Question

The sum of two side of triangle is 28 cm and difference between the two is 8 cm. If the third side is third side measure 10 cm. Then find the area of triangle​

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Answer 1

~Answer :

• The area of triangle is 9√19 cm.

Given : The sum of two side of triangle is 28 cm and difference between the two is 8 cm. If the third side is third side measure 10 cm.

To Find : The area of triangle ?

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Solution : Let the area of triangle be 'x'.

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• a + b = 28 cm⠀⠀⠀⠀⠀⠀ (i)
• a - b = 8 cm⠀⠀⠀⠀⠀⠀ ⠀(ii)

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Then, add equation (i) & (ii), we get ;

• a + b + a - b = 28 cm + 8 cm

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After cancelling the unlike terms,

${\sf:\implies{\cancel{a~+~b}~+~\cancel{a~-~b}~-~28 cm~+~8 cm}}$

${\sf:\implies{2a~=~36 cm}}$

${\sf:\implies{a~=~\cancel\dfrac{36}{2}cm}}$

• $\boxed{\sf{\pmb{\pink{a~=~18 cm}}}}$★

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Then, from equation (i), we get ;

• a + b = 28 cm

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${\bf{\underline{Substituting~ the~ given ~values~in~ equation~\bf{(i)},~we~get~;}}}$

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${\sf:\implies{a~+~b~=~28 cm}}$

${\sf:\implies{18 cm~+~b~=~28 cm}}$

${\sf:\implies{b~=~28 cm~-~18 cm}}$

• $\boxed{\sf{\pmb{\purple{b~=~10 cm}}}}$★

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~Formula Used :

• ${\underline{\boxed{\pmb{\sf{ Area~of~∆~=~\sqrt{s(s~-~a)(s~-~b)(s~-~c)} }}}}}$

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${\bf{\underline{Putting~the~values~in~formula}}}$

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${\sf:\implies{Area~of~∆~=~\sqrt{\dfrac{18 cm~+~10 cm~+~10 cm}{2} \bigg(\dfrac{18 cm~+~10 cm~+~10 cm}{2}~-~18 cm\bigg) \bigg(\dfrac{18 cm~+~10 cm~+~10 cm}{2}~-~10 cm\bigg) \bigg(\dfrac{18 cm~+~10 cm~+~10 cm}{2}~-~10 cm\bigg)}}}$

${\sf:\implies{Area~of~∆~=~\sqrt{\dfrac{38 cm}{2} \bigg(\dfrac{38 cm}{2}~-~18 cm\bigg) \bigg(\dfrac{38 cm}{2}~-~10 cm\bigg) \bigg(\dfrac{38 cm}{2}~-~10 cm\bigg)}}}$

${\sf:\implies{Area~of~∆~=~\sqrt{\cancel\dfrac{38 cm}{2} \bigg(\cancel\dfrac{38 cm}{2}~-~18 cm\bigg) \bigg(\cancel\dfrac{38 cm}{2}~-~10 cm\bigg) \bigg(\cancel\dfrac{38 cm}{2}~-~10 cm\bigg)}}}$

${\sf:\implies{Area~of~∆~=~\sqrt{19 cm(19 cm~-~18 cm) (19 cm~-~10 cm) (19 cm~-~10 cm)}}}$

${\sf:\implies{Area~of~∆~=~\sqrt{19 cm~×~1 cm~×~9 cm~×~9 cm}}}$

${\sf:\implies{Area~of~∆~=~\sqrt{19 cm~×~1 cm~×~\underline{9 cm~×~9 cm}}}}$

• ${\underline{\boxed{\pmb{\sf{\pink{Area~of~∆~=~9\sqrt{9 cm}}}}}}} \; {\red{\bigstar}}$

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Hence,

• The area of triangle is 9√19 cm.