Answer:
The area of the triangle is 18 square units.
Step-by-step explanation:
First, we determine the lengths of segments AB, BC and AC by Pythagorean Theorem:
AB
![AB = \sqrt{(5-2)^{2}+[6-(-1)]^{2}}](https://tex.z-dn.net/?f=AB%20%3D%20%5Csqrt%7B%285-2%29%5E%7B2%7D%2B%5B6-%28-1%29%5D%5E%7B2%7D%7D)
BC
AC
![AC = \sqrt{(-1-2)^{2}+[4-(-1)]^{2}}](https://tex.z-dn.net/?f=AC%20%3D%20%5Csqrt%7B%28-1-2%29%5E%7B2%7D%2B%5B4-%28-1%29%5D%5E%7B2%7D%7D)
Now we determine the area of the triangle by Heron's formula:
(1)
(2)
Where:
- Area of the triangle.
- Semiparameter.
If we know that , and , then the area of the triangle is:
The area of the triangle is 18 square units.
SLope = (2 + 2)/(2 + 2) = 4/4 = 1
Answer
1
Answer:
∠1 ≅ ∠2 ⇒ proved down
Step-by-step explanation:
#12
In the given figure
∵ LJ // WK
∵ LP is a transversal
∵ ∠1 and ∠KWP are corresponding angles
∵ The corresponding angles are equal in measures
∴ m∠1 = m∠KWP
∴ ∠1 ≅ ∠KWP ⇒ (1)
∵ WK // AP
∵ WP is a transversal
∵ ∠KWP and ∠WPA are interior alternate angles
∵ The interior alternate angles are equal in measures
∴ m∠KWP = m∠WPA
∴ ∠KWP ≅ ∠WPA ⇒ (2)
→ From (1) and (2)
∵ ∠1 and ∠WPA are congruent to ∠KWP
∴ ∠1 and ∠WPA are congruent
∴ ∠1 ≅ ∠WPA ⇒ (3)
∵ WP // AG
∵ AP is a transversal
∵ ∠WPA and ∠2 are interior alternate angles
∵ The interior alternate angles are equal in measures
∴ m∠WPA = m∠2
∴ ∠WPA ≅ ∠2 ⇒ (4)
→ From (3) and (4)
∵ ∠1 and ∠2 are congruent to ∠WPA
∴ ∠1 and ∠2 are congruent
∴ ∠1 ≅ ∠2 ⇒ proved
Answer:
2/3 of n +5>12
Step-by-step explanation:
<h2>
Answer:</h2>
<u><em>The answer has to be 30. </em></u>
<h2>
Step-by-step explanation:</h2>
<em>Correct me if I am incorrect but I would believe I am correct. Have a good one!</em>
<h2>
✍(◔◡◔)</h2>
