Draw the quadrilateral ABCD gives in the figure. Draw a triangle having some area that of quadrilateral
Answers
Answer:
in pic
Step-by-step explanation:
Solution
Draw line AB=5.5 cm
Draw an arc of radius 6.5 cm from point A and an other arc of radius 4.5 cm from point B both meet at point C
Join C to A and B
Draw an angle of 40
0
at point A and an other angle of 40
0
at point C. Both cut at point D
Join D to C and A
Hence ABCD is the required quadrilateral.
In triangle ABC,
s=
2
6.5+5.5+4.5
=8.25
Then area of ABC=
8.25(8.25−6.5)(8.25−5.5)(8.25−4.5)
=
8.25×1.75×2.75×3.75
=
148.89
=12.20
In triangle ACD
sin40
0
AD
=
sin100
0
6.5
⇒AD=
0.98
6.5×0.64
=4.25cm
Then area triangle ACD =
2
1
absinθ=6.5×4.25sin40
0
=27.63×0.64=17.68cm
2
So area of ABCD quadrilateral =12.20+17.68=29.88 sq cm
Answer:
Draw a triangle having same base but double the height of quadrilateral. It will have equal area.
Take quadrilateral Parallelogram.
Step-by-step explanation:
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