Question

problem 25
a semicircle is inscribed in an isosceles triangle with base 16 and height 15 so that the diameter of
the semicircle is contained in the base of the triangle as shown. what is the radius of the semicircle?
如图所示，一个底边为16，高为15的等腰三角形内有一个半圆与两条腰相切，并且半圆的
直径与三角形的底边重合，问这个半圆的半径是多少？
(2016年)
4√3
120/17
10
17√2/2
17√3/2

Explanation:

Step1: Find the length of the waist of the isosceles triangle

The base of the isosceles triangle is \(16\), so half of the base is \(\frac{16}{2} = 8\). The height is \(15\). According to the Pythagorean theorem, the length of the waist \(l=\sqrt{8^{2}+15^{2}}=\sqrt{64 + 225}=\sqrt{289} = 17\).

Step2: Set up the equation using the area method

Let the radius of the semicircle be \(r\). The area of the isosceles triangle can be calculated in two ways. One way is \(\frac{1}{2}\times16\times15=120\). Another way is to consider the area of the large triangle as the sum of the area of the two small right - angled triangles (formed by the radius and the waist) and the area of the semicircle? No, actually, we can also use the fact that the area of the triangle can be expressed as the sum of the area of the triangle above the semicircle and the area of the semicircle? Wait, a better way: draw a line from the vertex of the triangle to the center of the semicircle. The distance from the vertex to the center is \(15 - r\). The two small right - angled triangles (with the waist as the hypotenuse, the radius as one leg, and the distance from the center to the vertex as the other leg? No, the correct way is: the area of the isosceles triangle is equal to the sum of the areas of two right - angled triangles with the waist as the base and the radius as the height. The area of each right - angled triangle is \(\frac{1}{2}\times17\times r\), and there are two of them, so the sum of their areas is \(2\times\frac{1}{2}\times17\times r=17r\). But also, the area of the isosceles triangle is \(\frac{1}{2}\times16\times15 = 120\). Wait, no, actually, the correct model is: the large isosceles triangle can be thought of as composed of two congruent right - angled triangles (when we draw the height) and the semicircle? No, let's use the formula of the area of a triangle in terms of in - radius (but this is a semicircle, not a circle). Another approach: the distance from the center of the semicircle to each waist is equal to the radius \(r\). The area of the isosceles triangle \(S=\frac{1}{2}\times\) perimeter of the triangle (related to the tangent) \(\times r\)? No, for a circle inscribed in a triangle, the area \(S = r\times s\), where \(s\) is the semi - perimeter. But here it is a semicircle. Wait, let's consider the similar triangles. The large isosceles triangle has height \(15\) and base \(16\). The small triangle above the semicircle is also isosceles, with height \(15 - r\) and base \(2x\) (where \(x\) is the horizontal distance from the center of the semicircle to the waist). Since the two triangles are similar, we have the ratio of similarity: \(\frac{15 - r}{15}=\frac{x}{8}\) (because the base of the large triangle is \(16\), half - base is \(8\), and the base of the small triangle is \(2x\), half - base is \(x\)). Also, from the right - angled triangle formed by the radius \(r\), the segment of the waist, and the line from the center to the vertex, we know that by the Pythagorean theorem in the right - angled triangle with legs \(r\) and \(x\) and hypotenuse equal to the distance from the center to the vertex? No, the correct similar - triangle approach: the large triangle has height \(h = 15\), base \(b = 16\), waist \(l=17\). The small triangle (above the semicircle) has height \(h'=15 - r\), and its base \(b'\) is related to \(h'\) by the similarity of triangles. Since the two triangles are similar, \(\frac{h'}{h}=\frac{b'}{b}\), so \(b'=\frac{16(15 - r)}{15}\). Also, the distance from the center of the semicircle to the waist is \(r\), and we can use the formula for the distance from a point (center of the semicircle) to a line (the waist of the triangle). The equation of the waist: let's place the triangle in the coordinate system, with the base on the \(x\) - axis, center of the semicircle at the origin \((0,0)\), the left - hand waist passes through the points \((8,15)\) and \((- 8,0)\)? No, the left - hand waist passes through \((- 8,0)\) and \((0,15)\)? Wait, no, the base is from \((- 8,0)\) to \((8,0)\), and the vertex is at \((0,15)\). The equation of the left - hand waist (from \((- 8,0)\) to \((0,15)\)): the slope \(m=\frac{15 - 0}{0+8}=\frac{15}{8}\), so the equation is \(y=\frac{15}{8}(x + 8)\), that is \(15x-8y + 120 = 0\). The distance from the point \((0,r)\) (wait, no, the center of the semicircle is at \((0,0)\), and the semicircle is above the \(x\) - axis, so the center is \((0,0)\), and the top of the semicircle is at \((0,r)\)? No, the diameter is on the \(x\) - axis from \((- 8,0)\) to \((8,0)\), so the center is \((0,0)\), and the radius is \(r\), so the semicircle is \(y=\sqrt{r^{2}-x^{2}}\) (upper half). The distance from the center \((0,0)\) to the line \(15x - 8y+120 = 0\) is equal to the radius \(r\). The formula for the distance from a point \((x_0,y_0)\) to a line \(Ax+By + C = 0\) is \(d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^{2}+B^{2}}}\). So \(r=\frac{\vert15\times0-8\times0 + 120\vert}{\sqrt{15^{2}+(-8)^{2}}}=\frac{120}{17}\). Wait, no, the distance from the center of the semicircle (which is at \((0,0)\)) to the waist (the line \(15x - 8y+120 = 0\)) should be equal to the radius? Wait, the center of the semicircle is on the \(x\) - axis (since the diameter is on the \(x\) - axis), so the center is \((0,0)\), and the semicircle is \(y=\sqrt{r^{2}-x^{2}}\) with \(y\geq0\). The distance from the point \((0,0)\) to the line \(15x-8y + 120 = 0\) is \(d=\frac{\vert120\vert}{\sqrt{15^{2}+8^{2}}}=\frac{120}{17}\). But the distance from the center of the semicircle to the waist should be equal to the radius \(r\), so \(r=\frac{120}{17}\). Wait, let's check with the area method. The area of the isosceles triangle is \(\frac{1}{2}\times16\times15 = 120\). The area of the triangle can also be calculated as the sum of the areas of two triangles with the waist as the base and the radius as the height. The length of the waist is \(17\), and the sum of the areas of the two triangles is \(2\times\frac{1}{2}\times17\times r=17r\). But also, the area of the isosceles triangle is equal to the area of the triangle above the semicircle plus the area of the semicircle? No, that's wrong. The correct area method: when we draw the radius to the point of tangency on the waist, the radius is perpendicular to the waist. So the area of the isosceles triangle is equal to the sum of the areas of two right - angled triangles (each with the waist as the hypotenuse, the radius as one leg, and the segment from the vertex to the point of tangency as the other leg) plus the area of the semicircle? No, a simpler way: the area of the isosceles triangle is \(\frac{1}{2}\times\) base \(\times\) height \(=\frac{1}{2}\times16\times15 = 120\). Also, the area of the isosceles triangle can be expressed as the sum of the areas of two triangles with the waist as the base and the radius as the height. The length of each waist is \(17\), and there are two waists, so the area is \(2\times\frac{1}{2}\times17\times r=17r\). Wait, this is wrong because the two triangles with the waist as the base and the radius as the height do not cover the entire isosceles triangle. The correct model is: the large isosceles triangle is composed of a smaller isosceles triangle (above the semicircle) and the region of the semicircle? No, the correct way is to use the similarity of triangles. The smaller triangle above the semicircle has a height of \(15 - r\) and a base of \(2x\). Since the two triangles (the large one and the small one above the semicircle) are similar, we have \(\frac{15 - r}{15}=\frac{x}{8}\). Also, the distance from the center of the semicircle to the waist is \(r\), and in the right - angled triangle formed by the radius, the segment from the center to the vertex, and the segment from the center to the waist, we have \(\frac{r}{15 - r}=\frac{8}{17}\) (because in the large right - angled triangle with legs \(8\) and \(15\) and hypotenuse \(17\), the ratio of the leg \(8\) to the hypotenuse \(17\) is equal to the ratio of the radius \(r\) to the distance from the vertex to the center \(15 - r\)). Cross - multiplying gives \(17r=8\times(15 - r)\).

Step3: Solve the equation \(17r = 120-8r\)

Add \(8r\) to both sides of the equation: \(17r+8r=120\), that is \(25r = 120\)? No, wait, \(17r=120 - 8r\), then \(17r + 8r=120\), \(25r=120\)? No, that's wrong. Wait, from the similar triangles: the large triangle has height \(15\), base \(16\), the small triangle above the semicircle has height \(15 - r\), and the base of the small triangle: the distance from the center of the semicircle to the waist is \(r\), and using the formula for the distance from a point to a line, we have the correct equation from the distance formula. The line of the waist: passing through \((- 8,0)\) and \((0,15)\), the equation is \(y=\frac{15}{8}(x + 8)\), or \(15x-8y + 120 = 0\). The center of the semicircle is at \((0,0)\), and the distance from \((0,0)\) to the line \(15x-8y + 120 = 0\) is \(d=\frac{\vert15\times0-8\times0 + 120\vert}{\sqrt{15^{2}+(-8)^{2}}}=\frac{120}{17}\). Since the distance from the center of the semicircle to the waist is equal to the radius \(r\) (because the semicircle is tangent to the waist), so \(r=\frac{120}{17}\).

Answer:

\(\frac{120}{17}\) (or \(120/17\))