the parallel sides of trapezium are 20 cm and 13 cm its non parallel sides are 10 cm find the area of trapezium give me answer
who give me answer I will mark Brainliest answer
Answers
Given :-
parallel sides of a trapezium are 20cm and 13cm
non-parallel side are 10cm
I named the trapezium ABCD ( in the attached image)
two parallel sides = AB = 13cm and DC =20cm
non-parallel side = AD = BC = 10cm each
( draw a line parallel to BC ) ,BC = AE = 10cm
draw another line which is perpendicular to DC
now,
AB = EC = 13cm
DE = 20 - 13 = 7cm
(as AF is perpendicular to DC it divides DF, and FE equally )
so,
DF = FE = 7/2
DF = 3.5cm and FE = 3.5cm
AFE triangle and AFD triangle are right angled triangle.
triangle AFE
\begin{gathered} = {(AF )}^{2} + {(FE )}^{2} = {( AE)}^{2} \\ \\ = {(AF )}^{2} + {(3.5)}^{2} = {(10)}^{2} \\ \\ = {(AF )}^{2} + 12.25 = 100 \\ \\ = {(AF )}^{2} = 100 - 12.25 \\ \\ = {(AF )}^{2} = 87.75 \\ \\ = AF = \sqrt{87.75 } \\ \\ = AF = 9.36\end{gathered}
=(AF)
2
+(FE)
2
=(AE)
2
=(AF)
2
+(3.5)
2
=(10)
2
=(AF)
2
+12.25=100
=(AF)
2
=100−12.25
=(AF)
2
=87.75
=AF=
87.75
=AF=9.36
the height (AF) = 9.3cm
area of trapezium
\begin{gathered} = \frac{1}{2} \times (sum \: \: of \: \: two \: \: parallel \: \: sides) \times distance \: \: between \: them \\ \\ = \frac{1}{2} \times (13 + 20) \times 9.3 \\ \\ = \frac{1}{2} \times 33 \times 9.3 \\ \\ = \frac{33 \times 9.3}{2} \\ \\ = \frac{306.9}{2} \\ \\ = 153.45 \end{gathered}
=
2
1
×(sumoftwoparallelsides)×distancebetweenthem
=
2
1
×(13+20)×9.3
=
2
1
×33×9.3
=
2
33×9.3
=
2
306.9
=153.45
Area \: \: of \: \: trapezium = 153.45 {cm}^{2}Areaoftrapezium=153.45cm
2
Explanation:
hope it will help you buddy
