Question

An angle bisector AC divides a trapezoid ABCD into two similar triangles △ABC and △ACD. Find the perimeter of this trapezoid if the leg AB=9 cm and the leg CD=12 cm.

Answer 1

Answer:

46 cm

Step-by-step explanation:

It is given that the angle bisector AC divides the trapezoid ABCD into two similar triangles Δ ABC and Δ ACD.

Let us first find the corresponding sides of triangles Δ ABC and Δ ACD.

Since AC is the angle bisector of ∠A,

∠DAC = ∠CAB --- (1)

Also, since ∠DAC and ∠ACB are alternate interior angles,

∠DAC = ∠ACB --- (2)

From (1) and (2),

∠CAB = ∠ACB

Therefore, in Δ ABC,

BC = AB (sides opposite to equal angles are equal)

So, BC = 9 cm.

Now, since ∠DAC = ∠ACB and ∠DAC = ∠CAB, we can take either ∠ACB or ∠CAB as the angle corresponding to ∠DAC. Let us take ∠ACB as the corresponding angle of ∠DAC.

So, the side opposite to ∠DAC in Δ ACD is the corresponding side to the side opposite to ∠ACB in Δ ABC.

That is, the side CD in Δ ACD is the corresponding side to the side AB in Δ ABC.

Now, suppose if the side BC in Δ ABC corresponds to side AD of Δ ACD, then the remaining side AC of Δ ABC should correspond to side AC of Δ ACD which is not possible since they are congruent.

So, the side BC in Δ ABC should correspond to side AC of Δ ACD and the remaining side AC of Δ ABC should correspond to side AD of Δ ACD.

Therefore, we have,

$\frac{AB}{CD} =\frac{BC}{AC} =\frac{AC}{AD}$

But, $\frac{AB}{CD} =\frac{9}{12} =\frac{3}{4}$

Therefore, $\frac{BC}{AC} =\frac{3}{4}$

$\frac{9}{AC} =\frac{3}{4}$

$AC=(\frac{4}{3} )(9)$

AC = 12 cm

Also,

$\frac{AC}{AD} =\frac{3}{4}$

$\frac{12}{AD} =\frac{3}{4}$

$AD=(\frac{4}{3} )(12)$

AD = 16 cm

Now, the perimeter of the trapezoid = AB + BC + CD + AD

= 9 + 9 + 12 + 16

= 46 cm