Question

An isosceles triangle has the perimeter 30 cm and each of the equal sides is 12 cm. Find its area.​

Answer 1

$\large {\dag \; {\underline{\underline{\orange{\pmb{\sf{ \; Given \; :- }}}}}}}$

• Equal sides of the Triangle = 12 cm
• Perimeter of the Triangle = 30 cm

$\large {\dag \; {\underline{\underline{\color{darkblue}{\pmb{\sf{ \; To \; Find \; :- }}}}}}}$

• Area of the Triangle = ?

$\large {\dag \; {\underline{\underline{\pink{\pmb{\sf{ \; Solution \; :- }}}}}}}$

✭ Formula Used :

• ${\underline{\boxed{\red{\sf{ S = \dfrac{ a + b + c}{2} }}}}}$

• ${\underline{\boxed{\red{\sf{ Perimeter = a + b + c }}}}}$

• ${\underline{\boxed{\red{\sf{ Area = \sqrt{s (s - a)(s - b)(s - c)} }}}}}$

Where :

• ➟ s = Semi - Perimeter
• ➟ a = Side 1
• ➟ b = Side 2
• ➟ c = Side 3

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✭ Calculating the 3rd Side :

${\longmapsto{\qquad{\sf{ Perimeter = a + b + c }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ 30 = 12 + 12 + c }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ 30 = 24 + c }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ 30 - 24 = c }}}} \\ \\ \\ \ {\qquad \; {\therefore \; {\underline{\boxed{\color{maroon}{\pmb{\frak{ 3rd \; Side = 6 \; cm }}}}}}}}$

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✭ Calculating the Semi - Perimeter :

${\implies{\qquad{\sf{ S = \dfrac{a + b + c}{2} }}}} \\ \\ \\ \ {\implies{\qquad{\sf{ S = \dfrac{12 + 12 + 6}{2} }}}} \\ \\ \\ \ {\implies{\qquad{\sf{ S = \dfrac{30}{2} }}}} \\ \\ \\ \ {\implies{\qquad{\sf{ S = \cancel\dfrac{30}{2} }}}} \\ \\ \\ \ {\qquad \; {\therefore \; {\underline{\boxed{\orange{\pmb{\frak{ Semi - Perimeter = 15 \; cm }}}}}}}}$

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✭ Calculating the Area :

${\dashrightarrow{\qquad{\sf{ Area = \sqrt{s (s - a)(s - b)(s - c)} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = \sqrt{15 (15 - 12)(15 - 12)(15 - 6)} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = \sqrt{15 \times 3 \times 3 \times 9} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = \sqrt{3 \times 5 \times 3 \times 3 \times 3 \times 3} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = 3 \times 3 \times \sqrt{3} \times \sqrt{5} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = 9 \times \sqrt{15} }}}} \\ \\ \\ \ {\qquad \; {\therefore \; {\underline{\boxed{\purple{\pmb{\frak{ Area = 9 \sqrt{15} \; {cm}^{2} }}}}}}}}$

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✭ Therefore :

❛❛ Area of the given Triangle is 9√15 cm² . ❜❜

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Answer 2

Step-by-step explanation:

Given

The perimeter of an isosceles triangle is 30 cm

The length of the sides which are equal is12 cm.

Let the third side length be 'a cm'.

Then,

⇒30=12+12+x

⇒30=24+x

⇒24+x=30

⇒x=30−24

⇒ x = 6

$\boxed{ \red{\pmb{So, \: the \: length \: of \: the \: third \: side \: is \: 6 cm.}}}$

Thus,the semi perimeter of the isosceles triangle (s) = 30/2 cm =15 cm

By using Heron's Formula,

Area of the triangle,

$\bf= \sqrt{s(s - a)(s - b)(s - c)} \\ \bf= \sqrt{15(15 - 12)(15 - 12)(15 - 6)} \: {cm}^{2} \\ \bf= \sqrt{15 \times 3 \times 3 \times 9} \: {cm}^{2} \\ \bf= 9 \sqrt{15} \: {cm}^{2}$

$\boxed{ \red{ \pmb{Hence, \: the \: area \: of \: the \: triangle \: is \: 9 \sqrt 15 cm^ 2 .}}}$