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Sagot :
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### Problem 7: Triangle [tex]\( \triangle KLM \)[/tex] with vertices [tex]\( K(-5,18) \)[/tex], [tex]\( L(10,-2) \)[/tex], and [tex]\( M(-5,-10) \)[/tex].
#### (a) Find the length of each side.
To find the lengths of the sides of the triangle, we will use the distance formula:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
1. Length of side [tex]\( KL \)[/tex]:
- Coordinates of [tex]\( K \)[/tex] are [tex]\((-5, 18)\)[/tex].
- Coordinates of [tex]\( L \)[/tex] are [tex]\( (10, -2) \)[/tex].
- [tex]\( \text{Length of } KL = \sqrt{(10 - (-5))^2 + (-2 - 18)^2} \)[/tex].
- [tex]\( KL = \sqrt{(10 + 5)^2 + (-2 - 18)^2} \)[/tex].
- [tex]\( KL = \sqrt{15^2 + (-20)^2} \)[/tex].
- [tex]\( KL = \sqrt{225 + 400} \)[/tex].
- [tex]\( KL = \sqrt{625} \)[/tex].
- [tex]\( KL = 25 \)[/tex].
2. Length of side [tex]\( LM \)[/tex]:
- Coordinates of [tex]\( L \)[/tex] are [tex]\( (10, -2) \)[/tex].
- Coordinates of [tex]\( M \)[/tex] are [tex]\((-5, -10)\)[/tex].
- [tex]\( \text{Length of } LM = \sqrt{(10 - (-5))^2 + (-2 - (-10))^2} \)[/tex].
- [tex]\( LM = \sqrt{(10 + 5)^2 + (-2 + 10)^2} \)[/tex].
- [tex]\( LM = \sqrt{15^2 + 8^2} \)[/tex].
- [tex]\( LM = \sqrt{225 + 64} \)[/tex].
- [tex]\( LM = \sqrt{289} \)[/tex].
- [tex]\( LM = 17 \)[/tex].
3. Length of side [tex]\( MK \)[/tex]:
- Coordinates of [tex]\( M \)[/tex] are [tex]\((-5, -10)\)[/tex].
- Coordinates of [tex]\( K \)[/tex] are [tex]\((-5, 18)\)[/tex].
- [tex]\( \text{Length of } MK = \sqrt{(-5 - (-5))^2 + (18 - (-10))^2} \)[/tex].
- [tex]\( MK = \sqrt{(0)^2 + (18 + 10)^2} \)[/tex].
- [tex]\( MK = \sqrt{0 + 28^2} \)[/tex].
- [tex]\( MK = \sqrt{784} \)[/tex].
- [tex]\( MK = 28 \)[/tex].
#### (b) Find the perimeter of [tex]\( \triangle KLM \)[/tex].
The perimeter of the triangle is the sum of the lengths of its sides:
[tex]\[ \text{Perimeter} = KL + LM + MK \][/tex]
[tex]\[ \text{Perimeter} = 25 + 17 + 28 \][/tex]
[tex]\[ \text{Perimeter} = 70 \][/tex]
#### (c) Find the area of [tex]\( \triangle KLM \)[/tex].
To find the area of the triangle given its vertices, we can use the Shoelace formula:
[tex]\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \][/tex]
For [tex]\( \triangle KLM \)[/tex] with vertices [tex]\( K(-5,18) \)[/tex], [tex]\( L(10,-2) \)[/tex], and [tex]\( M(-5,-10) \)[/tex]:
[tex]\[ \text{Area} = \frac{1}{2} \left| (-5)((-2) - (-10)) + 10((-10) - 18) + (-5)(18 - (-2)) \right| \][/tex]
Substituting the values:
[tex]\[ \text{Area} = \frac{1}{2} \left| (-5)(8) + 10(-28) + (-5)(20) \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| (-40) - 280 - 100 \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| -420 \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \times 420 \][/tex]
[tex]\[ \text{Area} = 210 \][/tex]
### Summary of results:
(a) The lengths of the sides are: [tex]\( KL = 25 \)[/tex], [tex]\( LM = 17 \)[/tex], [tex]\( MK = 28 \)[/tex].
(b) The perimeter of [tex]\( \triangle KLM \)[/tex] is: 70.
(c) The area of [tex]\( \triangle KLM \)[/tex] is: 210.
### Problem 7: Triangle [tex]\( \triangle KLM \)[/tex] with vertices [tex]\( K(-5,18) \)[/tex], [tex]\( L(10,-2) \)[/tex], and [tex]\( M(-5,-10) \)[/tex].
#### (a) Find the length of each side.
To find the lengths of the sides of the triangle, we will use the distance formula:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
1. Length of side [tex]\( KL \)[/tex]:
- Coordinates of [tex]\( K \)[/tex] are [tex]\((-5, 18)\)[/tex].
- Coordinates of [tex]\( L \)[/tex] are [tex]\( (10, -2) \)[/tex].
- [tex]\( \text{Length of } KL = \sqrt{(10 - (-5))^2 + (-2 - 18)^2} \)[/tex].
- [tex]\( KL = \sqrt{(10 + 5)^2 + (-2 - 18)^2} \)[/tex].
- [tex]\( KL = \sqrt{15^2 + (-20)^2} \)[/tex].
- [tex]\( KL = \sqrt{225 + 400} \)[/tex].
- [tex]\( KL = \sqrt{625} \)[/tex].
- [tex]\( KL = 25 \)[/tex].
2. Length of side [tex]\( LM \)[/tex]:
- Coordinates of [tex]\( L \)[/tex] are [tex]\( (10, -2) \)[/tex].
- Coordinates of [tex]\( M \)[/tex] are [tex]\((-5, -10)\)[/tex].
- [tex]\( \text{Length of } LM = \sqrt{(10 - (-5))^2 + (-2 - (-10))^2} \)[/tex].
- [tex]\( LM = \sqrt{(10 + 5)^2 + (-2 + 10)^2} \)[/tex].
- [tex]\( LM = \sqrt{15^2 + 8^2} \)[/tex].
- [tex]\( LM = \sqrt{225 + 64} \)[/tex].
- [tex]\( LM = \sqrt{289} \)[/tex].
- [tex]\( LM = 17 \)[/tex].
3. Length of side [tex]\( MK \)[/tex]:
- Coordinates of [tex]\( M \)[/tex] are [tex]\((-5, -10)\)[/tex].
- Coordinates of [tex]\( K \)[/tex] are [tex]\((-5, 18)\)[/tex].
- [tex]\( \text{Length of } MK = \sqrt{(-5 - (-5))^2 + (18 - (-10))^2} \)[/tex].
- [tex]\( MK = \sqrt{(0)^2 + (18 + 10)^2} \)[/tex].
- [tex]\( MK = \sqrt{0 + 28^2} \)[/tex].
- [tex]\( MK = \sqrt{784} \)[/tex].
- [tex]\( MK = 28 \)[/tex].
#### (b) Find the perimeter of [tex]\( \triangle KLM \)[/tex].
The perimeter of the triangle is the sum of the lengths of its sides:
[tex]\[ \text{Perimeter} = KL + LM + MK \][/tex]
[tex]\[ \text{Perimeter} = 25 + 17 + 28 \][/tex]
[tex]\[ \text{Perimeter} = 70 \][/tex]
#### (c) Find the area of [tex]\( \triangle KLM \)[/tex].
To find the area of the triangle given its vertices, we can use the Shoelace formula:
[tex]\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \][/tex]
For [tex]\( \triangle KLM \)[/tex] with vertices [tex]\( K(-5,18) \)[/tex], [tex]\( L(10,-2) \)[/tex], and [tex]\( M(-5,-10) \)[/tex]:
[tex]\[ \text{Area} = \frac{1}{2} \left| (-5)((-2) - (-10)) + 10((-10) - 18) + (-5)(18 - (-2)) \right| \][/tex]
Substituting the values:
[tex]\[ \text{Area} = \frac{1}{2} \left| (-5)(8) + 10(-28) + (-5)(20) \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| (-40) - 280 - 100 \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| -420 \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \times 420 \][/tex]
[tex]\[ \text{Area} = 210 \][/tex]
### Summary of results:
(a) The lengths of the sides are: [tex]\( KL = 25 \)[/tex], [tex]\( LM = 17 \)[/tex], [tex]\( MK = 28 \)[/tex].
(b) The perimeter of [tex]\( \triangle KLM \)[/tex] is: 70.
(c) The area of [tex]\( \triangle KLM \)[/tex] is: 210.
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