Geometry Problems with Solutions for Grade 12

Triangle $ABC$ has two sides with equal length and therefore is an isosceles triangle with base AC. The measure of angle $\mathrm{\angle }BAC$ is given by:

$$\mathrm{\angle }BAC={\displaystyle \frac{{180}^{\circ }-{36}^{\circ }}{2}}={72}^{\circ }$$ Angle $\mathrm{\angle }BOC$ is a central angle and $\mathrm{\angle }BAC$ is an inscribed angle and both angles intercept the same arc; hence: $$\text{Measure of }\mathrm{\angle }BOC=2\times \text{Measure of }\mathrm{\angle }BAC={144}^{\circ }$$

Let $R_1,R_2$ and $R_3$ be the radii of circles C1, C2 and C3 with $R_1=R_2=R$.

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Let $C3O$ be the distance from C3 to the segment C1, C2. Hence $$h=C3O+R(I)$$

Apply Pythagoras' theorem applied to triangle C3 O C1 $$C3O^2+{\left({\displaystyle \frac{x}{2}}\right)}^2=(R+R_3{)}^2$$

which gives $$C3O=\sqrt{(R+R_3{)}^2-{\left({\displaystyle \frac{x}{2}}\right)}^2}$$

Using (I), we obtain $$h=R+C_{3}O=R+\sqrt{(R+R_3{)}^2-{\left({\displaystyle \frac{x}{2}}\right)}^2}$$

Let $r$, $R_2$, and $R_3$ be the radii of circles $C1$, $C2$, and $C3$ respectively, with $R_2=R_3=R$.

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Pythagoras' theorem applied to triangle $MC1C3$ $$C1C{3}^2=MC{3}^2+MC{1}^2(I)$$ Substitute: $C1C3=r+R,MC3=R,MC1=R-r$ in $(I)$ Expand, simplify and solve for $R$ $$R=4r=40\text{ cm}$$

Angles $\mathrm{\angle }BtA$ and $\mathrm{\angle }CtD$ are vertical and therefore

$$\mathrm{\angle }BtA={90}^{\circ }$$ Since $\mathrm{\angle }BtA={90}^{\circ }$, $AB$ is the diameter of the circle using the converse of Thales' theorem.

Since $CD$ is parallel to $AB$ Triangle $BtA$ and triangle $CtD$ are similar and hence the proportionality of the corresponding sides gives: $${\displaystyle \frac{3}{5}}={\displaystyle \frac{AB}{x}}$$ Solve for the diameter $AB$ $$AB={\displaystyle \frac{3x}{5}}$$ $$\text{radius}={\displaystyle \frac{AB}{2}}={\displaystyle \frac{3x}{10}}$$ $$\text{area}=\pi \times {\text{radius}}^2=\pi {\left({\displaystyle \frac{3x}{10}}\right)}^2=0.09\pi x^2$$

Let us split the large square into four small squares as shown in the image.

Let us consider one small square, the bottom left one for example. The square below has a side length of $2x$, which is half of the given square. Part of this square is shaded, while the other part is not shaded. Let us find the area of the non-shaded (white) part. The shaded part is a quarter of a disk (circle).

. $$\text{Area of non-shaded region}=\text{area of the square - 1/4 of the area of the circle}=(2x{)}^2-{\displaystyle \frac{1}{4}}\pi (2x{)}^2$$

If we go back to the given shape in problem 5, the total area of the non-shaded part is $8$ times the non-shaded area in the image of the small square above, which is given by $(2x{)}^2-{\displaystyle \frac{1}{4}}\pi (2x{)}^2$.

The area A of the shaded part in the shape of problem 5 is given by: $$A=\text{Total area of the large square}-\text{Total non-shaded area}$$ $$=(4x{)}^2-8[(2x{)}^2-{\displaystyle \frac{1}{4}}\pi (2x{)}^2]$$ $$=16x^2({\displaystyle \frac{\pi }{2}}-1)$$

The right triangle in the image has a hypotenuse whose length is $R$, one side $r$ and the second side half of the cord whose length is $20/2$ mm.

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Use the Pythagorean theorem

$$R^2=r^2+{10}^2(I)$$ The area $A$ of the ring is found by suntracting the area of the small circle from that of the large circle: $$A=\pi (R^2-r^2)(II)$$ \] The first equation $I$ gives $$R^2-r^2={10}^2=100$$ Substitute $(II)$ to obatin the area of the ring $$A=100\pi$$ \]

The area of a parallelogram may be calculated using the cross product of vectors $\mathbf{AB}$ and $\mathbf{AD}$ as follows: $$\text{Area}=|\mathbf{AB}\times \mathbf{AD}|$$ where $\mathbf{AB}$ and $\mathbf{AD}$ are 3-dimensional vectors with the $z$ component equal to zero so that we can perform the cross product $\mathbf{AB}\times \mathbf{AD}$.

Each vectors is calculated using the coordinates of the points defining the vector as follows: $$\mathbf{AB}=⟨2-(-2),b-(-2),0-0⟩=⟨4,b+2,0⟩$$ $$\mathbf{AD}=⟨4-(-2),2-(-2),0⟩=⟨6,4,0⟩$$ Computing the magnitude of the cross product: $$|⟨4,b+2,0⟩\times ⟨6,4,0⟩|=80$$ The determinant form of the cross product in three dimensions is: $$\mathbf{AB}\times \mathbf{AD}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 4& b+2& 0\\ 6& 4& 0\end{array}\right|$$ Expanding along the first row: $$=\mathbf{i}\left|\begin{array}{cc}b+2& 0\\ 4& 0\end{array}\right|-\mathbf{j}\left|\begin{array}{cc}4& 0\\ 6& 0\end{array}\right|+\mathbf{k}\left|\begin{array}{cc}4& b+2\\ 6& 4\end{array}\right|$$ which gives $$\mathbf{AB}\times \mathbf{AD}=(4-6b)\mathbf{k}$$ Taking the magnitude: $$|4-6b|=80$$ Solving for $b$: $$4-6b=80\text{or}4-6b=-80$$ $$b=-{\displaystyle \frac{38}{3}},b=14$$ Since point $B(2,b)$ is in the first quadrant, we choose $b=14$. Since $ABCD$ is a parallelogram, we have: $$\mathbf{AB}=\mathbf{DC}$$ Thus, $$\mathbf{AB}=⟨4,16,0⟩$$ $$\mathbf{DC}=⟨c-4,d-2,0⟩$$ Equating corresponding components: $$c-4=4,d-2=16$$ Solving: $$c=8,d=18$$ Hence : $b=14$ , $c=8$ and $d=18$

There are 3 right triangles. Use the Pythagorean theorem to write an equation for each triangle: $$y^2+z^2={12}^2(I)$$ $$x^2+z^2=9^2(II)$$ $$(y+x{)}^2={12}^2+9^2=225(III)$$ Solve equation $(III)$ by extracting the square root $$x+y=15$$ Subtract equation $(II)$ form equation $(I)$, we obtain: $$y^2-x^2={12}^2-9^2=63(IV)$$ Factor the left side of the above equation: $$(y-x)(y+x)=63$$ Substitute $x+y=15$ in the above equation and simplify to obtain: $$y-x={\displaystyle \frac{21}{5}}$$ Solve the system made up of equations $$\{\begin{array}{l}x+y=15\\ y-x={\displaystyle \frac{21}{5}}\end{array}$$

to find $x={\displaystyle \frac{27}{5}}$ and $y={\displaystyle \frac{48}{5}}$

Use equation $(I)$ to find $z$ $${\left({\displaystyle \frac{48}{5}}\right)}^2+z^2={12}^2$$ $$z^2={12}^2-{\left({\displaystyle \frac{48}{5}}\right)}^2={\displaystyle \frac{1296}{25}}$$ Hence $$z={\displaystyle \frac{36}{5}}$$ The values of $x,y$ and $z$ are $$x={\displaystyle \frac{27}{5}},y={\displaystyle \frac{48}{5}},z={\displaystyle \frac{36}{5}}$$

Split the given rectangle into 4 other rectangles as shown.

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Pythagoras' theorem applied to the top-left right triangle of side $a$ and hypotenuse $4$ gives: $$a^2+c^2=4^2(I)$$ Pythagoras' theorem applied to the bottom-left right triangle gives: $$b^2+c^2=x^2(II)$$ Pythagoras' theorem applied to the bottom-right right triangle. $$b^2+d^2=5^2(III)$$ Pythagoras' theorem applied to the top-right right triangle. $$a^2+d^2=6^2(IV)$$ Subtracting equations $(II)$ from $(I)$ $$a^2-b^2=4^2-x^2$$ Subtracting equations $(III)$ from $(IV)$ $$a^2-b^2=6^2-5^2$$ Substitute $a^2-b^2=4^2-x^2$ in the above equation to obatin: $$4^2-x^2=6^2-5^2$$ Solve for $x$. $$x=\sqrt{5}$$

Because of the symmetry, the shaded region may be considered as made up of two equal (in area) regions. The area of the left half of the shaded region is given by the area of the sector $BOC$ minus the area of the triangle $BOC$.

Since the distance between the centers is $6$, the length of $OM$ is given by: $$OM=6/2=3$$ $BOM$ is a right triangle, hence $$\cos \mathrm{\angle }BOM={\displaystyle \frac{OM}{OB}}={\displaystyle \frac{3}{4}}$$ $$\mathrm{\angle }BOM=\arccos \left({\displaystyle \frac{3}{4}}\right)$$ The area of sector $BOC$ is: $$\text{Area of sector }BOC={\displaystyle \frac{1}{2}}(2\mathrm{\angle }BOM)r^2$$ The area of triangle $BOC$ is: $$\text{Area of triangle }BOC={\displaystyle \frac{1}{2}}\sin (2\mathrm{\angle }BOM)r^2$$ Thus, the area of the shaded region is: $$2[{\displaystyle \frac{1}{2}}(2\mathrm{\angle }BOM)r^2-{\displaystyle \frac{1}{2}}\sin (2\mathrm{\angle }BOM)r^2]$$ Substitute $\mathrm{\angle }BOM$ by $\arccos \left({\displaystyle \frac{3}{4}}\right)$ and $r=4$ $$=(2\arccos \left({\displaystyle \frac{3}{4}}\right)-\sin (2\arccos \left({\displaystyle \frac{3}{4}}\right)))4^2$$ $$\approx 7.25\text{ square units}\text{(rounded to 3 decimal places)}$$