We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
623
4
avatar+1434 

Two right triangles share a side as follows:

What is the area of triangle ACE?

 

Thanks so much!

 #2
avatar+100571 
+1

Triangle ABD is isosceles.....and angle BAD   = 45

So angle DAC = 45....so....angle CAB is bisected....which sets up the following relationship :

 

CA / AB  = CE / BE

 

But triangle ABC is a right triangle   ...so   BC   =  sqrt (10^2 + 6^2)   =  sqrt (136)  = 2 sqrt (34)

 

So

 

10  / 6  = CE / BE

 

So....CB  is divided into 16 equal parts....and  CE  is 10 of these....so  CE  = (10 / 16) * 2 sqrt (34)  =

 

(5/4) sqrt (34)  = 1.25 sqrt (34)

 

The measure of angle  ACB  =   arcsin (6/(2sqrt (34)) 

 

So....the area of triangle ACE  = (1/2)(10)(1.25)sqrt (34)sin (arcsin (6/(2sqrt (34)) )  =

 

(1/4) * 10 * 1.25 * 6   =     

 

18.75  units^2

 

 

cool cool cool

 Mar 2, 2018
 #3
avatar+1434 
+2

Thank you!

 #4
avatar
0

Here's an alternative method.

 

Call the distance of E from the line AC, x, then the distance of E from the line BD will be 6 - x.

Call the distance of E from the line AB, y.

 

We can now write down the areas of the three smaller triangles.

Area ACE = 5x, area AEB = 3y and area BED = 3(6 - x).

 

So, area ACB = 30 = 5x + 3y, and area ABD = 18 = 3(6 - x) + 3y.

From the second equation, x = y, so substituting into the first equation 8x = 30, x = 15/4.

 

So, area ACE = 5*(15/4) = 75/4, (which is the one we are asked to find), area AEB = 3*(15/4) = 45/4

and area BED = 3(6 - 15/4) = 27/4.

 Mar 3, 2018

12 Online Users

avatar