The two sides of a triangle are in the ratio of 2:3 and it’s third side is 5cm . if the perimeter of the triangle is 15cm , what i

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1. [tex]\large\underline{\sf{Solution-}}[/tex]

   Let us consider a triangle ABC such that AB, BC and CA is represented by a, b, c respectively.

   According to statement,

   Let assume that,

   • a : b = 2 : 3 and c = 5 cm.

   Let a = 2x and b = 3x

   Further given that, Perimeter of Triangle = 15 cm

   ⟼ a + b + c = 15

   ⟼ 2x + 3x + 5 = 15

   ⟼ 5x + 5 = 15

   ⟼ 5x = 15 – 5

   ⟼ 5x = 10

   ⇛ x = 2

   Hence,

   The sides of triangle ABC are

   • ⟼ a = 2x = 2 × 2 = 4 cm

   • ⟼ b = 3x = 3 × 2 = 6 cm

   • ⟼ c = 5 cm

   We know that,

   [tex]\underline{\boxed{\sf Semi \: Perimeter \ of \ a \ triangle \: (s)= \dfrac{1}{2} (a+b+c)}}[/tex]

   [tex]\rm :\longmapsto\:s = \dfrac{1}{2}(4 + 6 + 5)[/tex]

   [tex]\rm :\longmapsto\:s = \dfrac{1}{2}(15)[/tex]

   [tex]\bf\implies \:s = \dfrac{15}{2} \: cm[/tex]

   We know,

   [tex]\underline{\boxed{\bf Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)} }}[/tex]

   [tex] \rm \:= \: \: \sqrt{\dfrac{15}{2}\bigg(\dfrac{15}{2} – 4\bigg)\bigg(\dfrac{15}{2} – 6 \bigg)\bigg(\dfrac{15}{2} – 5\bigg) } [/tex]

   [tex] \rm \:= \: \: \sqrt{\dfrac{15}{2}\bigg(\dfrac{15 – 8}{2}\bigg)\bigg(\dfrac{15 – 12}{2} \bigg)\bigg(\dfrac{15 – 10}{2}\bigg) } [/tex]

   [tex] \rm \:= \: \: \sqrt{\dfrac{15}{2}\bigg(\dfrac{7}{2}\bigg)\bigg(\dfrac{3}{2} \bigg)\bigg(\dfrac{5}{2}\bigg) } [/tex]

   [tex] \rm \:= \: \:\dfrac{15}{4} \sqrt{7} \: {cm}^{2} [/tex]

   Additional Information :-

   [tex]\green{\boxed{\bf{Area_{(rectangle)} = l \times b}}}[/tex]

   [tex]\green{\boxed{\bf{Area_{(square)} = 4 \times side}}}[/tex]

   [tex]\green{\boxed{\bf{Area_{(circle)} = \pi \: {r}^{2} }}}[/tex]

   [tex]\green{\boxed{\bf{Area_{(rhombus)} = h \times b}}}[/tex]

   [tex]\green{\boxed{\bf{Area_{(parallelogram)} = h \times b}}}[/tex]

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