Find the area of a triangle two sides of which are 18cm and 10cm and t

1.	Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.
Answer» ${\bold{\pink{\underline{\green{G}\purple{iv}\orange{en}\red{:-}}}}}$ $\bf\:perimeter=42\:<klux>CM</klux>$ $\bf\:First\:side\:(a)=18\:cm$ $\bf\:second\:side\:(<klux>B</klux>)=10\:cm$ $\bf\:Third\:side\:(<klux>C</klux>)=perimeter-(a+b)$ $\bf\:Third\:side=42-(18+10)\:cm$ $\bf\:Third\:side=42-28\:cm$ $\bf\:Third\:side=<klux>14</klux>\:cm$ ${\bold{\blue{\underline{\red{To}\:\pink{Fi}\green{nd}\purple{:-}}}}}$ $\bf\green{{Area\:of\:triangle=?}}$ ${\bold{\pink{\underline{\red{So}\purple{lut}\green{ion}\orange{:-}}}}}$ Now we have, a = 18 cm ,b = 10 cm and c = 14 cm $\bf\:Semi\:perimeter=\dfrac{a+b+c}{2}$ $\bf\:Semi\:perimeter=\dfrac{18+10+14}{2}$ $\bf\:Semi\:perimeter=\dfrac{42}{2}$ $\bf\red{{Semi\:perimeter=21\:cm}}$ $\bf\boxed\star\pink{\underline{\underline{{Using\:heron's\:formula:-}}}}$ $\bf\:Area=\sqrt{s(s-a)(s-b)(s-c)}$ $\bf\:Area=\sqrt{21(21-18)(21-10)(21-14)}$ $\bf\:Area=\sqrt{21\times\:3\times\:11\times\:7}$ $\bf\:Area=\sqrt{3\times\:7\times\:3\times\:11\times\:7}$ $\bf\:Area=3\times\:7\sqrt{11}\:c{m}^{2}$ $\bf\orange{{Area=21\sqrt{11}\:c{m}^{2}}}$ Hence,area of triangle will be 21√11 cm²