Rectangle Perimeter With Limited Information
What is the perimeter of rectangle A B C D ?
The answer is 38.
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8 solutions
The right angle there tempts us to do pythagoras :P Another way would be to realize the Δ E B A ∼ Δ D C E and we easily compute C E = 9 from it :)
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I just feel silly seeing @Paola Ramírez and your solution now after solving this problem using trigonometry (arctangents)! Ahahaha! Very beautiful solutions (:
I wasn't even thinking of the triangles being similar when I wrote the problem, but that is an interesting approach!
That's a nice way of ssolve the problem
I used the Euclid's Theorem: let's think like CD Would be the height on the hyphotenuse of the triangle.. Se we have: CD^2=BE•CE =>CE=36/4=9 so BC=9+4=13 and 2p=2(CD+BC)=2(6+13)=38
I got 37.5 I just used different namings (x-4) for in place of x and x instead of (x+4) Was my calculations too precise of a little wrong please reply someone
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Define A D = x . Then E C = x − 4 . Solve for E D 2 using the Pythagorean Theorem:
E C 2 + C D 2 = E D 2
( x − 4 ) 2 + 6 2 = E D 2
x 2 − 8 x + 1 6 + 3 6 = E D 2
x 2 − 8 x + 5 2 = E D 2
A E 2 = 5 2 , just as Paola calculated it. Now solve for x using the Pythagorean Theorem and Δ A E D :
A E 2 + E D 2 = A D 2
5 2 + x 2 − 8 x + 5 2 = x 2
x 2 − 8 x + 1 0 4 = x 2
1 0 4 = 8 x
x = 1 3
Now you have A D = 1 3 , and you can solve for the perimeter from there.
how do you get √2 ?
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By Phythagorean theorem
A B = C D = 6 ⇒ A E = A B 2 + B E 2 = 6 2 + 4 2 = 5 2
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yes I got the same too but in your answer you wrote √2^2 above ,is that a typo?
Can be done by proving the triangles similar like this, will make it more simpler.
You can use the rule that two perpendicular lines have slopes that are negative reciprocals. Since AE has slope 4 6 , ED has slope 6 − 4 . To find length x, divide length 6 by the slope 6 4 . The required horizontal distance is 9 for a vertical distance of 6 at the given slope.
Per Nihar Mahajan's comment:
m ∠ C E D + 9 0 ∘ + m ∠ A E B = 1 8 0 ∘
m ∠ C E D + 9 0 ∘ + m ∠ E D C = 1 8 0 ∘
Thus, ∠ A E B ≅ ∠ E D C .
Δ D C E ∼ Δ E B A because of AA similarity.
In fact, all three triangles in the figure are similar to each other!
You can solve for E C using a proportion:
6 4 = E C 6
My approach when creating the problem was the same as Paola's. I think both approaches are good!
Aha easy solution. TLDR your solutions sorry.
So, using the leg rule, let x be the rest of the length from 4
4/6 = 6/x, x = 9
That means our length is 13 and width is 6. 2(13+9) = 38. Tada!
I used Euclid's Theorem.. Let s look like CD is the height on the hyphotenuse of the triangle.. So we know that BE•CE=CD^2 so CE=36/4=9=> BC=9+4=13 and so 2p=2(BC+CD)=2(13+6)=38
Let E C = x . 6 is the geometric mean of BE and EC. Therefore, x = 9 , and the perimeter of rectangle ABCD is 3 8
The exact answer is 38.9~39
Relevant wiki: Pythagorean Theorem
By Pythagorean theorem ,
( 2 ) 2 + ( x 2 + 3 6 ) 2 = ( x + 4 ) 2 ⇒ x = 9
The perimeter is 2 x + 2 0 = 2 ( 9 ) + 2 0 = 3 8 .