Homework question | Math step-by-step solution

Answer

The radius of the inscribed sphere $R_{sh}$ is equal to 10 units. This value is determined by calculating the radius of the inscribed circle of the base and using the trigonometric relationship derived from the angle between the lateral face and the base.

Image Analysis

The image displays a regular triangular pyramid $SABC$ with a sphere inscribed inside it.

• Points: $S$ is the vertex, $O$ is the center of the equilateral base $\triangle ABC$, and $O_1$ is the center of the inscribed sphere.
• Lines: $SD$ is the apothem (height of the lateral face), and $OD$ is the radius of the circle inscribed in the base.
• Angle: $\alpha$ (angle $\angle SDO$) represents the dihedral angle between the lateral face and the base.
• Point of Tangency: $M$ is the point where the sphere touches the apothem $SD$.

Explanation

1. Calculate the side length of the base The base area $S_{ABC}$ of an equilateral triangle with side $a$ is given by the formula: $S_{ABC} = \frac{a^2\sqrt{3}}{4}$ This formula relates the area of an equilateral triangle to its side length. Substituting the given area: $300\sqrt{3} = \frac{a^2\sqrt{3}}{4} \implies a^2 = 1200 \implies a = \sqrt{400 \cdot 3} = 20\sqrt{3}$ ⚠️ This step is required on exams to establish the dimensions of the pyramid's base.

2. Calculate the inradius of the base ($r_{base}$) In a regular triangle, the distance from the center $O$ to the midpoint of the side $D$ (which is the radius of the inscribed circle of the base) is: $r_{base} = OD = \frac{a}{2\sqrt{3}}$ This represents the distance from the base center to the boundary of the base. Substituting $a = 20\sqrt{3}$: $OD = \frac{20\sqrt{3}}{2\sqrt{3}} = 10$ The distance $OD$ is exactly 10 units.

3. Analyze the cross-section $SOD$ We consider the right triangle $\triangle SOD$. The center of the sphere $O_1$ lies on the altitude $SO$. The radius of the sphere $R_{sh}$ is the segment $O_1O$. Since the sphere is tangent to the face at $M$, $O_1M \perp SD$ and $O_1M = R_{sh}$. The segment $DO_1$ is the bisector of the angle $\alpha$ ($\angle SDO$), because $O_1$ is equidistant from the base and the lateral face. $\angle O_1DO = \frac{\alpha}{2}$ This bisector property is fundamental for finding the center of an inscribed sphere.

4. Determine the trigonometric ratio for $\frac{\alpha}{2}$ We are given $\cos \alpha = \frac{8}{17}$. We use the half-angle formula for tangent: $\tan \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}}$ The tangent of the half-angle relates the sphere's radius to the base inradius. Substituting the values: $\tan \frac{\alpha}{2} = \sqrt{\frac{1 - 8/17}{1 + 8/17}} = \sqrt{\frac{9/17}{25/17}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$ The calculated ratio is 0.6.

5. Calculate the sphere radius ($R_{sh}$) In the right triangle $\triangle O_1OD$: $R_{sh} = O_1O = OD \cdot \tan\left(\frac{\alpha}{2}\right)$ This formula uses the definition of tangent in a right triangle to find the opposite side. $R_{sh} = 10 \cdot \frac{3}{5} = 6$ Note: Re-evaluating the geometry. In a regular pyramid, if $r$ is the inradius of the base, the sphere radius $R$ is $r \cdot \tan(\alpha/2)$. Let's double check the calculation. $10 \times 0.6 = 6$.

Final Answer

The radius of the inscribed sphere is: $\boxed{6}$

Common Mistakes

• Confusing angles: Students often use $\alpha$ instead of $\alpha/2$ in the final calculation. The center of the inscribed sphere always lies on the bisector of the dihedral angle.
• Radius definitions: Mixing up the radius of the circumscribed circle ($R_{base} = \frac{a}{\sqrt{3}}$) with the radius of the inscribed circle ($r_{base} = \frac{a}{2\sqrt{3}}$) for the base. Remember that $OD$ refers to the small radius ($r$).