ABCD is a square. P is a point within ABCD such that PA=1,PB=29 and PC=41. What is area of square ABCD ?
I started by placing a square of sides k so that one vertex is at B(0,0), and letting sides fall on the x and y axes
so the other vertices are C(k,0), D(k,k) and A(0,k)
I let point P be (x,y)
PO = 29 ----> x^2 + y^2 = 29^2
PA = 1 -----> x^2 + (y-k)^2 = 1^2
PC = 41 ----> (x-k)^2 + y^2 = 41^2
so I had 3 messy equations in 3 unknowns.
After a page of frustrating algebraic manipulations,
(which I really don't feel like typing here)
I finally found
x = 1/√2 , y = 41/√2 , and k = 21√2
but it was really the k value we are after,
since the area of the square is simply k^2
if k = 21√2, then
k^2 = (21√2)^2 = 882
So the area is 882 units^2
btw, you can test the distances PO, PA, and PC with your calculator, they work
Hmmm. I don't buy it. You can't have
x = 1/√2 , y = 41/√2 , and k = 21√2
and
x^2 + (y-k)^2 = 1^2
Maybe I'll visit the figure.
x^2 + (y-k)^2
= (1/√2)^2 + (41/√2 - 21√2)^2 ,,,,,, 21√2 = 42/√2
= 1/2 + (-1/√2)^2
= 1/2+1/2
= 1
My bad. I misread it. Even after looking at it several times, I still saw 21/√2 rather than 21√2.
To find the area of square ABCD, we need to use the properties of squares and apply the Pythagorean theorem.
Let's denote the side length of the square as "s". Since ABCD is a square, all sides are of equal length.
We are given that PA = 1, PB = 29, and PC = 41. To find the side length "s", we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
Applying the Pythagorean theorem to triangle PAB, we have:
PA² + PB² = AB²
1² + 29² = AB²
1 + 841 = AB²
842 = AB²
Similarly, applying the Pythagorean theorem to triangle PAC, we have:
PA² + PC² = AC²
1² + 41² = AC²
1 + 1681 = AC²
1682 = AC²
Since ABCD is a square, AB = AC. Therefore, we can equate AB² and AC²:
842 = 1682
AB² = AC²
Since both sides of the equation are equal to the same value, we can conclude that AB = AC.
Now, considering triangle PBC, we know that PB = 29 and PC = 41. To find BC, we can again use the Pythagorean theorem:
PB² + PC² = BC²
29² + 41² = BC²
841 + 1681 = BC²
2522 = BC²
By substituting either AB or AC with BC, we can see that AB = AC = BC. Therefore, all sides of the square are equal, and we can write this as s = AB = AC = BC.
Finally, to find the area of the square ABCD, we can use the formula A = s², where A denotes the area and s denotes the side length:
A = s² = (AB)² = (AC)² = (BC)²
A = (AB)² = 842
A = 842
The area of the square ABCD is 842 square units.