A geometry problem by Viknes Selvam
Geometry
Level
4
The vertices of a triangle are P ( 5 , 3 ) , Q ( 8 , h ) , and R ( -1 , - 1 ) . Given the area of the triangle PQR is 15 unit square , find the possible value of h .
The answer is 10.
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1 solution
I don't know if I am right, because this problem seems to have two solutions, that is 10 and 0.
We can use the formula below to find the area of a triangle. 2 1 5 3 8 h − 1 − 1 5 3 = 1 5 2 1 ∣ ( 5 h − 8 − 3 + 1 5 ) − ( 2 4 − h − 5 + 1 5 ) ∣ = 1 5 ∣ 6 h − 3 0 ∣ = 3 0 As LHS has the absolute value sign, we need to consider two cases. Case 1: 6 h − 3 0 > 0 We will have 6 h − 3 0 = 3 0 ⟹ h = 0 Case 2: 6 h − 3 0 < 0 We will have 3 0 − 6 h = 3 0 ⟹ h = 1 0
Therefore 0 and 10 are the possible values of h, like what I've stated above. And I think the one who come out with this problem wants a positive value of h, is it true?
Edit 27/08/2017: The final answer is 10, it was confirmed.