Question

Suppose given an isosceles triangle with a leg measuring 5 in. Two lines are drawn through some point on the base, each parallel to one of the legs. Find the perimeter of the constructed parallelogram.

Answer 1

Let ΔABC be isosceles triangle with legs AB=BC=5 in. Draw an arbitrary point D on the base AC and spend two lines DE and DF such that DE || BC and DF || AB.

You get the parallelogram BFDE.

1. Consider ΔAED. This triangle is isosceles, because ∠ADE≅∠ACB as corresponding angles at parallel lines. The legs of this triangle are AE=ED.

2. Consider ΔDFC. This triangle is isosceles, because ∠FDC≅∠BAC as corresponding angles at parallel lines. The legs of this triangle are DF=FC.

3. Find the perimeter of parallelogram BFDE.

$P_{BFDE}=BF+FD+ED+BE.$

Since AE=ED and DF=FC, you have that

$P_{BFDE}=BF+FC+AE+BE=AB+BC=5+5=10\ in.$

Answer: 10 in.