the point a 2, 9 b a, 5 c 5, 5 are the vertices of a triangle ABC right angled at B find the value of a and hence find the area of triangle ABC
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Apply Pythagoras theorem
AB² + BC² = CA²
[(a-2)² +(5-9)²] + [(a-5)²+ (5-5)²] =[(5-2)² + (5-9)²]
a²-4a +4 +16 +a²-10a +25 =9 +16
2a²- 14a +45= 25
2a²- 14a +20=0
a²-7a +10=0
by making factors we will get
(a-5)(a-2)=0
so a= 5,2
by taking a=2
area of triangle= 1/2 x AB x BC
= 1/2 x 4 x 3= 6
AB² + BC² = CA²
[(a-2)² +(5-9)²] + [(a-5)²+ (5-5)²] =[(5-2)² + (5-9)²]
a²-4a +4 +16 +a²-10a +25 =9 +16
2a²- 14a +45= 25
2a²- 14a +20=0
a²-7a +10=0
by making factors we will get
(a-5)(a-2)=0
so a= 5,2
by taking a=2
area of triangle= 1/2 x AB x BC
= 1/2 x 4 x 3= 6
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