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.

* 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.

slicergiza slicergiza · Answer 2 · 2019-08-27T09:49:59+02:00

Answer:

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

Step-by-step 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,

[tex]\because AM^2+DM^2 = AD^2[/tex]

[tex]AB^2+MB^2+MC^2+CD^2=AD^2[/tex]

[tex]AB^2 + MB^2 + MB^2 + AB^2 = AD^2[/tex] ( from equation (2) )

[tex]2AB^2 + 2MB^2 = AD^2[/tex]

[tex]2AB^2 + 2(\frac{BC}{2})^2 = AD^2[/tex]

[tex]2AB^2 + 2\times \frac{BC^2}{4} = AD^2[/tex]

[tex]2AB^2 + \frac{BC^2}{2}= AD^2[/tex]

[tex]2AB^2 + \frac{AD^2}{2} = AD^2[/tex] ( from equation (1) )

[tex]\frac{4AB^2+AD^2}{2}=AD^2[/tex]

[tex]4AB^2+AD^2 = 2AD^2[/tex]

[tex]4AB^2 = AD^2[/tex]

[tex]\implies 2AB= AD-----(3)[/tex]

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.

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.

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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.