## Questions

- A rectangular room has a length of 12.0 meters and width 8.0 meters. Find the maximum percentage error in estimating the perimeter of the room.
- In this question, mathematical tables or calculators should not be used. The base and perpendicular height of a triangle measured to the nearest centimeters are 12cm and 8cm respectively;

Find;- The absolute error in calculating the are of the triangle
- The percentage error in the area, giving the answer to 1 decimal place

- A rectangular plate has a perimeter of 28cm. determine the dimensions of the plate that give the maximum area
- A wire of length 5.2m is cut into two pieces without wastage. One of the pieces is 3.08m long. What is the shortest possible length of the second piece?
- The dimensions of a rectangle are 10cm and 15cm. If there is an error of 5% in each of the Measurements. Find the percentage error in the area of the rectangle.
- Find the products of 17.3 and 13.8. Find also the percentage error in getting the product.
- The mass of a metal is given as 14kg to the nearest l0g. Find the percentage error in this measurement.
- Complete the table below for the functions y = cos x and y = 2 cos (x +30
^{°}) for 0^{°}≤ X ≤ 360^{°}- On the same axis, draw the graphs of y = cosx and y = 2 cos (x + 30
^{°}) for 0^{°}≤ X≤ 360^{°} - State the amplitude of the graph y = cos x
^{°} - State the period of the graph y = 2cos (x + 30
^{°})

- State the amplitude of the graph y = cos x
- Use your graph to solve

cos x = 2 cos (x + 30^{°})

- On the same axis, draw the graphs of y = cosx and y = 2 cos (x + 30
- Given that 8≤ y ≤ 12 and 1 ≤ x ≤ 6, find the maximum possible value of:
^{(y + x)}/_{(y – x) }

## Answers

- Maximum perimeter = 2(12.05 + 8.05) = 40.2cm

Actual perimeter = 2(12.0 + 18.0)= 40.0cm

Error = 40.2cm – 40.0cm = 0.2cm

%error =^{(0.2 x100)}/_{40}

= 0.5% - A = ½ x 12 x 8 = 48
- Absolute error

=^{[ ½ x12.5 x 8.5 – ½ x 11.5 x 7.5 ]}/_{2}= 5 - % error =
^{5}/_{24}X 100%

= 10.4%

- Absolute error
- A = L x W

A = x (14−x) = 14x − x^{²}^{dA}/_{dx}= 14 – 2x = 0

14 = 2x, »x = 7

Maximum area = 7(14 -7)

= 7x 7 = 49cm^{²}

Shortest possible length of 2^{nd}piece

= 5.15 – 3.085= 2.065m- Absolute error 10 ± 0.05 and 15 ± 0.05

Max area = 10..5 x 15.05

Min area = 9.95 x 14.95 = 148.7525

a.e =^{(150.2525 – 15 + 150 – 148.7525)}/_{2}

= 1.25

% error =^{1.25}/_{150}x 100

= 0.8333% - 17.35 X 13.85 = 240.3

17.35 X 13.75 = 237.2

»17.3 X 13.8 = 238.7

Max err 240.3 – 238.7 = 1.5

Min err 238.7 – 237.2 = 1.6

Max err =^{(1.6 + 1.5)}/_{2}=^{3.1}/_{2}= 1.55

Product 238.7 ± 1.55

Last product 240

Max err = 1.55

Relative err = 1.55

»28.1%

error = 1.55 x 100 = 0.6% 28.1

Relative err =^{1.55}/_{238.7} - 14 Kg to the nearest
^{10}/_{1000}Kg

A.E = 0.01

% E =^{0.01}/_{14}x 100

=0.07 - Amplitude of y = cos x is 1 unit

And Y = 2cos (x +30) 2 units - period of y = 2 cos (x + 30
^{°})

=330^{°}

- Amplitude of y = cos x is 1 unit
- Cos x = 2 cos (x + 30
^{°})

x = 40^{°}± 1

x = 219^{°}± 1

^{(y + x)}/_{(y – x)}=^{(12 + 6)}/_{(8 – 6)}=^{18}/_{2}

= 9

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