Challenge 1 Spring of 2017: Trapezoid Solution Algebra

Promoting the study of Geometry

Joining forces with institut-delbol.com, mouctar.org is publishing the first challenge.

The callenge second part is now solved. No more answers will be accepted. There was no winner for this challenge.

The following is the Algebraic method.

 

PROBLEM 2: THE TRAPEZOID

An agency is assigning the area bounded by figure ABCD to a city population.

Each member will receive an equal area of a land plot.

Calculations have shown that out of the 5160 members of the population, 2000 will receive their plots from the area covered by triangle ADC

The remaining people will be assigned plots from ABC, (See graph).

DC=1050\;m and AC=1810\:m

1. Find the area of ABCD (30\; points)

2. Verify the area of ADC using the Heron’s Formula (10\; points)

3.  What is the area, in square meters, assigned to each person? (10\; points)

 

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SOLUTION: METHOD BY ALGEBRA

Let AB=x

BC=d_{1}

From the right triangle \triangle ABC We have the hypothenuse AC=1810.

We have:

d_{1}=\sqrt{(1810)^{2}-x^{2}}

We also have the right triangle \triangle DFC

FC=\sqrt{(1050)^{2}-x^{2}}

Another equality:

AD=d_{2}=BF=BC-FC

We can then say:

d_{2}=\sqrt{(1810)^{2}-x^{2}}-\sqrt{(1050)^{2}-x^{2}}

 

The area of \triangle ABC contains 3160 plots of land out of 5160.

For \triangle ABC:

\mathcal{A} \triangle ABC

The area of the \triangle ABC=\frac{1}{2}\cdot x \cdot BC

OR

Area \triangle ABC=\frac{1}{2}\cdot x \cdot d_{1}

Area \triangle ABC=\frac{1}{2}x \sqrt{(1810)^{2}-x^{2}}

 

Now let’s look for the total area:

Area \square ABCD=\square ABFD+\triangle DFC

Area \square ABFD:

Area \square ABFD=x\cdot d_{2}

We plugin:

Area \square ABFD=x\cdot (\sqrt{(1810)^{2}-x^{2}}-\sqrt{(1050)^{2}-x^{2}})

Area \square ABFD=x\sqrt{(1810)^{2}-x^{2}}-x\sqrt{(1050)^{2}-x^{2}}

 

For \triangle DFC:

The area of the \triangle DFC=\frac{1}{2}\cdot x \cdot FC

OR

Area \triangle DFC=\frac{1}{2}\cdot x \cdot d_{1}

Area \triangle DFC=\frac{1}{2}\cdot x \cdot \sqrt{(1050)^{2}-x^{2}}

TOTAL AREA :

Area \square ABCD=\square ABFD+\triangle DFC

We plugin:

Area \displaystyle{\square ABCD=x\sqrt{(1810)^{2}-x^{2}}-x\sqrt{(1050)^{2}-x^{2}}+\frac{1}{2}x \sqrt{(1050)^{2}-x^{2}}}

Area \square ABCD=x\sqrt{(1810)^{2}-x^{2}}-\frac{1}{2}x \sqrt{(1050)^{2}-x^{2}}

Comparing Areas:

Area \triangle ABC=3160p

Area \square ABCD=5160p

This yields:

\frac{Area\; \square ABCD}{Area \; \triangle ABC}=\frac{5160p}{3160p}

OR

Area \; \triangle ABC=\frac{3160p}{5160p}Area\; \square ABCD

We simplify:

Area \; \triangle ABC=\frac{79}{129}Area\; \square ABCD

We plugin:

\displaystyle{\frac{1}{2} x  \sqrt{(1810)^{2}-x^{2}}=\frac{79}{129}(x\sqrt{(1810)^{2}-x^{2}}-\frac{1}{2}x \sqrt{(1050)^{2}-x^{2}})}

We can write, for simplicity:

\displaystyle{\frac{129}{158}x  \sqrt{(1810)^{2}-x^{2}}=x\sqrt{(1810)^{2}-x^{2}}-\frac{1}{2}x \sqrt{(1050)^{2}-x^{2}}}

We divide both sides by x

\displaystyle{\frac{129}{158}  \sqrt{(1810)^{2}-x^{2}}=\sqrt{(1810)^{2}-x^{2}}-\frac{1}{2} \sqrt{(1050)^{2}-x^{2}}}

By moving we get:

\frac{1}{2} \sqrt{(1050)^{2}-x^{2}}=(1-\frac{129}{158}) \sqrt{(1810)^{2}-x^{2}}

\frac{1}{2} \sqrt{(1050)^{2}-x^{2}}=\frac{29}{158} \sqrt{(1810)^{2}-x^{2}}

Multiply both sides by 2:

\sqrt{(1050)^{2}-x^{2}}=\frac{29}{79} \sqrt{(1810)^{2}-x^{2}}

 We can also write:

\frac{0.79}{0.29}\sqrt{(1050)^{2}-x^{2}}=\sqrt{(1810)^{2}-x^{2}}

Let k=\frac{0.79}{0.29}

We get:

\sqrt{(1810)^{2}-x^{2}}=k\sqrt{(1050)^{2}-x^{2}}

Let’s square both sides:

(1810)^{2}-x^{2}=k^{2}((1050)^{2}-x^{2})

(1810)^{2}-x^{2}=k^{2}(1050)^{2}-k^{2}x^{2}

(k^{2}-1)x^{2}=k^{2}\cdot (1050)^{2}-(1810)^{2}

x^{2}=\frac{k^{2}\cdot (1050)^{2}-(1810)^{2}}{k^{2}-1}

x=\sqrt{\frac{k^{2}\cdot (1050)^{2}-(1810)^{2}}{k^{2}-1}}

We plugin the values:

\displaystyle{x=\sqrt{\frac{(\frac{0.79}{0.29})^{2}\cdot (1050)^{2}-(1810)^{2}}{(\frac{0.79}{0.29})^{2}-1}}}

x=874.0605963

Rounded Value here:

x=874\; m

 

Areas:

The other sides:

d_{1}=\sqrt{(1810)^{2}-x^{2}}

d_{1}=\sqrt{(1810)^{2}-(874.0605963)^{2}}

d_{1}=1584.966269

d_{1}=1585\;m

d_{2}=\sqrt{(1810)^{2}-x^{2}}-\sqrt{(1050)^{2}-x^{2}}

\displaystyle{d_{2}=\sqrt{(1810)^{2}-(874.0605963)^{2}}-\sqrt{(1050)^{2}-(874.0605963)^{2}}}

d_{2}=1003.143208

d_{2}=1003\;m

Area \triangle ABC=\frac{1}{2}x\cdot d_{1}

Area \triangle ABC=\frac{1}{2}\times  874.0605963\times 1584.966269

Area \triangle ABC=692678.2811

Rounding:

Area \triangle ABC=692678 \;m^{2}

 

TOTAL\; AREA:

Trapezoid:

 

AREA\; \square ABCD=\frac{d_{1}+d_{2}}{2}\cdot x

AREA\; \square ABCD=\frac{1584.966269+1003.143208}{2}\times 874.0605963

AREA\; \square ABCD=1131082.256

 

AREA\; ABCD=1131082\;m^{2}

 

SIZE\; OF \;THE\; PLOT:\; 5160\; plots

AREA \;PLOT=\frac{1131082}{5160}

AREA \;PLOT=219\;m^{2}

 Using the heron Formula for Area ADC

The sides are: 1050, 1810, 1003.143208

Calculating s:

s=\frac{1050+1810+1003.143208}{2}

s=1931.571604

s-1050=1931.571604-1050=881.571604

s-1810=1931.571604-1810=121.571604

s-1003.143208=1931.571604-1003.143208=928.428396

Area \triangle ADC=\sqrt{1931.571604 \times 881.571604 \times 121.571604 \times 928.428396 }

Area \triangle ADC=438403.9751

ROUNDING:

Area \triangle ADC=438404\; m^{2}

 

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