Question

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

Answer 1

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

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.

`(VW)/(YZ) =(WX)/(ZX) =(VX)/(YX)\ \textrm{corresponding sides of similar triangles are in proportion}\\`

On substituting the given values we get

`(100)/(l) =(60)/(30)\\\\l=(3000)/(60)=50\ ft`

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