+0

# Help

0
65
2

A square with sides 6 inches is shown. If P is a point such that the segment PA, PB, and PC are equal in length, and segment PC is perpendicular to segment FD, what is the area, in square inches, of triangle APB?

Jan 12, 2020

#1
0

The area of triangle APB is 13/2.

Jan 12, 2020
#2
+18828
0

C will be the midpoint of FD; therefore, FC = 3"

Call the three equal distances PA, PB and PC all equal to x".

Draw a line segment from P perpendicular to the side AF. Call the point of intersection X.

This makes the length of PX = 3" and the length of XF = x".

Since AF = 6" and XF = x", AX = (6 - x)"

Look at triangle AXP. This is a right triangle. The hypotenuse (PA) has value x"; side AX = (6 - x)", and side XP = 3".

Using the Pythagorean Theorem:  AX2 + XP2 =  AP2

--->                                               (6 - x)2 + 32  =  x2

--->                                       36 - 12x + x2 + 9  =  x2

--->                                                     45 - 12x  =  0

--->                                                            12x  =  45

--->                                                                x  =  3.75"

Since x = 3.75", CP also equals 3.75".

Since each side of the square is 6", the height of triangle(ABP) = 6" - 3.75"  = 2.25".

So, the area of triangle(ABP)  =  .5 x 6" x 2.25"  =  6.75".

Jan 12, 2020