Question

26.The sides of a triangular plot are in the ratio of 3: 5: 7 and its perimeter is 300 m.
Find its area.

Answer 1

Given: Ratio of sides of the triangle is 3:5:7 & perimeter of triangle is 300 m.

Need to find: Dimensions of triangle?

⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's consider side as 'x'

Angles of the triangle :-

• 3x
• 5x
• 7x

⠀⠀⠀⠀⠀

As we know that,

⠀⠀⠀⠀⠀

$\begin{gathered}\star\:{\underline{\boxed{\frak{Perimeter_{\:(triangle)} = a + b + c}}}}\\\\\\ \bf{\dag}\:{\underline{\frak{Putting\:given\:values\:in\:formula,}}}\\\\\\ :\implies\sf 300 = 3x + 5x + 7x \\\\\\ :\implies\sf 10x + 5x = 850\\\\\\ :\implies\sf 15x = 300 \\\\\\ :\implies\sf x = \cancel{\dfrac{300}{15}}\\\\\\ :\implies{\underline{\boxed{\frak{\purple{x = 20 ^{ \circ}}}}}}\:\bigstar\\\\\end{gathered}$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Therefore,

⠀⠀⠀⠀⠀

Side, x = 20°

⠀⠀⠀⠀⠀

$\therefore\:{\underline{\sf{Hence,\:Dimensions\:of\:triangle\:are \: \bf{60\:m}\:,\bf{100\: m} \: \sf{and}\: \bf{140\:m}\: \sf{respectively}.}}}$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Now, Area :-

$\begin{gathered}\star\:{\underline{\boxed{\frak{Area_{\:(triangle)} = \sqrt{s(s - a)(s - b)(s - c)} }}}}\\\\\\ \bf{\dag}\:{\underline{\frak{Putting\:given\:values\:in\:formula,}}}\\\\\\ :\implies\sf \sqrt{150(150 - 60)(150 - 100)(150 - 140) {m}^{2} } \\\\\\ :\implies\sf \sqrt{150 \times 90 \times 50 \times 10 {m}^{2} } \\\\\\ :\implies\sf \sqrt{30 \times 5 \times 30 \times 3 \times 5 \times 10 \times 10}\\\\\\ :\implies\sf 30 \times 5 \times 10 \sqrt{3} \\\\\\ :\implies{\underline{\boxed{\frak{\purple{1500 \sqrt{3} }}}}}\:\bigstar\\\\\end{gathered}$

Hence, Area = 1500√3 m²

Answer 2

Given : The sides of Triangle are in ratio 3:5:7 & the Perimeter of Triangle is 300 m .

Need To Find : Area of Triangle.

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's Consider 3x , 5x & 7x be the three side of triangle.

$\dag\underline {\frak{As \:We \:know\:that\:: }}$

$\star\underline{\boxed {\sf{ Perimeter _{(Triangle)} = A + B + C }}}$

Where ,

• A , B & C are three sides of Triangle.

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$\qquad \quad:\implies \sf{ 300 = 3x + 5x + 7x }$

$\qquad \quad:\implies \sf{ 300 = 8x + 7x }$

$\qquad \quad:\implies \sf{ 300 = 15x }$

$\qquad \quad:\implies \sf{ \dfrac{\cancel {300}}{\cancel {15}} = x }$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  x = 20\: m}}}}\:\bf{\bigstar}$

Therefore

• First Side or A is 3x = 3 × 20 = 60 m

• Second Side or B is 5x = 5 × 20 = 100 m

• Third Side or C is 7x = 7 × 20 = 140 m

$\therefore\:{\underline{\sf{Hence,\:Dimensions\:of\:triangle\:are \: \bf{60\:m}\:,\bf{100\: m} \: \sf{and}\: \bf{140\:m}\: \sf{respectively}.}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Finding Area of Triangle :

$\dag\underline {\frak{As \:We \:know\:that\:: }}$

$\star\:{\underline{\boxed{\sf{Area_{\:(triangle)} = \sqrt{s(s - a)(s - b)(s - c)} }}}}$

Where,

• a , b & c are three sides of Triangle & s is the Semi-Perimeter of Triangle.

$\sf{Semi-Perimeter}\begin {cases}\sf{Semi-Perimeter \:= \dfrac{Perimeter}{2}}\\\\\sf{Then,}\\\\\sf{Semi-Perimeter \:=\cancel {\dfrac{300}{2}}}\\\\ \sf{ s \:or\: Semi-Perimeter \:= \:150m\:}\end{cases}$

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$:\implies\sf \sqrt{150(150 - 60)(150 - 100)(150 - 140) {m}^{2} } \\\\\\ :\implies\sf \sqrt{150 \times 90 \times 50 \times 10 {m}^{2} } \\\\\\ :\implies\sf \sqrt{30 \times 5 \times 30 \times 3 \times 5 \times 10 \times 10}\\\\\\ :\implies\sf 30 \times 5 \times 10 \sqrt{3} \\\\\\ :\implies{\underline{\boxed{\frak{\purple{1500 \sqrt{3} }}}}}\:\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm { Hence ,\:Area \:of\:Triangle \:is\:\bf{1500\sqrt{3}\: m^{2}}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀