The task is to find the area of the large, blue rectangle. Attached to the blue rectangle are two other rectangles, per the picture. The one on the upper right has an area of 168 square cm; the one on the lower left has an area of 100 square cm. The sides of each smaller rectangle leave a little space, as you see. For the yellow rectangle, the exposed segment has a length of 4; for the green rectangle, the exposed segment has a length of 6.
Obviously, to calculate the blue rectangle's area, we need to know the lengths of its sides. All we know right now is that the sides are 6 + something and 4 + something.
Let's start with the yellow rectangle. 12 times x is 168, so x must be 168/12, which is 14. That means the vertical side of the blue rectangle is 14 + 4 = 18 cm.
Let's move on to the green rectangle. Here, 18x = 100, so x = 50/9. This makes the long side of the rectangle (50/9 + 6) cm, or 104/9 cm.
So, we multiply 104/9 times 18, and we get 208 square cm for the blue rectangle's area.
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1 comment:
The task is to find the area of the large, blue rectangle. Attached to the blue rectangle are two other rectangles, per the picture. The one on the upper right has an area of 168 square cm; the one on the lower left has an area of 100 square cm. The sides of each smaller rectangle leave a little space, as you see. For the yellow rectangle, the exposed segment has a length of 4; for the green rectangle, the exposed segment has a length of 6.
Obviously, to calculate the blue rectangle's area, we need to know the lengths of its sides. All we know right now is that the sides are 6 + something and 4 + something.
Let's start with the yellow rectangle. 12 times x is 168, so x must be 168/12, which is 14. That means the vertical side of the blue rectangle is 14 + 4 = 18 cm.
Let's move on to the green rectangle. Here, 18x = 100, so x = 50/9. This makes the long side of the rectangle (50/9 + 6) cm, or 104/9 cm.
So, we multiply 104/9 times 18, and we get 208 square cm for the blue rectangle's area.
QED.
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