+30 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?
+9481 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*
+9481 Oh actually this is wrong...I didn't read the question right...it asks for area of triangle ABC not area of triangle ABD!!
+30 Which is always why you read the problem right!
It's ok, people make mistakes
+9673 \(\text{AD}^2 + \text{DC}^2 = \text{AC}^2\\ \text{AD}^2 = 13^2 - 5^2 = 144\\ \text{AD} = 12\)
\(\text{BD}^2 + \text{DA}^2 = \text{AB}^2\\ \text{BD}^2 = 15^2 - 12^2 = 81\\ \text{BD} = 9\\ \text{BC} + \text{CD} = 9\\ \text{BC} = 9 - 5 = 4\)
Area = \(\dfrac{4\times 12}{2} \) = \(24 \;\text{sq. units}\)
Yay!
