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(1) In triangle JKL, we have JK = JL = 25 and KL = 30. Find the area.

 

(2) In triangle ABC, we have AB = 5, AC = 12, BC = 13.  Find the circumradius.

 May 31, 2021
 #1
avatar+121055 
+2

(1) In triangle JKL, we have JK = JL = 25 and KL = 30. Find the area.

 

Let  J  be  the  apex

And the triangle  is isosceles

Let the  height  = JM

 

Then  we  have right triangle  JML   such  that

JM   =sqrt  ( JL^2  -  (KL/2)^2  )    =    sqrt ( 25^2  - 15^2)   =  sqrt (625 - 225)   = sqrt 400  = 20

 

So....the  area  =   (1/2)KL * JM =  (1/2)(30) (20)  =    300 units^2

 

 

cool cool cool

 May 31, 2021
 #2
avatar+121055 
+2

(2) In triangle ABC, we have AB = 5, AC = 12, BC = 13.  Find the circumradius.

 

This  is a right  triangle.......the  circumradius = (1/2)   (hypotenuse)  =  (1/2)(13)  =  6.5

 

 

cool cool cool

 May 31, 2021
 #3
avatar+34416 
+2

1)   you could also use Heron's formula

         semi perimeter =  s = (25 + 25 + 30)/2 = 40

             area = sqrt (s (s-JK)(s-JL)(s-KL) )

      

                area = sqrt ( 40(15)(15)(10)) = 300 units2

 May 31, 2021

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