The sides of a triangle are in the ratio of 12: 17: 25 and its perimet

1.	The sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540cm. Find its area.
Answer» align="absmiddle" alt="\HUGE{\bf{\green{\mathfrak{\MALTESE{Question:-}}}}}" class="latex-formula" ID="TexFormula1" src="https://tex.z-dn.net/?f=%20%5Chuge%7B%5Cbf%7B%5Cgreen%7B%5Cmathfrak%7B%5Cmaltese%7BQuestion%3A-%7D%7D%7D%7D%7D%20" title="\huge{\bf{\green{\mathfrak{\maltese{Question:-}}}}}"> $\bullet{\mapsto}$ The sides of a triangle are in the ratio of 12:17:25 and its perimeter is 540cm. Find its area. $\huge {\bf{\orange{\mathfrak{\maltese{Answer:-}}}}}$ $\bullet{\leadsto} \: \textsf{Sides of triangle are in the ratio 12:17:25.}$ $\bullet{\leadsto} \: \textsf{Perimeter of the triangle = 540 cm.}$ $\bullet{\leadsto} \: \textsf{Let the sides of the triangle be x.}$ $\bullet{\leadsto} \: \therefore{\textsf{1st side = a = 12x.}}$ $\bullet{\leadsto} \: \therefore{\textsf{2nd side = b = 17x.}}$ $\bullet{\leadsto} \: \therefore{\textsf{3rd side = c = 25x.}}$ $\bullet{\leadsto} \: \textsf{Perimeter of the triangle = 540 cm.}$ $\bullet{\leadsto} \: \textsf{We know that,}$ $\bullet{\leadsto} \: \boxed{\textsf{Perimeter of a triangle = a + b + c.}}$ $\bullet{\leadsto} \: \textsf{Substituting the values,}$ $\bullet{\leadsto} \: \sf 540 \: = \: 12x \: + \: 17x \: + \: 25x$ $\bullet{\leadsto} \: \sf 540 \: = \: 54x$ $\bullet{\leadsto} \: \sf 54x \: = \: 540$ $\bullet{\leadsto} \: \sf x \: = \: \dfrac{540}{54}$ $\bullet{\leadsto} \: \boxed{\sf x \: = \: 10 \: cm.}$ $\bullet{\leadsto} \: \sf 1st \: side \: = \: a \: = \: 12x \: = \: 12 \: * \: 10 \: = \: 120 \: cm.$ $\bullet{\leadsto} \: \sf 2nd \: side \: = \: b \: = \: 17x \: = \: 17 \: * \: 10 \: = \: 170 \: cm.$ $\bullet{\leadsto} \: \sf 3rd \: side \: = \: c \: = \: 25x \: = \: 25 \: * \: 10 \: = \: 250 \: cm.$ $\bullet{\leadsto} \: \textsf{Using Heron's Formula,}$ $\bullet{\leadsto} \: \sf Semi-Perimeter \: = \: ?$ $\bullet{\leadsto} \: \sf Perimeter \: = \: 540 \: cm.$ $\bullet{\leadsto} \: \boxed{\sf Semi-Perimeter \: = \: \dfrac{Perimeter}{2}}$ $\bullet{\leadsto} \: \sf S \: = \: \dfrac{540}{2}$ $\bullet{\leadsto} \: \sf S \: = \: 270 \: cm.$ $\bullet{\leadsto} \: \boxed{\sf Heron's \: Formula \: = \: \sqrt{s(s \: - \: a)(s \: - \: b)(s \: - \: c)}.}$ $\bullet{\leadsto} \: \textsf{Substituting the values,}$ $\bullet{\leadsto} \: \sf \sqrt{270(270 \: - \: 120)(270 \: - \: 170)(270 \: - \: 250)}$ $\bullet{\leadsto} \: \sf \sqrt{270(150)(100)(20)}$ $\bullet{\leadsto} \: \sf \sqrt{270 \: * \: 150 \: * \: 100 \: * \: 20}$ $\bullet{\leadsto} \: \sf \sqrt{27 \: * \: 15 \: * \: 1 \: * \: 2 \: * \: 10^{5}}$ $\bullet{\leadsto} \: \sf \sqrt{810 \: * \: 10^{5}}$ $\bullet{\leadsto} \: \sf \sqrt{81 \: * \: 10^{6}}$ $\bullet{\leadsto} \: \sf 9 \: * \: 10^{3}$ $\bullet{\leadsto} \: \sf 9 \: * \: 1000$ $\bullet{\leadsto} \: \boxed{\sf 9,000 \: cm^{2}.}$ $\huge{\bf{\red{\mathfrak{\maltese{Conclusion:-}}}}}$ $\bullet{\mapsto} \: \boxed{\therefore{\sf Area \: of \: the \: triangle \: = \: 9,000 \: cm^{2}.}}$ $\huge{\bf{\purple{\mathfrak{\maltese{Formulas \: Used:-}}}}}$ $\bullet{\leadsto} \: \sf Heron's \: Formula \: = \: \sqrt{s(s \: - \: a)(s \: - \: b)(s \: - \: c)}.$ $\bullet{\leadsto} \: \sf Semi-Perimeter \: = \: \dfrac{Perimeter}{2} \: or \: \dfrac{a \: + \: b \: + \: c}{2}$ $\huge{\bf{\pink{\mathfrak{\maltese{Note:-}}}}}$ $\bullet{\leadsto}$ Here, to find the area of the triangle we don't know the height (h) of the triangle but we can USE Heron's Formula to find out the area of the triangle as we know the Semi-Perimeter (Half of the given perimeter) and the measure of all 3 sides.

The sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540cm. Find its area.

Answer»

align="absmiddle" alt="\HUGE{\bf{\green{\mathfrak{\MALTESE{Question:-}}}}}" class="latex-formula" ID="TexFormula1" src="https://tex.z-dn.net/?f=%20%5Chuge%7B%5Cbf%7B%5Cgreen%7B%5Cmathfrak%7B%5Cmaltese%7BQuestion%3A-%7D%7D%7D%7D%7D%20" title="\huge{\bf{\green{\mathfrak{\maltese{Question:-}}}}}">

$\bullet{\mapsto}$ The sides of a triangle are in the ratio of 12:17:25 and its perimeter is 540cm. Find its area.

$\huge {\bf{\orange{\mathfrak{\maltese{Answer:-}}}}}$

$\bullet{\leadsto} \: \textsf{Sides of triangle are in the ratio 12:17:25.}$

$\bullet{\leadsto} \: \textsf{Perimeter of the triangle = 540 cm.}$

$\bullet{\leadsto} \: \textsf{Let the sides of the triangle be x.}$

$\bullet{\leadsto} \: \therefore{\textsf{1st side = a = 12x.}}$

$\bullet{\leadsto} \: \therefore{\textsf{2nd side = b = 17x.}}$

$\bullet{\leadsto} \: \therefore{\textsf{3rd side = c = 25x.}}$

$\bullet{\leadsto} \: \textsf{Perimeter of the triangle = 540 cm.}$

$\bullet{\leadsto} \: \textsf{We know that,}$

$\bullet{\leadsto} \: \boxed{\textsf{Perimeter of a triangle = a + b + c.}}$

$\bullet{\leadsto} \: \textsf{Substituting the values,}$

$\bullet{\leadsto} \: \sf 540 \: = \: 12x \: + \: 17x \: + \: 25x$

$\bullet{\leadsto} \: \sf 540 \: = \: 54x$

$\bullet{\leadsto} \: \sf 54x \: = \: 540$

$\bullet{\leadsto} \: \sf x \: = \: \dfrac{540}{54}$

$\bullet{\leadsto} \: \boxed{\sf x \: = \: 10 \: cm.}$

$\bullet{\leadsto} \: \sf 1st \: side \: = \: a \: = \: 12x \: = \: 12 \: * \: 10 \: = \: 120 \: cm.$

$\bullet{\leadsto} \: \sf 2nd \: side \: = \: b \: = \: 17x \: = \: 17 \: * \: 10 \: = \: 170 \: cm.$

$\bullet{\leadsto} \: \sf 3rd \: side \: = \: c \: = \: 25x \: = \: 25 \: * \: 10 \: = \: 250 \: cm.$

$\bullet{\leadsto} \: \textsf{Using Heron's Formula,}$

$\bullet{\leadsto} \: \sf Semi-Perimeter \: = \: ?$

$\bullet{\leadsto} \: \sf Perimeter \: = \: 540 \: cm.$

$\bullet{\leadsto} \: \boxed{\sf Semi-Perimeter \: = \: \dfrac{Perimeter}{2}}$

$\bullet{\leadsto} \: \sf S \: = \: \dfrac{540}{2}$

$\bullet{\leadsto} \: \sf S \: = \: 270 \: cm.$

$\bullet{\leadsto} \: \boxed{\sf Heron's \: Formula \: = \: \sqrt{s(s \: - \: a)(s \: - \: b)(s \: - \: c)}.}$

$\bullet{\leadsto} \: \textsf{Substituting the values,}$

$\bullet{\leadsto} \: \sf \sqrt{270(270 \: - \: 120)(270 \: - \: 170)(270 \: - \: 250)}$

$\bullet{\leadsto} \: \sf \sqrt{270(150)(100)(20)}$

$\bullet{\leadsto} \: \sf \sqrt{270 \: * \: 150 \: * \: 100 \: * \: 20}$

$\bullet{\leadsto} \: \sf \sqrt{27 \: * \: 15 \: * \: 1 \: * \: 2 \: * \: 10^{5}}$

$\bullet{\leadsto} \: \sf \sqrt{810 \: * \: 10^{5}}$

$\bullet{\leadsto} \: \sf \sqrt{81 \: * \: 10^{6}}$

$\bullet{\leadsto} \: \sf 9 \: * \: 10^{3}$

$\bullet{\leadsto} \: \sf 9 \: * \: 1000$

$\bullet{\leadsto} \: \boxed{\sf 9,000 \: cm^{2}.}$

$\huge{\bf{\red{\mathfrak{\maltese{Conclusion:-}}}}}$

$\bullet{\mapsto} \: \boxed{\therefore{\sf Area \: of \: the \: triangle \: = \: 9,000 \: cm^{2}.}}$

$\huge{\bf{\purple{\mathfrak{\maltese{Formulas \: Used:-}}}}}$

$\bullet{\leadsto} \: \sf Heron's \: Formula \: = \: \sqrt{s(s \: - \: a)(s \: - \: b)(s \: - \: c)}.$

$\bullet{\leadsto} \: \sf Semi-Perimeter \: = \: \dfrac{Perimeter}{2} \: or \: \dfrac{a \: + \: b \: + \: c}{2}$

$\huge{\bf{\pink{\mathfrak{\maltese{Note:-}}}}}$

$\bullet{\leadsto}$ Here, to find the area of the triangle we don't know the height (h) of the triangle but we can USE Heron's Formula to find out the area of the triangle as we know the Semi-Perimeter (Half of the given perimeter) and the measure of all 3 sides.

The sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540cm. Find its area.

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