Solve the math textbook exercise 5 Lesson: Review on calculating the perimeter and area of some shapes
1. Solve problem 1 page 166 Math textbook 5
A rectangular orchard has length \(120m\), width equal to \(\dfrac{2}{3}\) length.
a) Calculate the perimeter of that garden.
b) Calculate the area of that garden in square meters, hectares.
Solution method
– Calculate width = length \(\times \,\dfrac{2}{3}\).
– Calculate perimeter = (length + width) \(\times \,2\).
– Calculate area = length \(\times\) width.
Solution guide
a) The width of the rectangular garden is:
\(120 \times \dfrac{2}{3} = 80\;(m)\)
The perimeter of the rectangular garden is:
\((120 + 80) \times 2 = 400\;(m)\)
b) The area of the rectangular garden is:
\(120 \times 80 = 9600\;(m^2)\)
\(9600m^2= 0.96ha\)
Answer:
a) \(400m\)
b) \(9600m^2\); \(0.96ha\).
2. Solve problem 2 page 167 Math textbook 5
The figure below is a trapezoidal plot of land drawn on a map of scale \(1 : 1000\). Calculate the area of that land in square meters.
Solution method
Step 1: The length of the large bottom, the small bottom, the actual height of the land
Step 2: Calculate the actual area of the land
Solution guide
The length of the great base of the plot is:
\(5 \times 1000 = 5000 \, (cm) = 50m\)
The length of the bottom of the plot is:
\(3 \times 1000 = 3000 \, (cm) = 30m\)
The length and height of the plot is:
\(2 \times 1000 = 2000 \, (cm) = 20m\)
The actual area of the land is:
\(\dfrac{(50 + 30) \times 20}{2} = 800 \, (m^2)\)
Answer: \(800m^2\)
3. Solve problem 3 page 167 Math textbook 5
On the figure below, calculate the area:
a) Square \(ABCD\).
b) The colored part of the circle.
Solution method
– Area of square \(ABCD\) is equal to \(4\) times the area of triangle \(BOC\). Triangle \(BOC\) is a right triangle with the lengths of two right angles \(4cm\) and \(4cm\).
– The area of the colored part of the circle is equal to the area of the circle with radius \(4cm\) minus the area of the square \(ABCD\).
Solution guide
a) The area of \(ABCD\) square is equal to \(4\) times the area of triangle \(BOC\). Triangle \(BOC\) is a right triangle with the lengths of two right angles \(4cm\) and \(4cm\).
Area of right triangle \(BOC\) is:
\(\dfrac{{4 \times 4}}{2} = 8\;(cm^2)\)
The area of square ABCD is:
\(8 × 4 = 32\; (cm^2)\)
b) Observing the figure has shown that a circle with center \(O\) has a radius of \(OA = OB = OC = OD = 4cm\).
The area of the shaded part of the circle is equal to the area of the circle with radius \(4cm\) minus the area of the square \(ABCD\).
The area of a circle with center \(O\) is:
\(4 × 4 × 3.14 = 50.24\; (cm^2)\)
The area of the shaded part of the circle is:
\(50.24 – 32 = 18.24\;(cm^2)\)
Answer:
a) \(32cm^2\)
b) \(18.24cm^2\).
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