Calculate area of the Yellow shaded Quadrilateral | Important Geometry skills explained

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Calculate area of the Yellow shaded Quadrilateral | Important Geometry skills explained

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[danisco]

I love you PreMath, please come to brazil to teach various tricks.
Great video!!!

[pwmiles]

Call the central crossing-point O. The blue and green triangles have the same height measured from BE, so
OE/OB = 28/35

The white triangle is similar to the blue triangle, so the areas are in proportion to the square of the ratio of corresponding sides i.e.
Area of white triangle = 28(35/28)^2 = 35^2/28 = 5x35/4 = 175/4

The white and green triangles add to half the area of the rectangle so the yellow area and the blue triangle add to the same amount

Yellow area = 35 + 175/4 - 28 = 203/4 = 50.75 cm^2

[theoyanto]

This was great, didn't know about the a b c d thing with quadrilateral, new knowledge thanks, and you've already taught me same base and height area thing but didn't see it, needed the video for this one, another priceless Pearl of wisdom you've shared, thank you very much

[TheDHemant]

Time 3.39 mins was a new learning for me. and subsequent video suggestion was logical and useful. Very very good Sir

[irfanfaiz]

You explain well brother.
I am also a Mathematics Teacher,
Math with irfan faiz

[onpathak]

Please also provide proof that products of pairs of opposite triangles formed by diagonals in a quadrilateral are equal. This result is quiet interesting, which I did not know.

[gatxn]

hiii, I am a big fan from brazil! My teacher indicated your channel and I loved it. Ty very much for all your videos

[Ramkabharosa]

Let AC meet BE in the point F. Let g & h be the heights of triangles AEF and ABF with AF as base.
Then g & h will also be the heights of ΔCEF & ΔCBF with CF as base. So y/35 = |AF|/|CF| = 35/28
(triangles with same height have areas proportional to lengths of their bases). Thus y = 35.(35/28)
= 175/4. Hence area(ADEF) = area(ADC) - 28 = area(ABC) - 28 = (175/4) + 35 - 28 = 203/4 = 50 3⁄4.

[spiderjump]

Join A and E and let the point of intersection between AC and EB be F. Ratio of
Area of triangle AEF : area of triangle AFB= 4:5 since areas of EFC to BFC = 4:5
Area of AEC= area BCE hence area of AEF= area of BFC = 35
Area of ABF= 5/4x35= 43.75
Yellow area = 43.75 +35–28=50.75

[noahvale]

I look forward to your videos every day.

[abcdpqrs]

Let h be the of the triangle of area (35+28 = 63) units². The two triangles of areas 63 and 28 have same bases, so height ratio is the area ratio: 4h/9; the triangle in white is similar to triangle in blue. So, (4/9:5/9)² = 16:25: ratio of areas. From this, white area is 43.75. the total area of yellow and blue is (43.75+35 = 78.75 units²) since it is a rectangle. That leaves yellow area as 50.75 units²

[adiammu]

Really useful for my primary son thanks for sharing 😊

[alster]

Wow!! I learned something new here... Tricky at first but became easy at the end

[zawatsky]

В конце вовсе не обязательно было искать Z. Достаточно было от АВС (который равен ADC) отнять 28. S=43, 75+35-28=43, 75+7=50.75 см².

[Ramkabharosa]

Let ABCE be any quadrilateral and F be the point where the diagonals meet.
Let g and h be the heights of triangles ABC and AEC with AC as base. Then
area(ABF)= ½.g.|AF|, area(CBF)= ½.g.|FC|, area(AEF)= ½.h.|AF|, & area(CEF)= ½.h.|FC|.
So area(ABF).area(CEF) = (½).g.|AF|.(½).h.|FC| = area(CBF).area(AEF). Q.E.D.

[HappyFamilyOnline]

Awesome presentation👍
Thanks for sharing😊😊

[himo]

28 : 35 = 4 : 5 28*5/4*5/4=175/4
175/4+35 = x+28
area of Yellow Quadrilateral : 50.75

[misterenter-izrz]

A very brain-ached puzzle, I now have some idea, but I is difficult to imagine how can you create such terrific puzzle?
The area of the upper triangle is 28x5^2/4^2=28x25/16=43.75, so half of the rectangle area is 43.75+35=78.75, therefore our answer is 78.75-28=50.75, done.🙂

[TheDHemant]

Other possible measures for the rectangle, I have to check if L=35 cm & B=4.5 cm will also work. Will it work, Sir?

[vara]

Once again, you have an interesting problem with the answer.