In right triangle ABC, the hypotenuse AB is 13 units long and leg AC is 5 units long. Point D lies on side AB so that AD = \(17\) units. What is the area of triangle BCD?
AD lies on AB AB is only 13 so AD cannot be 17 ....the other similar question a few down the list from this states AD = 10
Assuming that side AB is extended to point D (past point B) so that AD = 17:
1) Find the size of BC: using the Pythagorean Theorem, BC = 12.
2) Find the size of angle(ABC): sin(ABC) = 5/13 ---> angle(ABC) = 22.62o.
3) Find the size of angle(CBD): angle(CBD) = 180o - 22.62o = 157.38o.
4) Find the area of triangle(BCD) using the formula
area = ½ ·a · b · sin(C) = ½ ·12 · 4 · sin(157.38o)
okay, so BC is 12, BD= 4
If we extend AC to E so that ADE is a right triangle, then point F is on DE so that triangle BDE is a smaller scale of ADE,
BF and FD are x and x^2+x^2=16, so x=\(\sqrt8\)
and so BCD is 12\(\times \sqrt8\) /2, which is 6sqrt8
and guest you are correct but if it is lie on, then i guess the problem is wrong?