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.
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
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.