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Question 1064005: ΔABC has vertices at A(5,2), B(−3,1), and C(−2,5). Point D is located on the intersection of the altitude and AB�, in such a way that D has coordinates at approximately (−1.52,1.18).
https://cds.flipswitch.com/tools/asset/media/607812
Match each question with the correct measurement, rounded to one decimal place.
What is the approximate area of the triangle?
What is the approximate length of the base of the triangle for the given altitude?
What is the approximate length of the altitude of the triangle?
16.1 units
3.9 units
8.1 units
7.6 units
15.8 units
8.0 units
Found 5 solutions by mananth, ikleyn, greenestamps, MathTherapy, math_tutor2020:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website!
Find AB by distance formula
A(5,2) B(-3,1)
AB = sqrt((2-1)^2+(5-(-3))^2) =sqrt(65) Base of triangle
C=(-2,5) , D(-1.52,1.18)
CD = sqrt((-2+1.52)^2+(5-(1.18))^2) =3.85
Area of triangle = 1/2 *8.06*3.85
15.5 unit^2
Answer by ikleyn(52776)  (Show Source):
You can put this solution on YOUR website! .
Here I will explain you how to find quickly the area of triangle ABC
with the vertices at the integer grid points in a coordinate plane.
Draw the minimal rectangle on the grid which concludes the given triangle.
Identify the triangles on the grid inside this rectangle that are outside the given triangle.
It is quite easy to find the area of each such a triangle.
After that, you will find the area of your triangle ABC mentally, by subtraction
the areas of the three other triangles from the area of rectangle.
Notice that doing this way, You will find the PRECISE area of triangle ABC mentally and quickly,
without doing boring calculations.
It is the best strategy and the most valuable piece of knowledge, which you can learn from this problem.
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
Interesting problem; and interesting responses....
Finding the area of the triangle using the coordinates of A, B, C, and D in the formula for the area of a triangle (one-half base times height) is not a particularly good way. For one thing the coordinates of D are only given approximately; for another thing, you end up doing ugly arithmetic with decimal approximations of the lengths of the base and height.
On the other hand, since the problem asks to find lengths rounded to the nearest tenth, you can get the answer that way.
But, wait -- using either that approximate method or any of at least two exact methods, the correct answer of (exactly) 15.5 is not one of the answer choices. So here the problem is faulty....
Finally, as a note regarding the response from the other tutor, you should never take it as the gospel truth when someone says that a particular method is the best way to work a problem. Enclosing the triangle in a rectangle and finding the area of the given triangle as the area of the rectangle minus the area of the three small triangles is A good way to solve the problem; but there are other equally good ways.
The length of the base AB is found simply using the Pythagorean Theorem -- although again, given the set of answer choices and the instruction to find the length to the nearest tenth, 8.1 is the obvious answer.
And, finally, only one of the answer choices is reasonable for the length of altitude AD.
Answer by MathTherapy(10551)  (Show Source):
You can put this solution on YOUR website!
ΔABC has vertices at A(5,2), B(−3,1), and C(−2,5). Point D is located on the intersection of the altitude and AB�, in such a way that D has coordinates at approximately (−1.52,1.18).
https://cds.flipswitch.com/tools/asset/media/607812
Match each question with the correct measurement, rounded to one decimal place.
What is the approximate area of the triangle?
What is the approximate length of the base of the triangle for the given altitude?
What is the approximate length of the altitude of the triangle?
16.1 units
3.9 units
8.1 units
7.6 units
15.8 units
8.0 units
The EASIEST and BEST method to determine the AREA of any POLYGON is the SHOELACE METHOD!
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
Answers:
Area = 15.8 square units
Base = 8.1 units
Altitude = 3.9 units
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Explanation
Use the distance formula to determine the length of side AB
A = (x1,y1) = (5,2) and B = (x2,y2) = (-3,1)
 approximately
Follow similar steps to find the distance from C = (-2,5) to D = (-1.52,1.18) is roughly 3.85004 which rounds to 3.9
Then,
area of triangle = 0.5*base*height
area = 0.5*AB*CD
area = 0.5*8.1*3.9
area = 15.795
area = 15.8 square units.
There's rounding error going on here.
The area of triangle ABC should be 15.5 square units exactly.
This can be determined in the next section below.
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When given the coordinates of the vertex points, we could use the shoelace formula to fairly quickly find the area of any polygon.
Given list of vertex points
(5,2)
(-3,1)
(-2,5)
Make a copy of the first point (5,2) and place it at the bottom of the list
(5,2)
(-3,1)
(-2,5)
(5,2)
This will help form a loop.
Space the x and y coordinates out. Each (x,y) point gets its own row.
Then draw in the diagonals as indicated below.
Multiply along the red diagonal pairs and add up those products.
5*1+(-3*5)+(-2*2) = 5-15-4 = -14
Do the same for the blue diagonal pairs.
5*5+(-2*1)+(-3*2) = 25-2-6 = 17
Subtract the results.
red - blue = -14 - 17 = -31
We get a negative result. Let's take the absolute value to get |-31| = 31
Lastly, take half of this to get the area = (1/2)*31 = 15.5
The area of the triangle is exactly 15.5 square units.
GeoGebra can be used to verify.
More practice on the shoelace formula is found here
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What's another way to see how the area is exactly 15.5?
My post is already quite lengthy so this section will be a brief overview rather than getting into the gritty details.
The distance formula will show that
AB = sqrt(65)
The equation of line AB is y = (1/8)x + 11/8
The equation of line CD is y = -8x - 11
Intersect those two equations to determine point D is at the exact location (-99/65, 77/65)
Note how -99/65 = -1.52 and 77/65 = 1.18 approximately.
So that's where your teacher is getting (-1.52, 1.18) for point D.
Use the distance formula to determine that CD = 31/sqrt(65)
Then lastly,
area = 0.5*base*height
area = 0.5*AB*CD
area = 0.5*sqrt(65)*( 31/sqrt(65) )
area = 0.5*31
area = 15.5
which is exact without any rounding done to it.
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