Find Area of the Shaded Triangle in a Rectangle | Important Geometry skills explained | Fun Olympiad

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Find Area of the Shaded Triangle in a Rectangle | Important Geometry skills explained | Fun Olympiad

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[anatoliy]

Superb colorful task! I tried to solve this problem using Heron's theorem, but your method is better. Thank you very much, Mr PreMath! Sir, do you have any math textbooks uploaded or in paper form of your authorship? God bless you!

[santiagoarosam]

It can be solved more easily by transforming the initial triangles into rectangles:
The horizontal and vertical lines, passing through the right and bottom vertices respectively of the green triangle, divide the original rectangle into four rectangles; the two in the left column add up to an area of ​​value 2x28=56, the two in the upper row 2x35=70 and the lower right 2x21=42, values ​​obtained from the surfaces of the original triangles. If we give the surface of the lower left rectangle a value “a”, then the surface of the upper left will be (56-a) and that of the upper right (70-56+a=14+a).
On the other hand, we know that the ratios between the areas of the rectangles in each column are constant, since the heights of each row and the widths of each column are. Considering this constant relationship and the previously obtained area values, the following equality can be established:
(56-a)/a=(14+a)/42 ⇒ 42x(56-a)=a(14+a) ⇒ a²+56a-2352=0 ⇒ a=28 ⇒
Total original rectangle area = (56-a)+(14+a)+a+42 = 28+42+28+42=140
Green triangle area= 140-35-28-21 = 56

[nirupamasingh]

Sir there is aprinting mistake in onefactorization step. It should be negative of five y square

[mikefochtman]

@6:02 when you broke -2xy into +3xy-5xy, you accidently changed the sign of -5y^2 to +5y^2. It still works out since when you then factored -5y from the last two terms, you factored it as -5y(x+y), which corrected the sign mistake. Just an FYI.

[ajbonmg]

All the numbers are multiples of 7, so I divided through by 7 to simplify it. Triangles now have areas 4, 3, 5.
Some straight guesses - the lengths of the left triangle could be 2 and 4.
The lengths of the bottom right triangle 2 and 3.
The lengths of the top triangle 2 and 5.
It works! So the rectangle is 5 by 4 (=20), subtract the triangles (3+4+5=12), giving green area 8. Then multiply up by the 7 again, and the answer is 56! :)

[tonekjansen]

Because of the particular given values for the surface areas of the triangles, there is a much easier way to see that the area of the rectangle is 140. Bring the top-left corner of the 28-triangle to the middle of the left side of the rectangle. This gives a new triangle with area 14. This together with the area of the 21-triangle equals 35, which is exactly equal to the area of the 35-triangle. This implies that the point where the two corners of the 21- and the 35-triangle meet must be exactly in the middle of right side of the rectangle. This is turn means that the area of the top half of the rectangle is twice the area of the 35-triangle, and the total area of the rectangle is 140.

If the areas of the triangles do not have such nice values, we can generalize as follows. Draw a horizontal line through the point on the right side of the rectangle where the 21- and the 35-triangle meet. This splits the rectangle into an upper and a lower part. The area of the upper part is 2*35=70. To determine the area of the lower part we move the top-left corner of the 28-triangle to the new horizontal line. This reduces its area to, let's say, 28*f. The area of the lower part of the rectangle is then 2*(28*f+21). We now need to determine the reduction factor f. We can do this by looking at the ratio between the areas of the upper and lower part of the rectangle. The latter is proportional to f, the former to 1-f, or (2*(28*f+21))/(2*35)=f/(1-f). This can be rearranged into a quadratic equation in f. The relevant solution is f=1/2, which gives, again, a value of 140 for the area of the rectangle.

[Sam_on_YouTube]

I did this a MUCH simpler way. My way is not generalizable, and presumes rather than proves a unique answer. So it is good for some purposes, like a multiple choice test, but it isn't as robust. But here's what I did:
Scale the whole area down by a factor of 2. That means that the areas of the triangles are now 35, 28, and 21 over 2. That means the triangles can be written as factors of those three numbers, if you can find a combination that works. I found one. In mine, the diagram is not at all to scale. But if there is only 1 answer, that's fine.

If the top is 5, and the left side is 14, then the bottm divides into 2 and 3 and the right side into 7 and 7. Take the area of each triangle and you'll get the right numbers. Multiply 14×5 and then scale the area back up by the same factor of 2 as the beginning and you get a total area of 140.

Then subtract the areas of the triangles and you've got 56.

EDIT:
WAY easier than what I did, divide the whole thing by a factor of 14 instead. The top is then 5 long and the height is 2 high. On the right side it splits into 1 and 1. The bottom remains 2 and 3.

Still not to scale, but closer. Regardless:
5×2×14=140. WAY easier to find that breakdown.

[TimBoulette]

@6:22 That should be 3x(x+y) - 5y(x-y)

[MarcelCox]

This kind of problem is easier to solve (the formulas stay simpler) if you remain with sirfaces instead of going down to side length. In fact, allinformation is in terms of surfaces and the result as well. This is an indicator that an elegant solution might exist in terms of calculating with surfaces only. In this case, break down the big rectangle in 4 smaller ones by drawning a horizontal and a vertical line where the triangles touch the borders. If you label the 4 areas of these rectangles as a (top left), b top right), c (bottom left), d (bottom right), you get a set of equations as follows:
a+b=2*35 (twice the 35 trangle), a+c=2"28, d=2"21. Furthermore, given the 4 rectangles are linked by the formula a*d=b*c. Solving this set of equations yields a=28, b=42, c=28 and d=42. The surface of the green triangle is (b+c+d)/2 which gives you 56, the solution. I leave the intermediate steps for you to work out.

[vidyadharjoshi]

Let H be the height, B be Base, x be base of 28, y be the height of 35. 1/2(Hx) = 28, 1/2(By) = 35, 1/2(B-x)*(H-y) = 21.
BHsq -168BH +3920 = 0 Solving for BH gives 140. So Area of Triangle = 140 - (28+21+35) = 56

[opytmx]

Nice solution! Thanks a lot. I just took 28 = 7 x 2 x 2?, primes, 35 = 5 x 7?, primes and 21 = 3 x 7?, primes, and the corresponding rectangles 4 x 14 = 56, 10 x 7 = 70 and 6 x 7 = 42. All opposite sides have to be the same, as well the sum of their parts resulting in: left side = 14, right side = 7 + 7 = 14, top = 10, bottom = 4 + 6 = 10, so the big rectangle must be 10 x 14 = 140, and the green one 140 - 84 = 56. Of course, the drawing would look very different, but the result is the same 🙂

[lybcxds]

تمرين جميل جيد. شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم. تحياتنا لكم من غزة فلسطين .

[buff]

Спасибо за видео
Это было тяжело но я смог решить

[renamila]

Without your video I wouldn`t have had any glue to solve. I tried it then a little bit different. First I named both side oft the Rectangle x and y. Then I got for the bottom length 56/y plus x - 56/y. On the right side I got: 70/x plus y-70/x. Then I concentrated on the 21 triangle. Then I set up following equation 1/2 * (x-56/y)*(y-70/x) equals 21=> (x-56/y)*((y-70/x) equals 42 =>
x*y - 70x/x - 56y/y + 56*70/x*y = x*y- 126 + 3920/x*y = 42 |-42 and |*x*y => x^2*y^2 -168x*y +3920 equals 0. As x*y equals A of the whole rectangle I substitute it. So A^2 - 168A +3920 = 0. Then simple solve the quadratic equation. The rest is simple.

[Ramkabharosa]

Let a√7 = length of rectangle & b√7 = height of rectangle. Put x√7 = width of the left
triangle & y√7 = height of the upper-right triangle. Lengths here are all in (√7)-units,
so we must multiply (ab -12) by 7 to get the area of the green triangle (in sq. units).
Now bx = 8, ay =10, and (a-x).(b-y) = 6 (twice the areas of corresponding triangles).

So x= 8/b, y=10/a, & ab - ay - bx + xy = 6. Thus ab - 10 - 8 - 6 + (80/ab) = 0. Hence
(ab)² - 24(ab) + 80 = 0. So (ab - 20)(ab - 4)=0. Since ab -12>0, we must have ab=20.
Thus ab -12=8 and so the area will be 8(7)=56 sq. units. Note a=20/b, x=8/b, &
y = 10/a = 10/(20/b) = b/2 always. Also b can be anything > 0.
.

[dvol]

In my opinion, not the best solution. I decided in another way.
I drew a vertical line to the bottom corner and a horizontal line to the right corner. I took the area of the upper-left rectangle as X. After that, I wrote down the equality of the ratio of the areas of the rectangles, which turned out to be a simple quadratic equation. And found out the area of X.
X/(28*2-x)=(35*2-x)/21*2 => x^2-168x+3920=0 => x=28
Next, adding up the areas of 3 rectangles and subtracting X, we get the total area.
28*2+35*2+21*2 -28 = 140
Then we subtract the known areas, and that's it.
140-28-21-35 = 56

[holyshit]

Quite easy, once we write equation (fe for area of bottom right triangle)
all we need to do is solve quadratic equation for area of rectangle and subtract areas of right triangles
I labeled x, y as side lengths of rectangle then I wrote equation (y-56/x)(x-70/y)=42

[williammcbrayer]

You flipped a sign at 5:54. It should read - 5y^2 = 0. You proceeded to factor it correctly, in spite of this error.

[uekxmpg]

Draw vertical and horizontal lines which pass bottom and right hand vertices respectively. Then the area of green triangle is 35+28+21- area of left top rectangle. Let x, y be lengths of vertical and horizontal sides of rectangle respectively. It follows that two orthogonal sides of area 21 triangle is 56/y - x and 70/x - y and multifly these two yields 42. Put xy=z and simplify the formula to get z^2 - 168z+ 56x70 = 0. Thus z=28, 140. Therefore the area of green triangle is 35+28+21- area of left top rectangle =35+28+21-28=56. Thank you for your Hints!

[HappyFamilyOnline]

Superb. Very well explained👍👍
Thanks for sharing😊😊