Find the distance between the two points. Round to the nearest tenth if necessary.
1 answer:
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<h3>Answer: C) 12</h3>
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Explanation:
Refer to the diagram below. I'm using the points A,B,C,D,E to help label the triangles, and to label the angles as well.
For now, let's focus on triangle DEC.
The top most angle is given as D = 65. Note how the angles D and E are opposite the congruent lines (marked with the red single tickmarks). This means that angles D and E are congruent to one another. So E = 65 as well. This is shown in the diagram below.
For now we don't know what angle C is, so we'll call it y. More specifically, we'll make y the measure of angle DCE.
For any triangle, the three angles always add to 180
y+65+65 = 180
y+130 = 180
y = 180-130
y = 50
So angle DCE is 50 degrees.
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Now we move onto triangle ABC.
Because angles DCE and ACB are vertical angles, this makes angle ACB 50 degrees also. Effectively, anywhere you see a 'y' in the diagram below, replace it with 50.
So triangle ABC has the three angles y = 50, y = 50 and 7x-4
We can solve for x like so
y+y+(7x-4) = 180
50+50+(7x-4) = 180
7x+96 = 180
7x = 180-96
7x = 84
x = 84/7
x = 12
This points to choice C as the final answer.
Answer:
The area of a parallelogram is 360 in.²
Step-by-step explanation:
Where DG = GH
GP = 12 in.
AB = 39 in.
∠DAB + ∠ABC = 180° (Adjacent angles of a parallelogram)
Whereby ∠DAB is bisected by AG and ∠ABC is bisected by BH
Therefore, ∠GAB + ∠HBA = 90°
Hence, ∠BPA = 90° (Sum of interior angles of a triangle)
We note that ∠AGD = ∠GAB (Alternate angles of parallel lines)
∴ ∠AGD = ∠AGD since ∠AGD = ∠GAB (Bisected angle)
Hence AD = DG (Side length of isosceles triangle)
The bisector of ∠ADG is parallel to BH and will bisect AG at point Q
Hence ΔDAQ ≅ ΔDGQ ≅ ΔGPH and AQ = QG = GP
Hence, AP = 3 × GP = 3 × 12 = 36
∠GAB = 22.62°
GH = 13 in.
∴ AD 13 in.
BP = 39 × sin(22.62°) = 15 in.
GH = √(GP² + HP²)
∠DAB = 2 × 22.62° = 45.24°
The height of the parallelogram = AD × sin(∠DAB) = 13 × sin(45.24°)
The height of the parallelogram = 120/13 = 9.23 in.
The area of a parallelogram = Base × Height = (120/13) × 39 = 360 in.²
