Question

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

Answer 1

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\tt Find \:the\: area\: of \:a \:triangle.\:Its\: two \:sides \:are\: 18\:cm\\\tt and\: 10\:cm, \:perimeter\: is\: 42\: cm.$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

• $\bf First \:side\:of\:the \:triangle(a) = 18\:cm$
• $\bf Second\:side\:of\:the \:triangle(b) = 10\:cm$
• $\bf Perimeter \:the \:triangle(P) = 42\:cm$

$\sf \bf {\boxed {\mathbb {TO\:FIND}}}$

• $\bf Area\:of\:the \:triangle(A)$

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

${\pink {\underline {\bf {\pmb {Third \:side\:of\:the \:triangle(c)}}}}}$

${\blue {\boxed {\boxed {\boxed {\green {\pmb {P_{(Triangle)}=a+b+c}}}}}}}$

• $\sf P= perimeter\:of\:the \:triangle$
• $\sf a=first \:side \:of\:the \:triangle$
• $\sf b=second\:side \:of\:the \:triangle$
• $\sf c=third \:side \:of\:the \:triangle$

${\underbrace {\overbrace {\orange {\pmb {Substitute \:the \:values}}}}}$

$\bf \implies 42=18+10+c$

$\bf \implies 42=28+c$

$\bf \implies c=42-28$

$\implies {\blue {\boxed {\boxed {\purple {\sf c=14\:cm}}}}}$

——————————————————————————————

${\pink {\underline {\bf {\pmb {Semiperimeter \:of\:the \:triangle(S)}}}}}$

${\blue {\boxed {\boxed {\boxed {\green {\pmb {S=\dfrac{P}{2}}}}}}}}$

• $\sf S=semiperimeter \:of\:the \:triangle$
• $\sf P=perimeter \:of\:the \:triangle$

${\underbrace {\overbrace {\orange {\pmb {Substitute \:the \:values}}}}}$

$\bf \implies S=\dfrac{42}{2}$

$\bf \implies S=\dfrac{\cancel{42}}{\cancel{2}}$

$\implies {\blue {\boxed {\boxed {\purple {\sf S=21\:cm}}}}}$

——————————————————————————————

${\pink {\underline {\bf {\pmb {Area \:of\:the \:triangle(A)}}}}}$

${\blue {\boxed {\boxed {\boxed {\green {\pmb {A_{(Triangle)}=\sqrt{S\Big(S-a\Big) \Big(S-b\Big) \Big(S-c\Big)}}}}}}}}$

• $\sf A=area\:of \:the \:triangle$
• $\sf S=semiperimeter \:of\:the \:triangle$
• $\sf a=first \:side \:of\:the \:triangle$
• $\sf b=second\:side \:of\:the \:triangle$
• $\sf c=third \:side \:of\:the \:triangle$

${\underbrace {\overbrace {\orange {\pmb {Substitute \:the \:values}}}}}$

$\bf \implies A=\sqrt{21\Big(21-18\Big) \Big(21-10\Big) \Big(21-14\Big)}$

$\bf \implies A=\sqrt{21\Big(3\Big) \Big(11\Big) \Big(7\Big)}$

$\bf \implies A=\sqrt{21\times 3\times 11\times 7}$

$\bf \implies A=\sqrt{4851}$

$\bf \implies A=\sqrt{441\times 11}$

${\blue {\boxed {\boxed {\purple {\mathfrak {A=21\sqrt{11}\:{cm}^{2}}}}}}}$

${\underbrace {\red {\overline {\red {\underline {\red {\pmb {\sf {{\therefore} The\:area \:of \:the \:triangle \:is\:21\sqrt{11}\:{cm}^{2}}}}}}}}}}$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

____________________________________________

$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

$\sf Area \:of \:the \:triangle = \dfrac{1}{2}bh$

$\sf Perimeter\:of \:the \:triangle =a+b+c$

Answer 2

Given :

• First side of the triangle = a = 18 cm
• Second side of the triangle = b = 10 cm
• Third side of the triangle = c = 42 cm

A N S W E R :

• The Area of a triangle is ${\sf{21 \sqrt{11} cm^2}}$

To find :

• Find Area of the triangle ?

Solution :

• a + b = 18 + 10 = 28 cm
• Perimeter of the rectangle = a + b + c = 42 cm
• c = 42 cm - 28 cm = 14 cm

Also,

${\sf{S \:=\; \dfrac{a + b + c}{2} = \cancel{\dfrac{42}{2} }}}$${\sf{=\;21}}$

Now, By Using Heron's Formula,

• $\boxed{\bf{Area\: of\: the\; ∆ \;=\;A\:=\; \sqrt{s(s - a)(s - b)(s - c)} }}$

${\sf{A\:=\; \sqrt{21(21 - 18)(21 - 10)(21 - 14)} }}$

:$\implies$${\sf{A\:=\; \sqrt{21\:×\:3\:×\:11\;×\;7} }}$

$~~~~~$:$\implies$${\sf{A\:=\; \sqrt{21\:×\:21\:×\:21} }}$

$~~~~~~~~~~$:$\implies$${\underline{\boxed{\frak{\pink{A\;=\;21\sqrt{11} cm^2}}}}}$★

Hence,

• ${\underline{\sf{The \: Area\: of\:a\: rectangle\;is\; \bf{21\sqrt{11\:cm^2}}}}}$

$~~~~$$\qquad\quad\therefore{\underline{\textsf{\textbf{Hence, Proved!}}}}$

$~~~~~~~~~~~~~~~$ ____________________