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ABC and CDE are equilateral triangles with side lengths 3 and 5, respectively. Find the perimeter of ADE. May 23, 2021

#1
#2
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P = (√19) + 5 + 7 Guest May 23, 2021
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FD = 1/2

AF = 5(cos30º)

AD = sqrt(FD2 + AF2) = √19

AE = sqrt(AF2 + FE2) = 7 Guest May 24, 2021
#3
+1 The diagram got crowded but I guess it could balance the deserted feeling of the previous two posts!! The answers given are correct, but here is one way we can get those answers. We are asked to find the perimeter of the triangle ADE. We know DE has length 5, so we must find the lengths of the remaining two sides.

length of AE: the triangle AFH is a 30-60-90 triangle with hypothenuse equal to 5 units; so HF has length $$\frac{1}{2}(5)$$=2.5, and AH has length $$\frac{\sqrt{3}}{2}(5)$$. The triangle AHE, therefore, has legs with lengths 5.5 and $$\frac{5\sqrt{3}}{2}$$$${5.5}^{2}+(\frac{5\sqrt{3}}{2})^2=49$$, so AE has length 7.

Length of AD. triangle DHB is also a 30-60-90 triangle with the lengthof the hypothenuse equal to 2. Thus HB has length 1 and DH length$$\frac{\sqrt{3}}{2}(2)=\sqrt{3}$$. Triangle ADH , therefore, has legth 4 and $$\sqrt{3}$$; $$4^2+(\sqrt{3})^2=19$$, so length of AD is $$\sqrt{19}$$.

ADE, therefore, has perimeter equal to 5+7+$$\sqrt{19}$$= 12 +$$\sqrt{19}$$.

May 24, 2021