1) In triangle ABC, AB=AC and D is a point on line AC so that line BD bisects angle ABC. If BD=BC, what is the measure, in degrees, of angle A?

2) In a certain iscoceles right triangle, the altitude to the hypotenuse has length 4sqrt2. What is the area of the triangle?

3) In triangle ABC, AB=17, AC=*, and BC=15. Let D be the foot of the altitude from C to AB. Find the area of triangle ACD.

Please help soon, thanks!

Guest Oct 13, 2018

#1**+1 **

2) In a certain iscoceles right triangle, the altitude to the hypotenuse has length 4sqrt2. What is the area of the triangle?

Area = (1/2)hypotenuse * (altidude drawn to the hypotenuse) (1)

Area (1/2) product of the leg lengths (2)

Equate (1) and (2) and let the leg lengths = L and we have

(1/2) hypotenuse * ( 4sqrt 2 ) = (1/2) L^2 multiply through by 1/2

hypotenuse * ( 4sqrt(2) = L^2

The hypotenuse = sqrt ( L^2 + L^2) = sqrt (2 L^2 ) = Lsqrt 2

L sqrt 2 * 4* sqrt 2 = L^2 divide through by L

4 sqrt 2 * sqrt 2 = L

4*2 = L

8 = L

So...the area = (1/2) L^2 = (1/2) * 8^2 = 32 units^2

CPhill Oct 13, 2018

#2**+1 **

1) In triangle ABC, AB=AC and D is a point on line AC so that line BD bisects angle ABC. If BD=BC, what is the measure, in degrees, of angle A?

Call the each bisected angle. a

And call angle BCA, b

And since BD = BC...then angle BDC = angle BCD = angle BCA = b

So....since AB = AC....then angle ABC = angle BCA ⇒ 2a = b ⇒ 4a = 2b

So..in triangle BDC....

a + 2b =180

a + 4a = 180

5a = 180

a = 36

And b = 2a = 72

So...in triangle ABC

Angle ABC + Angle BCA + Angle A = 180

2a + b + Angle A = 180

72 + 72 + Angle A = 180

144 + Angle A = 180

Angle A = 180 - 144 = 108°

CPhill Oct 13, 2018