Only reals accepted ; )

Geometry Level 5

$P(3 , 1)$ , $Q(6 , 5)$ and $R(x,y)$ are three points such that $\angle PRQ=90^\circ$ . Also, area of $\triangle PQR$ is $7 \text{ units}^2$ . Find the number of points $R$ possible.

Note: Only real values of $x,y$ accepted.

The answer is 0.

2 solutions

Niranjan Khanderia
Sep 1, 2015

Since $\angle~PRQ=90^o$ , R is on he circumference of a circle with PQ as a diameter. The maximum area with this is if triangle PRQ is 45-45-90. The area of this triangle is 6.25. So there an be NO real triangle with given conditions.

Fast one !!

Akshat Sharda - 5 years, 7 months ago

Nelson Mandela
Apr 4, 2015

Correct me if I am wrong.

There can be only two points possible (6,1) and (3,5) for angle PRQ to be 90 degrees.(square)

If point is (6,1) area is 1/2 x 3 x 4 = 6 units.

If point is (3,5) also area is 1/2 x 3 x 4 = 6 units.

So, I think answer is 0.

actually, i didn't understand the question. where did triangle ABC come from? i thought it was meant to confuse so i didn't take into account the area at all.

Lipsa Kar - 5 years, 9 months ago

I think that ABC is PQR.

Nelson Mandela - 5 years, 9 months ago

There are infinite points R on the circumference of the circle with diameter PQ, such that angle at P is 90 . The maximum area possible is 6.25.

Niranjan Khanderia - 5 years, 9 months ago

Only reals accepted ; )

The answer is 0.

2 solutions

0 pending reports