Question

Question
Find The Area Of A ∆ Triangle Two sides Of Which Are 18cm And 10Cm And The Perimeter Is 42Cm?

${\pink}{\rule}{1cm}{0cm}}$
​

Answer 1

Step-by-step explanation:

★Given

Find The Area Of A ∆ Triangle Two sides Of Which Are 18cm And 10Cm And The Perimeter Is 42Cm?

★To Find

$\bold { \rule{666cm}{0.1cm}}$

• Area Of A Triangle?

$\bold { \rule{666cm}{0.1cm}}$

Solution

${ \qquad { \sf \pink{\longmapsto \: Let A=18Cm}}}$

${ \qquad { \sf \red{\longmapsto \:B=10Cm}}}$

${ \qquad { \sf \blue{\longmapsto \: Let \: Perimeter=42Cm}}}$

So,

${ \qquad { \sf \purple{\longmapsto \: a + b + c=42Cm}}}$

${ \qquad { \sf {\longmapsto \:18 + 10 + c = 42}}}$

${ \qquad { \sf {\longmapsto \: = 28 + c = 42}}}$

${ \qquad { \sf {\longmapsto \: = c = \sf \pink{42 - 28}}}}$

${ \qquad { \sf {\longmapsto \:c = \sf \red{14cm} }}}$

We Know,

$\large{\purple{\bigstar}} \: \</u></strong><strong><u>:</u></strong><strong><u>{\underline{\boxed{\pink{\sf{S =\sf\green{\frac{A+B+C}{2}} }}}}}</u></strong><strong><u>$

~Putting Terms According To It

${ \qquad { \sf \blue {\longmapsto \: S= \sf \red{ \frac{18 + \sf \pink{10} + \sf \green{14}}{ \sf \purple{2}} } }}}$

$\qquad\longrightarrow\quad\sf \dfrac{\cancel{ \sf \green{42}}}{\cancel{ \sf \red{2}}} \sf \blue{ = 21}$

We Know,

$\bold { \rule{666cm}{0.1cm}}$

$\sf \blue{Area \: of ∆= }\large{\pink{\bigstar}} \: \</u></strong><strong><u>:</u></strong><strong><u>{\underline{\boxed{\green{\sf{ \sqrt{s(s - a)(s - b</u></strong><strong><u>)</u></strong><strong><u>(s-c)} }}}}}$

$\blue \bigstar{ \qquad { \sf {\longmapsto \sf \pink{ = } \sf \green{ \sqrt{21(21 - 18)(21 - 10)(21 - 14)} }}}}$

${ \qquad { \sf \red {\longmapsto \: \sf \green{ = }\: \sf \pink{ \sqrt{21 \times 3 \times 11 \times 7} }\: \: }}}$

${ \qquad { \sf \red {\longmapsto = \sf \pink{ 21} \sf \green{\sqrt[]{11cm {}^{2} } }}}}$

Therefore,

$\bold { \rule{666cm}{0.1cm}}$

Area Of Triangle=${ \qquad { \sf \red {\longmapsto = \sf \pink{ 21} \sf \green{\sqrt[]{11cm {}^{2} } }}}}$

$\bold { \rule{666cm}{0.1cm}}$

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More Information:

${\longmapsto{\qquad{\sf \blue{ Perimeter{\small_{(Rectangle)}} = \sf \pink{Length +Breadth} }}}}$

$\large{\blue{\bigstar}} \: \: {\underline{\boxed{\blue{\sf{ Area{\small_{(Rectangle)}} \sf \pink{= Length \times Breadth }}}}}} \: \: {\blue{\bigstar}}★$

Answer 2

Question:-

Find The Area Of A ∆ Triangle Two sides Of Which Are 18cm And 10Cm And The Perimeter Is 42Cm?

$\color{cyan}{\rule{45pt}{7pt}}\color{cyan}{\rule{45pt}{7pt}}\color{cyan}{\rule{45pt}{7pt}}\color{cyan}{\rule{45pt}{7pt}}$

Answer:-

$\large{\underbrace{\leadsto{\boxed{\sf\color{blue}{21\sqrt{11}cm^2}}}}}$

$\tt{\underline{Step-by-step \: explanation}}$

As per the data given in the question, we have to find the area of the triangle for the two sides.

The sides of triangle given: a = 18 cm, b = 10 cm

Perimeter is 42 cm

As we know that,

The perimeter of the triangle => (a + b + c)

=> 42 = 18 + 10 + c

=> 42 = 28 + c

=> c = 42 - 28

=> c = 14 cm

So, Semi Perimeter

$\rm{s = (a + b + c)}$

=> $\sf{\dfrac{42}{2} = 21 \:cm}$

Now, by using Heron's Formula

Area of triangle = $\sqrt{s \:(s - a)(s - b)(s - c)}$

$\sf :\implies{\sqrt{21 \: (21 - 18)(21 - 10)(21 - 14)}}$

$\sf :\implies{21 × 3 × 11 × 7}$

$\dashrightarrow{\boxed{\bf\color{magenta}{21\sqrt{11}cm^{2}}}}$

$\color{cyan}{\rule{45pt}{7pt}}\color{cyan}{\rule{45pt}{7pt}}\color{cyan}{\rule{45pt}{7pt}}\color{cyan}{\rule{45pt}{7pt}}$