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Sagot :
Absolutely, let's go through the detailed steps to solve this problem.
(i) Construct the parallelogram ABCD:
1. Draw AB = 6 cm: Start by drawing a horizontal line segment AB that is 6 cm long.
2. Draw angle ABC = 60°: At point B, construct a 60° angle with AB. This will be the line BC.
3. Draw BC = 7.5 cm: Measure 7.5 cm from B along the 60° angle line to point C.
4. Draw line parallel to AB from point C: From point C, draw a line parallel to AB using a ruler and a set square.
5. Draw line parallel to BC from point A: From point A, draw a line parallel to BC.
6. Intersection point D: These two parallel lines will intersect at point D. Connect D to A, forming the parallelogram ABCD.
(ii) Construct a triangle with equal area to the parallelogram ABCD:
1. Area of the parallelogram ABCD: The area of parallelogram is given by
[tex]\[ \text{Area of parallelogram} = AB \times BC \times \sin(\angle ABC) \][/tex]
Plugging in the values,
[tex]\[ \text{Area of parallelogram} = 6 \, \text{cm} \times 7.5 \, \text{cm} \times \sin(60^\circ) \][/tex]
The sine of 60° is [tex]\(\frac{\sqrt{3}}{2}\)[/tex], so
[tex]\[ \text{Area of parallelogram} \approx 6 \times 7.5 \times 0.866 = 38.97114317029974 \, \text{cm}^2 \][/tex]
2. Area of the triangle: A triangle with an area equal to half of the parallelogram’s area.
[tex]\[ \text{Area of triangle} = \frac{38.97114317029974 \, \text{cm}^2}{2} = 19.48557158514987 \, \text{cm}^2 \][/tex]
(iii) Construct the height of the triangle and measure it:
1. Height of the triangle: We’ll use the area formula for triangles. The area of a triangle is given by:
[tex]\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]
Given that the base is [tex]\(BC = 7.5 \, \text{cm}\)[/tex] and the area is [tex]\(19.48557158514987 \, \text{cm}^2\)[/tex], we can solve for the height:
[tex]\[ 19.48557158514987 = \frac{1}{2} \times 7.5 \times \text{height} \][/tex]
Therefore,
[tex]\[ \text{height} = \frac{2 \times 19.48557158514987}{7.5} = 5.196152422706632 \, \text{cm} \][/tex]
(iv) Calculate the area of the triangle so constructed:
The area of the triangle, based on the calculations above, is:
[tex]\[ \text{Area of triangle} = 19.48557158514987 \, \text{cm}^2 \][/tex]
To summarize:
- The area of the parallelogram ABCD is approximately [tex]\(38.97114317029974 \, \text{cm}^2\)[/tex].
- The area of the triangle with equal area is [tex]\(19.48557158514987 \, \text{cm}^2\)[/tex], as the area of the triangle equals half the area of the parallelogram.
- The height of the constructed triangle from the base BC is [tex]\(5.196152422706632 \, \text{cm}\)[/tex].
(i) Construct the parallelogram ABCD:
1. Draw AB = 6 cm: Start by drawing a horizontal line segment AB that is 6 cm long.
2. Draw angle ABC = 60°: At point B, construct a 60° angle with AB. This will be the line BC.
3. Draw BC = 7.5 cm: Measure 7.5 cm from B along the 60° angle line to point C.
4. Draw line parallel to AB from point C: From point C, draw a line parallel to AB using a ruler and a set square.
5. Draw line parallel to BC from point A: From point A, draw a line parallel to BC.
6. Intersection point D: These two parallel lines will intersect at point D. Connect D to A, forming the parallelogram ABCD.
(ii) Construct a triangle with equal area to the parallelogram ABCD:
1. Area of the parallelogram ABCD: The area of parallelogram is given by
[tex]\[ \text{Area of parallelogram} = AB \times BC \times \sin(\angle ABC) \][/tex]
Plugging in the values,
[tex]\[ \text{Area of parallelogram} = 6 \, \text{cm} \times 7.5 \, \text{cm} \times \sin(60^\circ) \][/tex]
The sine of 60° is [tex]\(\frac{\sqrt{3}}{2}\)[/tex], so
[tex]\[ \text{Area of parallelogram} \approx 6 \times 7.5 \times 0.866 = 38.97114317029974 \, \text{cm}^2 \][/tex]
2. Area of the triangle: A triangle with an area equal to half of the parallelogram’s area.
[tex]\[ \text{Area of triangle} = \frac{38.97114317029974 \, \text{cm}^2}{2} = 19.48557158514987 \, \text{cm}^2 \][/tex]
(iii) Construct the height of the triangle and measure it:
1. Height of the triangle: We’ll use the area formula for triangles. The area of a triangle is given by:
[tex]\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]
Given that the base is [tex]\(BC = 7.5 \, \text{cm}\)[/tex] and the area is [tex]\(19.48557158514987 \, \text{cm}^2\)[/tex], we can solve for the height:
[tex]\[ 19.48557158514987 = \frac{1}{2} \times 7.5 \times \text{height} \][/tex]
Therefore,
[tex]\[ \text{height} = \frac{2 \times 19.48557158514987}{7.5} = 5.196152422706632 \, \text{cm} \][/tex]
(iv) Calculate the area of the triangle so constructed:
The area of the triangle, based on the calculations above, is:
[tex]\[ \text{Area of triangle} = 19.48557158514987 \, \text{cm}^2 \][/tex]
To summarize:
- The area of the parallelogram ABCD is approximately [tex]\(38.97114317029974 \, \text{cm}^2\)[/tex].
- The area of the triangle with equal area is [tex]\(19.48557158514987 \, \text{cm}^2\)[/tex], as the area of the triangle equals half the area of the parallelogram.
- The height of the constructed triangle from the base BC is [tex]\(5.196152422706632 \, \text{cm}\)[/tex].
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