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 $\angle BAC$ is given by:

$$ \angle BAC = \frac{180^\circ - 36^\circ}{2} = 72^\circ $$

Angle $\angle BOC$ is a central angle and $\angle BAC$ is an inscribed angle, and both angles intercept the same arc $BC$. According to the inscribed angle theorem:

$$ \text{Measure of } \angle BOC = 2 \times \text{Measure of } \angle BAC = 2 \times 72^\circ = \mathbf{144^\circ} $$

Let $R_1, R_2$ and $R_3$ be the radii of circles $C_1, C_2$ and $C_3$ with $R_1 = R_2 = R$.

Let $C_3O$ be the vertical distance from the center of $C_3$ to the horizontal line segment connecting the centers $C_1$ and $C_2$. Hence: $$ h = C_3O + R \quad (I) $$

Apply Pythagoras' theorem to the right triangle $C_3 O C_1$: $$ C_3O^2 + \left(\frac{x}{2}\right)^2 = (R + R_3)^2 $$

Solving for $C_3O$: $$ C_3O = \sqrt{(R + R_3)^2 - \left(\frac{x}{2}\right)^2} $$

Using equation (I), we substitute $C_3O$ to obtain the final height $h$: $$ \mathbf{h = R + \sqrt{(R + R_3)^2 - \left(\frac{x}{2}\right)^2}} $$

Let $r, R_2$, and $R_3$ be the radii of circles $C_1, C_2$, and $C_3$ respectively, with $R_2 = R_3 = R$. Note that $r = 10$ cm.

Apply Pythagoras' theorem to the right triangle $M C_1 C_3$: $$ C_1C_3^2 = MC_3^2 + MC_1^2 \quad (I) $$

From the geometry of the figure, we can substitute the side lengths with their radii equivalents: $C_1C_3 = r + R$ $MC_3 = R$ $MC_1 = R - r$

Substitute these into equation (I): $$ (r + R)^2 = R^2 + (R - r)^2 $$

Expand and simplify the equation: $$ r^2 + 2rR + R^2 = R^2 + R^2 - 2rR + r^2 $$ $$ 2rR = R^2 - 2rR $$ $$ 4rR = R^2 $$

Since $R > 0$, divide both sides by $R$: $$ R = 4r $$

Substitute $r = 10$ cm: $$ \mathbf{R = 4(10) = 40 \text{ cm}} $$

Angles $\angle BtA$ and $\angle CtD$ are vertical angles, and therefore: $$ \angle BtA = 90^\circ $$

Because $\angle BtA = 90^\circ$ and it is an inscribed angle, $AB$ must be the diameter of the circle (by the converse of Thales' theorem).

Since $CD$ is parallel to $AB$, triangle $BtA$ and triangle $CtD$ are similar triangles. The proportionality of their corresponding altitudes and bases gives: $$ \frac{3}{5} = \frac{AB}{x} $$

Solve for the diameter $AB$: $$ AB = \frac{3x}{5} $$

The radius is half of the diameter: $$ \text{Radius} = \frac{AB}{2} = \frac{3x}{10} $$

Finally, calculate the area of the circle: $$ \text{Area} = \pi \times \text{Radius}^2 = \pi \left(\frac{3x}{10}\right)^2 = \mathbf{0.09\pi x^2} $$

Let us split the large square into four smaller squares by drawing horizontal and vertical lines through the center.

Consider one small square (e.g., the bottom left one). This square has a side length of $2x$ (half of $4x$). Part of this square is shaded, while the other part is white (non-shaded). The shaded part is bounded by a quarter circle. Let's focus on the white part first.

The area of the non-shaded (white) corner in one small square is the area of the small square minus the area of the quarter circle: $$ \text{Area of one white region} = (2x)^2 - \frac{1}{4}\pi(2x)^2 = 4x^2 - \pi x^2 $$

Looking at the entire large square, there are 8 identical white regions (2 in each of the 4 small squares).

The area $A$ of the shaded part in the whole shape is the total area of the large square minus the 8 non-shaded areas: $$ A = (4x)^2 - 8 \left[ (2x)^2 - \frac{1}{4}\pi(2x)^2 \right] $$ $$ A = 16x^2 - 8 \left[ 4x^2 - \pi x^2 \right] $$ $$ A = 16x^2 - 32x^2 + 8\pi x^2 $$ $$ A = 8\pi x^2 - 16x^2 $$ $$ \mathbf{A = 16x^2 \left( \frac{\pi}{2} - 1 \right)} $$

Draw a radius to the point of tangency (which creates a right angle) and a radius to the end of the chord. This forms a right triangle.

The right triangle has a hypotenuse of length $R$, one leg of length $r$, and the second leg is exactly half of the chord's length ($20 / 2 = 10$ mm). Use the Pythagorean theorem:

$$ R^2 = r^2 + 10^2 \quad (I) $$ $$ R^2 - r^2 = 100 $$

The area $A$ of the ring is found by subtracting the area of the small circle from that of the large circle: $$ A = \pi R^2 - \pi r^2 = \pi (R^2 - r^2) \quad (II) $$

Substitute $R^2 - r^2 = 100$ into the area formula: $$ \mathbf{A = 100\pi \text{ mm}^2} $$

The area of a parallelogram can be calculated using the magnitude of the cross product of vectors $\mathbf{AB}$ and $\mathbf{AD}$:

$$ \text{Area} = | \mathbf{AB} \times \mathbf{AD} | $$

First, calculate the vectors from the given points $A(-2, -2), B(2, b), D(4, 2)$: $$ \mathbf{AB} = \langle 2 - (-2), b - (-2), 0 \rangle = \langle 4, b + 2, 0 \rangle $$ $$ \mathbf{AD} = \langle 4 - (-2), 2 - (-2), 0 \rangle = \langle 6, 4, 0 \rangle $$

Compute the determinant form of the cross product: $$ \mathbf{AB} \times \mathbf{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & b+2 & 0 \\ 6 & 4 & 0 \end{vmatrix} = \mathbf{k}((4)(4) - (6)(b+2)) $$ $$ \mathbf{AB} \times \mathbf{AD} = (16 - 6b - 12)\mathbf{k} = (4 - 6b)\mathbf{k} $$

Taking the magnitude and setting it equal to 80: $$ |4 - 6b| = 80 $$ $$ 4 - 6b = 80 \quad \text{or} \quad 4 - 6b = -80 $$ $$ -6b = 76 \implies b = -\frac{38}{3} $$ $$ -6b = -84 \implies b = 14 $$

Since point $B(2, b)$ is clearly in the first quadrant on the graph, we choose $b = 14$.

Since $ABCD$ is a parallelogram, opposite vectors are equal: $\mathbf{AB} = \mathbf{DC}$. $$ \mathbf{AB} = \langle 4, 14 + 2, 0 \rangle = \langle 4, 16, 0 \rangle $$ $$ \mathbf{DC} = \langle c - 4, d - 2, 0 \rangle $$

Equating corresponding components: $c - 4 = 4 \implies c = 8$ $d - 2 = 16 \implies d = 18$

Conclusion: $$ \mathbf{b = 14, \; c = 8, \; d = 18} $$

There are 3 right triangles. Use the Pythagorean theorem to write an equation for each:

$$ y^2 + z^2 = 12^2 \quad (I) $$ $$ x^2 + z^2 = 9^2 \quad (II) $$ $$ (x + y)^2 = 12^2 + 9^2 = 144 + 81 = 225 \quad (III) $$

Solve equation (III) by extracting the square root (since lengths must be positive): $$ x + y = 15 $$

Subtract equation (II) from equation (I) to eliminate $z^2$: $$ (y^2 + z^2) - (x^2 + z^2) = 144 - 81 $$ $$ y^2 - x^2 = 63 $$

Factor the difference of squares: $$ (y - x)(y + x) = 63 $$

Substitute $x + y = 15$ into the factored equation: $$ (y - x)(15) = 63 \implies y - x = \frac{63}{15} = \frac{21}{5} $$

Now, solve the system of linear equations: $$ y + x = 15 = \frac{75}{5} $$ $$ y - x = \frac{21}{5} $$ Adding them gives: $2y = \frac{96}{5} \implies \mathbf{y = \frac{48}{5}}$. Subtracting them gives: $2x = \frac{54}{5} \implies \mathbf{x = \frac{27}{5}}$.

Use equation (I) to find $z$: $$ \left(\frac{48}{5}\right)^2 + z^2 = 144 \implies z^2 = 144 - \frac{2304}{25} = \frac{3600 - 2304}{25} = \frac{1296}{25} $$ $$ \mathbf{z = \frac{36}{5}} $$

Split the rectangle into 4 smaller rectangles by drawing vertical and horizontal lines through the interior point.

Apply Pythagoras' theorem to the four right triangles formed at the corners (let the legs be $a, b, c, d$ as shown in the diagram):

$$ a^2 + c^2 = 4^2 \quad (I) \text{ (Top-left)} $$ $$ b^2 + c^2 = x^2 \quad (II) \text{ (Bottom-left)} $$ $$ b^2 + d^2 = 5^2 \quad (III) \text{ (Bottom-right)} $$ $$ a^2 + d^2 = 6^2 \quad (IV) \text{ (Top-right)} $$

Subtract equation (II) from (I): $$ a^2 - b^2 = 4^2 - x^2 $$

Subtract equation (III) from (IV): $$ a^2 - b^2 = 6^2 - 5^2 $$

Set the two expressions for $(a^2 - b^2)$ equal to each other: $$ 4^2 - x^2 = 6^2 - 5^2 $$ $$ 16 - x^2 = 36 - 25 $$ $$ 16 - x^2 = 11 $$ $$ x^2 = 5 $$ $$ \mathbf{x = \sqrt{5}} $$

Because of symmetry, the shaded region can be split into two equal left and right halves. Let's calculate the area of the left half.

The left half of the overlapping region is the area of the circular sector $BOC$ minus the area of the triangle $BOC$. Since the distance between the centers is $6$, the length of $OM$ (half the distance) is $3$.

Triangle $BOM$ is a right triangle, so we can find angle $BOM$: $$ \cos(\angle BOM) = \frac{OM}{OB} = \frac{3}{4} $$ $$ \angle BOM = \arccos(0.75) $$

The full central angle $BOC$ is twice $\angle BOM$. Therefore:

1. Area of sector $BOC = \frac{1}{2} (2 \angle BOM) r^2$ (in radians)

2. Area of triangle $BOC = \frac{1}{2} \sin(2 \angle BOM) r^2$

The total shaded area is twice the left half: $$ \text{Total Area} = 2 \left[ \frac{1}{2}(2 \angle BOM)r^2 - \frac{1}{2}\sin(2 \angle BOM)r^2 \right] $$ $$ \text{Total Area} = r^2 \left( 2 \angle BOM - \sin(2 \angle BOM) \right) $$

Substitute $r = 4$ and $\angle BOM = \arccos(0.75)$: $$ \text{Total Area} = 4^2 \left( 2\arccos(0.75) - \sin(2\arccos(0.75)) \right) $$ $$ \text{Total Area} = 16 \left( 1.445 - 0.992 \right) \approx \mathbf{7.25 \text{ square units}} $$

1. Identify the Theorem:
This problem requires the Intersecting Chords Theorem. It states that when two chords intersect inside a circle, the product of the segments of one chord is equal to the product of the segments of the other chord.

2. Set up the Equation:
$$ AP \times PB = CP \times PD $$

3. Substitute and Solve:
$$ 4 \times 6 = 3 \times PD $$ $$ 24 = 3 \times PD $$ $$ \mathbf{PD = 8} $$

1. Break Down the Hexagon:
A regular hexagon inscribed in a circle can be divided into exactly 6 identical equilateral triangles by drawing lines from the center of the circle to the 6 vertices. Because it's a regular hexagon, the side length of the hexagon is exactly equal to the radius of the circumscribed circle ($s = R = 10$).

2. Area of One Equilateral Triangle:
The formula for the area of an equilateral triangle with side $s$ is: $$ A_{\text{triangle}} = \frac{\sqrt{3}}{4}s^2 $$ Substitute $s = 10$: $$ A_{\text{triangle}} = \frac{\sqrt{3}}{4}(100) = 25\sqrt{3} $$

3. Total Area:
Multiply by the 6 triangles that make up the hexagon: $$ A_{\text{total}} = 6 \times 25\sqrt{3} = \mathbf{150\sqrt{3}} $$

1. Identify the Theorem:
To find the length of a median given all three sides of a triangle, we use Apollonius's Theorem, which relates the length of a median to the lengths of its sides: $$ AB^2 + AC^2 = 2(AM^2 + BM^2) $$

2. Determine the Segments:
Since $AM$ is a median, $M$ splits $BC$ perfectly in half. $$ BM = \frac{BC}{2} = \frac{9}{2} = 4.5 $$

3. Substitute and Solve:
$$ 7^2 + 8^2 = 2\left(AM^2 + 4.5^2\right) $$ $$ 49 + 64 = 2(AM^2 + 20.25) $$ $$ 113 = 2AM^2 + 40.5 $$ $$ 72.5 = 2AM^2 $$ $$ AM^2 = 36.25 = \frac{145}{4} $$ $$ \mathbf{AM = \frac{\sqrt{145}}{2}} \approx 6.02 $$