Want this question answered?
You should break it down in to smaller shapes. Two rectangles. Then figure out all the lengths. Multiply to find the area of the two rectangles. then add the products to get the final area.
They are all rectangles (or 2 squares and 4 rectangles).They are all rectangles (or 2 squares and 4 rectangles).They are all rectangles (or 2 squares and 4 rectangles).They are all rectangles (or 2 squares and 4 rectangles).
It is 36m + 72n which can be factorised to 36*(m + 2n)
You could consider the cross as two intersecting rectangles. Calculate the area of both rectangles and the area of the intersection (overlap). Then area of cross = sum of the areas of the rectangles minus the area of the overlap.
Lemma 2 says that if the curved area is approximated by inscribed and also by exscribed rectangles, where all rectanlges are of equal width, then the ratio of the exscribed areas summed up, to the curved area will be 1 as the number of rectangles tends to infinity. Also, the ratio of the sum of the area of the inscribed rectangles to the curved area to the curved area will be equal to 1 as the number of rectangles tends to infinity. In less technical language, the rectangles approximate the area under the curve, and as we use more and more (and thus thinner and thinner) rectangles, the approximation will get better and better; in fact we can make it as accurate as we want to. Lemma 3 says the same as the above, but drops the requirement that the rectangles all be of equal width.
45 percent of -- 36m = 80m45% of x = 36m(45% * x)/45% = 36m/45%x = 36m/45%x = 36m/0.45x = 80m
Some rectangles don't have equal sides.
You should break it down in to smaller shapes. Two rectangles. Then figure out all the lengths. Multiply to find the area of the two rectangles. then add the products to get the final area.
They are all rectangles (or 2 squares and 4 rectangles).They are all rectangles (or 2 squares and 4 rectangles).They are all rectangles (or 2 squares and 4 rectangles).They are all rectangles (or 2 squares and 4 rectangles).
It is 36m + 72n which can be factorised to 36*(m + 2n)
You could consider the cross as two intersecting rectangles. Calculate the area of both rectangles and the area of the intersection (overlap). Then area of cross = sum of the areas of the rectangles minus the area of the overlap.
Yes, all Squares are rectangles, but not all rectangles are squares because it needs to have all equal sides.
Yes, all rectangles are parallelograms. However, not all parallelograms are rectangles.
Lemma 2 says that if the curved area is approximated by inscribed and also by exscribed rectangles, where all rectanlges are of equal width, then the ratio of the exscribed areas summed up, to the curved area will be 1 as the number of rectangles tends to infinity. Also, the ratio of the sum of the area of the inscribed rectangles to the curved area to the curved area will be equal to 1 as the number of rectangles tends to infinity. In less technical language, the rectangles approximate the area under the curve, and as we use more and more (and thus thinner and thinner) rectangles, the approximation will get better and better; in fact we can make it as accurate as we want to. Lemma 3 says the same as the above, but drops the requirement that the rectangles all be of equal width.
Each side will be 6. i.e 62 = 36
In fact, some are rectangles, but not all are.
An L-shaped area can be divided into two rectangles. The total area is the sum of the areas of the two rectangles.