Exercise 11 Page 611

Let's begin with drawing a right isosceles triangle ABC. We know that its hypotenuse has a length of h.

We are given that DE is a midsegment that is not parallel to the hypotenuse. This means, according to the Triangle Midsegment Theorem, that DE is parallel to one of the legs of this triangle. Since the legs of an isosceles triangle are congruent, we can choose to which leg this midsegment is parallel.

Now, let's recall that the midsegment has a length that is one half the length of the side to which it is parallel. This means that DE= 12BC. Since we know that the length of DE is 4x, we can find the length of BC.

DE=1/2BC

Substitute

DE= 4x

4x=1/2BC

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Solve for BC

MultEqn

LHS * 2=RHS* 2

2*4x=2*1/2BC

DenomMultFracToNumber

2 * a/2= a

2*4x=1* BC

Multiply

Multiply

8x=BC

RearrangeEqn

Rearrange equation

BC=8x

The length of BC is 8x. Since △ ABC is isosceles, the length of AB is also 8x.

Finally, we will evaluate the perimeter of △ ABC by adding all the side lengths. 8x+8x+ h=16x+h The perimeter of △ ABC is 16x+h.