Question

Find the area of a triangle whose two sides are 18 cm and 10 cm and
perimeter is 42 cm.​

Answer 1

Question:

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

perimeter is 42 cm.​

Given:

• AB = 18 cm

• BC = 10 cm

• Perimeter - 42 cm

To Find:

• Area

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf A }\put(0.5,-0.3){ \bf C }\put(5.2,-0.3){ \bf B }\end{picture}$

Solution:

$\mathcal{PERIMETER}\\\\\\\textsf{Perimeter=ab+bc+ca}\\\\\\42=18+10+ca\\\\\\42=28+ca\\\\\\42-28=ca\\\\\\ca=42-28\\\\\\ca=14cm$

$\mathcal{AREA}\\\\\\\sqrt{s(s-a)(s-b)(s-c)}\\\\\\\textsf{Where s is semiperimeter}\\\\\\s=\frac{a+b+c}{2}\\\\\\s=\frac{18+10+14}{2}\\\\\\s=\frac{42}{2}\\\\\\s=21cm\\\\\\\sqrt{s(s-a)(s-b)(s-c)}\\\\\\\sqrt{21(21-18)(21-10)(21-14)}\\\\\\\sqrt{21(3)(11)(7)}\\\\\\\sqrt{3\times3\times7\times7\times11} \\\\\\3\times7\sqrt{11} \\\\\\21\sqrt{11} \\\\\\\boxed {=21\sqrt{11}}$

Answer:

$\boxed {=21\sqrt{11}}$