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Triangle $ABC$ has sides of length $11$ inches, $15$ inches and $20$ inches. What is the length of the altitude to the side of length $15$ inches? Express your answer in simplest radical form. 

 Mar 18, 2024

Best Answer 

 #2
avatar+128794 
+1

Area =  12sqrt (46) =  (1/2) (15) * altitude

 

12sqrt (46)   / 7.5  =  altitude ≈  10.85

 

cool cool cool

 Mar 18, 2024
 #1
avatar+394 
+1

By Heron's formula, the area of the triangle is 

\(\sqrt{23*(23-11)*(23-15)*(23-20)}=\sqrt{23*12*8*3}=\sqrt{6624}=12\sqrt{46}\).

Because side * altitude / 2 = area, then.

\(\frac{20x}{2}=12\sqrt{46}\)

\(10x=12\sqrt{46}\)

\(x=\frac{6\sqrt{46}}{5}\).

 Mar 18, 2024
 #2
avatar+128794 
+1
Best Answer

Area =  12sqrt (46) =  (1/2) (15) * altitude

 

12sqrt (46)   / 7.5  =  altitude ≈  10.85

 

cool cool cool

CPhill Mar 18, 2024

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