Question

Let ABCD be a square and P be a point inside the square such that PB = 23 and PD = 29. Find the area of triangle APC. PLEASE ANSWER FASTTTTT
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Answer 1

Answer:

2

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Accepted

Let us work in a coordinate plane such that :

A(0,0),B(a,0),C(a,a),D(0,a),P(x,y)

(where a is the square's sidelength).

Distance constraints give :

{PB2PD2==(x−a)2+y2x2+(y−a)2==232292(1)

(Geometrical interpretation : P is any of the two intersection points P1 and P2 of circles C1 and C2 with resp. centers B and D and radii 23 and 29.)

Subtracting equations in (1), we get:

2xa−2ya=312.(2)

Therefore, the area of triangle APC is:

12det(AP→,AC→)=12∣∣∣xyaa∣∣∣=12a(x−y)=78.(3)

(using (2)).

Answer 2

Answer:

Rasam kittum ethu cheythal. so thanne cheyyam. Area of ∆APC = 27cm.