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1. Point  X is on AC such that AX=3CX=12 . If $\angle ABC = \angle BXA = 90, then what is BX?

 

2. Find the length of the shortest altitude of a triangle with sides of lengths 10, 24, and 26.

 

3.In the diagram below, we have DE=2EC and AB=DC=20 . Find the length of FG.

4.In the diagram below, we have $\angleABC= \angleACB=\angleDEC=\angleCDE$, BC=8, and DB=2. Find AB.

 

5.Right  triangle ABC has , AB=3,BC=4 and AC=5. Square XYZW is inscribed in triangle ABC  with X and Y on AC ,  W on AB , and  Z on BC. What is the side length of the square?

 

6. In triangle ABC, we have\angleBAC= 60 and \angle ABC = 45. The bisector of \angle A intersects BC at point T, and AT=24. What is the area of triangle ABC?

Guest Jul 11, 2018
edited by Guest  Jul 11, 2018
 #1
avatar+25 
+3

1: 4√3

2: 120/13

3: 12

4: 6*(√3)/3

5: 60/37

6: 216 + 72√3

 

PS.  I am not sure about all of these answers

CluelesssPersonnn  Jul 11, 2018
 #2
avatar
+1

4 is wrong

Guest Jul 11, 2018
 #3
avatar+88848 
+2

2. Find the length of the shortest altitude of a triangle with sides of lengths 10, 24, and 26.

 

The area  of the triangle will be  (1/2)(10) (24)  = 120  units^2

 

So....if we let  10  be one altitude....the shortest altitude  will be  drawn  from the intersection of the legs  to the hypotenuse

 

And since the areas  are equal....we have that

 

(1/2) (length of the shortest altitude) (26)  = 120    

 

(13) (length of shortest  altitude)  = 120      divide both sides  by 13

 

length of the shortest altitude  = 120 / 13   units  .....just as CluelesssPersonn  found  !!!!

 

 

cool cool cool

CPhill  Jul 11, 2018
 #4
avatar+88848 
+2

3.

 

Since DE  = 2EC....then

EC + 2EC  = DC

3EC  = 20

EC  = 20/3

And DE  = 2EC  = 2(20/3)  = 40/3

 

Note that triangle  AFB  is similar to  triangle EFD

So

AB/ED  = FB / FD

So

20/(40/3)  = FB / FD

60/40 =  FB / FD

6/4  = FB /FD

3/2 = FB / FD

 

Then BF  consists of 5 equal parts.....FB is 3  of them and FD  is 2 of  them

 

And in triangle BDC, FG  is drawn parallel to DC....and we have the following relationship :

 

BF / BD  = FG / DC

 

3 / 5   = FG / 20      multiply both sides by 20

 

20 * 3  / 5    = FG

 

60 / 5  = FG

 

12  = FG

 

 

cool cool cool

CPhill  Jul 11, 2018
 #5
avatar+88848 
+1

4.In the diagram below, we have \(angleABC= angleACB=angleDEC=angleCDE\), BC=8, and DB=2. Find AB.

 

Since angle DEC  = angle BEC,    and angle DEC  = angle ABC = angle ECB.....then angles  BEC  and ECB  are equal....then in triangle BEC,   BC  = BE 

Then

BC  = BD  + DE

8  = 2 + DE

6  = DE

 

And angle ABC  = angle ACB...so...AC  = AB

 

 

Note that since angles ACB, ACB, DEC and CDE  are all equal, we have the following similar triangles :

 

ΔABC  ~  Δ BEC   ~  ΔCED          and  because  angles DEC and CDE are equal, then   CD  = EC

 

So

 

AC/BC  = BC /EC   

AC / 8  = 8 / EC

AC * EC   = 8*8

AC * EC  = 64

AC  = 64/EC    (1)

 

And

AC/BC  = CD/ED

AC/8  = CD / 6

AC/8 = EC/ 6

AC = (8/6)EC

AC = (4/3)EC     (2)

 

Equating (1)  and (2), we have that

 

64/EC  = (4/3)EC

(64)(3/4)  = EC^2

48  = EC^2

√48  = EC

4√3  = EC

 

And  AC  = (4/3)EC   = (4/3) * 4√3   =     16/√3  =   AB

 

 

cool cool cool

CPhill  Jul 12, 2018

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