Question

What is the distance, in feet, across the patch of swamp water?

Answer 1

Answer:

Therefore the distance across the patch of swamp water is 50 ft

Step-by-step explanation:

Given:

VW = 100 ft

WX = 60 ft

XZ = 30 ft

To Find:

ZY = l = ?

Solution:

In  Δ VWX and Δ YZX

∠W ≅ ∠ Z    …………..{measure of each angle is 90° given}

∠VXW ≅ ∠YXZ      ..............{vertically opposite angles  are equal}

Δ ABC ~ Δ DEC ….{Angle-Angle Similarity test}

If two triangles are similar then their sides are in proportion.

$\frac{VW}{YZ} =\frac{WX}{ZX} =\frac{VX}{YX}\ \textrm{corresponding sides of similar triangles are in proportion}$  

On substituting the given values we get

$\frac{100}{l} =\frac{60}{30}\\\\l=\frac{3000}{60}=50\ ft$

Therefore the distance across the patch of swamp water is 50 ft