Question

What is AE and the perimeter of BCDE? Round your answer to the nearest tenth.

Answer 1

Answer:

AE = 43.2 units

Perimeter = 229.2 units

Step-by-step explanation:

Let the side AE be 'x'.

Consider triangles AEB and ADC

Statements                                                     Reasons

1. ∠ ABE ≅ ∠ ACD                                        Right angles are congruent.

2. ∠A ≅ ∠A                                                   Common angle

Therefore, the two triangles are similar by AA postulate.

Now, for similar triangles, the ratio of their corresponding sides are also proportional to each other. Therefore,

$\frac{AE}{AD}=\frac{AB}{AC}=\frac{BE}{DC}\\\\\frac{AE}{AE+ED}=\frac{AC-BC}{AC}$

Now, plug in the given values and solve for 'x'. This gives,

$\frac{x}{x+72}=\frac{88-55}{88}\\\\88x=33(x+72)\\\\88x=33x+2376\\\\88x-33x=2376\\\\55x=2376\\\\x=\frac{2376}{55}=43.2$

Therefore, AE = 43.2 units

Now, from right angled triangle ABE,

$(AB)^2+(BE)^2=(AE)^2...(Pythagoras\ Theorem)\\\\(33)^2+(BE)^2=(43.2)^2\\\\BE=\sqrt{(43.2)^2-(33)^2}=27.879$

Similarly from right angled triangle ACD,

$CD=\sqrt{AD^2-AC^2}\\\\CD=\sqrt{(72+43.2)^2-88^2}=74.344$

Now, perimeter is the sum of all the sides of a figure. Therefore, the perimeter of BCDE is given as:

Perimeter = BE + ED + CD + BC

Perimeter = 27.879 + 72 + 74.344 + 55 = 229.223 ≈ 229.2 (Nearest tenth)

Therefore, the perimeter = 229.2 units