Question

the perimeter of a triangular fields is 1080m and it's side are in the ratio 25:17:12.find the area of the triangle also find the cost of ploughing the field rate rs 18.80 per 20 so.metre​

Answer 1

$\dag \: \underline{\sf AnsWer :}$

We are provided a field which is in a triangular shape whose perimeter is 1080 m and it's sides are in the ratio 25:17:12. We are asked to find the the area of the triangle also find the cost of ploughing the field rate rs 18.80 per 20 m². To solve this sum first we have to assume the sides of a traingle as :

• a = 25x m
• b = 17x m
• c = 12x m

$\dashrightarrow\:\:\sf a + b + c = Perimeter$

$\dashrightarrow\:\:\sf 25x+ 17x+ 12x= 1080$

$\dashrightarrow\:\:\sf 54x= 1080$

$\dashrightarrow\:\:\sf x= \dfrac{1080}{54}$

$\dashrightarrow\:\: \underline{ \boxed{\sf x= 20}}$

$\: \: \: \qquad\dag\:\underline{\textbf{Sides of a triangle :}}$

$\bullet\:\:\textsf{a = 25x = 25(20) = \textbf{500 m}}$

$\bullet\:\:\textsf{b = 17x = 17(20) = \textbf{340 m}}$

$\bullet\:\:\textsf{c = 12x = 12(20) = \textbf{240 m}}$

• Here, we have find the sides of ∆ and now we will find the semi perimeter of ∆ :

$\: \: \: \qquad\dag\:\underline{\textbf{Semi perimeter of a triangle :}}$

$:\implies \sf Semi \: Perimeter = \dfrac{Sum \: of \: sides \: of \: triangle}{2}$

$:\implies \sf Semi \: Perimeter = \dfrac{a + b + c}{2}$

$:\implies \sf Semi \: Perimeter = \dfrac{500+ 340 + 240}{2}$

$:\implies \sf Semi \: Perimeter = \dfrac{1080}{2}$

$:\implies \sf Semi \: Perimeter = \dfrac{1080}{2}$

$:\implies \underline{ \boxed{\sf Semi \: Perimeter = 540 \: m}}$

• Now, by using heron's formula we can find the area of ∆ :

$\longrightarrow\:\:\sf Area_{(of \: \Delta)} = \sqrt{S(S - a) (S - b) (S - c)}$

$\longrightarrow\:\:\sf Area_{(of \: \Delta)} = \sqrt{540(540- 500) (540- 340) (540 - 240)}$

$\longrightarrow\:\:\sf Area_{(of \: \Delta)} = \sqrt{540(40) (200) (300)}$

$\longrightarrow\:\:\sf Area_{(of \: \Delta)} = \sqrt{54 \times 10 \times 2 \times 2 \times 10 \times 2 \times 100 \times 3 \times 100}$

$\longrightarrow\:\:\sf Area_{(of \: \Delta)} = 10 \times 2 \times 100\sqrt{54 \times 2 \times 3}$

$\longrightarrow\:\:\sf Area_{(of \: \Delta)} = 2000\sqrt{27 \times 2 \times 2 \times 3}$

$\longrightarrow\:\:\sf Area_{(of \: \Delta)} = 2000 \times 2\sqrt{3 \times 3\times 3 \times 3}$

$\longrightarrow\:\:\sf Area_{(of \: \Delta)} = 4000 \times 3 \times 3$

$\longrightarrow\:\: \underline{ \boxed{\sf Area_{(of \: \Delta)} = 36000\: m^2}}$

$\: \: \: \qquad\dag\:\underline{\textbf{Cost of ploughing :}}$

$\leadsto\:\sf Cost \: of \: 20 \: m^2 = Rs. \: 18.80$

$\leadsto\:\sf Cost \: of \: 36000 \: m^2 = 36000 \times \dfrac{18.80}{20}$

$\leadsto\:\sf Cost \: of \: 36000 \: m^2 = 3600\times \dfrac{18.80}{2}$

$\leadsto\: \underline{ \boxed{\sf Cost \: of \: 36000 \: m^2 =Rs. \: 33840}}$

Answer 2

$\huge \fbox{Answer}$

$\large \underline{ \underline{ \bf{ \color{indigo}ɢɪᴠᴇɴ::}}}$

${ \mapsto \sf{Sides \: of \: triangle_{(in \: ratio)}=25:17:12}}$

${ \mapsto \sf{Perimeter \: of \: triangle=1,080m}}$

$\large \underline{ \underline{ \bf{ \color{orange}ᴛᴏ \: ғɪɴᴅ::}}}$

$\leadsto \sf{Area \: of \: triangle. }$

${ \leadsto \sf{Cost \: of \: ploughing \: the \: field \: { @} \: ₹ 18.80 \: per \: 20m².}}$

$\large \underline{ \underline{ \bf{ \color{cyan}ғᴏʀᴍᴜʟᴀ \: ʀᴇϙᴜɪʀᴇᴅ::}}}$

$\dag \underline\red{ \boxed {\blue{ \rm{Perimeter \: of \: triangle=sum \: of \: all \: side}}}}$

$\dag \underline \red {\boxed{ \blue{ \rm{Area \: of \: triangle= {\sqrt{s(s - a)( s- b)(s -c )}} }}}}$

Where,

• s is semi-perimeter of triangle.
• a, b and c is the sides of triangle.

$\large \underline{ \underline{ \bf{ \color{peru}ᴀᴄᴄᴏʀᴅɪɴɢ \: ᴛᴏ \: ϙᴜᴇsᴛɪᴏɴ::}}}$

${\dashrightarrow \sf{Perimeter \: of \: triangle=sum \: of \: all \: side}}$

${\dashrightarrow \sf{1,080m=25x + 17x + 12x}}$

${\dashrightarrow \sf{1,080m=54x}}$

${\dashrightarrow \sf{x = \frac{1,080}{54} }}$

${\dashrightarrow \sf{ x= 20}}$

$\bf \underline{Hence,sides \: of \: triangle \: is}$

$\rightarrowtail \sf{a = 25x = 25 \times 20 = 500m}$

$\rightarrowtail \sf{b = 17x = 17 \times 20 = 340m}$

$\rightarrowtail \sf{c= 12x = 12 \times 20 = 240m}$

$\bf \underline{And,}$

${ \rightarrowtail \sf{Semi-perimeter= \frac{1,080}{2} m=540m}}$

$\bf \underline{Again,}$

${ \dashrightarrow{ \sf{Area \: of \: triangle= {\sqrt{s(s - a)( s- b)(s -c )}}}}}$

${ \dashrightarrow{ \sf{Area \: of \: triangle= {\sqrt{540(540 - 500)( 540- 340)(540-240)}}}}}$

${ \dashrightarrow{ \sf{Area \: of \: triangle= {\sqrt{540 \times 40 \times 200 \times \times 300}}}}}$

${ \dashrightarrow{ \sf{Area \: of \: triangle= {\sqrt{54 \times 10 \times 10 \times 10 \times 2 \times 10 \times 2 \times 2 \times 10 \times 10 \times 3}}}}}$

${ \dashrightarrow{ \sf{Area \: of \: triangle= {\sqrt{54 \times \underline{ 10 \times 10} \times \underline{10 \times 2 }\times \underline{10 \times 2 }\times 2 \times \underline {10 \times 10} \times 3}}}}}$

${ \dashrightarrow{ \sf{Area \: of \: triangle= 10 \times 10 \times 10 \times 2{\sqrt{54 \times 2 \times 3}}}}}$

${ \dashrightarrow{ \sf{Area \: of \: triangle= 2,000{\sqrt{27 \times 2 \times 2 \times 3}}}}}$

${ \dashrightarrow{ \sf{Area \: of \: triangle= 2,000{\sqrt{27 \times \underline{2 \times 2 }\times 3}}}}}$

${ \dashrightarrow{ \sf{Area \: of \: triangle= 2,000 \times 2{\sqrt{27 \times 3}}}}}$

${ \dashrightarrow{ \sf{Area \: of \: triangle= 4,000{\sqrt{3 \times 3 \times 3\times3}}}}}$

${ \dashrightarrow{ \sf{Area \: of \: triangle= 4,000{\sqrt{ \underline{3 \times 3} \times \underline{3\times3}}}}}}$

${ \dashrightarrow{ \sf{Area \: of \: triangle= 4,000 \times 3 \times 3}}}$

${ \dashrightarrow{ \sf{Area \: of \: triangle= 36,000 {m}^{2} }}}$

$\bf \underline{Then,}$

$\dashrightarrow \sf{Cost \: of \: 20m²=18.80}$

${ \dashrightarrow \sf{Cost \: of \: 36,000m²=36,000 \times \frac{18.80}{20} }}$

${ \dashrightarrow \sf{Cost \: of \: 36,000m²=36,000 \times \frac{18 \cancel.80}{2,000} }}$

${ \dashrightarrow \sf{Cost \: of \: 36,000m²=36 ,\cancel{000} \times \frac{1,880}{2, \cancel{000}} }}$

${ \dashrightarrow \sf{Cost \: of \: 36,000m²=36 \times \frac{1,880}{2} }}$

${ \dashrightarrow \sf{Cost \: of \: 36,000m²= \cancel{36} \times \frac{1,880} {\cancel{2}} }}$

${ \dashrightarrow \sf{Cost \: of \: 36,000m²= 18 \times {1,880} }}$

${ \dashrightarrow \sf{Cost \: of \: 36,000m²=₹ 33,840 }}$

$\bf \underline{Hence,}$

$\: \: \: \: \: \star \underline \blue {\boxed{ \red{ \rm{Area \: of \: triangle= 36,000 {m}^{2} } }}}$

$\: \: \: \: \: \star \underline \blue {\boxed{ \red{ \rm{Cost \: of \: ploughing \: triangular \: field=₹33,840 } }}}$

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