Thanks sir to teaching me out the box💪

@PreMath

Жыл бұрын

Happy to help Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

The interesting thing about your solution is that the key step is multiplying a x b x c x d -- something that has no *geometric* meaning here, but is crucial for generating a solvable equation. Neat.

@e1woqf

Жыл бұрын

No, it's not crucial. Indeed there is no need to introduce a,b,c and d. Since both the upper and the lower rectangles share one side respectively, we can work with the ratio of areas, which leads us to the equation 56 (x+2) = (x-4)(4x-3)

@murdock5537

Жыл бұрын

@@e1woqf = (x - 4)(4x - 3) 🙂

If I multiply the diagonal areas and then equalize those products [56(x+2)] = [(4x-3)(x-4)], I get the same quadratic equation and of course the same values for x.

@e1woqf

Жыл бұрын

That's exactly what I did as well. No need to introduce extra variables.

@alster724

Жыл бұрын

The ratio-and-proportion technique

@krzysz5023

Жыл бұрын

Yeah I noticed that the radio of the area of the boxes would be proportional to the ratio of the corresponding sides (after doing a bunch of other work) and got that too! 😅

That was a challenging problem. I had to really think through the logic and then it occurred to me that essentially you solved it as a proportion (x+2)/(4x-3) x (x-4)/56. Only that you did it by labeling the sides and then cross multiplying. That was was excellent. Your problems really get me thinking outside of the box.

Nice problem. You can do it by ratios of the areas green/purple = brown/blue (x+2)/(4x - 3) = (x - 4)/56 Multiply out denominators and simplify 4 x^2 - 75x -100 = 0 x = (75 +/- sqrt(75^2 + 1600))/8 x = (75 +/- 5 sqrt(15^2 + 64))/8 x = (75 +/- 5 sqrt(289))/8 x = (75 +/- 85) / 8 x = -5/4 or 20 etc

@e1woqf

Жыл бұрын

Exactly.

@geraldillo

Жыл бұрын

It's the same; Premath skipped that step (unfortunately)

Your channel is a treasure of knowledge

Thanks! Great problem - like the way it mixes geometry with the algebra

Nice problem, excellent way to solve it, many thanks! fast lane: (x + 2)/(4x - 3) = (x - 4)/56 → (x + 2)56 = (4x - 3)(x - 4) → x = 20 btw: a = 2; b = 11; c = 7 ; d = 8 🙂 56 = 7(8) → 77 = 7(11) → 22 = 2(11) → 16 = 2(8)

The ratio between areas of the same column or row is constant throughout the array of rectangular cells → (X+2)/(4X-3=(X-4)/56 → X=20 → X+2=22 ; X -4=16 ;4X-3=77 If the dimensions of the purple cell are B*H and we put the dimensions of the other column and the other row as a function of B and H, we obtain a column width =B*b and a row height =H*h → Result some areas: Green=BHh ; Brown=BbHh ; Purple=BH ; Blue=BbH → With these algebraic expressions we can check the constancy of the relationships between areas of the same column or row: Green/Purple=BHh/BH=h = Brown/Blue=BbHh/BbH=h ; Green/Brown=BHh/BbHh=1/b = Purple/Blue=BH/BbH=1/b. Thanks and greetings to all.

I went with 56(x+2) = (4x-3)(x-4) , formed the quadratic from that, and calculated x with quadratic formula.

Another nice one, the best part is that you can also calculate the individual lengths of the small segments: 2 and 7 on the vertical axis, 11 and 8 on the horizontal axis, and it all works out. In fact, 7 • 8 was my first guess for the blue rectangle even before you started your procedure, because 56 doesn't have many other suitable divisors.

Amazing👍 Thanks for sharing😊

For 2 rectangles with a common length (A = l*w, & B = l*y), the ratio of their areas is equal to the ratio of their widths (A/B = lw/ly = w/y). We see that the 2 rectangles in the 1st column have the same lengths (as labelled in diagram: b), so the ratio of their areas is a/c. Likewise for the rectangles in the 2nd column, the ratios of their areas is also a/c. Therefore the two ratios of areas are equal to each other. Therefore, (x+2)/(4x-3) = (x-4)/56. Then solve for x, then get the areas of the 3 unknown areas. ^^Notice: As a mnenomic, you just set up the ratios EXACTLY in the same position as the subrectangles are positioned in the big, overall rectangle!! If you remember this, you can just write the equal ratios, and off you go!! (If you aren't sure how the ratios go, you can always derive this from scratch, which should only take 30-60 secs.)

The manipulation part of abcd was a bit tricky but the rest are easy. The ratio-proportion technique of the labelled areas would've been easier AB/AC=AD/CD (x+2)/(4x-3)=(x-4)/56

Very interesting sir

Yeah, these are the ones that cause me trouble in getting started.... So more like this please 🤓👍🏻

Thanks for video.Good luck sir!!!!!!!!!!

Good Morning Master 🇧🇷

Vvv nice explanation

Thats good idea

That was a fun problem. Thanks professor!😄

@PreMath

Жыл бұрын

Glad to hear that! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

Amazing sir

@PreMath

Жыл бұрын

Many many thanks You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

4:15 step 3, why did you multiply the area of rectangle ab by the area of rectangle cd?

Divide (1)by (2) and (3) by (4) each of which =b\d. Then cross multiply etc.

@e1woqf

Жыл бұрын

👍

@johnbrennan3372

Жыл бұрын

Thank you

Задачка простая, пропорция составляется сразу же из подобия прямоугольников.

16 , 22, 77, 56 (given) and 171

Super

@PreMath

Жыл бұрын

Thank you! Cheers! You are awesome, Mahalakshmi. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

We dont need to calculate the sides if we notice that blue to brown = purple to green: 56/(x-4) = (4x-3)/(x+2) then cross multiplication and simplification give: 4xx - 75x - 100 =0 and the positive solution X=20

I got the Idea: 56 / (x-4) = (4x-3) / (x+2) => x = 20 kind regards

@PreMath

Жыл бұрын

Excellent! Glad you think so! You are awesome, Michael. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

Thnku

@PreMath

Жыл бұрын

Thank you! Cheers! You are awesome, Pranav. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

Answer =16 , 22, 56, 77 and total area =171 or 19 x 9 A different approach. let the blue rectangle area = np (n= the horizontal line and p= the vertical line) let the vertical line of the beige rectangle= r; hence its area = nr let the horizontal line of the green rectangle = y; hence its area = ry let the vertical line of the pink rectangle = p ; hence its area= py Since np =56 and nr = x-4, then p/r = 56/(x-4) [ divide np by nr] Since py= 4x-3 and ry =x+2, then p/r = (4x-3)/x+2) [divide py by ry] Therefore 56/(x-4) = (4x-3)/(x+2) since both = p/r 56x + 112 = 4x^2 -19x +12 [ Crossmultiply] 0 = 4x^2 -75x -100 [ Simplify] quadratic equation. Using the quadratic formulae x =20 hence area of each is 16 (20-4) 22 ( 20 + 2) 77 (4 x 20 -3) 56 (given) Area of the large triangle = 171 or 19 x 9

S(orange)/S(blue) = S(green)/S(purple) => 56(x+2)=(4x-3)*(x-4) => x=20

Why does abcd = (ab)(cd) etc?

Real lateral thinking. I was nowhere near this trick.

I didn't get Step 3 at all. Are you multiplying the area of 2 rectangles ? (ab)*(cd) ? Why ? Thank you.

@kurtecaranum3047

Жыл бұрын

It's because Commutative Property comes into play to help us solve x easier (Green)(Blue) = (Orange)(Pink) (ab)(cd) = (ad)(bc) = abcd (x+2)(56) = (x-4)(4x-3) = abcd

Nice, X + 2 - X - 4 4X-3 - 56 56X + 112 = 4X² - 16X - 3X + 12 4X² - 75X - 100 = zero X = 20 Total area = 171 Bingo !!!!!!!!!

A=6x+51...trovo dai dati x=20,per cui A=171

What is the meaning of abcd?

@e1woqf

Жыл бұрын

a times b times c times d

x=20

I found a solution via the area ratios of 2 neighbouring rectangles respectively. The variables a, b, c and d are not required in this case: (x + 2) : (4x - 3) = (x - 4) : 56 56 (x + 2) = (x - 4) (4x - 3) 56x + 112 = 4x² - 3x - 16x + 12 4x² - 75x - 100 = 0 x = 20 (x = -1.25 => negative areas!) A = (x + 2) + (x - 4) + (4x - 3) + 56 A = (20 + 2) + (20 - 4) + (80 - 3) + 56 A = 22 + 16 + 77 + 56 = 171 Or if just the complete area is asked: A = 6x + 51 = 6 * 20 + 51 = 120 + 51 = 171