.5
WE NEED ONE MORE DATA TO SOLVE THE OTHER QUESTIONS.A QUADRILATERAL NEEDS 5 DATA..TRAPEZIUM NEEDS 4 DATA... WE ARE GIVE ONLY 3 DATA..IF BOTH BASE ANGLES ARE 63 THEN WE CN SOLVE..CHECK BACK YOUR QUESTION.
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SORRY ..IT IS AN ISICELLES TRAPEZOID...I MISSED THAT...SEE THE ANSWER BELOW
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.