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Triangle ABC has AB=BC=5 and AC=6. Let E be the foot of the altitude from B to \(\overline{AC}\) and let D be the foot of the altitude from A to BC. Compute the area of triangle DEC.

 

Suppose ABC is a scalene right triangle, and P is the point on hypotenuse \(\overline{AC}\) such that \(\angle{ABP} = 45^{\circ}\). Given that AP=1 and CP=2, compute the area of ABC.

 Dec 17, 2018
edited by Guest  Dec 17, 2018
 #1
avatar+94526 
+2

Triangle ABC has AB=BC=5 and AC=6. Let E be the foot of the altitude from B to AC  and let D be the foot of the altitude from A to BC. Compute the area of triangle DEC.

 

 

Here's an image :

 

 

 

 

 

Let A = (0,0)   C = (6,0)

The triangle is isoceles.....so E will bisect AC....so  ....EC = 3

And....triangle AEB is a 3-4-5  right triangle with AE = 3, EB = 4  and AB = 5

 

So....B = (3,4)

Sine angle BCE = 4/5 = sin DCA = sin DCE

And we can find AD  using the Law of Sines

 

AD/sin DCA = AC/sin ADC

 

AD/ (4/5) = 6/sin90

 

AD = (4/5)6  =  24/5 =  4.8

 

And...using the Pythagorean Theorem, 

 

DC =  sqrt (6^2 - 4.8^2 )   = 3.6

 

And  EC = 3

 

So....the area of triangle DEC  =

 

(1/2) (EC) (DC) sin DCE   =

 

(1/2)(3)(3.6) (4/5)  =

 

4.32 units^2

 

 

cool cool cool

 Dec 17, 2018
 #2
avatar+94526 
+2

Suppose ABC is a scalene right triangle, and P is the point on hypotenuse AC  such that angle ABP = 45° . Given that AP=1 and CP=2, compute the area of ABC.

 

Because angle ABP = 45, then the right angle ABC is bisected....this sets up the following relationship in the triangle

 

AP/CP = AB/BC

 

1/2  = AB/BC

 

Let AB = x

So BC = 2x

And AC = 5

 

So....by the Pythagorean Theorem, 

 

AB^2 + BC^2  = AC^2

 

x^2 + (2x)^2  =  5^2

 

x^2 + 4x^2  = 25

 

5x^2  =  25

 

x^2  =  5

 

x =  sqrt (5)  = AB

2x  = 2sqrt(5)  = BC

 

So.....the area of ABC =  (1/2)AB * BC  = (1/2)sqrt(5) * 2sqrt (5)  =  5 units^2

 

 

cool cool cool

 Dec 17, 2018

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