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Three lots with parallel side boundaries extend from the avenue to the boulevard as shown answer

User Tjarbo
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Three lots with parallel side boundaries extend from the avenue to the boulevard as shown in the diagram attached.

Given;
x + y + z = 140 meters


According to the proportion theorem,


(x)/(40) = (y)/(30)

Solving for x, using cross multiplication:


(x)/(1) = (40y)/(30)

Similarly,

x/40 = z/35

solving for z by cross multiplication,

z = 35x/40
z = 7x/8

Now, the third proportions,
y/30 = z/35

Solving for y,

y = 30z/35
y = 6z/7

x + y + z = 140

4y/3 + 6z/7 + 7x/8 = 140

(224y + 144z + 147x)/168 = 140

224y + 144z + 147x = 23520

Substitutoing z = 7x/8,

224y + (144 × 7x/8) + 147x = 23520

224y + 126x + 147x = 23520

224y + 273x = 23520
Now,
Substitute y = 3x/4

(224×3x/4) + 273x = 23520

168x + 273x = 23520

441x = 23520

x = 53.33

Now,
We know that
y = 3x/4
so,
y = 40

And,
z = 7x/8
Z = 46.66
Three lots with parallel side boundaries extend from the avenue to the boulevard as-example-1
User Javier Neyra
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