The video provides a clear algebraic solution to a basic geometry problem, though framing such a foundational concept as a "Harvard-level" challenge is a bit of an overstatement. It is a solid tutorial for beginners, even if the academic prestige is mostly marketing.
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Can You Solve This?Added:
This is an interesting math question. We have a big rectangle which is divided into four different portions using vertical and horizontal lines like this.
The area of this portion is 15 units square. Then the area of this portion is six units square and the area of this portion is 14 units square. Now the question says, "What will be the area of this portion?"
Before we begin, note that this figure is not drawn to scale. First of all, let us label this portion as one, this as two, this as three, and this as four.
Now we know that in the case of any rectangle of side length A and B, its area is given as A * B.
But we don't know anything about the side lengths of these rectangles and we only know about the areas. So you should keep in mind that for such questions, it's often helpful to assume side lengths as some variables and express the areas in terms of these variables.
Let me show you how. Let us assume this side length is equal to A. So if this is A, then this will also be equal to A.
Then assume this is B and thus this will also be equal to B.
Now if this is C, then this will also be equal to C. And finally, if this is D, then this will also be equal to D.
After we are done with labeling side lengths, we've turned a confusing picture into a simple set of variables.
So now we should write the area of each rectangle as the product of its side lengths.
For portion one or this rectangle, what will be its area? Yes, right. It will be A * C because the side lengths of this rectangle are A and C. But area is given as 15 and thus we have AC = 15. Now for portion two, we have BC = 6 because the side lengths of this rectangle are B and C and its area is given as six. Similarly, for portion three, we have BD = 14 because the sides are B and D and the area is 14. Now for portion four, we need to find its area which will be equal to A * D. Now what to do? Let us call this product AD as X.
Let us multiply this equation with this equation. So from here, we get AC multiplied by this BD which equals 15 * 14. So this gives A * B * C * D = 15 * 14, right? Next, multiply this equation with this equation. We get BC from here * AD from here = 6 * X. So this also gives A * B * C * D = 6 * X.
But hey, both of them represent the same quantity as this one. So we can equate them. 15 * 14 = 6 * X. Now to solve for X, divide by six on both sides to get X = 15 * 14 over six. This 15 is 3 * 5 and 14 is 2 * 7. Also, six is 2 * 3. So both two and three get canceled out and we are left with 5 * 7 which means X = 35 and that's it.
The area of this portion is 35 square units.
So good.
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