Find the area of a triangle two side of which are 18cm and 10cm and the perimeter is 42 cm
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Let a=18 cm, b=10 cmPerimeter =42cm{tex}\\therefore a+b+c=42cm{/tex}So, C=14 cm{tex}\\therefore \\text{ S=}\\frac{a+b+c}{2}\\text{ =}\\frac{18+10+14}{2}{/tex} = 21 cmnew area of triangles {tex}=\\sqrt{21(21-18)\\text{ (21}-\\text{10) (21}-\\text{14)}}\\\\ \\text{= }\\sqrt{21\\times 3\\times 11\\times 7}{/tex}{tex}\\text{=21}\\sqrt{11}\\text{ }sq\\text{ cm}{/tex}