Question

Suppose ABCD is a rectangle. Find AB and AD if point M is the midpoint of

BC
,
AM
⊥
MD
, and the perimeter of ABCD is 34 in.

Answer 1

If AMD is an isosceles right triangle, its height is half its width*. Then

AB = (1/2)AD

2(AB + AD) = 34 in

(1/2)AD +AD = (3/2)AD = 17 in

AD = 34/3 = 11 1/3 inches . . . . . multiply by 2/3

AB = 5 2/3 inches . . . . . . . . . . . = (1/2)AD

_____

* The altitude of an isosceles right triangle divides the triangle into two isosceles right triangles. One leg of each of these smaller triangles is half the length of the base, and the other leg is the altitude. Since the two legs of an isosceles right triangle are congruent, the altitude of AMD is half the base AD.

Answer 2

AB is 5.67 inches and AD is 11.33 inches ( approx )

Explanation:

Given,

ABCD is a rectangle,

So, AB = CD and AD = BC -----(1)

In which M is the midpoint of the side BC,

That is, BM = MC ----(2),

Also, AM ⊥ MD,

In triangle AMD,

`\because AM^2+DM^2 = AD^2`

`AB^2+MB^2+MC^2+CD^2=AD^2`

`AB^2 + MB^2 + MB^2 + AB^2 = AD^2` ( from equation (2) )

`2AB^2 + 2MB^2 = AD^2`

`2AB^2 + 2((BC)/(2))^2 = AD^2`

`2AB^2 + 2* (BC^2)/(4) = AD^2`

`2AB^2 + (BC^2)/(2)= AD^2`

`2AB^2 + (AD^2)/(2) = AD^2` ( from equation (1) )

`(4AB^2+AD^2)/(2)=AD^2`

`4AB^2+AD^2 = 2AD^2`

`4AB^2 = AD^2`

`\implies 2AB= AD-----(3)`

Now, the perimeter of the rectangle ABCD = 34 in,

⇒ 2(AB+AD) = 34

⇒ 2AB + 2AD = 34

From equation (3),

AD + 2AD = 34

3AD = 34

⇒ AD ≈ 11.33 inches

Again from equation (3),

AB ≈ 5.67 inches.