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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?

 Jun 1, 2019
 #1
avatar+8829 
+3

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*

 Jun 1, 2019
edited by hectictar  Jun 1, 2019
 #2
avatar+30 
+2

laughYay!

CuriousDude  Jun 1, 2019
edited by CuriousDude  Jun 1, 2019
 #3
avatar+8829 
+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
avatar+30 
+1

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

CuriousDude  Jun 1, 2019
 #5
avatar+7747 
+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 = \(\dfrac{4\times 12}{2} \) = \(24 \;\text{sq. units}\)

.
 Jun 1, 2019

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