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.

Guest Dec 17, 2018

edited by
Guest
Dec 17, 2018

#1**+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

CPhill Dec 17, 2018

#2**+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

CPhill Dec 17, 2018