What is the area, in square units, of triangle ABC in the figure shown if points A, B, C, and D are coplanar, angle D is a right angle, AC = 13, AB = 15 and DC = 5?

CuriousDude Jun 1, 2019

#1**+3 **

By the Pythagorean Theorem,

(DC)^{2} + (DA)^{2} = (AC)^{2}

5^{2} + (DA)^{2} = 13^{2}

(DA)^{2} = 13^{2} - 5^{2}

(DA)^{2} = 144

DA = 12

By the Pythagorean Theorem,

(DA)^{2} + (BD)^{2} = (AB)^{2}

12^{2} + (BD)^{2} = 15^{2}

(BD)^{2} = 15^{2} - 12^{2}

(BD)^{2} = 81

BD = 9

area of triangle ABD = (1/2)(base)(height)

area of triangle ABD = (1/2)( DA )( BD )

area of triangle ABD = (1/2)(12)(9)

area of triangle ABD = 54

area of triangle ACD = (1/2)(base)(height)

area of triangle ACD = (1/2)( DA )( DC )

area of triangle ACD = (1/2)( 12 )( 5 )

area of triangle ACD = 30

area of triangle ABC = area of triangle ABD - area of triangle ACD

area of triangle ABC = 54 - 30

area of triangle ABC = 24

*edited to fix error*

hectictar Jun 1, 2019

#3**+3 **

Oh actually this is wrong...I didn't read the question right...it asks for area of triangle ABC not area of triangle ABD!!

hectictar
Jun 1, 2019

#4**+1 **

Which is always why you read the problem right! It's ok, people make mistakes

CuriousDude
Jun 1, 2019

#5**+2 **

\(\text{AD}^2 + \text{DC}^2 = \text{AC}^2\\ \text{AD}^2 = 13^2 - 5^2 = 144\\ \text{AD} = 12\)

\(\text{BD}^2 + \text{DA}^2 = \text{AB}^2\\ \text{BD}^2 = 15^2 - 12^2 = 81\\ \text{BD} = 9\\ \text{BC} + \text{CD} = 9\\ \text{BC} = 9 - 5 = 4\)

Area = \(\dfrac{4\times 12}{2} \) = \(24 \;\text{sq. units}\)

.MaxWong Jun 1, 2019