# ABCD is a square with AB=12. Point P is in interior and the distances to A,B and to the side CD are equal. Find the distance.

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We are given that ABCD is a square with AB=12. Point P is inside the square and the distances to A,B and to the side CD are equal. We have to find this distance.

Now the distance of P from A and B is equal. Therefore we know that the perpendicular from P to AB cuts it halfway. Let this point be X, so that we have AX = BX = 12/2 = 6.

Now take the distance that we have to determine as Z. From the triangle AXP right angled at P we have Z^2 - 6^2 = (PX)^2. Also the distance from P to CD is Z. So PX + Z = 12

=> sqrt ( Z^2 - 6^2) + Z = 12

=> sqrt ( Z^2 - 6^2)  = 12 - Z

square both the sides

=> Z^2 - 36 = 144 + Z^2 - 24Z

=> 24Z = 180

=> Z = 180/24

=> Z = 7.5

Therefore the required distance is 7.5

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Let P be  the point which is equidistant from A, B and CD.

Then the point P being equidistant from A and B, the perperpendicular bisector of AB shoud pass through P  and bisect AB at E and CD at F respectively.

Now  the triangel AFE is a right angled triangle. Therefore AF^2 = AE^2+EF^2.

AF^2= (1/2)AB^2 + EF^2,  as E is mid point of AB.

AF^2 = 6^2 + 12 ^2 , as EF = BC = CD = 12 , sincw ABCD is a square.

Therefore AF^2 =  36+144 = 180.

Therefore AF = sqrt(180) = 6sqrt5.

Similarly BF = 6sqrt5.

AB = 12 given

The area of triangle ABF = (1/2)AB*FF = (1/2)12*12 = 72.

Therefore AP = BP = FP = R the circumradius  R of the triangle ABF , where F is the mid point of CD.

We know that in a triangle circumradius R =   abc/4*area of trangle .

Therefore the circum radius of ABF is R = AB*AF*BF/4*area of triangle = 12*(6sqrt5)(6sqrt5)/(4*72) = 12*36*5/(4*72) = 7.5cm.

Therefore the  distance AP= BP = FP = 7.5cm.