Question

plz answer this in detail
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Answer 1

AS TRIANGLE DCB IS RIGHT ANGLED DC^2+BC^2=BD^2

5^2+12^=BD^2

25+144=BD^2

169=BD^2

BD=13

FOR AREA OF TRIANGLE BCD WE HAVE

1/2×BASE×HEIGHT

1/2×5×12

30m^2

AND FOR TRIANGLE ABD WE HAVE HERONS FORMULA

S=(8+9+13)/2

30/2

15

=sqrt[s (s-a)×(s-b)×(s-c)]

sqrt [15×7×6×2]

6 sqrt 35 m^2

adding both areas we get

6 (5+sqrt 35) m^2

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