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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?

 Jan 16, 2022
 #1
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0

17 or 1.7 ?!?  cheeky

 Jan 16, 2022
 #2
avatar+36435 
+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

 Jan 16, 2022
edited by Guest  Jan 16, 2022
 #3
avatar+23200 
+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) = 22.62o.

 

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

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

 Jan 16, 2022
 #4
avatar+36435 
+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
avatar+361 
+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 \(6\sqrt8\)?

 Jan 17, 2022
 #6
avatar+361 
+1

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

 Jan 17, 2022
 #7
avatar+361 
+2

tell me if i am wrong

 Jan 17, 2022

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