Question: Estimate
8.7 \times 20.3
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(8.898−6.01)^2
A circle has a radius of 11.2m
(a) Estimate the area of the circle
You must show your working
(b) State whether your answer to (a) an underestimate or an
overestimate.
You must explain your answer.
A number p is rounded to 1 decimal place. The result is 15.4
Write down error interval for p.
A number h is truncated to 2 significant figures. The result is 6.2.
Write the error interval for h.
The length of a rectangle is measured to be 18cm to the nearest cm.
The width of the rectangle is 20cm to the nearest 10cm.
18cm
20cm
Find the lower bound for the area of the rectangle to 2d.p.
F = 25.14 N correct to 2 decimal places
A = 4.29 m² correct to 3 significant figures
By considering bound, find the value of p to a suitable degree of accuracy.
p = F
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A
p = pressure
F = force
A = area
1. **Estimate** $$\frac{8.7 \times 20.3}{(8.898 - 6.01)^2}$$
- Calculate numerator: $$8.7 \times 20.3 = 176.61$$
- Calculate denominator: $$8.898 - 6.01 = 2.888$$
- Square denominator: $$2.888^2 = 8.341\ (approx.)$$
- Divide: $$\frac{176.61}{8.341} \approx 21.18$$
2. **Estimate area of circle with radius 11.2 m**
- Formula for area: $$A = \pi r^2$$
- Use $$\pi \approx 3.14$$ for estimation
- Calculate: $$A \approx 3.14 \times (11.2)^2 = 3.14 \times 125.44 = 393.8\ m^2$$
3. **Is the answer an underestimate or overestimate?**
- Using $$\pi \approx 3.14$$ underestimates $$\pi$$ (actual $$\pi \approx 3.14159$$)
- So, area estimate is an **underestimate** because $$3.14 < \pi$$
4. **Error interval for p rounded to 1 decimal place with result 15.4**
- Rounding to 1 decimal place means error is $$\pm 0.05$$
- Interval: $$15.35 \leq p < 15.45$$
5. **Error interval for h truncated to 2 significant figures with result 6.2**
- Truncation means $$h$$ is at least 6.2 but less than next number at 2 sig figs
- Next number after 6.2 with 2 sig figs is 6.3
- Interval: $$6.2 \leq h < 6.3$$
6. **Lower bound for area of rectangle**
- Length measured to nearest cm: lower bound $$= 18 - 0.5 = 17.5\ cm$$
- Width measured to nearest 10 cm: lower bound $$= 20 - 5 = 15\ cm$$
- Area lower bound: $$17.5 \times 15 = 262.5\ cm^2$$
- Rounded to 2 d.p.: $$262.50\ cm^2$$
7. **Find pressure p with bounds**
- Force $$F = 25.14$$ N correct to 2 d.p., so bounds:
$$25.135 \leq F < 25.145$$
- Area $$A = 4.29$$ m² correct to 3 sig figs, so bounds:
$$4.285 \leq A < 4.295$$
- Pressure $$p = \frac{F}{A}$$
- Calculate upper bound for $$p$$ (max $$F$$, min $$A$$):
$$p_{max} = \frac{25.145}{4.285} \approx 5.87$$
- Calculate lower bound for $$p$$ (min $$F$$, max $$A$$):
$$p_{min} = \frac{25.135}{4.295} \approx 5.85$$
- So, $$p \approx 5.86 \pm 0.01$$ (to 2 decimal places)