can you explain why this is x2, idk why h1 h2 h3 become x2

Question

here it is :(((((((((

Answer 1

There is no h₃ (there can only be two heights associated with a parallelogram).

Let's take the analysis step by step.

We have two ways of finding the area of the parallelogram, which encloses all the other shapes:

1) take the side length x₁+x₂ and the height h₁ (perpendicular distance between the opposite sides of the parallelogram), then the area is (x₁+x₂)h₁;

2) take the side length x₃ and the height h₂, then the area is x₃h₂.

So these differently calculated areas must be the same: (x₁+x₂)h₁=x₃h₂.

Now we consider the combination of areas of shapes inside the big parallelogram:

We use congruency of certain shapes to work out the area of the big parallelogram in terms of known and unknown areas. The embedded parallelogram is split into two congruent triangles, so they have equal areas. Inside this embedded parallelogram we take the combined areas of the rightmost triangle, that is, 72+s₁+8. The other triangle forming the embedded parallelogram has the same area. (Note that we only have one given area of 79 and the little triangles are not marked.) So the area of the embedded parallelogram is 2(72+s₁+8).

We have two more congruent triangles. The areas of these, together with the area of the embedded parallelogram, make up the total area of the big parallelogram. We use the area of the leftmost triangle: Δs+s₂ to determine the area of its counterpart right (the big triangle on the other side of the big parallelogram). So the combined area is 2(Δs+s₂).

We now have the total area of the big parallelogram:

2(72+s₁+8)+2(Δs+s₂)

The next line in the analysis looks wrong or incomplete to me. The solver seems to have assumed (rather than proved) that there are two more congruent triangles. The areas s₂+79+s₁+10 form the area of a large triangle, and no other triangle in the figure is congruent to this triangle. However, the area of this triangle is the same as the area of a triangle on the same base x₃ formed by drawing a diagonal of the big parallelogram, because it would have the same height h₂. So the area of the big parallelogram is, in fact, double the area of the big triangle, as the solver has indicated in the logic. The rest of the logic now falls into place:

2(72+s₁+8)+2(Δs+s₂)=2(s₂+79+s₁+10), 80+s₁+Δs+s₂₌s₂+89+s₁ (adding numbers and dividing both sides by 2), 80+Δs=89 (s₁ and s₂ cancel out), making Δs=9.

Comments

ohhhh thank you so much so the way to solve this problem is wrong?

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No, my bad. The solution was correct, but I feel the logic was incomplete and needed further explanation. I've amended my response so that you can see the logic that was missing. If you don't understand why the triangles are congruent I can expand further. When I first wrote my response I missed the point. I apologise if I confused you.

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can u show me which triangles are congruent and why :( i dont really understand it :((( sorry 

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First, you might want an easy definition of geometrical congruency. Two triangles are congruent if one is a copy or a reflection of the other. That means they're the same shape and size. If you draw a diagonal of a parallelogram two congruent triangles are formed. In the problem, there is a smaller parallelogram inside the big one. In your diagram, x₁ is shown as the length of one side of this embedded parallelogram, so the opposite parallel side is the same length. One of the other two sides separates area s₂ from area 79, and its opposite parallel side has the same length. The diagonal starts at the left end of x₁ (or the right end of x₂) and cuts across the embedded parallelogram till it meets the bottom right-hand corner. To prove the two triangles formed by the diagonal are congruent:

the diagonal is common to both triangles,

the diagonal is a transverse of the parallelogram and creates two pairs of equal angles on alternate sides of the transverse.

We only need three conditions to proved that the triangles are congruent: we have identified three equalities (or congruencies): one side (common) and two angles. If two triangles are congruent, their areas are the same. We could have used three other conditions such as the lengths of the sides of the triangles all being of equal length (one common side and the fact that opposite sides of a parallelogram have equal length).

The two other congruent triangles are not as obvious, but let's find three conditions which tell us that the triangles are congruent (and have equal area). First, identify the triangles. One, which we can call triangle A, is on the left and has x₂ and x₃ as two of its sides, and its area is s₂+Δs. The other triangle, triangle B, is on the right. Let's see if we can find the length of its sides. One side has length x₃ because it's the opposite side of the big parallelogram.

The big parallelogram has its long side marked by x₁ and x₂, so its length is x₁+x₂. The opposite long side must have the same length. The embedded parallelogram has side length x₁ and so its opposite parallel side is the same length, and forms part of the long side of the big parallelogram (length x₁+x₂). The remaining part of this long side is another side of triangle B, so this side must have length x₁+x₂-x₁=x₂. This length matches the length of the side in triangle A. The angle between the two sides of triangle A has the same measure as the two sides of angle B because the opposite angles of a parallelogram (the big one) have the same measure. So we have satisfied 3 conditions (two sides and the included angle) and therefore proved triangles A and B to be congruent, so they have the same area.

Another rule you might need to remember is that all triangles on the same base length and between the same two parallel lines have equal area (because they have the same height).

I hope this lengthy explanation helps.