Don't Get This One Wrong on the Test - It's Actually EASY
8.7K views · Sep 28, 2026 · Education
Comments · 7
@paparmar · 1 day ago (edited)
The largest semicircle that "just fits" inside a right triangle (sides a, b, hypotenuse c) has radius r = ab/(a+b) if its diameter lies along side c, ab/(a+c) if along side b, and ab/(b+c) if along side a. In this case, it's along side c (hypotenuse) so r = 6*8/(6+8) = 24/7.
@bernardbrinette5388 · 2 days ago
Exactly what I found.
@HamidZandi-xo6bl · 2 days ago
If you connect center of circle to verex of right angle, you make two triangles. <br>Now, the area of right angle triangle is equal to the sum of two triangles areas, and you calculate radius of the circle....
@Claudio_Bruzzone · 2 days ago
I reflect the figure across the hypotenuse to obtain a quadrilateral with an inscribed circle. <br>The radius of a circle inscribed in a polygon is: <br>r = 2 × Area / Perimeter. <br>In our case: <br>Area of the quadrilateral = 6 × 8 = 48 cm². <br>Perimeter of the quadrilateral = 6 + 8 + 6 + 8 = 28 cm. <br>r = 2 × 48 / 28 = 24/7 cm. <br>Area of the semicircle = ½(24/7)²π = 288π/49 cm².
@ロプノールs · 2 days ago
考え方（how to solve) 半径（radius)をXとして三角形の相似（∽）を使って半円の面積（area)を求めます。<br>大の△＝△ABC。中の△＝△ODCとします。２つの△は∠が等しく相似（∽）<br>Xを求める。６：８＝X:（８－X）８X＝４８－６X １４X＝４８ X＝２４/７<br>半円の面積（area)=（２４/７）²×π×１/２＝５７６/４９π×１/２＝２８８/４９π answer=288/49π㎠
@HamidZandi-xo6bl · 2 days ago
If you connect center of circle to verex of right angle, you make two triangles. <br>Now, the area of right angle triangle is equal to the sum of two triangles areas, and you calculate radius of the circle....
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