find the area of atriangle two sides of which are 18cm 10 cm and perimetre is 42 cm

Dear Student,

Please find below the solution to the asked query:

Given : Two sides of given triangle are 18 cm 10 cm and perimeter is 42 cm , So

Third side =  42 - 18 - 10 = 14 cm

We know Heron's formula :

Area of triangle  = $\sqrt{s\left(s - a\right)\left(s - b\right)\left(s - c\right)}$ , Here s  = $\frac{a + b\hspace{0.17em}+ c}{2}$

We have :  a= 18 , b = 10  and c  =  14  , So

s  = $\frac{18 + 10\hspace{0.17em}+ 14}{2} = \frac{42}{2}$ =  21 cm

Therefore,

Area of given triangle = $\sqrt{21\left(21 -18\right)\left(21 - 10\right)\left(21 - 14\right)} = \sqrt{21 \times 3 \times 11 \times 7 }=\mathbf{ }\mathbf{21}\sqrt{\mathbf{11}}\mathbf{ }\mathbf{=}\mathbf{21}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{3}\mathbf{.}\mathbf{316}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{69}\mathbf{.}\mathbf{636}\approx \mathbf{69}\mathbf{.}\mathbf{64}\mathbf{ }{\mathbf{cm}}^{\mathbf{2}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$

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