Question

Calculate the area of shaded region in the above figure.

1X4=4 M

120 m

122 m

24 m

26 m

22 m
​

Answer 1

Answer:

314

Step-by-step explanation:

120

+122

=242

+26

+24

+22

=314

Answer 2

Step-by-step explanation:

semiperimeter of triangle ABC (s)

semiperimeter of triangle ABC (s)= 120+122+22/2

semiperimeter of triangle ABC (s)= 120+122+22/2=132

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)=√132*12*10*110

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)=√132*12*10*110=√12*11*12*11*10*10

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)=√132*12*10*110=√12*11*12*11*10*10=12*11*10

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)=√132*12*10*110=√12*11*12*11*10*10=12*11*10=1320m^2

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)=√132*12*10*110=√12*11*12*11*10*10=12*11*10=1320m^2Semiperimeter of triangle OBC

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)=√132*12*10*110=√12*11*12*11*10*10=12*11*10=1320m^2Semiperimeter of triangle OBC=24+26+22/2=36

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)=√132*12*10*110=√12*11*12*11*10*10=12*11*10=1320m^2Semiperimeter of triangle OBC=24+26+22/2=36Area of triangle OBC

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)=√132*12*10*110=√12*11*12*11*10*10=12*11*10=1320m^2Semiperimeter of triangle OBC=24+26+22/2=36Area of triangle OBC=√36(36-24)(36-26)(36-22)

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)=√132*12*10*110=√12*11*12*11*10*10=12*11*10=1320m^2Semiperimeter of triangle OBC=24+26+22/2=36Area of triangle OBC=√36(36-24)(36-26)(36-22)=√36*14*10*14

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)=√132*12*10*110=√12*11*12*11*10*10=12*11*10=1320m^2Semiperimeter of triangle OBC=24+26+22/2=36Area of triangle OBC=√36(36-24)(36-26)(36-22)=√36*14*10*14=254m^2

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)=√132*12*10*110=√12*11*12*11*10*10=12*11*10=1320m^2Semiperimeter of triangle OBC=24+26+22/2=36Area of triangle OBC=√36(36-24)(36-26)(36-22)=√36*14*10*14=254m^2Area of shaded region

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)=√132*12*10*110=√12*11*12*11*10*10=12*11*10=1320m^2Semiperimeter of triangle OBC=24+26+22/2=36Area of triangle OBC=√36(36-24)(36-26)(36-22)=√36*14*10*14=254m^2Area of shaded regionarea of triangle ABC- area of triangle OBC

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)=√132*12*10*110=√12*11*12*11*10*10=12*11*10=1320m^2Semiperimeter of triangle OBC=24+26+22/2=36Area of triangle OBC=√36(36-24)(36-26)(36-22)=√36*14*10*14=254m^2Area of shaded regionarea of triangle ABC- area of triangle OBC=1320-254

semiperimeter of triangle ABC (s)= 120+122+22/2=132Area of triangle ABC=√s(s-a)(s-b)(s-c)=√132(132-120)(132-122)(132-22)=√132*12*10*110=√12*11*12*11*10*10=12*11*10=1320m^2Semiperimeter of triangle OBC=24+26+22/2=36Area of triangle OBC=√36(36-24)(36-26)(36-22)=√36*14*10*14=254m^2Area of shaded regionarea of triangle ABC- area of triangle OBC=1320-254=1065.45m^2

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