How to find the area of a triangle inside of a rectangle

We are given the area of the rectangle is 96 and we know two parts are 5 and 8, as indicated on the picture. We need to find the area of the shaded triangle. I will present the easiest solution to solve this tricky geometry puzzle!

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#math #algebra #mathbasics

[Itstoearly]

I completed the 2 sides of the rectangle using x and y, then wrote out the area of the 3 white triangles, and using the 2 formulas: Area of shaded = 96 - (area of 3 white triangles), and (y+8)*(x+5)=96, all the variables cancelled out. Not as elegant as your solution, but it worked!

[dlevi]

No idea what your first solution was like, however, if we call a the shorter side of the rectangle, b the longer side, and x the area of the shaded triangle, we know that:

ab = 96

but also, since the rectangle is divided into 4 triangles (one of base b, height 5 at the top, another one of base b-8 and height a-5 bottom left, a third one of base 8 and height a bottom right, and the shaded one in the middle), we also have:

5b/2 + (b-8)(a-5)/2 + 8a/2 + x = 96

multiplying out
5/2b + ab/2 +40/2 -5/2b -8a/2 + 8a/2 + x = 96

the terms in 'a' and 'b' cancel out
ab/2 + 40/2 + x = 96

but we know ab = 96
48 + 20 + x = 96

from which
x = 28

It's possibly slightly longer, but it doesn't need any auxiliary construction.

[MinhTran-flqg]

An algebraic solution to this is:



Let the two sides of the rectangle be 5+x and 8+y

Given that the product of two sides is 96, 8x+5y+xy=56 (i)

As the rectangle can be divided into four smaller triangles, including the one with the area A we are looking for, A+8(5+x)/2+5(8+y)/2+xy/2=96, or 2A+80+(8x+5y+xy)=192 (ii)

From equations (i) and (ii), A=(192-80-56)/2=28

[krischan]

There's a simpler solution: If that's enough information, then any rectangle with area 96 and the horizontal lenght being at least 8 and the vertical at least 5 will do. So I will choose a rectangle with the horizontal length 8, so the vertical length is 12. That leads to a triangle with a base of 12-5=7 and a height of 8, having an area of 7*8/2=28.

[L_Ratio_]

What I did is assigned the horizontal unknown distance be x and vertical unknown distance be y. One equation we can get is that (8+x)(5+y) =96. Now I have the formula memorised for area of triangle through coordinates of vertices (x, 0), (0,y), (8+x, 5+y), which is basically a determinant to solve, and on substituting y, it just beautifully cancels out the x term and gives the area as 28.

[ausaramun]

Let 96 = (8+x)(5+y) = 40 + 5x + 8y + xy
Therefore 56 = 5x + 8y + xy

Let 96 = (5)(8+x)/2 + (8)(5+y)/2 + xy/2 + A
Therefore 96 = 20 + 2.5x + 20 + 4y + 0.5xy + A
Therefore 56 = 2.5x + 4y + 0.5xy

Let 56 = 56
5x + 8y + xy = 0.5(5x+8y+xy) + A
A = 0.5(5x+8y+xy) = 0.5(56) = 28 QED

[Qermaq]

One cool thing I noticed is that if the unknown length below 5 is 0, the one next to 8 is 11.2, and if the one next to 8 is 0, the one below 5 must be 7. There are infinite possibilities. Unless you count 7, 0 then the only nice numbers are 3, 4.

[Mnaughten]

I first found the prime factorization of 96, then found the combinations that had both values that were bigger than 8 and 5, (small list to make). So I knew that the sides must be 8x12 or 6x16. Then I found the area of unshaded triangles of initial picture. Then I subtracted those areas, 96-6-32-30=28 and 96-24-4-40=28

Both rectangle sizes got me to the same answer.

Edit: this solution only takes into account that the sides are whole numbers, so unless the problem says with only whole numbers, I don’t think it is absolutely valid.

[JJ_TheGreat]

0:31 Area of shaded triangle is 28 - based on finding areas of the other (non-shaded) triangles, adding them up and subtracting from 96. 96-68 is 28…
And the dimensions for 96 have to be a width of 12 and a height of 8, so you can subtract to finish the other sizes of the rectangle, they have to add up to those sizes!

[Robi]

I went with "hunch" solution, but got the exact same result: 28.
If the rectangle's area is 96, that means it has be factored into its' sides.That's where I assumed it's 8 and 12. Rest was just computing the areas of the three smaller triangles (6, 30 and 32) and subtract them from the whole area and voila: 28
Took me less than a minute in my head.
It's not the rigoristic solution but on a test where you're given the answers it would be enough ;)

[pauldorman]

It took me a little while to understand, so for others still scratching your heads, read on.

Pause the video at the 2 minute mark so you can see all the shaded triangles. First thing to observe is that the larger rectangle is divided in two, and obviously the area of the top rectangle plus the area of the bottom rectangle equals 96. Our excellent teacher has shown us that we can construct an additional triangle which allows us to "fill out" the top and bottom rectangles so that they each contain a single triangle with he same base and height, and therefore half the respective area. The area of all the the shaded triangles together is therefore half the area of the large rectangle, or 48, and all we need to do is subtract the area of the red triangle that spans the top and bottom rectangles, which is simply half its base (5) times height (8): 48 - 20 = 28. QED!

I haven't been a student for over 30 years, but I love coming to channels like this to learn different ways of solving problems that I really couldn't appreciate when I was a kid. This was an excellent video. I love a bit of mystery!

[raymundbelleza]

I did this through measuring and got 28.6.

[professorvatcraft]

I solved this using algebra :]
1. Give the right side of the rectangle length x
-> The left side also has length x
-> Top and bottom sides have length 96/x
2. On the left side, we know part of it is length 5
-> The other part of it will be x - 5
3. Do the same for the bottom side
-> the other part of it will be 96/x - 8
4. The area of the marked triangle will be the area of the rectangle - the sum area of the 3 small unmarked triangles
-> A(MT) = A(R) - A(T1) - A(T2) - A(T3) (*)
A: Area
MT: Marked triangle
R: Rectangle
T1: Triangle 1 (has base x and height 8)
T2: Triangle 2 (has base 5 and height 96/x)
T3: Triangle 3 (has base x - 5 and height 96/x - 8)
5. Area of triangle 1: A(T1) = 1/2 * x * 8 = 4x
6. Area of triangle 2: A(T2) = 1/2 * 96/x * 5 = 240/x
7. Area of triangle 3: A(T3) = 1/2 * (x-5)(96/x - 8) = 48 - 4x - 240/x + 20 = -4x - 240/x + 68
8. Filling the values in to (*)
A(MT) = 96 - 4x - 240/x - (-4x - 240/x + 68)
A(MT) = 96 - 4x - 240/x + 4x + 240/x - 68
A(MT) = 96 - 68
A(MT) = 28 ✅
I assume this is how most of us solve this problem :]
Your solution is really interesting! I did not think you can do that :D

[harishhimanshu]

My method was like a Hit and Trial
take the area 96 and make its factors 32 * 3 means it will have atleast a 3 in it and then
i got that Length would be 8+4 = 12

and Breadth is 8
and I got the all traingle area value as 6, 30, 32
and 96 - 68 = 28
I solved it in my mind that when I check the Ans I was so Happy

[turmericgarage]

For the first time I paused when asked and tried to do it in my head. Then got lazy and guessed if 96 is 12x8, then the bottom left white is a 3 4 5 of area 6. The other two white triangles are area 30 and 32. Thus 28 for the shaded area.

[LoGiX-wins]

I couldn't understand. Can you 0lease break down the steps and explain why you did them? (eg - How could you just create the division saying that it is at the half of the length 5+y)

[Gremriel]

I swear you did this one before, and I was about to ask AndyMath what he thought about it, but then the video got deleted.

[zachansen]

it's funny this one I find to be much simpler using algebra and can't follow your explanation at all.

Just find the area of the other three triangles in terms of X and Y, then substitute either x for y or y for x using the x*y=96 (x=96/y) or (y=96/x) given and then subtract from the total area.

[tawney]

I don’t understand how this works. If the sides of the rectangle were more than 10 wouldn’t that throw the whole thing off?

Was it said that 5 was half of the length? Isn’t this also just assuming the shaded triangles equal half of the rectangle?

[shreyas]

how do we know the area of the bottom red and black triangle is equal to half the area of the bottom rectangle?