Find (AB)^2 in the figure below

No response or is it a case of local man cannot can?

AB^2= 96 unit^2

My work, using coordinate geometry.

1) Fix the diagram on a rectangular coordinate system such that point O is at the origin and CD lies on the x-axis. Doing this gives the following coordinates:
O(0, 0), F(4, 0), D(6, 0), E(-3, 0), C(-6, 0)

2) The equation of the circle that yields the bigger semicircle is

x² + y² = 6² ————(1)

and the equation of the circle that yields the smaller semicircle is

(x - h)² + (y - k)² = r² ————(2)

where (h, k) is the center of the circle and r = |AB|/2 is it's radius

3) the smaller semicircle contains the points F(4, 0) and E(-3, 0) therefore the coordinates of both points satisfies eq. (2)

(-3 - h)² + k² = r² ————(3)
and
(4 - h)² + k² = r² ————(4)

Solving (3) & (4) simultaneously, we will discover that h = ½ and substitution into either (3) or (4) yields k² + (49/4) = r²

4) Observe that point (h, k), O(0, 0) and point B are vertices of a right triangle of sides a, b, and c such that

a² = h² + k² = ¼ + k²,
b² = 6², and
c² = r² = k² + (49/4)
and by the Pythagorean theorem, c² + a² = b².

Substituting and solving for k², we find that k² = 47/4 and therefore r² = k² + (49/4) = 24

5) r² = |AB|²/4
|AB|² = 4r² = 96 unit²

|AB|² = 96 unit²

Boringdraw1:
AB^2= 96 unit^2

96 is correct show ur workings please.

Martinez39:
My work, using coordinate geometry.

1) Fix the diagram on a rectangular coordinate system such that point O is at the origin and CD lies on the x-axis. Doing this gives the following coordinates:
O(0, 0), F(4, 0), D(6, 0), E(-3, 0), C(-6, 0)

2) The equation of the circle that yields the bigger semicircle is

and the equation of the circle that yields the smaller semicircle is


3) the smaller semicircle contains the points F(4, 0) and E(-3, 0) therefore the coordinates of both points satisfies eq. (2)

Solving (3) & (4) simultaneously, we will discover thath = ½ and substitution into either (3) or (4) yields k² + (49/4) = r²

4) Observe that point (h, k), O(0, 0) and point B are vertices of a right triangle of sidea, b, and c such that


5) r² = |AB|²/4
|AB|² = 4r² = 96 unit²

|AB|² = 96 unit²

Good one.

Awaiting more shapes tomorrow.

Martinez39:
Awaiting more shapes tomorrow.

Sure sir

Mrshape:

Sure sir

In that your meme, what's the meaning of "CAN"?

Martinez39:
In that your meme, what's the meaning of "CAN"?

I think it means I will not bother attempting, I am not doing again, this is beyond me, in fact local man can not can.

G'day Mrshape what about yesterday's shape answer

Kkings11:
G'day Mrshape what about yesterday's shape answer

ok sir I will post it soon

Martinez39:
My work, using coordinate geometry.

1) Fix the diagram on a rectangular coordinate system such that point O is at the origin and CD lies on the x-axis. Doing this gives the following coordinates:
O(0, 0), F(4, 0), D(6, 0), E(-3, 0), C(-6, 0)

2) The equation of the circle that yields the bigger semicircle is

and the equation of the circle that yields the smaller semicircle is


3) the smaller semicircle contains the points F(4, 0) and E(-3, 0) therefore the coordinates of both points satisfies eq. (2)

Solving (3) & (4) simultaneously, we will discover that h = ½ and substitution into either (3) or (4) yields k² + (49/4) = r²

4) Observe that point (h, k), O(0, 0) and point B are vertices of a right triangle of sides a, b, and c such that


5) r² = |AB|²/4
|AB|² = 4r² = 96 unit²

|AB|² = 96 unit²

Nice solution

Martinez39:
My work, using coordinate geometry.

1) Fix the diagram on a rectangular coordinate system such that point O is at the origin and CD lies on the x-axis. Doing this gives the following coordinates:
O(0, 0), F(4, 0), D(6, 0), E(-3, 0), C(-6, 0)

2) The equation of the circle that yields the bigger semicircle is

and the equation of the circle that yields the smaller semicircle is


3) the smaller semicircle contains the points F(4, 0) and E(-3, 0) therefore the coordinates of both points satisfies eq. (2)

Solving (3) & (4) simultaneously, we will discover that h = ½ and substitution into either (3) or (4) yields k² + (49/4) = r²

4) Observe that point (h, k), O(0, 0) and point B are vertices of a right triangle of sides a, b, and c such that


5) r² = |AB|²/4
|AB|² = 4r² = 96 unit²

|AB|² = 96 unit²

Sweeet