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# geometry

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157
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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
0

17 or 1.7 ?!?

Jan 16, 2022
#2
+36435
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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
+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
+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
+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
+361
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and guest you are correct but if it is lie on, then i guess the problem is wrong?

Jan 17, 2022
#7
+361
+2

tell me if i am wrong

Jan 17, 2022