The sides of a quadrilateral taken in order are 20 cm, 27 cm, 7 cm and 24m. The angle between the last two sides is a right angle, find the area of quadrilateral.
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Answer:
375.85 m
Step-by-step explanation:
Answer
The sides of a quadrilateral field taken order as
AB=26 m
BC=27 m
CD=7 m
and DA=24 m
Diagonal AC is joined
Now △ADC
AC
2
=AD
2
+CD
2
AC=
24
2
+7
2
AC=
625
AC=25 m
Now area of △ABC
S=
2
1
(AB+BC+CA)
=
2
1
(26+27+25)
=
2
78
=39 m
By using heron's formula
Area of △ABC=
S(S−AB)(S−BC)(S−CA)
=
39(39−26)(39−27)(39−25)
=
39×13×12×14
=
85179
=291.85 m
2
Now area of △ADC
S=
2
1
(AD+CD+AC)
=
2
1
(25+24+7)
=
2
56
=28 m
By using heron's formula
Area of △ADC=
S(S−AD)(S−DC)(S−CA)
=
28(28−24)(28−7)(28−25)
=
28×4×21×3
=
7056
=84 m
2
hence, total area is375.85 m
2
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