## February 2023

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In the Math Corner of my February 2023 Diary I asked
for the area of this semicircle.

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**• Solution**

This one's too easy. First, label up key points P, Q, A, and B. Then, since you know this is a semicircle, that bottom horizontal line
is a diameter, so halfway along it is the circle's center O. Not knowing precisely where that is in relation to P and Q, let's just suppose it's a
distance *x* to the right of P.

Construct radii, both of course length *r*, from O to A and B.

With two right-angled triangles there, sides (*r*, 4, *x*) and (*r*, 2, 6−*x*), you could now chew
through the algebra, applying Pythagoras' Theorem to the two triangles APO and BQO, then eliminating *r* and solving for *x*.

It's hardly worth the trouble, though. Plainly *x* = 2 and the two triangles are congruent. So
*r*² = 4² + 2², which is 20. A circle radius *r* has area π*r*², so our semicircle
area is 10π.