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?
By the Pythagorean Theorem,
(DC)2 + (DA)2 = (AC)2
52 + (DA)2 = 132
(DA)2 = 132 - 52
(DA)2 = 144
DA = 12
By the Pythagorean Theorem,
(DA)2 + (BD)2 = (AB)2
122 + (BD)2 = 152
(BD)2 = 152 - 122
(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*
Oh actually this is wrong...I didn't read the question right...it asks for area of triangle ABC not area of triangle ABD!!
Which is always why you read the problem right! It's ok, people make mistakes
\(\text{AD}^2 + \text{DC}^2 = \text{AC}^2\\ \text{AD}^2 = 13^2 - 5^2 = 144\\ \text{AD} = 12\)
\(\text{BD}^2 + \text{DA}^2 = \text{AB}^2\\ \text{BD}^2 = 15^2 - 12^2 = 81\\ \text{BD} = 9\\ \text{BC} + \text{CD} = 9\\ \text{BC} = 9 - 5 = 4\)
Area = \(\dfrac{4\times 12}{2} \) = \(24 \;\text{sq. units}\)