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A market gardener has to fertilize a triangular field with sides of lengths 90m, 45m, and 65m. The fertilizer is to be spread so that 1 kg covers 10m squared. One bag of fertilizer has a mass of 9.1 kg. How many bags of fertilizer will he need?

micheala950831  Mar 28, 2017

Best Answer 

 #1
avatar+19620 
+1

A market gardener has to fertilize a triangular field with sides of lengths 90m, 45m, and 65m.

The fertilizer is to be spread so that 1 kg covers 10m squared. One bag of fertilizer has a mass of 9.1 kg.

How many bags of fertilizer will he need?

 

Let A = the area of the triangular field

Let a = 90 m
Let b = 45 m
Let c = 65 m

 

Heron:

\(\begin{array}{|rcll|} \hline A &=& \sqrt{s(s-a)(s-b)(s-c)} \quad & | \quad s = \frac{a+b+c}{2}= \frac{90+45+65}{2}=100 \\ A &=& \sqrt{100\cdot(100-90)\cdot(100-45)\cdot(100-65)} \\ A &=& \sqrt{100\cdot(10)\cdot(55)\cdot(35)} \\ A &=& 10\cdot\sqrt{10\cdot 55\cdot 35} \\ A &=& 10\cdot\sqrt{19250} \\ A &=& 10\cdot 138.744369255 \\ A &=& 1387.44369255\ m^2 \\ \hline \end{array} \)

 

The area of the triangular field is 1387.44 m2

 

Let x = bags of fertilizer

\(\begin{array}{|rcll|} \hline x &=& A \cdot \frac{1\ kg}{10\ m^2} \cdot \frac{1\ \text{bag}}{9.1\ kg} \quad & | \quad A &=& 1387.44369255\ m^2 \\ x &=& 1387.44369255\ m^2 \cdot \frac{1\ kg}{10\ m^2} \cdot \frac{1\ \text{bag}}{9.1\ kg} \\ x &=& \frac{1387.44369255 }{10\cdot 9.1 } \ \text{bags} \\ x &=& \frac{1387.44369255 }{91} \ \text{bags} \\ x &=& 15.2466339841\ \text{bags} \\ \hline \end{array} \)

 

We need \(\approx\)16 bags

 

laugh

heureka  Mar 28, 2017
 #1
avatar+19620 
+1
Best Answer

A market gardener has to fertilize a triangular field with sides of lengths 90m, 45m, and 65m.

The fertilizer is to be spread so that 1 kg covers 10m squared. One bag of fertilizer has a mass of 9.1 kg.

How many bags of fertilizer will he need?

 

Let A = the area of the triangular field

Let a = 90 m
Let b = 45 m
Let c = 65 m

 

Heron:

\(\begin{array}{|rcll|} \hline A &=& \sqrt{s(s-a)(s-b)(s-c)} \quad & | \quad s = \frac{a+b+c}{2}= \frac{90+45+65}{2}=100 \\ A &=& \sqrt{100\cdot(100-90)\cdot(100-45)\cdot(100-65)} \\ A &=& \sqrt{100\cdot(10)\cdot(55)\cdot(35)} \\ A &=& 10\cdot\sqrt{10\cdot 55\cdot 35} \\ A &=& 10\cdot\sqrt{19250} \\ A &=& 10\cdot 138.744369255 \\ A &=& 1387.44369255\ m^2 \\ \hline \end{array} \)

 

The area of the triangular field is 1387.44 m2

 

Let x = bags of fertilizer

\(\begin{array}{|rcll|} \hline x &=& A \cdot \frac{1\ kg}{10\ m^2} \cdot \frac{1\ \text{bag}}{9.1\ kg} \quad & | \quad A &=& 1387.44369255\ m^2 \\ x &=& 1387.44369255\ m^2 \cdot \frac{1\ kg}{10\ m^2} \cdot \frac{1\ \text{bag}}{9.1\ kg} \\ x &=& \frac{1387.44369255 }{10\cdot 9.1 } \ \text{bags} \\ x &=& \frac{1387.44369255 }{91} \ \text{bags} \\ x &=& 15.2466339841\ \text{bags} \\ \hline \end{array} \)

 

We need \(\approx\)16 bags

 

laugh

heureka  Mar 28, 2017

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