MedVision ad

Area of a Triangle (1 Viewer)

cossine

Well-Known Member
Joined
Jul 24, 2020
Messages
628
Gender
Male
HSC
2017
We need to find the area of the colored triangle (image is given below) with coordinates (0, 0), (20, 0), (20, 9). This will mean that the area of the triangle is 90 units^2 since 20*9/2 = 90. However if we follow the steps from b to e we will see the area of the triangle is 90.5.

(b) Show that the yellow (lower) right triangle has area 27.5.
(c) Show that the red rectangle has area 45.
(d) Show that the blue (upper) right triangle has area 18.
(e) Add the results of parts (b), (c), and (d), showing that the area of the colored
region is 90.5.

Question: Why do we get a different result?


Screenshot 2024-01-03 014843.png
 

bruhmoment.

Active Member
Joined
Jun 19, 2023
Messages
238
Gender
Male
HSC
2024
Okay i think i got it, if you look at both the axis, the numbers aren't in a uniform manner for both axis, meaning for the x - axis, its increasing by 3, whereas for the y axis, its increasing by 2. so thats why you would get different results from the two ways you did. hope that helps😀
 

Luukas.2

Well-Known Member
Joined
Sep 21, 2023
Messages
445
Gender
Male
HSC
2023
Hint: The diagram is deliberately misleading.

It encourages the reader to make an assumption that is not justified.

Just because something appears to be true, doesn't mean it is true... what is the diagram implying is true that isn't actually known for sure from the information given?

If the yellow and blue triangles and the red rectangle make up a triangle, then the vertex where all three meet cannot be at (11, 5)... in that case, all the calculations in (b), (c), and (d) are wrong.

Alternatively, if the vertex common to the yellow, blue, and red regions is (11, 5), then the three regions combine to form a quadrilateral rather than a triangle because the hypotenuses of the two triangles meet at at an angle that is not 180 degrees.

In other words:
  • EITHER (0, 0) to (20, 9) is a straight line, in which case the rectangle's corner is at (11, 4.95) rather than at (11, 5) and the calculations should be:
    • yellow = 0.5 x 11 x 4.95 = 27.225 sq units
    • red = 9 x 4.95 = 44.55 sq units
    • blue = 0.5 x 9 x 4.05 = 18.225 sq units
    • Combined area of triangle = 27.225 + 44.55 + 18.225 = 90 sq units, as expected
  • OR (0, 0) to (20, 9) to (11, 5) to (0, 0) forms a triangle of area 0.5, which is the extra area found by the above method as it is outside the area of the triangle from (0, 0) to (20, 0) to (20, 9) to (0, 0)

There is a third option, as well...

The vertex where the yellow and blue triangles and the red rectangle meet if they do make up the larger triangle, could be at y = 5, rather than at x = 11 (as was the first case above). In that case, what appears to be the point (11, 5) will actually be at (100/9 , 5), and the calculations in (b), (c), and (d) are again wrong:
  • yellow = 0.5 x 100/9 x 5 = 250/9 sq units
  • red = (20 - 100/9) x 5 = 80/9 x 5 = 400/9 sq units
  • blue = 0.5 x (20 - 100/9) x 4 = 0.5 x 80/9 x 4 = 160/9 sq units
  • Combined area of triangle = 250/9 + 400/9 + 160/9 = (250 + 400 + 160)/9 = 810/9 = 90 sq units, as expected
 
Last edited:

WeiWeiMan

Well-Known Member
Joined
Aug 7, 2023
Messages
1,021
Location
behind you
Gender
Male
HSC
2026
We need to find the area of the colored triangle (image is given below) with coordinates (0, 0), (20, 0), (20, 9). This will mean that the area of the triangle is 90 units^2 since 20*9/2 = 90. However if we follow the steps from b to e we will see the area of the triangle is 90.5.

(b) Show that the yellow (lower) right triangle has area 27.5.
(c) Show that the red rectangle has area 45.
(d) Show that the blue (upper) right triangle has area 18.
(e) Add the results of parts (b), (c), and (d), showing that the area of the colored
region is 90.5.

Question: Why do we get a different result?


View attachment 41997
The thing isn't a triangle so you can't just do the area like 1/2 bh
 

cossine

Well-Known Member
Joined
Jul 24, 2020
Messages
628
Gender
Male
HSC
2017
Hint: The diagram is deliberately misleading.

It encourages the reader to make an assumption that is not justified.

Just because something appears to be true, doesn't mean it is true... what is the diagram implying is true that isn't actually known for sure from the information given?

If the yellow and blue triangles and the red rectangle make up a triangle, then the vertex where all three meet cannot be at (11, 5)... in that case, all the calculations in (b), (c), and (d) are wrong.

Alternatively, if the vertex common to the yellow, blue, and red regions is (11, 5), then the three regions combine to form a quadrilateral rather than a triangle because the hypotenuses of the two triangles meet at at an angle that is not 180 degrees.

In other words:
  • EITHER (0, 0) to (20, 9) is a straight line, in which case the rectangle's corner is at (11, 4.95) rather than at (11, 5) and the calculations should be:
    • yellow = 0.5 x 11 x 4.95 = 27.225 sq units
    • red = 9 x 4.95 = 44.55 sq units
    • blue = 0.5 x 9 x 4.05 = 18.225 sq units
    • Combined area of triangle = 27.225 + 44.55 + 18.225 = 90 sq units, as expected
  • OR (0, 0) to (20, 9) to (11, 5) to (0, 0) forms a triangle of area 0.5, which is the extra area found by the above method as it is outside the area of the triangle from (0, 0) to (20, 0) to (20, 9) to (0, 0)

There is a third option, as well...

The vertex where the yellow and blue triangles and the red rectangle meet if they do make up the larger triangle, could be at y = 5, rather than at x = 11 (as was the first case above). In that case, what appears to be the point (11, 5) will actually be at (100/9 , 5), and the calculations in (b), (c), and (d) are again wrong:
  • yellow = 0.5 x 100/9 x 5 = 250/9 sq units
  • red = (20 - 100/9) x 5 = 80/9 x 5 = 400/9 sq units
  • blue = 0.5 x (20 - 100/9) x 4 = 0.5 x 80/9 x 4 = 160/9 sq units
  • Combined area of triangle = 250/9 + 400/9 + 160/9 = (250 + 400 + 160)/9 = 810/9 = 90 sq units, as expected
Thank you.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top