ABCD is a rectangle with AB=50 and BC=120. E is a point on CD (possibly extended) such that AE=EC. What is the length of DE?
119
thanks
Let ABCD be a rectangle such that AB=5 and BC=12. There exist two distinct points X1 and X2 on BC such that ∠AX1D=∠AX2D=90∘. Suppose that d is the distance from X1 to X2. What is d2?
To find the length of DE, we need to use the properties of a rectangle and apply the Pythagorean theorem.
Step 1: Draw ABCD rectangle with given dimensions.
A__________B
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|____E_____C
D
Step 2: Since AE = EC, we can divide CD into two equal parts, making CE = ED.
A__________B
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________E__|
D
C
Step 3: Apply the Pythagorean theorem to find the length of DE.
In the right triangle AED, we have AE = ED, AD = 50, and AD perpendicular to AE.
Using the Pythagorean theorem, where the hypotenuse (AD) is squared and equal to the sum of the squares of the other two sides:
AD^2 = AE^2 + ED^2
Since AE = ED, we can substitute AE for ED in the equation:
AD^2 = AE^2 + AE^2
Simplifying the equation:
AD^2 = 2 * AE^2
Now we can substitute the known values into the equation:
50^2 = 2 * AE^2
2500 = 2 * AE^2
Divide both sides of the equation by 2:
AE^2 = 1250
Taking the square root of both sides:
AE = sqrt(1250)
Therefore, the length of DE is equal to AE, so DE = sqrt(1250).