IB myp 4-5 MATHEMATICS – Practice Questions- All Topics
Topic :Geometry–Enlargement around a given point
Topic :Geometry- Weightage : 21 %
All Questions for Topic : Volume and capacity (additional shapes),Enlargement around a given point,Enlargement by a rational factor,Gradients of perpendicular lines,Identical representation of transformations
Question
The following question about the requirements for a reliable mobile phone signal. Mobile phones are also known as cell phones.
In Diagram 1, identical transmitters are positioned at the vertices of an equilateral triangle. The equilateral triangle has sides of length $20 \mathrm{~km}$ and area $173 \mathrm{~km}^2$. The transmitters are placed at $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$, and they emit a frequency in a circular radius up to $10 \mathrm{~km}$.
Question (a)
The unshaded area inside the triangle represents the part of the triangle that receives no signal. Find the area of the triangle that does not receive a signal.
▶️Answer/Explanation
Ans:
To find the area of the triangle that does not receive a signal, we can use the given information and calculations:
The shaded region represents the part of the triangle that receives no signal. We can calculate the area of this shaded region by subtracting the area of the sectors from the area of the triangle.
The area of a circle with a radius of $10$ units is given by $\pi \times (10^2)$ or $100 \pi$ (approximately $314$ square units).
One possible method to calculate the area of the shaded region is by considering the area of the sectors. Each sector forms $\frac{1}{6}$ of the area of the circle.
The area of one sector is $\frac{1}{6}$ of the area of the circle, which is $\frac{1}{6} \times 100 \pi = \frac{50}{3} \pi$ (approximately $52.359$ square units).
Since there are three sectors in the shaded region, the total area of the shaded region (the part of the triangle that does not receive a signal) is $3 \times \frac{50}{3} \pi = 50 \pi$ (approximately $157.08$ square units).
Therefore, the area of the triangle that receive a signal is approximately $157.08$ square units.
If we subtract the area of the sectors ($50 \pi$) from the total area of the triangle (given as $173$ square units), we get $173 – 50 \pi \approx 15.92$ square units.
Hence, the area of the triangle that does not receive a signal is approximately $15.92$ square units or $16$ square units (rounded to the nearest whole number).
Question (b)
In Diagram 2, identical transmitters are positioned at the vertices of a square. The square has sides of length $20 \mathrm{~km}$ and area $400 \mathrm{~km}^2$. The transmitters are placed at $A, B, C$ and $D$, and they emit a frequency in a circular radius up to $10 \mathrm{~km}$.
The unshaded area inside the square represents the part of the square that receives no signal. Find the area of the square that does not receive a signal.
▶️Answer/Explanation
Ans:
The area covered by each transmitter’s circular signal is $\frac{1}{4} \pi \times (10 \mathrm{~km})^2 = 25 \pi \mathrm{~km}^2$.
Since there are four transmitters positioned at the vertices of the square, the total area covered by the circular signals is $4 \times (25 \pi \mathrm{~km}^2) = 100 \pi \mathrm{~km}^2$.
Therefore, the area of the square that does not receive a signal is $400 \mathrm{~km}^2 – 100 \pi \mathrm{~km}^2 \approx 400 – 314 \approx 86 \mathrm{~km}^2$.