Look at the figure below: triangle abc has measure of angle abc equal to 90 degrees. e is a point on the hypotenuse ac such that the segment be is perpendicular to ac. d is a point outside the triangle on the right of e so that triangle edc has the measure of angle edc equal to 90 degrees. which triangle is similar to triangle aeb?

δaeb δabc δ bec

step-by-step explanation:

given:   ∠abc= , => ∠aeb =   , and ∠edc =  

let ∠bac = x , then by angle sum property in δabc, ∠bca = 90°-x

now, we can see, in δaeb, ∠bae = x (∵∠bac = x)

so, again by angle sum property in δaeb , ∠abe = 180°-x - 90° = 90°-x

in δabc & δaeb

∠abc=∠aeb (∵each 90°)

∠bac = ∠bae (∵ each equals to x)

∠acb = ∠abe = 90°-x

so, by aaa criteria of similarity

δaeb is similar to δabc

by using transitive property,

δaeb δabc δ bec

here, neither δedc nor δcde is similar to δaeb, beause ∠edc and ∠cde are arbitrary.

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240

x= 3^y

step-by-step explanation: