Using the slope formula find the slope of the line that passes through these points (2,5) and (8,-3)
2 answers:
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Answer:
6 square units
Step-by-step explanation:
The attachment shows the figure redrawn to scale, along with helpful semicircle O with diameter EB. Trig relations can be used to find the area of triangle ADB. The relation between an inscribed angle and its corresponding central angle is also helpful.
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<h3>angles</h3>Any right triangle can be inscribed in a semicircle whose diameter is the hypotenuse. Accordingly, we have constructed semicircle O whose center is the midpoint of EB. Hence its radius is EB/2 = 4/2 = 2.
Then angle EBD is an inscribed angle, hence half the measure of central angle EOD. Since ∠EOD is double the measure of ∠EBD, it is congruent to ∠EBC, which is also double the measure of ∠EBD. We have called the measure of this double angle θ. Its complement is the angle marked β.
<h3>segments</h3>The fact that angles EOD and EBC are congruent means that segments OD and BC are parallel. Segment BC is the altitude of ΔADB with respect to base AD.
The tangent ratio is ...
Tan = Opposite/Adjacent
Applying that to angle β, we have
tan(β) = OD/AD [eqn 1]
tan(β) = BC/AC [eqn 2]
These two relations can be solved for the measures of AD and BC, which we need in order to find the area of ΔADB.
<h3>area</h3>Equation [eqn 1] tells us ...
AD = OD/tan(β)
Equation [eqn 2] tells us ...
BC = AC·tan(β)
The area of ΔADB is ...
Area = 1/2bh
Area = 1/2(AD)(BC) = 1/2(OD/tan(β))·(AC·tan(β))
Area = 1/2(OD)(AC) . . . . . . factors tan(β) cancel
The length of AC is given as 6, and we know OD = 2, so the area is ...
Area = 1/2(2)(6) = 6
The area of triangle ADB is 6 square units.
