+0  
 
0
59
2
avatar

A triangle has vertices at A(6,8), B(0,0), and C(16,0).  A square is inscribed in triangle ABC so that one side of the square aligns with side AB.  Find the side length of the square.

 Dec 9, 2020
 #1
avatar+846 
+2

A triangle has vertices at A(6,8), B(0,0), and C(16,0).  A square is inscribed in triangle ABC so that one side of the square aligns with side AB.  Find the side length of the square.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

BC = 16

 

AB = sqrt(82 + 62) = 10

 

AC = sqrt(82 + 102) = √164

 

Use the law of cosine to calculate  ∠ABC and ∠BAC

 

Use this formula to find "a"               (a / tan∠BAC) + a + (a / tan∠ABC) = AB

 

 Dec 9, 2020
 #2
avatar+114221 
+2

See the following image  :

 

 

AB  = 10    BC  = 16

 

Since  FG is parallel to AB.....then triangle  GFC  is similar to  triangle  ABC

 

Therefore....

 

AB/BC  = GF/FC

10/16  = GF/ FC

5/8  = GF/FC

8/5 = FC / GF

Let  GF = the side  of the square  = s

So....

(8/5)s  = FC

 

Draw  altitude  AH

Triangle BHA is a right triangle

sin angle B  = 8/10  = 4/5

But triangle BEF  is also a right triangle  with  hypotenuse BF

Therefore  sin angle B   =  EF/ BF = s/BF

4/5 = s/BF

5/4  = BF/s

(5/4)s   =BF

 

But   BF  +  BC  = 16

 

Therefore

 

(8/5)s  + (5/4)s   =16

 

[( 32 + 25) / 20 ] s  = 16

 

(57/20)s  =16

 

s =  (20)(16)/57  =  320 / 57

 

 

cool cool cool

 Dec 10, 2020

25 Online Users