Can you find area of the Purple Trapezoid? | Trapezium | (Important skills explained) |

[@miguelgnievesl6882]

The measurement of AB is not needed. The trapezoid has a major base BC = 10√2 and a minor base DE = 8. We obtain the height CD by drawing a segment parallel to CD from E to BC. Then CD = 10√2 -8 and the Purple Area = (1/2)(BC + DE)(CD) = (1/2)(10√2+8)(10√2-8) = 68 cm²

[@maurypb]

I found it the same way.

[@cleiberrocha1449]

Me too.

[@ganymed1236]

Yes, that's what I did too.👏

[@ОльгаСоломашенко-ь6ы]

Согласна!

[@quigonkenny]

Working literally inside the box...

[@misterenter-iz7rz]

(10sqrt(2)+8)×(10sqrt(2)-8)/2=(200-64)/2=136/2=68.😊

[@PreMath]

Thanks ❤️

[@jhill4874]

This method seemed overly complicated. Like others did, I used all the 45d angles with the extended triangle and eventually subtracted 32 from 100 and got 68 sq cm. Thanks for another great problem.

[@Waldlaeufer70]

Let's have a close look: - Complete the purple area to get the same triangle as the blue one: - A(purple-white triangle) = A(blue) = 100 cm² - A(white addition) = 1/2 * 8² = 32 cm² A(purple) = A(blue) - A(white) = 100 cm² - 32 cm² = 68 cm²

[@PreMath]

Super job! Thanks ❤️

[@Copernicusfreud]

That is how I solved the problem.

[@habichmeyer]

takes more time to write it down, than to calculate

[@hcgreier6037]

Best answer! No need to calculate all the a's, b's, x's and roots....

[@prossvay8744]

1/2(x^2)=100 x=10√2cm area of the purple region=1/2(10√2+8)(10√2-8)={200-64)/2=68cm^2 Thanks

[@PreMath]

Great job! Thanks ❤️

[@johnjones8580]

We have all the necessary information to calculate the area once we establish the length of BC. The angle of the trapezoid is at 45 degrees, meaning that the difference between the parallel lines of the trapezoid is also its height. When applying the formula for the area of the trapezoid, we will discover that the area is half of a difference of squares. IE, [(BC-8)(BC+8)]/2 = (BC^2 - 8^2)/2 = 68

[@ybaggi]

Finally one of your puzzles I could do in my head.

[@silvaadamyan9098]

The triangle ABC = the triangle BCF, hence their areas are equal to 100 each. The area of the rectangular DFE with angles of 45 degrees is equal to (8^2)/2= 32. So the area of the trapezoid BCDE = 100 - 32 = 68.

[@juanalfaro7522]

I extended the figure to include EDF as done, but after finding DEF = 45 I saw that triangle BCF is congruent to triangle BCA. Therefore [BCF] = [BCA] = 100 and [EDF] = 8^2 /2 = 32, so [BCDE] = 100-32 = 68

[@quigonkenny]

Let t be the sides of the blue triangle AC and BC. Area of blue triangle ∆ACB: A = bh/2 100 = AC(BC)/2 = t²/2 t² = 200 t = √200 = 10√2 Let F be a point on BC where EF is perpendicular to BC. As ∆ACB is right isosceles, ∠ABC is 45°, therefore ∠FBE is also 45° (90°-45°) and ∆EFB is right isosceles and similar to ∆BCA. As DC and EF are parallel to each other and perpendicular to CB, CF is 8 units and thus BF is (10√2)-8 units. As ∆EFB is isosceles, EF and thus DC are also (10√2)-8 units. Area of purple trapezoid: A = (b₁+b₂)h/2 = (10√2+8)(10√2-8)/2 A = [(10√2)² - 8²]/2 = (200-64)/2 A = 136/2 = 68 cm²

[@cyruschang1904]

If we extend the two sides of the pink trapezoid, we get a white isosceles right triangle with its two short sides being 8 cm. The combined area of the white triangle + the pink trapezoid = the area of the green triangle = 100 cm^2 The pink area = 100 cm^2 - 100 cm^2 (8/✓200)^2 = 68 cm^2 (8/✓200) is the ratio of the short side of the white triangle to that of the green triangle, these two triangles are similar to each other, both being 45°−45°−90° triangles

[@MegaSuperEnrique]

I created a point F along CB such that CF=8, FB=10√2-8, therefore height=FE=FB, then area of rhombus=0.5* diff of squares. =68

[@PreMath]

Bravo! Thanks ❤️

[@cleiberrocha1449]

This value can also be confirmed by calculating the hypotenuse of triangle ABF. It must be reduced by eight plus ten root two. 

[@MrPaulc222]

The route I took was a little different. I extended line DE to a point F and made a rectangle where lengths BF and EF were both 10*sqrt(2) - 8. This gave a rectangle of (10*sqrt(2) - 8) by 10*sqrt(2), so 86.863... The triangle ((10*sqrt(2))^2)/2 = 18.863... Subtract one from the other for 68 sq cm. Your way was cleaner and better.

[@williamwingo4740]

Did it mentally. Construct triangle DEF which has area (8^2)/2 = 64/2 = 32. Since triangle ABC is congruent to triangle BCF, the purple area is 100 -- 32 = 68. Per Ardua Stabilis. 🤠

[@PreMath]

Bravo! Thanks ❤️

[@Copernicusfreud]

Yay! I solved the problem. I saw that triangle ABC and triangle BCF were congruent triangles and worked the problem that way.

[@PreMath]

Bravo! Keep rocking ❤️🌹

[@santiagoarosam430]

Prolongamos AD y BE hasta su intersección en F→ ABC=BCF→ BCDE=BCF-DEF =100 -(8²/2) =100-32 =68 Gracias y un saludo cordial.

[@PreMath]

Thanks ❤️

[@phungpham1725]

1/We have BC= 10sqrt 2. Drop the height EH to BC, the triangle EHB is a right isosceles so EH=EB= BC-DE=10sqrt2 -8 Area of the purple region= 1/2 .(10sqrt2+8).(10sqrt2-8) =1/2 .(200-64) = 68 sq cm

[@vpexmc]

Tried this problem without video and was right just didnt know trapezoid formula had all measurements but didnt know how to find area. Also where is point H?

[@marioalb9726]

A₁ = ½ b.h = ½ 8² = 32 cm² A = A₂ - A₁ A = 100 - 32 A = 68 cm² ( Solved √ )

[@paul.njihia]

We can also solve by finding height of the trapezium which is 10√2-8.

[@Ibrahimfamilyvlog2097l]

Very nice sar❤❤❤❤

[@giuseppemalaguti435]

l=BM=10√2...h=10√2-8...A=(8+10√2)(10√2-8)/2=(200-64)/2=68

[@PreMath]

Super! Thanks ❤️

[@tontonbeber4555]

It took me 5 seconds ... 100-32=68. BC-(missing summit) is identical to ABC so 100, then missing summit triangle 8x8/2 = 32

[@anantamurthy5525]

You are assuming the segment parallel to CD forms a square.It could be a rectangle also i think.

[@AndreasPfizenmaier-y7w]

CDE -Area also ist 0.5*8*8 and CBE 10* sqrt 128*0.5 and finished. We dont Need to think outside the box!

[@ОльгаСоломашенко-ь6ы]

Достаточно опустить перпендикуляр из точки E. Это и будет высота трапеции. Треугольник прямоугольный с углом 45 градусов, значит высота равнв разности CB- DE.

[@PreMath]

Thanks dear ❤️

[@ccmplayer87]

I see, that makes sense 👍

[@raya.pawley3563]

Thank you

[@amarseth1]

It's easy to see ABC and BCF are congruent. Area of DEF is 32 so area of BCDE is 68.

[@AmirgabYT2185]

S=24+30√2≈66,3 cm²

[@MegaSuperEnrique]

Heart racing from another Monday morning pop quiz ...

[@PreMath]

All the best 🌹

[@MrPaulc222]

Yes, I got there in the end, but I overcomplicated it.

[@martinrosenau478]

Just one comment about the question itself: The area (100cm^2) is given with a unit while the length (8) is given without a unit. This is not good - especially because some people have conventions about what a number without a unit means. An example: In engineering, a length of "8" (without unit) would mean "8mm" which is 0.8cm. But even the students at school might calculate: 100cm^2 = 1dm^2 to simplify calculations (which is perfectly valid) and then assume that "8" means "8dm" which is 80cm...

[@terakhanthis]

Using just the thumbnail: aqua triangle is 100cm^2. Area of a triangle is base x height / 2. So each of its equal sides adjacent to the 90 degree corner will be equal to the square root of 200, or 14.14 approximately. The trapezoid has a 45 degree angle, with the opposite angles to that being right angles, so the height of it will be equal to the difference between the base length and the top length, or 6.14. Break it into a rectangle of 8 x 6.14 and a right triangle with a base and height of 6.14. This gives an area of 49.12 + 18.85, for a total of 68, rounded.

[@vpexmc]

Was always taught you should never round until the end well done though gg

[@ybodoN]

Let's finish our initial task which was to calculate h, the height of the purple trapezoid. To do this, let's use BC as a fold line to fold the purple trapezoid over the blue triangle. Now, D'E' forms an isosceles right triangle with A ⇒ AD' = 8 ⇒ C'D' = CD = h = 10√2 − 8.

[@mohammedaboal3z932]

9451=1 7532=0 0190=3 8822=4 0008=5 3060=? Please help me to understand this 🙏🏼

[@PreMath]

Please send the question to [e-mail] Thanks 🌹

[@n.662]

100-(8²/2)

[@mikepeters8132]

My initial rhought was to split the purple trapezoidal shape into two right angle triangles one CDE the other CEB. Is that a flawed approach because we do not know if the line CE would make 45° angles?

[@PreMath]

Thanks for the feedback ❤️

[@mikepeters8132]

​@PreMath I love your videos, I found your solution to be very elegant. I'm still not sure if mine would have worked. Thanks for all the time you put in to your chanel.

[@ybodoN]

Remember: _this diagram may NOT be 100% true to the scale!_ Actually ∠CEB ≈ 97.48°, ∠DEC ≈ 37.52° and ∠DCE ≈ 52.48° Your approach would only work in the case where DE = 5√2.

[@ChatGPTvzlomaetMir]

Geometry 7 class. Easy.

[@raffaeleguerrieri5482]

Se unisco C ad E non è una parallela con A B? Se si, CD =8?

[@alokranjan4149]

68 cm2

[@DayanandPandey-ns5qc]

Vish

[@김상수-v2f9l]

68 in 15 second.

[@amarendrasingh7327]

Are you a calculator

[@laxmikatta1774]

Day 3 of asking premath sir to reveal his face

[@PreMath]

Pretty soon! Thanks dear ❤️🌹

[@laxmikatta1774]

​@@PreMath👍👍🙌🙏🙏

[@MrKwabenaadu]

i have a very simple method to find the area of the the purple region.This your method is too complicated.

[@SinaRa-yj1xo]

The amount of "as well" and "therefore" this guy said is just too much 😂 Expand your paraphrasing, bro!