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?

Guest Jan 16, 2022

#2**+1 **

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

ElectricPavlov Jan 16, 2022

edited by
Guest
Jan 16, 2022

#3**+1 **

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) = 22.62^{o}.

3) Find the size of angle(CBD): angle(CBD) = 180^{o} - 22.62^{o} = 157.38^{o}.

4) Find the area of triangle(BCD) using the formula

area = ½ ·a · b · sin(C) = ½ ·12 · 4 · sin(157.38^{o})

= ...........

geno3141 Jan 16, 2022

#4**+1 **

Assuming that side AB is extended to point D (past point B) so that AD = 17:

.....but then point "D" would not lie * on *side AB ...

ElectricPavlov
Jan 16, 2022

#5**+2 **

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

so \(6\sqrt8\)?

XxmathguyxX Jan 17, 2022

#6**+1 **

and guest you are correct but if it is lie on, then i guess the problem is wrong?

XxmathguyxX Jan 17, 2022