Question

the sides of a triangle field are un the ratio 5:7:8 and its perimeter is 400 m find the length of each sides
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Answer 1

$\large{\underline{\underline{\red{\sf{GIVEN:}}}}}$

• There is a triangular field.

• Ratio of sides of a triangular field is 5:7:8.

$\large{\underline{\underline{\red{\sf{TO\:FIND:}}}}}$

• Length of three sides of the field.

$\large{\underline{\underline{\red{\sf{CONCEPT\:USED:}}}}}$

• We will first multiply each ratio by a variable as their Highest Common Factor since their HCF is the only number which will convert them into their simplest ratios.

$\large{\underline{\underline{\red{\sf{ANSWER:}}}}}$

Here the ratio of sides is 5:7:8.

Now Let us take the given ratio be 5x:7x:8x .

So ,sum of sides of triangle i.e. their perimeter.

= 5x+7x+8x = 20x .

But it is given 400m . So ,

Atq ,

$\sf{\implies 20x=400m}$

$\sf{\implies x=\dfrac{400m}{20}}$

$\sf{\red{\leadsto x=20m}}$

Hence ,

• Measure of first side =5x = 5×20m=100m.
• Measure of second side = 7x = 7×20m = 140m
• Measure of third side = 8x =8×20m =160m

Answer 2

Solution :

Diagram :

$\setlength{\unitlength}{0.8cm}\begin{picture}(6,5)\thicklines\put(1,0.5){\line(2,1){3}}\put(4,2){\line(-2,1){2}}\put(2,3){\line(-2,-5){1}}\put(0.7,0.3){ \bf A }\put(4.05,1.9){ \bf B }\put(1.7,2.95){ \bf C }\put(3.1,2.5){ \sf 8r }\put(1.3,1.7){ \sf \:\:\:\:7r }\put(2.5,1.05){ \sf \:\:5r }\end{picture}$

$\bigstar$ We have sides of triangle in ratio are 5:7:8 & perimeter is 400 m.

$\boxed{\bf{perimeter\:of\:\triangle=Side+Side+Side}}}$

$\longrightarrow\sf{5r+7r+8r=400}\\\\\longrightarrow\sf{20r=400}\\\\\longrightarrow\sf{r=\cancel{400/20}}\\\\\longrightarrow\bf{r=20\:m}$

Now;

$\underline{\boldsymbol{The\:length\:of\:each\:side\:of\:triangle\::}}}}$

$\bullet\:\sf{1^{st}\:side\:of\:\triangle=5r=5\times 20m=\boxed{\bf{100\:m}}}\\\\\bullet\sf{2^{nd}\:side\:of\:\triangle=7r=7\times 20m=\boxed{\bf{140\:m}}}\\\\\bullet\sf{3^{rd}\:side\:of\:\triangle=8r=8\times 20m=\boxed{\bf{160\:m}}}$