Explore detailed solutions for Grade 11 geometry problems. These examples cover coordinate geometry, circle theorems, and area/volume calculations for complex shapes.
Find all points of intersection of the circle \( x^2 + 2x + y^2 + 4y = -1 \) and the line \( x - y = 1 \).
Solve the linear equation for \( x \): \( x = 1 + y \).
Substitute into the circle equation:
\[ (1 + y)^2 + 2(1 + y) + y^2 + 4y = -1 \]Expand and simplify to standard quadratic form:
\[ 2y^2 + 8y + 4 = 0 \]Solving for \( y \) yields \( y = -2 \pm \sqrt{2} \). Using \( x = 1 + y \), the intersection points are:
\[ (-1 + \sqrt{2},\ -2 + \sqrt{2}) \quad \text{and} \quad (-1 - \sqrt{2},\ -2 - \sqrt{2}) \]Find the area of the triangle enclosed by the x-axis and the lines \( y = x \) and \( y = -2x + 3 \).
Solving the system \( y = x \) and \( y = -2x + 3 \) gives the intersection (vertex) at \( (1, 1) \). The height is \( h = 1 \).
The base lies on the x-axis from \( x = 0 \) to the intercept of \( y = -2x + 3 \), which is \( x = 1.5 \).
\[ \text{Area} = \frac{1}{2} \times 1.5 \times 1 = 0.75 \text{ or } \frac{3}{4} \]Find the length of the third side of a triangle if the area is 18 and two sides have lengths of 5 and 10.
Using the sine formula for area:
\[ 18 = \frac{1}{2} \times 5 \times 10 \times \sin(A) \Rightarrow \sin(A) = \frac{18}{25} \]Now apply the cosine law for side \( x \):
\[ x^2 = 5^2 + 10^2 - 2(5)(10)\cos(A) \]Since \( \cos(A) = \sqrt{1 - (18/25)^2} \approx 0.692 \):
\[ x = \sqrt{125 - 100(0.692)} \approx 7.46 \]In a circle, chords AC and BD intersect at O. If Area(BOC) = 15, AO = 10, and OB = 5, find the area of triangle AOD.
By the intersecting chords theorem, \( AO \cdot OC = BO \cdot OD \). This gives the ratio:
\[ \frac{AO}{BO} = \frac{OD}{OC} = \frac{10}{5} = 2 \]The triangles share vertical angles. The ratio of their areas is the product of their side ratios:
\[ \text{Area}_{\triangle AOD} = \text{Area}_{\triangle BOC} \times 2 \times 2 = 15 \times 4 = 60 \]
Two concentric circles have radii of 10 and 6. What is the length of chord AB of the larger circle that is tangent to the smaller circle?
Drawing a radius to the point of tangency forms a right triangle with the chord.
Using the Pythagorean Theorem for half-chord \( x \):
\[ 6^2 + x^2 = 10^2 \Rightarrow x = \sqrt{100 - 36} = 8 \]Total length \( AB = 2x = 16 \).
Point A is inside square BCDE (side = 20). AB = 9 and AE = 13. Find length AC.
Use cosine law in \(\triangle ABE\):
\[ 13^2 = 20^2 + 9^2 - 2(20)(9)\cos(T) \]Use cosine law in \(\triangle ACB\):
\[ x^2 = 20^2 + 9^2 - 2(20)(9)\cos(90^\circ - T) \]
Note that \( \cos(90^\circ - T) = \sin(T) \). Solving the first equation gives \( \cos(T) = \frac{13}{15} \).
Using the identity \( \sin(T) = \sqrt{1 - (13/15)^2} = \frac{2\sqrt{14}}{15} \):
\[ AC = \sqrt{481 - 48\sqrt{14}} \approx 17.4 \]A metal object combines a hemisphere atop a cylinder. Radius \( r = 5 \text{ cm} \), cylinder height \( h = 12 \text{ cm} \). Density is \( 7.8 \text{ g/cm}^3 \).
a) Total Volume:
\[ V_{cyl} = \pi(5)^2(12) = 300\pi \] \[ V_{hemi} = \frac{2}{3}\pi(5)^3 = \frac{250}{3}\pi \] \[ V_{total} = \frac{1150}{3}\pi \text{ cm}^3 \]b) Mass:
\[ \text{Mass} = 7.8 \times \text{Volume} \approx 9396.66 \text{ g} \]A garden is a \( 12 \times 8 \text{ m} \) rectangle with a semicircle attached to one 8m side.
a) Total Area:
\[ A_{rect} = 12 \times 8 = 96, \quad A_{semi} = \frac{1}{2}\pi(4)^2 = 8\pi \] \[ A_{total} = 96 + 8\pi \approx 121.13 \text{ m}^2 \]b) Perimeter: We exclude the internal boundary.
\[ P = 12 + 8 + 12 (\text{rectangle sides}) + \pi(4) (\text{arc}) \approx 44.57 \text{ m} \]Explore more on Intersecting Chords, the Equation of a Circle, and High School Math.