Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Certainly! Let's solve this step-by-step.
### Step-by-Step Solution
#### Step 1: Constructing the Triangle
We are given the sides of a triangle:
- [tex]\( a = 7 \)[/tex] cm
- [tex]\( b = 6 \)[/tex] cm
- [tex]\( c = 5 \)[/tex] cm
You can construct this triangle using a compass and a ruler. Here is how you can construct it:
1. Draw a line segment [tex]\( AB = 7 \)[/tex] cm.
2. Using a compass, draw an arc of radius 6 cm from point [tex]\( A \)[/tex].
3. Using the same compass, draw another arc of radius 5 cm from point [tex]\( B \)[/tex].
4. The intersection of these two arcs will be point [tex]\( C \)[/tex].
5. Connect [tex]\( C \)[/tex] to [tex]\( A \)[/tex] and [tex]\( C \)[/tex] to [tex]\( B \)[/tex].
Now you have triangle [tex]\( ABC \)[/tex] with the specified sides.
#### Step 2: Finding the Area of Triangle [tex]\( ABC \)[/tex]
To find the area of the triangle [tex]\( ABC \)[/tex], we use Heron's formula:
- First, we calculate the semi-perimeter [tex]\( s \)[/tex]:
[tex]\[ s = \frac{a + b + c}{2} = \frac{7 + 6 + 5}{2} = 9 \, \text{cm} \][/tex]
- Then, the area [tex]\( A \)[/tex] of the triangle is given by:
[tex]\[ A = \sqrt{s(s-a)(s-b)(s-c)} \][/tex]
Substituting the values:
[tex]\[ A = \sqrt{9 \cdot (9 - 7) \cdot (9 - 6) \cdot (9 - 5)} \][/tex]
[tex]\[ A = \sqrt{9 \cdot 2 \cdot 3 \cdot 4} \][/tex]
[tex]\[ A = \sqrt{216} \][/tex]
[tex]\[ A \approx 14.696938456699069 \, \text{cm}^2 \][/tex]
#### Step 3: Constructing the Parallelogram
We are to construct a parallelogram whose area is equal to the area of triangle [tex]\( ABC \)[/tex] and one of whose angles is [tex]\( 60^\circ \)[/tex]. Let the base of the parallelogram be one of the sides of the triangle, say [tex]\( 7 \)[/tex] cm (same as [tex]\( a \)[/tex]).
#### Step 4: Calculating the Height of the Parallelogram
The area of the parallelogram is given by:
[tex]\[ \text{Area}_{\text{parallelogram}} = \text{base} \times \text{height} \][/tex]
Given:
- Area of parallelogram [tex]\( = 14.696938456699069 \, \text{cm}^2 \)[/tex]
- Base [tex]\( = 7 \, \text{cm} \)[/tex]
- Angle [tex]\( = 60^\circ \)[/tex]
We need to find the height. The height [tex]\( h \)[/tex] can be calculated as follows:
From the area formula:
[tex]\[ h = \frac{\text{Area}}{\text{base}} \][/tex]
Substituting the values:
[tex]\[ h = \frac{14.696938456699069}{7} \][/tex]
[tex]\[ h = 2.424366106925306 \, \text{cm} \][/tex]
Thus, the height of the parallelogram, when one of the angles is [tex]\( 60^\circ \)[/tex], is approximately [tex]\( 2.424 \, \text{cm} \)[/tex].
### Reason
The construction follows from the provided side lengths and the angle constraint, with the area being conserved throughout the transformations. This ensures that the derived properties hold true geometrically as well as mathematically.
By following these steps, you will have both constructed a triangle [tex]\( ABC \)[/tex] and a parallelogram with the same area and a specified angle of [tex]\( 60^\circ \)[/tex].
### Step-by-Step Solution
#### Step 1: Constructing the Triangle
We are given the sides of a triangle:
- [tex]\( a = 7 \)[/tex] cm
- [tex]\( b = 6 \)[/tex] cm
- [tex]\( c = 5 \)[/tex] cm
You can construct this triangle using a compass and a ruler. Here is how you can construct it:
1. Draw a line segment [tex]\( AB = 7 \)[/tex] cm.
2. Using a compass, draw an arc of radius 6 cm from point [tex]\( A \)[/tex].
3. Using the same compass, draw another arc of radius 5 cm from point [tex]\( B \)[/tex].
4. The intersection of these two arcs will be point [tex]\( C \)[/tex].
5. Connect [tex]\( C \)[/tex] to [tex]\( A \)[/tex] and [tex]\( C \)[/tex] to [tex]\( B \)[/tex].
Now you have triangle [tex]\( ABC \)[/tex] with the specified sides.
#### Step 2: Finding the Area of Triangle [tex]\( ABC \)[/tex]
To find the area of the triangle [tex]\( ABC \)[/tex], we use Heron's formula:
- First, we calculate the semi-perimeter [tex]\( s \)[/tex]:
[tex]\[ s = \frac{a + b + c}{2} = \frac{7 + 6 + 5}{2} = 9 \, \text{cm} \][/tex]
- Then, the area [tex]\( A \)[/tex] of the triangle is given by:
[tex]\[ A = \sqrt{s(s-a)(s-b)(s-c)} \][/tex]
Substituting the values:
[tex]\[ A = \sqrt{9 \cdot (9 - 7) \cdot (9 - 6) \cdot (9 - 5)} \][/tex]
[tex]\[ A = \sqrt{9 \cdot 2 \cdot 3 \cdot 4} \][/tex]
[tex]\[ A = \sqrt{216} \][/tex]
[tex]\[ A \approx 14.696938456699069 \, \text{cm}^2 \][/tex]
#### Step 3: Constructing the Parallelogram
We are to construct a parallelogram whose area is equal to the area of triangle [tex]\( ABC \)[/tex] and one of whose angles is [tex]\( 60^\circ \)[/tex]. Let the base of the parallelogram be one of the sides of the triangle, say [tex]\( 7 \)[/tex] cm (same as [tex]\( a \)[/tex]).
#### Step 4: Calculating the Height of the Parallelogram
The area of the parallelogram is given by:
[tex]\[ \text{Area}_{\text{parallelogram}} = \text{base} \times \text{height} \][/tex]
Given:
- Area of parallelogram [tex]\( = 14.696938456699069 \, \text{cm}^2 \)[/tex]
- Base [tex]\( = 7 \, \text{cm} \)[/tex]
- Angle [tex]\( = 60^\circ \)[/tex]
We need to find the height. The height [tex]\( h \)[/tex] can be calculated as follows:
From the area formula:
[tex]\[ h = \frac{\text{Area}}{\text{base}} \][/tex]
Substituting the values:
[tex]\[ h = \frac{14.696938456699069}{7} \][/tex]
[tex]\[ h = 2.424366106925306 \, \text{cm} \][/tex]
Thus, the height of the parallelogram, when one of the angles is [tex]\( 60^\circ \)[/tex], is approximately [tex]\( 2.424 \, \text{cm} \)[/tex].
### Reason
The construction follows from the provided side lengths and the angle constraint, with the area being conserved throughout the transformations. This ensures that the derived properties hold true geometrically as well as mathematically.
By following these steps, you will have both constructed a triangle [tex]\( ABC \)[/tex] and a parallelogram with the same area and a specified angle of [tex]\( 60^\circ \)[/tex].
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
