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.
+130071 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.
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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
+130071 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
