the parallel side of a trapezium are 21 cm and 30 cm while its is 6cm find the numbers
Answers
Answer:
The answer is 144 cm².
Step-by-step explanation:
As shown in the attached figure, ABCD is a trapezium with AB = 20 cm, CD = 30 cm, AD = 6 cm and BC = 8 cm.
We are to find the area of the trapezium.
The side BE is drawn parallel to AD so that ABED is a parallelogram. Also, in ΔBEC, BE = 6 cm, EC = 10 cm and BC = 8 cm.
So,
s=\dfrac{BE+EC+BC}{2}=\dfrac{6+10+8}{2}=12~\textup{cm}.s=
2
BE+EC+BC
=
2
6+10+8
=12 cm.
Therefore, the area of ΔBEC is
\triangle=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{12(12-6)(12-10)(12-8)}=24~\textup{cm}^2.△=
s(s−a)(s−b)(s−c)
=
12(12−6)(12−10)(12−8)
=24 cm
2
.
We know that
\begin{gathered}\triangle=\dfrac{1}{2}\times BF\times EC\\\\\Rightarrow 24=\dfrac{1}{2}\times BF\times 10\\\\\Rightarrow BF=4.8.\end{gathered}
△=
2
1
×BF×EC
⇒24=
2
1
×BF×10
⇒BF=4.8.
The area of the parallelogram ABED will be
A=\dfrac{1}{2}(AB+CD)\times BF=\dfrac{1}{2}(20+30)\times 4.8=25\times 4.8=120~\textup{cm}^2.A=
2
1
(AB+CD)×BF=
2
1
(20+30)×4.8=25×4.8=120 cm
2
.
Thus, the area of the trapezium ABCD = Δ + A = 24 + 120 = 144 cm².
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Answer:
here
we know
Step-by-step explanation:
area of trapezium = 1/2(sum of parallel side) *height
=1/2(21+31)*6
= 26*6
=156cm^2
that 's all
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