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10.067. Inside a circle with a radius of 15 cm, a point $M$ is taken at a distance of 13 cm from the center. Through point $M$, a chord of length 18 cm is drawn. Find the lengths of the segments into which point $M$ divides the chord. | Solution.
Draw $O C \perp A B$ (Fig. 10.66). Then $C B=\frac{1}{2} A B=9$ cm. From $\triangle O B C$
Fig. 10.67
, we have $A C^{2}: x^{2}=S:(S / 2)$, where $S$ is the area of $\triangle A B C$. Hence, $A C^{2}=2 x^{2}$ and $x=18 \sqrt{2}$ (cm).
Answer: $18 \sqrt{2}$ cm. | 18\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,187 |
10.069. The radius of the circle circumscribed around a right-angled triangle is $15 \mathrm{cm}$, and the radius of the circle inscribed in it is 6 cm. Find the sides of the triangle. | ## Solution.
By the condition $A B=2 R=2 \cdot 15=30$ (cm) (Fig. 10.68). The radius of the inscribed circle $r=\frac{a+b-c}{2}$. Hence, $2 r=a+b-c$. Solving the system
$$
\left\{\begin{array}{l}
2 r = a + b - c, \\
a^{2} + b^{2} = c^{2}
\end{array} \Leftrightarrow \left\{\begin{array}{l}
12 = a + b - 30 \\
a^{2} + b^... | 18;24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,188 |
10.070. A circle is inscribed in a circular sector with a central angle of $120^{\circ}$. Find the radius of the inscribed circle if the radius of the given circle is $R$. | Solution.
$\Delta O_{2} A O_{1}$ is a right triangle, $\angle O_{1} O_{2} A=60^{\circ}, \sin 60^{\circ}=\frac{O_{1} A}{O_{2} O_{1}}=\frac{r}{R-r} ;$ $\frac{\sqrt{3}}{2}=\frac{r}{R-r}$ (Fig. 10.69). From this, $r=\frac{\sqrt{3} R}{2+\sqrt{3}}=\sqrt{3} R(2-\sqrt{3})$.
Answer: $\sqrt{3} R(2-\sqrt{3})$. | \sqrt{3}R(2-\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,189 |
10.072. In an isosceles triangle, the base is $30 \mathrm{~cm}$, and the lateral side is 39 cm. Determine the radius of the inscribed circle. | Solution.
Draw $B D \perp A C$ (Fig. 10.70); since $\triangle A B C$ is isosceles, $B D$ is also a median. We have $B D^{2}=A B^{2}-A D^{2}$, from which $B D=\sqrt{39^{2}-15^{2}}=36$ (cm) and, therefore, $S=\frac{1}{2} \cdot 30 \cdot 36=540\left(\mathrm{~cm}^{2}\right)$. But $S=p r=$ $=54 r$, from which $r=10 \mathrm{... | 10\mathrm{~} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,190 |
10.073. In a square with a side of 12 cm, the midpoints of its adjacent sides are connected to each other and to the opposite side of the square. Find the radius of the circle inscribed in the resulting triangle. | Solution.
Let's find the area of triangle $NMD$ (Fig. 10.71); $S_{\triangle NMD}=$ $=S_{ABCD}-2 S_{\triangle MCD}-S_{\triangle NBM} ; S_{\triangle NBM}=0.5 \cdot 6 \cdot 6=18\left(\mathrm{~cm}^{2}\right) ; S_{\triangle MCD}=0.5 \cdot 6 \cdot 12=$ $=36\left(\mathrm{~cm}^{2}\right) ; S_{\triangle NMD}=144-2 \cdot 36-18=... | 2\sqrt{5}-\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,191 |
10.074. One of the two parallel lines is tangent to a circle of radius $R$ at point $A$, while the other intersects this circle at points $B$ and $C$. Express the area of triangle $A B C$ as a function of the distance $x$ between the lines. | ## Solution.
Let the distance between the parallel lines be $x$ (Fig. 10.72); then the area $S$ of triangle $ABC$ is $0.5 BC \cdot x$. Since $AM \perp BC$, then $BM = MC$ and $BM \cdot MC = AM \cdot MD = x(2R - x)$. Therefore, $0.25 BC^2 = x(2R - x)$, from which $S = 0.5 x \cdot 2 \sqrt{2Rx - x^2} = x \sqrt{2Rx - x^2}... | x\sqrt{2Rx-x^2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,192 |
10.076. Find the ratio of the radius of the circle inscribed in an isosceles right triangle to the height drawn to the hypotenuse. | ## Solution.
Since $\triangle A B C$ is an isosceles right triangle, the height $C D$ is also the bisector, i.e., $\angle D C A=\angle A=45^{\circ}$ (Fig. 10.74); therefore, $A D=D C$ and $A C=\sqrt{2 D C}$. But $A C=A K+K C=D C+r(A K=A D$ as tangents drawn from the same point), hence $r=\sqrt{2} D C-D C$, i.e., $r / ... | \sqrt{2}-1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,194 |
10.079. Find the bisectors of the acute angles of a right triangle with legs of 24 and $18 \mathrm{~cm}$. | Solution.
$$
\left.\frac{DC}{AC}=\frac{BD}{AB} \text { (fig. } 10.77\right) ; AC=\sqrt{24^{2}+18^{2}}=30(\mathrm{~cm}), \frac{x}{30}=\frac{24-x}{18} \cdot \text { Hence }
$$
here $x=15, CD=15($ cm $), BD=24-15=9$ (cm).
The bisector of $\angle BAC: AD=\sqrt{AC \cdot AB-CD \cdot BD}=\sqrt{30 \cdot 18-15 \cdot 9}=9 \sq... | 9\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,197 |
10.081. The area of a rectangle is 9 cm $^{2}$, and the measure of one of the angles formed by the diagonals is $120^{\circ}$ (Fig. 10.79). Find the sides of the rectangle. | ## Solution.
$C D=x \cdot \triangle O C D$ - equilateral; $A O=O C=C D=x ; A C=2 x$. The area of the rectangle $S=\frac{1}{2} d_{1} d_{2} \sin 60=\frac{1}{2} 2 x \cdot 2 x \cdot \frac{\sqrt{3}}{2}, 9=x^{2} \sqrt{3}$. Therefore, $x=\frac{3}{\sqrt[4]{3}}=\frac{3^{4 / 4}}{3^{1 / 4}}=3^{3 / 4}=\sqrt[4]{27}$ (cm). $A C=2 \... | \sqrt[4]{27} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,199 |
10.083. The sum of the lengths of the diagonals of a rhombus is $m$, and its area is $S$. Find the side of the rhombus. | Solution.
$A C+B D=m ; A C=x, B D=y . S=\frac{1}{2} d_{1} d_{2}$ (Fig. 10.081). Solving the system $\left\{\begin{array}{l}x+y=m, \\ 2 S=x y,\end{array}\right.$ we get $x_{1}=\frac{m+\sqrt{m^{2}-8 S}}{2}, y_{1}=\frac{m-\sqrt{m^{2}-8 S}}{2}$ or $x_{2}=\frac{m-\sqrt{m^{2}-8 S}}{2}, y_{2}=\frac{m+\sqrt{m^{2}-8 S}}{2}$.
... | \frac{\sqrt{^{2}-4S}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,201 |
10.084. The perimeter of a rhombus is 2 m, the lengths of its diagonals are in the ratio $3: 4$. Find the area of the rhombus. | Solution.
The desired area will be found using the formula $S=\frac{1}{2} A C \cdot B D=2 A O \cdot O B$ (Fig.
10.81). In $\triangle A O B, A B^{2}=A O^{2}+B O^{2}$, where $O B=\frac{3}{4} A O, A B=\frac{2}{4}=\frac{1}{2}$ (m). Then we get the equation $\frac{1}{4}=A O^{2}+\frac{9}{16} A O^{2}$, from which $A O^{2}=\... | 0.24\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,202 |
10.085. A circle of radius $R$ is inscribed in an isosceles trapezoid. The upper base of the trapezoid is half the height of the trapezoid. Find the area of the trapezoid. | Solution.
$C D=y ; E D=x$ (Fig. 10.82). We solve the system $\left\{\begin{array}{l}2 y=(R+2 x)+R, \\ x^{2}+4 R^{2}=y^{2} .\end{array}\right.$ From this, we get $x=\frac{3}{2} R$, then $A D=R+2 x=R+3 R=4 R$. The required area $S=\frac{1}{2}(A D+B C) \cdot E C=\frac{1}{2}(R+4 R) \cdot 2 R=5 R^{2}$.
Answer: $5 R^{2}$. | 5R^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,203 |
10.087. A square of unit area is inscribed in an isosceles triangle, one side of which lies on the base of the triangle. Find the area of the triangle, given that the centers of mass of the triangle and the square coincide (the center of mass of the triangle lies at the intersection of its medians). | Solution.
$B D$ is the median, $B O: O D=2: 1 ; O D=0.5$ (cm), $B O=1($ cm $), B D=$ $=O D+B O=1.5$ (cm) (Fig.10.84). $\triangle B E F$ and $\triangle B D C$ are similar, $\frac{B D}{B E}=\frac{D C}{E F}$. Therefore, $D C=\frac{B D \cdot E F}{B E}, D C=\frac{1.5 \cdot 0.5}{0.5}=1.5(\mathrm{~cm}), A C=2 \times$ $\times... | \frac{9}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,205 |
10.088. A trapezoid is inscribed in a circle of radius $R$, with the lower base being twice as long as each of the other sides. Find the area of the trapezoid. | ## Solution.
$A D=2 x=2 R$ (Fig.10.85). From this, $x=R$. The required area $S=\frac{B C+A D}{2} \cdot C E, C E=\sqrt{C D^{2}-E D^{2}}=\sqrt{x^{2}-\frac{x^{2}}{4}}=\frac{\sqrt{3} x}{2}=\frac{\sqrt{3}}{2} R ;$ $S=\frac{R+2 R}{2} \cdot \frac{\sqrt{3}}{2} R=\frac{3 \sqrt{3}}{4} R^{2}$.
Answer: $\frac{3 \sqrt{3}}{4} R^{2... | \frac{3\sqrt{3}}{4}R^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,206 |
10.089. Find the area of the circle circumscribed around an isosceles triangle if the base of this triangle is 24 cm and the lateral side is $13 \mathrm{~cm}$. | Solution.
The area of a circle $S=\pi R^{2}$ (Fig.10.86). The radius of the circumscribed circle $R=\frac{a b c}{4 S_{\triangle A B C}}$, where $S_{\triangle A B C}=\frac{1}{2} A C \cdot B D$.
From $\triangle B D C$ we get $B D=\sqrt{B C^{2}-D C^{2}}=\sqrt{13^{2}-12^{2}}=5$ (cm). Then $S_{\triangle A B C}=\frac{1}{2}... | 285.61\pi\mathrm{}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,207 |
10.090. The distance from the center of the circle to a chord of length 16 cm is 15 cm. Find the area of the triangle circumscribed around the circle, if the perimeter of the triangle is $200 \mathrm{~cm}$. | ## Solution.
Radius of the circle from $\triangle O C B: R=\sqrt{15^{2}+8^{2}}=17$ (cm). The required area of the triangle $S_{\triangle D K L}=R \cdot p=17 \cdot \frac{200}{2}=1700$ (cm²) (Fig. 10.87).
Answer: $1700 \mathrm{~cm}^{2}$. | 1700\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,208 |
10.091. Find the area of a circle inscribed in an isosceles trapezoid, if its larger base is equal to $a$, and the angle at the smaller base is $120^{\circ}$. | Solution.
In $\triangle C E D, \angle C D E=60^{\circ}$ and $\cos 60=\frac{a-b}{2 c}$ (Fig. 10.88). Hence, $\frac{a-b}{2 c}=\frac{1}{2}$, $a-b=c$. For the circumscribed isosceles trapezoid, $a+b=2 c$.
Solving the system $\left\{\begin{array}{l}a-b=c, \\ a+b=2 c,\end{array}\right.$ we get $c=\frac{2}{3} a . \sin 60=\f... | \frac{\pi^{2}}{12} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,209 |
10.092. A triangle with angles of 15 and $60^{\circ}$ is inscribed in a circle of radius $R$ (Fig. 10.89). Find the area of the triangle. | Solution.
$$
\angle ABC = 105^{\circ}; \angle ABC = 180^{\circ} - \frac{1}{2} \angle AOC. \text{ Hence, } \angle AOC = 150^{\circ}. \text{ Then, }
$$
Fig. 10.89
; $A B=x, B C=y$. It is known that $2 p=x+y+c$. Therefore, $x+y=2 p-c$. The radius of the inscribed circle $r=\frac{x+y-c}{2}=\frac{2 p-c-c}{2}=p-c$. The area of the circle is found using the formula $S=\pi r^{2}=\pi(p-c)^{2}$.
Answer: $\pi(p-c)^{2}$. | \pi(p-)^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,211 |
10.094. Find the area of a circle inscribed in a right triangle (Fig. 10.90), if the projections of the legs on the hypotenuse are 9 and 16 m. | Solution.
$$
A C=A D+D C=25\left(\text { m); } \triangle A D B \text { and } \triangle A B C \text { - similar: } \frac{A D}{A B}=\frac{A B}{A C} .\right) \text { Hence }
$$
from $\frac{9}{x}=\frac{x}{25}, x^{2}=9 \cdot 25=225, x=\sqrt{225}=15($ m), $A B=15$ (m). $\triangle B D C$ and $\triangle A B C$ are similar: $... | 25\pi\mathrm{M}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,212 |
10.095. The area of an isosceles triangle is equal to $1 / 3$ of the area of the square constructed on the base of the given triangle. The lengths of the lateral sides of the triangle are shorter than the length of its base by 1 cm. Find the lengths of the sides and the height of the triangle, drawn to the base. | ## Solution.
According to the condition, $B C^{2}=3 \cdot \frac{1}{2} B C \cdot A H$ (Fig. 10.91) or $A H=\frac{2}{3} B C$. But $A H^{2}=A B^{2}-\left(\frac{1}{2} B C\right)^{2}$, so $\frac{4}{9} B C^{2}=A B^{2}-\frac{1}{4} B C^{2}$ or $A B^{2}=\frac{25}{36} B C^{2}$, i.e., $A B=\frac{5}{6} B C$. Then we get $A B=\fra... | 5;6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,213 |
10.096. The area of an isosceles trapezoid circumscribed around a circle is $32 \sqrt{3} \mathrm{~cm}^{2}$ (Fig. 10.92). Determine the lateral side of the trapezoid, given that the acute angle at the base is $\pi / 3$. | Solution.
$\triangle C E D-$ is a right triangle and $\sin 60^{\circ}=\frac{2 R}{C D}$. Therefore, $2 R=\frac{\sqrt{3}}{2} C D$, $C D=x \Rightarrow 2 R=\frac{\sqrt{3}}{2} x$. The area of the trapezoid $S=\frac{a+b}{2} h=2 R x=\frac{\sqrt{3}}{2} x^{2}$. According to the condition $\frac{\sqrt{3}}{2} x^{2}=32 \sqrt{3} \... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,214 |
10.097. The area of a right-angled triangle (Fig. 10.93) is $2 \sqrt{3}$ cm $^{2}$. Determine its height drawn to the hypotenuse, if it divides the right angle in the ratio $1: 2$. | Solution.
$\angle A B D=30^{\circ}, \angle D B C=60^{\circ}$. The area of $\triangle A B C$ is found using the formula $S=\frac{1}{2} A B \cdot B C$. From $\triangle B D C$ we get $B C=\frac{D C}{\sin 60^{\circ}}$. From $\triangle A D B$ we find $A B=\frac{A D}{\sin 30^{\circ}}$. Then $S=\frac{1}{2} \frac{D C \cdot A ... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,215 |
10.098. A line parallel to the base of a triangle divides it into parts whose areas are in the ratio of $2: 1$. In what ratio, counting from the vertex, does it divide the lateral sides? | Solution.
1. $S_{\triangle A B C}=S_{A M N C}+S_{\triangle M B N}=x+2 x=3 x ; \frac{S_{\triangle A B C}}{S_{\triangle B M N}}=\frac{3}{2}$ (Fig. 10.94). From this, $\left(\frac{M B}{A B}\right)^{2}=\frac{2}{3}, \frac{M B}{A B}=\sqrt{\frac{2}{3}}, A B=\sqrt{3} x, M B=\sqrt{2} x ; A M=(\sqrt{3}-\sqrt{2}) x$, $\frac{M B}... | (\sqrt{6}+2):1or(\sqrt{3}+1):2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,216 |
10.099. The area of an isosceles trapezoid (Fig. 10.95) circumscribed about a circle is $8 \mathrm{~cm}^{2}$. Determine the sides of the trapezoid if the angle at the base is $30^{\circ}$. | Solution.
From $\triangle C E D$ we get $\sin 30^{\circ}=\frac{2 R}{c}$. Hence $2 R=\frac{c}{2}$. The area of the trapezoid $S=2 R c=\frac{c^{2}}{2}=8$. From this, $c^{2}=16, c=4$ (cm), $C D=A B=4$ (cm). We have $\cos 30^{\circ}=\frac{E D}{C D}, E D=2 \sqrt{3}$ (cm). Solving the system $\left\{\begin{array}{l}a+b=4, \... | 4-2\sqrt{3};4+2\sqrt{3};4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,217 |
10.100. A regular hexagon $A B C D E F$ consists of two trapezoids sharing a common base $C F$ (Fig. 10.96). It is known that $A C=13$ cm, $A E=10$ cm. Find the area of the hexagon. | Solution.
$\triangle C K A$ is a right triangle, $A K=\frac{1}{2} A E=5$ (cm), $C K=\sqrt{A C^{2}-A K^{2}}=$ $=\sqrt{13^{2}-5^{2}}=12(\text{cm}), \cos \varphi=\frac{12}{13} ; \sin \varphi=\frac{5}{13}$. Consider $\triangle C O B(\angle C O B=$ $\left.=90^{\circ}\right): \sin 2 \varphi=\frac{5}{x}$ and $\sin 2 \varphi=... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,218 |
10.101. Find the area of a regular triangle (Fig. 10.97) inscribed in a square with side $a$, given that one of the vertices of the triangle coincides with a vertex of the square. | Solution.
$CD=a; BE=FD=x; AF=AE=a-x; CE=EF=CF=y. \mathrm{Pe}-$ solving the system $\left\{\begin{array}{l}y^{2}=a^{2}+x^{2}, \\ y^{2}=2(a-x)^{2} ;\end{array} a^{2}+x^{2}=2(a-x)^{2}\right.$. Solving this equation, we get $x=a(2-\sqrt{3})$. Let's find $y^{2}=a^{2}+a^{2}(2-\sqrt{3})^{2}=a^{2}(8-4 \sqrt{3})$. The area of ... | ^{2}(2\sqrt{3}-3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,219 |
10.102. The diagonal of an isosceles trapezoid bisects its obtuse angle. The smaller base of the trapezoid is 3 cm, and the perimeter is $42 \mathrm{~cm}$. Find the area of the trapezoid. | Solution.
By the condition $\angle B C A=\angle A C D$ (Fig. 10.98). But $\angle B C A=\angle C A D$, and so
Fig. 10.99
.
Solve the system $\left\{\begin{array}{l}p=x+y+2 R, \\ S=\frac{1}{2} x y, \\ x^{2}+y^{2}=4 R^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}24=x+y+2 R, \\ 24=\frac{1}{2} x y, \\ x^{2}+y^{2}=4 R^{2},\end{array}\right.\right.$ we get $x=6$ (cm), $... | 25\pi\mathrm{}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,221 |
10.105. Find the area of an isosceles triangle (Fig. 10.101) with an angle of $120^{\circ}$, if the radius of the inscribed circle is $\sqrt[4]{12}$ cm. | ## Solution.
From $\triangle B D C$, we find $B D=B C \cdot \sin 30^{\circ}=\frac{B C}{2} ; D C=B C \cdot \sin 60^{\circ}=\frac{\sqrt{3}}{2} B C$. The semiperimeter of $\triangle A B C \quad p=\frac{1}{2}(x+x+\sqrt{3} x)=x\left(1+\frac{\sqrt{3}}{2}\right)$.
Area of $\triangle A B C: S=p r=x\left(1+\frac{\sqrt{3}}{2}\... | 2(7+4\sqrt{3})(\mathrm{}^{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,222 |
10.106. On the sides of an isosceles right triangle with hypotenuse outside this triangle, squares are constructed. The centers of these squares are connected to each other. Find the area of the resulting triangle. | Solution.
$B E=K C=\frac{c}{2} ; E F=2 B E=c$ (Fig. 10.102). Quadrilateral $A B C D$ is a square. Therefore, $A C=B D=c$. The area of the desired triangle is
$$
S=\frac{1}{2} B D \cdot E F=\frac{1}{2} c^{2}=\frac{c^{2}}{2}
$$
Answer: $\frac{c^{2}}{2}$.
. Then the area of the given square $S_{1}=(a+b)^{2}=a^{2}+2 a b+b^{2}=c^{2}+2 a b$. Since $\angle B L K=60^{\circ}$, then $\angle B K L=30^{\circ}$, from which $c=2 b, a=b \sqrt{3}$ and, therefore, $S_{1}=4 b^{2}+2 b^{2} \sqrt{3}$. The area of the inscribed squ... | 4-2\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,224 |
10.108. Find the area of a square inscribed in an equilateral triangle with side $a$. | ## Solution.
$\triangle B K C$ and $\triangle E M C$ are similar $\Rightarrow \frac{B C}{E C}=\frac{B K}{E M}, B K=\sqrt{B C^{2}-K C^{2}}=$. $=\sqrt{a^{2}-\frac{a^{2}}{4}}=\frac{a \sqrt{3}}{2} ; E M=x$ and $x=\frac{E C \cdot B K}{B C}=\frac{(a-x) a \sqrt{3}}{2 a}$. Therefore, $2 a x=$ $=a \sqrt{3}(a-x)$. Solving the e... | 3^{2}(7-4\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,225 |
10.109. On the sides of an equilateral triangle, squares are constructed outside it. Their vertices, lying outside the triangle, are sequentially connected. Determine the area of the resulting hexagon if the side of the given triangle is $a$. | Solution.
The area of the desired hexagon (Fig. 10.105)
$S_{H K D E F G}=S_{\triangle A B C}+3 S_{A H K B}+3 S_{\triangle K B D}=\frac{a^{2} \sqrt{3}}{4}+3 a^{2}+3 \cdot \frac{1}{2} a^{2} \sin 120^{\circ}=$ $=\frac{a^{2} \sqrt{3}}{4}+3 a^{2}+\frac{3}{2} \frac{a^{2} \sqrt{3}}{2}=a^{2}(3+\sqrt{3})$.
Answer: $\quad a^{... | ^{2}(3+\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,226 |
10.110. A square with side a is cut at the corners so that a regular octagon is formed. Determine the area of this octagon. | ## Solution.
Let $A E=x$ (Fig. 10.106). Then $A B=2 A E+E F$ or $2 x+x \sqrt{2}=a$, from which $x=\frac{a}{2+\sqrt{2}}=\frac{a(2-\sqrt{2})}{2}$. Therefore, the desired area is
$$
S=S_{A B C D}-4 S_{\triangle A E N}=a^{2}-\frac{4 x^{2}}{2}=a^{2}-\frac{4 a^{2}(4-4 \sqrt{2}+2)}{8}=2 a^{2}(\sqrt{2}-1)
$$
Answer: $\quad ... | 2^{2}(\sqrt{2}-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,227 |
10.111. The side of a regular triangle inscribed in a circle is equal to $a$. Calculate the area of the square inscribed in the same circle. | ## Solution.
Let the radius of the circumscribed circle be denoted by $R$. Then $a=R \sqrt{3}$, from which $R=a / \sqrt{3}$. Since the side of the inscribed square is $R \sqrt{2}$, its area $S=2 R^{2}=2 a^{2} / 3$.
Answer: $2 a^{2} / 3$. | \frac{2a^2}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,228 |
10.112. Calculate the ratio of the areas of a square, an equilateral triangle, and a regular hexagon inscribed in the same circle. | Solution.
Let $R$ be the radius of the circle. Then the side of the inscribed regular triangle is $a_{3}=R \sqrt{3}$ and $S_{3}=\frac{a_{3}^{2} \sqrt{3}}{4}=\frac{3 R^{2} \sqrt{3}}{4}$. Next, the side of the square is $a_{4}=R \sqrt{2}$ and $S_{4}=a_{4}^{2}=2 R^{2}$, and finally, the side of the inscribed regular hexa... | 8:3\sqrt{3}:6\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,229 |
10.113. The side of an equilateral triangle inscribed in a circle is equal to $a$. Calculate the area of the segment cut off by it. | ## Solution.
The area of segment $A n B$ is the difference between the areas of sector $A O B$ and $\triangle A O B$ (Fig. 10.107). We find $S_{\text {sect } A O B}=\frac{\pi R^{2}}{3}, S_{\triangle A O B}=\frac{1}{2} a \cdot \frac{R}{2}=\frac{a R}{4}$, hence $S=\frac{\pi R^{2}}{3}-\frac{a R}{4}$. Since $R=\frac{a}{\s... | \frac{^{2}(4\pi-3\sqrt{3})}{36} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,230 |
10.115. On the diameter $2 R$ of a semicircle, a regular triangle is constructed, the side of which is equal to the diameter. The triangle is located on the same side of the diameter as the semicircle. Calculate the area of that part of the triangle which lies outside the circle. | Solution.
$\triangle A O D, \triangle D O E, \triangle O E C, \triangle D B E$ are equilateral triangles with side $R$ (Fig. 10.108). The desired area $S=S_{\triangle D B E}-S_{\text {segm }} . S_{\text {segm }}=\frac{\pi}{6} R^{2}-\frac{\sqrt{3}}{4} R^{2} ;$ $S_{\triangle D B E}=\frac{\sqrt{3}}{4} R^{2}, S=\frac{\sqr... | R^{2}(\frac{3\sqrt{3}-\pi}{6}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,231 |
10.118. Determine the side of the rhombus, knowing that its area is $S$, and the lengths of the diagonals are in the ratio $m: n$. | Solution.
Let's write $B D=m x ; A C=n x$ (Fig. 10.111). The area of the rhombus $S=\frac{1}{2} m n x^{2}$, from which $x=\sqrt{\frac{2 S}{m n}} \cdot$ Then $B D=m \sqrt{\frac{2 S}{m n}}=\sqrt{\frac{2 S m}{n}} ; A C=\sqrt{\frac{2 S n}{m}}$. The side of the rhombus $B C^{2}=\frac{1}{4} B D^{2}+\frac{1}{4} A C^{2}==\fra... | \sqrt{\frac{S(^{2}+n^{2})}{2n}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,233 |
10.119. The perimeter of a rhombus is $2 p$; the lengths of the diagonals are in the ratio $m: n$. Calculate the area of the rhombus. | Solution.
Using Fig. 10.111, we have $B D=m x$,
Fig. 10.112 $A C=n x . B C^{2}=\left(\frac{B D}{2}\right)^{2}+\left(\frac{A C}{2}\right)^{2}=\left(\frac{m x}{2}\right)^{2}+$ $+\left(\frac{n... | \frac{np^{2}}{2(^{2}+n^{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,234 |
10.120. Two circles of radius $R$ with centers at points $O_{1}$ and $O_{2}$ touch each other. They are intersected by a line at points $A, B, C$ and $D$ such that $A B=B C=C D$. Find the area of the quadrilateral $O_{1} A D O_{2}$. | Solution.
$\triangle A O_{1} B, \triangle O_{1} B O, \triangle B O C, \triangle O C O_{2}, \triangle C O_{2} D$ - equilateral with side $R$ (Fig. 10.112). The area of $\triangle A O_{1} B \quad S_{\triangle A O, B}=\frac{R^{2} \sqrt{3}}{4}$. The area of the desired quadrilateral $O_{1} A D O_{2} \quad S=5 S_{\triangle... | \frac{5R^{2}\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,235 |
10.121. Calculate the area of a rectangular trapezoid if its acute angle is $60^{\circ}$, the smaller base is $a$, and the larger lateral side is $b$. | Solution.
The height of the trapezoid is $\frac{b \sqrt{3}}{2}$, and the larger base is $a+\frac{b}{2}$.
Therefore, its area $S=\frac{1}{2}\left(a+a+\frac{b}{2}\right) \frac{b \sqrt{3}}{2}=\frac{(4 a+b) b \sqrt{3}}{8}$.
Answer: $(4 a+b) b \sqrt{3} / 8$. | \frac{(4a+b)b\sqrt{3}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,236 |
10.122. The larger base of the trapezoid has a length of 24 cm (Fig. 10.113). Find the length of its smaller base, given that the distance between the midpoints of the diagonals of the trapezoid is 4 cm. | Solution.
$A E=E C ; D F=F B, K E=D C / 2 ; F T=D C / 2 ; D C=x, K T=2 K E+E F=$ $=x+4$. On the other hand, $K T=\frac{x+24}{2}$. Solving the equation $x+4=\frac{x+24}{2}$, we get $x=16$ (cm).
Answer: $16 \mathrm{~cm}$. | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,237 |
10.123. The area of an isosceles trapezoid circumscribed around a circle is $S$. Determine the lateral side of the trapezoid, given that the acute angle at the base is $\pi / 6$. | ## Solution.
Let $x$ be the length of the lateral side; then the height of the trapezoid is $\frac{1}{2} x$. Since the trapezoid is circumscribed around a circle, the sum of its bases is equal to the sum of the lateral sides. Therefore, the area of the trapezoid $S=\frac{1}{2} \cdot 2 x \cdot \frac{1}{2} x$, from whic... | \sqrt{2S} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,238 |
10.124. A trapezoid is divided by its diagonals into four triangles (Fig. 10.114). Prove that the triangles adjacent to the lateral sides are equal in area.
Fig. 10.114
 and \( n \). Determine the area of the trapezoid. | Solution.
Let $A K=m, K B=n$ (Fig. 10.117). Then $K B=B M=M C=n$, $A K=A N=N D=m$.
We find the height of the trapezoid:
$h=B H=\sqrt{A B^{2}-A H^{2}}=\sqrt{(m+n)^{2}-(m-n)^{2}}=2 \sqrt{m n}$.
Thus, $S=\frac{1}{2}(B C+A D) h=(m+n) h=2 \sqrt{m n}(m+n)$.
Answer: $\quad 2 \sqrt{m n}(m+n)$ | 2\sqrt{n}(+n) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,243 |
10.130. The side of a square inscribed in a circle cuts off a segment, the area of which is equal to $(2 \pi-4) \mathrm{cm}^{2}$. Find the area of the square. | Solution.
Let $R$ be the radius of the circle: Then the area of the sector is $\frac{\pi R^{2}}{4}$, and
Fig. 10.118
the area of the triangle is $\frac{R^{2}}{2}$, and therefore, the area ... | 16\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,244 |
10.131. A circle with an area of $Q$ is inscribed in a rhombus with an acute angle of $30^{\circ}$. Find the area of the rhombus. | ## Solution.
Draw radii $O K, O L, O M, O N$ to the points of tangency (Fig. 10.118). $\angle A B C=180^{\circ}-\angle B A D=180^{\circ}-30^{\circ}=150^{\circ}$. Since the diagonals of a rhombus are angle bisectors, $\angle A B O=\frac{150^{\circ}}{2}=75^{\circ}$. Therefore, $B O=\frac{K O}{\sin 75^{\circ}}=\frac{K O}... | \frac{8Q}{\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,245 |
10.132. A circle is inscribed in a circular sector, the arc of which contains $60^{\circ}$. Find the ratio of the area of this circle to the area of the sector. | ## Solution.
Let $B, D, E$ be the points of tangency. Denote the radius of the circle from which the sector is cut as $R$, and the radius of the inscribed circle as $r$ (Fig. 10.119). Thus, $B O=D O=E O=r$, and $B O=R . O O=R-r$. $D O=O O \sin 30^{\circ}$. This means that $r=\frac{1}{2}(R-r)$, from which $R=3 r$. The ... | 2:3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,246 |
10.133. A tangent of length $2a$ is drawn from a point $M$, which is at a distance $a$ from the circle, to this circle. Find the area of a regular hexagon inscribed in the circle. | ## Solution.
Draw the radius $O A$ to the point of tangency (Fig. 10.120) and denote the radius of the circle by $r$. Then in $\triangle O A M$ we have $(2 a)^{2}+r^{2}=(a+r)^{2}$ or $4 a^{2}+r^{2}=a^{2}+2 a r+r^{2}$, from which $r=\frac{3 a}{2}$. Thus,
$$
\begin{aligned}
& S=\frac{6 r^{2} \sqrt{3}}{4}=\frac{27 a^{2}... | \frac{27^{2}\sqrt{3}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,247 |
10.134. In an isosceles trapezoid, one base is equal to $40 \mathrm{~cm}$, and the other is 24 cm. The diagonals of this trapezoid are perpendicular to each other. Find its area. | ## Solution.
The area of an isosceles trapezoid, whose diagonals are perpendicular to each other, is equal to the square of its height: $S=h^{2}$. On the other hand, $S=\frac{a+b}{2} h$, from which $h=\frac{a+b}{2}=\frac{40+24}{2}=32$ (cm). And the area
$S=32^{2}=1024\left(\mathrm{~cm}^{2}\right)$.
Answer: $1024 \ma... | 1024\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,248 |
10.137. A chord $AB$ of constant length slides with its ends along a circle of radius $R$. A point $C$ on this chord, located at distances $a$ and $b$ from the ends $A$ and $B$ of the chord, describes a circle when the chord completes a full revolution. Calculate the area of the annulus enclosed between the given circl... | ## Solution.
$\triangle A O B$ is isosceles (Fig. 10.123). Therefore, $\angle B A O = \angle A B O = \alpha$. Let $O C = r$. By the cosine theorem, $r^{2} = R^{2} + a^{2} - 2 a R \cos \alpha$. On the other hand, $r^{2} = R^{2} + b^{2} - 2 b R \cos \alpha$. Equating these, we get $a^{2} - b^{2} = 2 R \cos \alpha (a - b... | \pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,249 |
10.138. Three equal circles of radius $r$ touch each other pairwise. Calculate the area of the figure located outside the circles and bounded by their arcs between the points of tangency. | ## Solution.
The segments connecting the centers of the circles (Fig. 10.124) form an equilateral triangle with side length $2r$. Its area is: $S_{1}=\frac{4 r^{2} \sqrt{3}}{4}=r^{2} \sqrt{3}$. The area of one sector $S_{2}=\frac{\pi r^{2} \cdot 60^{\circ}}{360^{\circ}}=\frac{\pi r^{2}}{6}$. Since there are three sect... | \frac{r^{2}(2\sqrt{3}-\pi)}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,250 |
10.139. Semicircles are described on the sides of a rhombus, turned inward. Determine the area of the resulting rosette if the diagonals of the rhombus are equal to $a$ and $b$. | ## Solution.
The area of the obtained rosette will be equal to the sum of the areas of four shaded (Fig. 10.125) parts. The area of each shaded part is equal to the difference between the area of a semicircle and the area of a right triangle with sides $\frac{a}{2}$ and $\frac{b}{2}$. The side of the rhombus can be fo... | \frac{\pi(^{2}+b^{2})-4}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,251 |
10.140. Prove that if lines parallel to the diagonals of a quadrilateral are drawn through its vertices, then the area of the parallelogram determined by these lines is twice the area of the given quadrilateral. | Solution.
According to the condition (Fig. 10.126), $K L\|A C\| N M, K N\|B D\| L M$. The area of parallelogram $K L M N$ is
Fig. 10.126
. Draw $B L \perp A D, C M \perp A D$. $S_{A B C D}=\frac{B C+A D}{2} B L$. Therefore, $B L=\frac{2 S_{A B C D}}{B C+A D}=4$ (cm). According to the condition, $A B=C D, B L=C M$. Thus, $A L=M D=\frac{A D-L M}{2}=3$ (cm). Then $A B=\sqrt{\dot{B} L^{2}+A L^{2}}=\sqrt{16+9}=5(\m... | 5\mathrm{~} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,253 |
10.142. A circle is inscribed in an equilateral triangle, and a regular hexagon is inscribed in the circle. Find the ratio of the areas of the triangle and the hexagon. | ## Solution.
Let the side of the equilateral triangle be $a$. Then its area $S_{1}=\frac{a^{2} \sqrt{3}}{4} \cdot$ The radius of the circle inscribed in the triangle, $r=\frac{a \sqrt{3}}{6}$.
. By the cosine theorem,
$A B^{2}=R_{1}^{2}+R_{1}^{2}-2 R_{1} R_{1} \cos 60^{\circ}=R_{1}^{2}$.
On the other hand, $A B^{2}=R_{2}^{2}+R_{2}^{2}-2 R_{2} R_{2} \cos 120^{\circ}=3 R_{2}^{2}$. Therefore, $R_{1}^... | 3:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,255 |
10.145. The height of the rhombus is $12 \mathrm{~cm}$, and one of its diagonals is $15 \mathrm{~cm}$. Find the area of the rhombus. | Solution.
Let $B K=12$ (cm), $B D=15$ (cm) (Fig. 10.130). From $\triangle B K D$ $\sin \alpha=\frac{B K}{B D}=\frac{12}{15}=\frac{4}{5} ;$ since $0<\alpha<90^{\circ}$, then $\cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$.
From $\triangle A O D \cos \alpha=\frac{O D}{A D}=\frac{B D}{2 A D}$,... | 150\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,257 |
10.146. The length of the height dropped to the base of an isosceles triangle is 25 cm, and the radius of the inscribed circle is $8 \mathrm{~cm}$. Find the length of the base of the triangle.
. Then $B K=25$ (cm), $O K=O L=O M=8$ (cm). Therefore,
$$
\begin{aligned}
& B O=B K-O K=17(\mathrm{~cm}) \\
& B M=\sqrt{B O^{2}-O M^{2}}=\sqrt{289-64}=15(\mathrm{~cm})
\end{aligned}
$$
Then $\operatorname{tg} \alpha=\frac{O M}{B M}=\frac{8}{15}$. On the ... | \frac{80}{3}\mathrm{~} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,258 |
10.147. In a parallelogram with a perimeter of 32 cm, the diagonals are drawn. The difference between the perimeters of two adjacent triangles is 8 cm. Find the lengths of the sides of the parallelogram. | Solution.
Let the sides of the parallelogram be denoted by $a$ and $b$. Then its perimeter is $p=2(a+b)=32$. The perimeter of one triangle is $p_{1}=b+\frac{d_{1}}{2}+\frac{d_{2}}{2}$, and the second is $p_{2}=a+\frac{d_{1}}{2}+\frac{d_{2}}{2}$, where $d_{1}, d_{2}$ are the diagonals of the parallelogram. The differen... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,259 |
10.148. Find the area of an isosceles trapezoid if its height is equal to $h$, and the lateral side is seen from the center of the circumscribed circle at an angle of $60^{\circ}$. | Solution.
Since the central angle $C O D$ is equal to $60^{\circ}$ (Fig. 10.132), the inscribed angle $C A D$ is equal to $30^{\circ}$. Therefore, $h=\frac{1}{2} A C$ and from $\triangle A K C$ we get $A K=\sqrt{A C^{2}-C K^{2}}=h \sqrt{3}$. We find the area of the trapezoid:
$$
S=\frac{1}{2}(B C+A D) h=(A E+E K) h=A... | ^{2}\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,260 |
10.149. A circle with radius $R$ is divided into two segments by a chord equal to the side of the inscribed square. Determine the area of the smaller of these segments. | ## Solution.
Let the side of the inscribed square be denoted by $a$. The diameter of the circle is its diagonal. This means that $2 a^{2}=(2 R)^{2}$, from which $a=R \sqrt{2}$. The area of the circle $S_{1}=\pi R^{2}$. The area of the square $S_{2}=a^{2}=2 R^{2}$. The area of the smaller segment
$$
\begin{aligned}
& ... | \frac{R^{2}(\pi-2)}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,261 |
10.150. Determine the area of the circular ring enclosed between two concentric circles, the lengths of which are $C_{1}$ and $C_{2}\left(C_{1}>C_{2}\right)$. | ## Solution.
The circumference of the circles $C_{1}=2 \pi R_{1}, C_{2}=2 \pi R_{2}$, hence $R_{1}=\frac{C_{1}}{2 \pi}, R_{2}=\frac{C_{2}}{2 \pi}$. The area of the larger circle $S_{1}=\pi R_{1}^{2}=\frac{\pi C_{1}^{2}}{4 \pi^{2}}=\frac{C_{1}^{2}}{4 \pi} ; S_{2}=\frac{C_{2}^{2}}{4 \pi}$. The area of the ring $S=S_{1}-... | \frac{C_{1}^{2}-C_{2}^{2}}{4\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,262 |
10.151. A circle is divided into two segments by a chord equal to the side of an inscribed regular triangle. Determine the ratio of the areas of these segments. | Solution.
Let $r$ be the radius of the circle, and $S_{1}$ and $S_{2}$ be the areas of the segments. Then
$$
S_{1}=\frac{\pi r^{2}}{3}-\frac{1}{2} r \cdot \frac{r \sqrt{3}}{2}=\frac{1}{12} r^{2}(4 \pi-3 \sqrt{3}),
$$
$$
\begin{aligned}
& S_{2}=\pi r^{2}-S_{1}=\pi r^{2}-\frac{1}{12} r^{2}(4 \pi-3 \sqrt{3})=\frac{1}{1... | \frac{4\pi-3\sqrt{3}}{8\pi+3\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,263 |
10.152. A circle is inscribed in a regular hexagon with side length $a$, and another circle is circumscribed around the same hexagon. Determine the area of the circular ring enclosed between these two circles. | ## Solution.
The radius of the inscribed circle in the hexagon $r_{1}=\frac{a}{2 \operatorname{tg} \frac{180^{\circ}}{6}}=\frac{a \sqrt{3}}{2}$.
The radius of the circumscribed circle around the hexagon $r_{2}=a$. The area of the larger circle $S_{2}=\pi a^{2}$, the smaller one $-S_{1}=\pi \frac{3 a^{2}}{4}$.
The ar... | \frac{\pi^{2}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,264 |
10.153. A circle of radius $R$ is divided by two concentric circles into three figures of equal area. Find the radii of these circles. | ## Solution.
Let the radii of the inner circles be denoted by $R_{1}$ and $R_{2}$. Let $S$ be the area of the smallest circle; then $\pi R_{1}^{2}=S, \pi R_{2}^{2}=2 S, \pi R^{2}=3 S$. Therefore, $\pi R_{1}^{2}=\frac{\pi R^{2}}{3}$ and $\pi R_{2}^{2}=\frac{2 \pi R^{2}}{3}$, from which $R_{1}=\frac{R}{\sqrt{3}}$ and $R... | \frac{R}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,265 |
10.155. In a circle of radius $R$, two parallel chords are drawn on opposite sides of the center, one of which is equal to the side of an inscribed regular triangle, and the other to the side of an inscribed regular hexagon. Determine the area of the part of the circle contained between the chords. | Solution.
The length of the chord equal to the side of the inscribed regular hexagon, $a_{6}=R$, and the triangle $a_{3}=\frac{3 R}{\sqrt{3}}=\sqrt{3} R$ (Fig. 10.133). The area of the sector $O A B S_{1}$ is equal to the area of the segment plus the area of the triangle
$$
\begin{aligned}
& A O B: S_{1}=S_{\text {se... | \frac{R^{2}(\pi+\sqrt{3})}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,267 |
10.156. Two equilateral triangles are inscribed in a circle of radius $R$ such that their mutual intersection divides each side into three equal segments. Find the area of the intersection of these triangles. | ## Solution.
The figure formed by the intersection of such triangles will be a regular hexagon. The side
of the inscribed equilateral triangle is $a_{3}=\frac{3 R}{\sqrt{3}}=\sqrt{3} R$. Then the side of the regular hexagon is $a_{6}=\frac{1}{3} a_{3}=\frac{\sqrt{3}}{3} R$. The radius of the inscribed circle in it $r... | \frac{\sqrt{3}R^{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,268 |
10.157. Through points $R$ and $E$, belonging to sides $A B$ and $A D$ of parallelogram $A B C D$, and such that $A R=(2 / 3) A B, A E=(1 / 3) A D$, a line is drawn. Find the ratio of the area of the parallelogram to the area of the resulting triangle. | ## Solution.
Let $h$ be the height of parallelogram $ABCD$, and $h_{1}$ be the height of triangle $ARE$ (Fig. 10.134). Then $S_{ABCD} = AD \cdot h$, and $S_{\triangle ARE} = \frac{1}{2} AE \cdot h_{1}$. But $\frac{h}{h_{1}} = \frac{AB}{AR} = \frac{3}{2}$. Therefore, $\frac{S_{ABCD}}{S_{\triangle ARE}} = \frac{AD \cdot... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,269 |
10.158. Three circles with radii $R_{1}=6 \mathrm{~cm}, R_{2}=7 \mathrm{~cm}, R_{3}=8 \mathrm{~cm}$ touch each other pairwise. Determine the area of the triangle whose vertices coincide with the centers of these circles. | ## Solution.
The segments connecting the centers of these circles are: $a=R_{1}+R_{1}=6+7=13$ (cm), $b=R_{2}+R_{3}=7+8=15$ (cm), $c=R_{1}+R_{3}=$ $=6+8=14$ (cm). Then the area of this triangle can be found using Heron's formula, where $p=\frac{a+b+c}{2}=\frac{13+15+14}{2}=21$ (cm):
$$
S=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{... | 84\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,270 |
10.159. Find the ratio of the areas of an equilateral triangle, a square, and a regular hexagon, the lengths of whose sides are equal. | ## Solution.
Let the side of the equilateral triangle, square, and hexagon be $a$. Then $S_{3}=\frac{a^{2} \sqrt{3}}{4}, S_{4}=a^{2}, S_{6}=\frac{6 \sqrt{3} a^{2}}{4}$. Therefore,
$$
S_{3}: S_{4}: S_{6}=\frac{a^{2} \sqrt{3}}{4}: a^{2}: \frac{6 \sqrt{3}}{4} a^{2}=\sqrt{3}: 4: 6 \sqrt{3}
$$
Answer: $\sqrt{3}: 4: 6 \sq... | \sqrt{3}:4:6\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,271 |
10.160. In a trapezoid with an area of $594 \mathrm{~m}^{2}$, the height is 22 m, and the difference between the parallel sides is $6 \mathrm{~m}$. Find the length of each of the parallel sides. | ## Solution.
Since $S=\frac{1}{2}(a+b) h$, then $\frac{1}{2}(a+b) \cdot 22=594$, from which $a+b=54$. From the system of equations $\left\{\begin{array}{l}a+b=54, \\ a-b=6\end{array}\right.$ we find $a=30$ (m), $b=24$ (m).
Answer: $\quad 30$ m and 24 m. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,272 |
10.161. A perpendicular is drawn to the hypotenuse through the vertex of the right angle of a right-angled triangle with legs of 6 and 8 cm. Calculate the areas of the resulting triangles. | Solution.
Let $C K$ be the perpendicular to the hypotenuse $A B$ of $\triangle A B C$ (Fig. 10.135).
Fig. 10.135
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,273 |
10.162. Calculate the area of an isosceles triangle if the length of its height, drawn to the lateral side, is $12 \mathrm{~cm}$, and the length of the base is $15 \mathrm{cm}$. | ## Solution.
Let $A K$ be the perpendicular to the side $B C$ (Fig. 10.136). Then $A K=12, A C=15$ (cm); $K C=\sqrt{A C^{2}-A K^{2}}=\sqrt{225-144}=9$ (cm). Let $A B=B C=x$. Then $A B^{2}=A K^{2}+13 x^{2}$ or $x^{2}=144+(x-9)^{2}$, from which $x=\frac{15}{2}$ (cm). Then $S_{A B C}=\frac{1}{2} A K \cdot B C=\frac{1}{2}... | 75\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,274 |
10.163. The sides of a triangle are 13, 14, and 15 cm. Find the ratio of the areas of the circumscribed and inscribed circles of this triangle. | Solution.
Let $r$ and $R$ be the radii of the inscribed and circumscribed circles. Then $r=\frac{S}{p}, R=\frac{a b c}{4 S} \cdot$ We find
\[
\begin{aligned}
& S=\sqrt{p(p-a)(p-b)(p-c)}= \\
& =\sqrt{21 \cdot 8 \cdot 7 \cdot 6}=84\left(\mathrm{~cm}^{2}\right)
\end{aligned}
\]
Therefore, $r=4 \mathrm{~cm}, R=\frac{65}... | (\frac{65}{32})^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,275 |
10.165. A circle is inscribed in a regular triangle, and another circle is circumscribed around it. Find the area of the resulting ring if the side of the triangle is $a$. | Solution.
We have $S_{\Delta}=\frac{\sqrt{3}}{4} a^{2}, p=\frac{3}{2} a, r=\frac{S_{\Delta}}{p}=\frac{\sqrt{3}}{4} a^{2} \cdot \frac{2}{3 a}=\frac{\sqrt{3}}{6} a, R=\frac{a^{3}}{4 S_{\Delta}}=$ $=\frac{a^{3}}{\sqrt{3} a^{2}}=\frac{\sqrt{3}}{3} a$. Therefore, $S=\pi R^{2}-\pi r^{2}=\pi a^{2}\left(\frac{1}{3}-\frac{1}{1... | \frac{\pi^{2}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,276 |
10.166. One of the legs of a right triangle is 15 cm, and the radius of the circle inscribed in the triangle is 3 cm. Find the area of the triangle. | ## Solution.
Let $K, L, M$ be the points of tangency of the inscribed circle (Fig. 10.138). Then $K O=L O=M O=3$ (cm), $C A=15$ (cm). From this, $C M=K C=3$ (cm), $M A=C A-C M=15-3=12$ (cm). Therefore, $L A=$ $=12$ (cm). Let $B K=B L=x$. Then the area of the triangle $S=2 S_{B K O}+2 S_{O L A}+S_{K O M C}=$ $=2 \cdot ... | 60\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,277 |
10.168. Prove that if the diameter of a semicircle is divided into two arbitrary parts and on each of them a semicircle is constructed (inside the given semicircle), then the area enclosed between the three semicircles is equal to the area of a circle whose diameter is the length of the perpendicular from the point of ... | ## Solution.
Let $R$ be the radius of the given semicircle, and $r$ be the radius of one of the constructed semicircles (Fig. 10.139).
Fig. 10.139
. Then, from the right triangles $\triangle A B D$ and $\triangle O K B$, we get
$$
\sin \alpha=\frac{O K}{O B}=\frac{A D}{A B} \Leftrightarrow \frac{r}{R-r}=\frac{a}{R} \Leftrightarrow R r=R a-r a \Leftrightarrow r=\frac{R ... | \pi(\frac{R}{R+})^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,281 |
10.171. The bases of the trapezoid are equal to $a$ and $b$, and the angles at the larger base are $\pi / 6$ and $\pi / 4$. Find the area of the trapezoid. | ## Solution.
Let $B C \| A D$ and $\angle A=\pi / 6, \angle D=\pi / 4$. Drop perpendiculars $B K$ and $C L$ to the side $A D$ (Fig. 10.142). Let $C L=B K$ be denoted by $h$. Then $\operatorname{tg} \pi / 4=\frac{C L}{L D}$. Therefore, $L D=h$. Then $A K=b-a-h$, and $\operatorname{tg} \pi / 6=\frac{B K}{A K}=$ $=\frac{... | \frac{(\sqrt{3}-1)(b^{2}-^{2})}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,282 |
10.174. The area of an isosceles trapezoid circumscribed around a circle is $S$. Determine the radius of this circle if the angle at the base of the trapezoid is $30^{\circ}$. | ## Solution.
Let $B C \| A D, \angle A=30^{\circ}$ (Fig. 10.144). Denote $B C=a, A D=b$;
Fig. 10.145
. Drop perpendiculars $KM$ and $KN$ to $BC$ and $CD$.
From $\triangle KMC$, $KC=\frac{MK}{\sin \alpha}$. From $\triangle KNC$, $KC=\frac{KN}{\sin \beta}$. This means that $\frac{\sin \beta}{\sin \alpha}=\frac{KN}{MK}$. From $\triangle ACD$ by t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 51,285 |
10.178. Find the lengths of the sides of isosceles triangle $ABC$ with base $AC$, if it is known that the lengths of its heights $AN$ and $BM$ are equal to $\boldsymbol{n}$ and $m$ respectively. | Solution.
$\triangle C M B \sim \triangle C N A$ (Fig. 10.147), from which $\frac{B M}{A N}=\frac{B C}{A C}$. Therefore, $A C=\frac{n}{m} B C$, $M C=\frac{1}{2} A C=\frac{n}{2 m} B C$. From $\triangle B M C$, $B C^{2}=M B^{2}+M C^{2}$, $B C^{2}=m^{2}+B C^{2}\left(\frac{n}{2 m}\right)^{2}$, from which $B C=\frac{2 m^{2... | \frac{2^{2}}{\sqrt{4^{2}-n^{2}}},\frac{2n}{\sqrt{4^{2}-n^{2}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,287 |
10.179. A rhombus, in which the side is equal to the smaller diagonal (Fig. 10.148), is equal in area to a circle of radius $R$. Determine the side of the rhombus. | Solution.
From the condition $\triangle A B D$ is equilateral, therefore, $\angle B A D=60^{\circ}$.
Area of the rhombus: $S_{\mathrm{p}}=x^{2} \sin 60^{\circ}=\frac{\sqrt{3}}{2} x^{2}=\pi R^{2} \Rightarrow x=\sqrt{\frac{2 \pi R^{2}}{\sqrt{3}}}=R \sqrt{\frac{2 \pi}{\sqrt{3}}}$.
Answer: $\quad R \sqrt{\frac{2 \pi}{\s... | R\sqrt{\frac{2\pi}{\sqrt{3}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,288 |
10.181. In a square with side $a$, the midpoints of two adjacent sides are connected to each other and to the opposite vertex of the square. Determine the area of the inner triangle.
Fig. 10... | ## Solution.
Let $K, L$ be the midpoints of $A D$ and $C D$ (Fig. 10.150). Then $S_{1}=\frac{1}{2} a \cdot \frac{a}{2}=\frac{a^{2}}{4}$; $S_{2}=\frac{1}{2} \frac{a}{2} \cdot \frac{a}{2}=\frac{a^{2}}{8}$; $S_{3}=\frac{1}{2} a \cdot \frac{a}{2}=\frac{a^{2}}{4}$. The area of the square $S_{0}=a^{2}$. Therefore, $S=S_{0}-... | \frac{3^{2}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,289 |
10.182. A circle is circumscribed around a square with side $a$. A square is inscribed in one of the resulting segments. Determine the area of this square. | ## Solution.
The diagonal of the square is the diameter of the circle (Fig. 10.151). $(2 R)^{2}=2 a^{2}$, hence $R=\sqrt{2} a$. Let the side of the smaller square be $x$. Then $P M^{2}+P O^{2}=O M^{2}, P M=\frac{x}{2}, P O=\frac{a}{2}+x$. Therefore, $\frac{x^{2}}{4}+\left(\frac{a}{2}+x\right)^{2}=\frac{a^{2}}{2}$, fro... | \frac{^{2}}{25} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,290 |
10.184. Calculate the area of a trapezoid, the parallel sides of which are 16 and 44 cm, and the non-parallel sides are 17 and 25 cm. | Solution.
Let $B K$ and $C L$ be the perpendiculars dropped onto $A D$ (Fig. 10.152). $B C=16$ (cm), $A D=44$ (cm), $A B=17$ (cm), $C D=25$ (cm). Denote $A K=x, L D=y$. Then $A B^{2}-A K^{2}=B K^{2}, C L^{2}=C D^{2}-L D^{2}$. This means that $17^{2}-x^{2}=B K^{2}, C L^{2}=25^{2}-y^{2}$. We have the system $\left\{\beg... | 450\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,291 |
10.185. In an isosceles trapezoid, the length of the midline is 5, and the diagonals are perpendicular to each other. Find the area of the trapezoid. | Solution.
Let the bases of the trapezoid be $a$ and $b$, and the height be $h$. Then the area
Fig. 10.153
. Then $\frac{B C+A D}{2}=h=17$, $\frac{12 k+5 k}{2}=17$, from which $k=2$. Therefore, $B C=10, A D=24$. Let $C L \perp A D$, $B K \perp A D, A K=L D=\frac{A D-B C}{2}=7$ (cm); $C D=\sqrt{C L^{2}+L D^{2}}=\sqrt{17^{2}+7^{2}}=$ $=13 \sqrt{2}, \sin \alpha=\frac{C L}{C D}=\f... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,293 |
10.188. The radius of the circle circumscribed around a right triangle relates to the radius of the inscribed circle as 5:2. Find the area of the triangle if one of its legs is equal to $a$. | Solution.
Let $R$ and $r$ be the radii of the inscribed and circumscribed circles, $BC = a$ (Fig. 10.155). Let $BD = x$; then $BL = x$ (as tangents drawn from the same point), $LA = AK = 2R - x$ (since $\triangle ABC$ is a right triangle, then $AB = 2R$). We have $AC^2 + BC^2 = AB^2$ or $(r + 2R - x)^2 + a^2 = 4R^2$. ... | \frac{2a^2}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,295 |
10.189. A square is inscribed in a segment whose arc is $60^{\circ}$. Calculate the area of the square if the radius of the circle is $2 \sqrt{3}+\sqrt{17}$. | Solution.
Let $B C=C D=D A=A B=x$ (Fig. 10.156). Consider $\triangle O N C$,
Fig. 10.156
. A semicircle is inscribed in it with the diameter lying on the largest side. Find the ratio of the area of the semicircle to the area of the triangle. | ## Solution.
By the condition $A B: B C: A C=2: 3: 4$. Let $A B=2 x$, then $B C=3 x$, $A C=4 x, S_{\triangle A B C}=\frac{B C \cdot A C}{2} \cdot \sin \angle B C A, \cos \angle B C A=\frac{4 x^{2}-9 x^{2}-16 x^{2}}{-24 x^{2}}=\frac{7}{8}$, $\sin \angle B C A=\sqrt{1-\frac{49}{64}}=\frac{\sqrt{15}}{8} ; S_{\triangle A ... | \frac{3\pi\sqrt{15}}{50} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,297 |
11.001. The base of the pyramid is a right-angled triangle with a hypotenuse equal to \( c \) and an acute angle of \( 30^{\circ} \). The lateral edges of the pyramid are inclined to the base plane at an angle of \( 45^{\circ} \). Find the volume of the pyramid. | ## Solution.
By the condition $\angle A C B=90^{\circ}, \angle B A C=30^{\circ}$ (Fig. 11.3); therefore, $B C=\frac{c}{2}, A C=\frac{c \sqrt{3}}{2}$, from which $S_{\text {base }}=\frac{1}{2} A C \cdot B C=\frac{c \sqrt{3}}{8}$. Draw $S O$ such that $A \dot{O}=O B$. Then $O$ is the center of the circumscribed circle a... | \frac{\sqrt[3]{3}}{48} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,298 |
11.002. Calculate the volume of a regular tetrahedron if the radius of the circle circumscribed around its face is $R$. | Solution.
For a tetrahedron with equal edges, denote the edges by \(a\). The radius of the circle circumscribed around the base of the tetrahedron (which is an equilateral triangle) is \(R = \frac{a \sqrt{3}}{3}\), from which \(a = R \sqrt{3}\). The height of the tetrahedron projects onto the center of the circle circ... | \frac{R^{3}\sqrt{6}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,299 |
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