The key to solve the problem for students to learn is to use the tangent properties: 1. Radius to tangent is perpendicular. 2. 2 circles having a common tangent point give a straight line joining centres of the circles and the tangent point. This is the only construction needed for this problem. Then using Pythagoras theorem for the right-angled triangle AOC to set up an equation with the unknown radius of semicircle will solve the problem easily as shown by previous comments.

Very good. What's the name of the app/software that you are using to write ✍️ electronically on your computer screen?

Area of the yellow sermicircle=1/2(π)(8^2)-(π)(3^2)=23π square units.

In triangle ACO, CO=R-3, AO=R-4 and AC=3 so (R-3)^2 = (R-4)^2 +9 etc

@pk2712

3 ай бұрын

I did the same thing . I don't know why math booster did that difference of squares thing instead . The difference of squares thing seems like more work . I solved the problem before I watched the video , and did it basically the same as Math Booster did in the first method ; but , I used one leg as R-4 , the other as 3 , and the hypotenuse as R-3 in the Pythagorean to solve for R=8 .

@johnbrennan3372

3 ай бұрын

Once you realise that pts D,C and O are collinear, it’s easy.

@pk2712

2 ай бұрын

@@johnbrennan3372 You are absolutely right . This is the key to the problem .

@DB-lg5sq

2 ай бұрын

Yes,good

Let O be the center of the semicircle, and D be the point on the circumference where radius OD passes through C. By rule, D is also the point of tangency between the circle and the semicircle, so CD is a radius of the circle snd CD = 3. Triangle ∆CAO: CA² + OA² = OC² 3² + (r-4)² = (r-3)² 9 + r² - 8r + 16 = r² - 6r + 9 2r = 16 r = 8 Purple area: A = πr²/2 - π3² A = π8²/2 - 9π A = 32π - 9π = 23π

Without calculus: if you drop the perpendicular from D to BO and draw the line BC by comparing angles it results that triangles BAC and AOC are congruent, so AO=AB=4 --> R=8. The rest are formulas... 👌🏻

(R-3)^2=(R-4)^2+3^2...R=8

@RAG981

3 ай бұрын

Exactly.

@pk2712

2 ай бұрын

This is exactly the way I solved the problem . All that figuring out of CO-- OA and CO+OA , seems like a clumsy way to solve the problem . We simple have the right triangle with legs R-4 and 3 and hypotenuse R-3 .