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Home » Thus, the original radius is \(\boxed{\frac{3 + \sqrt141}3}\). - Midis

Thus, the original radius is \(\boxed{\frac{3 + \sqrt141}3}\). - Midis

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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A' = \frac{\sqrt{3}}{4} (14)^2 = \frac{\sqrt{3}}{4} \times 196 = 49\sq...
\Delta A = A' - A = 49\sqrt{3} - 36\sqrt{3} = 13\sqrt{3}
\boxed{13\sqrt{3} \text{ square centimeters}}
A right triangle has a hypotenuse of length \( z = 10 \) cm and an ins...
A_{\text{circle}} = \pi c^2 = \pi (2)^2 = 4\pi
For a right triangle, the radius \( c \) of the inscribed circle is gi...
where \( a \) and \( b \) are the legs. But we also know:
A_{\text{triangle}} = \frac{1}{2}ab = r \cdot s
where \( r = c = 2 \) and \( s = \frac{a + b + z}{2} \) is the semi-pe...
Using the identity for right triangles:

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