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,
[tex] \frac{x}{40} = \frac{y}{30} [/tex]
Solving for x, using cross multiplication:
[tex] \frac{x}{1} = \frac{40y}{30} [/tex]
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
