Answer :
Sure, let's go through each part step-by-step.
### Part (a)
To find the sum of the numbers 2,483.65, 701.532, and 102.7 and give the answer to one decimal place.
1. Add the three numbers:
[tex]\[ 2483.65 + 701.532 + 102.7 = 3287.882 \][/tex]
2. Round this result to one decimal place:
[tex]\[ 3287.9 \][/tex]
Therefore, the sum of the numbers to one decimal place is 3287.9.
### Part (b)
In this part, we will calculate the perimeter and area of the quadrilateral ABCD. Let's start with the given information and proceed with the calculations.
Given:
- AB = 3 cm
- BC = 4 cm
- CD = 12 cm
- Angle ABC = 90°
#### (i) Perimeter of ABCD
1. Calculate the length of BD using the Pythagorean theorem. Because triangle ABC is a right triangle:
[tex]\[ BD^2 = AB^2 + BC^2 \][/tex]
[tex]\[ BD = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \; \text{cm} \][/tex]
2. Add the lengths of AB, BC, CD, and BD to find the perimeter:
[tex]\[ \text{Perimeter of ABCD} = AB + BC + CD + BD = 3 + 4 + 12 + 5 = 24 \; \text{cm} \][/tex]
Therefore, the perimeter of the quadrilateral ABCD is 24.0 cm.
#### (ii) Area of ABCD
1. Calculate the area of triangle ABC. Since triangle ABC is a right triangle, its area can be calculated as:
[tex]\[ \text{Area of } \triangle ABC = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 3 \times 4 = 6 \; \text{cm}^2 \][/tex]
2. Calculate the area of triangle BCD. To simplify, assume CD is the base and BD is the height (which might not be exactly accurate if this were a more complex figure, but we use it here as given):
[tex]\[ \text{Area of } \triangle BCD = \frac{1}{2} \times CD \times BD = \frac{1}{2} \times 12 \times 5 = 30 \; \text{cm}^2 \][/tex]
3. Add the areas of triangles ABC and BCD to find the total area of quadrilateral ABCD:
[tex]\[ \text{Area of ABCD} = \text{Area of } \triangle ABC + \text{Area of } \triangle BCD = 6 + 30 = 36 \; \text{cm}^2 \][/tex]
Therefore, the area of the quadrilateral ABCD is 36.0 cm².
In summary:
- The sum of the numbers is 3287.9.
- The perimeter of ABCD is 24.0 cm.
- The area of ABCD is 36.0 cm².
### Part (a)
To find the sum of the numbers 2,483.65, 701.532, and 102.7 and give the answer to one decimal place.
1. Add the three numbers:
[tex]\[ 2483.65 + 701.532 + 102.7 = 3287.882 \][/tex]
2. Round this result to one decimal place:
[tex]\[ 3287.9 \][/tex]
Therefore, the sum of the numbers to one decimal place is 3287.9.
### Part (b)
In this part, we will calculate the perimeter and area of the quadrilateral ABCD. Let's start with the given information and proceed with the calculations.
Given:
- AB = 3 cm
- BC = 4 cm
- CD = 12 cm
- Angle ABC = 90°
#### (i) Perimeter of ABCD
1. Calculate the length of BD using the Pythagorean theorem. Because triangle ABC is a right triangle:
[tex]\[ BD^2 = AB^2 + BC^2 \][/tex]
[tex]\[ BD = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \; \text{cm} \][/tex]
2. Add the lengths of AB, BC, CD, and BD to find the perimeter:
[tex]\[ \text{Perimeter of ABCD} = AB + BC + CD + BD = 3 + 4 + 12 + 5 = 24 \; \text{cm} \][/tex]
Therefore, the perimeter of the quadrilateral ABCD is 24.0 cm.
#### (ii) Area of ABCD
1. Calculate the area of triangle ABC. Since triangle ABC is a right triangle, its area can be calculated as:
[tex]\[ \text{Area of } \triangle ABC = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 3 \times 4 = 6 \; \text{cm}^2 \][/tex]
2. Calculate the area of triangle BCD. To simplify, assume CD is the base and BD is the height (which might not be exactly accurate if this were a more complex figure, but we use it here as given):
[tex]\[ \text{Area of } \triangle BCD = \frac{1}{2} \times CD \times BD = \frac{1}{2} \times 12 \times 5 = 30 \; \text{cm}^2 \][/tex]
3. Add the areas of triangles ABC and BCD to find the total area of quadrilateral ABCD:
[tex]\[ \text{Area of ABCD} = \text{Area of } \triangle ABC + \text{Area of } \triangle BCD = 6 + 30 = 36 \; \text{cm}^2 \][/tex]
Therefore, the area of the quadrilateral ABCD is 36.0 cm².
In summary:
- The sum of the numbers is 3287.9.
- The perimeter of ABCD is 24.0 cm.
- The area of ABCD is 36.0 cm².
