Question

find the are of shaded region by herons formula​

Answer 1

Answer:

1074 m2

Step-by-step explanation:

perimeter of triangle ABC

= 120 + 122 + 22 = 264m

semi perimeter

s = 264/2 = 132 m

area of triangle ABC

= √[(132(132-122)(132-120)(132-22)]

= 1320m^2

perimeter of inner triangle BOC

= 24+26+22 = 72 m

s = 72/2 = 36m

area of Triangle BOC

= √[36(36-24)(36-26)(36-22)]

= 245.92 => 246m^2 (approx)

area of shaded region

= 1320 - 246 = 1074m^2 Ans

Answer 2

$\Large\frak{\underline{\red{Given :}}}$

• Dimensions of Triangle ABC :

AB = 120m, BC = 22m, CA = 122m.

• Dimensions of Triangle EBC :

EB = 24m, BC = 22m, EC = 26m.

$\Large\frak{\underline{\red{Need\:To\:Find :}}}$

• Area of Shaded region.

$\Large\frak{\underline{\red{ Complete\:Solution:}}}$

Firstly, we have to find the area of Triangle ABC.

$\longrightarrow{\boxed{\sf{\orange{ Semiperimeter (s) = \dfrac{sum\:of\:all\:sides}{2}}}}}$

• Substituting the values.

$\longrightarrow{\sf{ \dfrac{ 120m+122m+22m}{2}=264}}$

$\large\implies{\boxed{\sf{ 132}}}$

$\longrightarrow{\boxed{\sf{\orange{ Area_{triangle}\: (Heron's\:Formulae) = \sqrt{ s(s-a)(s-b)(s-c)}}}}}$

• Substituting the values.

$\longrightarrow{\sf{ \sqrt{132(132-122)(132-120)(132-22)}}}$

$\longrightarrow{\sf{ \sqrt{132(10)(12)(110)}}}$

• Factorise the numbers.

$\longrightarrow{\sf{ \sqrt{ 2\times 2\times 3\times 11\times 2\times 5 \times 2\times 2\times 3\times 2\times 5\times 11}}}$

• Simplify.

$\longrightarrow{\sf{ 2 \times 2\times 2\times 3 \times 5\times 11 }}$

$\large\implies{\boxed{\sf{ 1320m^{2}}}}$

Now, we need to find Area of Triangle EBC.

$\longrightarrow{\boxed{\sf{\orange{ Semiperimeter (s) = \dfrac{sum\:of\:all\:sides}{2}}}}}$

• Substituting the values.

$\longrightarrow{\sf{ \dfrac{ 26m+24m+22m}{2}=\dfrac{72m}{2}}}$

$\large\implies{\boxed{\sf{ 36m}}}$

$\longrightarrow{\boxed{\sf{\orange{ Area_{(triangle)}\: (Heron's\:Formulae) = \sqrt{ s(s-a)(s-b)(s-c)}}}}}$

• Substituting the values.

$\longrightarrow{\sf{ \sqrt{36(36-26)(36-24)(36-22)}}}$

$\longrightarrow{\sf{ \sqrt{36(10)(12)(14)}}}$

• Factorise the numbers.

$\longrightarrow{\sf{ \sqrt{ 2\times 2\times 3\times 3\times 2\times 5\times 2\times 2\times 3 \times 7\times 2} }}$

• Simplify.

$\longrightarrow{\sf{ 2\times 2\times 2 \times 3 \sqrt{ 3\times 7\times 5} }}$

$\large\implies{\boxed{\sf{ 24\sqrt{105}m^{2}}}}$

We should subtract Area of Triangle ABC by area of Triangle EBC to get the area of shaded region.

$\longrightarrow{\sf{ 1320m^{2} - 24\sqrt{105}m^{2}}}$

$\Large{\boxed{\sf{\red{ 1074cm^{2}approx.}}}}$

Hence, the area of shaded region will be 1074.08cm².

And we are done! :D

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