Question

lisa is looking to fill a part of a park with sand. the shape below represents the part of the park being covered in sand. how many square yards of the park will lisa cover in sand? the area of the park being covered in sand changed, as shown below. would this new area of the park be more than or less than the original area of the park, 130 square yards? what would be a good estimate for the new area of the park? 10 yd 6 yd 10 yd the new area of the park would be 130 square yards. a good estimate for the new area of the park would be about square yards.

Explanation:

Step1: Calculate the area of the rectangle

The formula for the area of a rectangle is \(A = l\times w\). Here, \(l = 10\) yd and \(w=10\) yd. So, \(A_{rectangle}=10\times10 = 100\) square yards.

Step2: Calculate the area of the triangle

The formula for the area of a triangle is \(A=\frac{1}{2}\times b\times h\). Here, \(b = 6\) yd and \(h = 10\) yd. So, \(A_{triangle}=\frac{1}{2}\times6\times10=30\) square yards.

Step3: Calculate the total area

The total area \(A = A_{rectangle}+A_{triangle}\). Substituting the values, \(A=100 + 30=130\) square yards. But wait, the original area was 130. However, if we assume some approximation (maybe the jagged part was a negligible error in the first - case calculation). But if we consider the new shape as a combination of rectangle and triangle as above.

Another way: If we compare to the original area of 130. The new area is composed of a \(10\times10\) rectangle (\(100\)) and a triangle with base \(6\) and height \(10\) (\(30\)). The sum is \(100 + 30=130\). But if we assume that in the original problem the shape might have had some complex calculation (maybe the jagged part was over - estimated in the original 130). But if we go by the new simple shape (rectangle + triangle):
The new area is more than if the original had some "missing" parts (but if we assume the new shape is a proper rectangle - triangle combination). Wait, no, actually, if we calculate the new area as \(A=(10 + 6)\times10-\frac{1}{2}\times6\times10\) (using the formula for the area of a trapezoid \(A=\frac{(a + b)h}{2}\), where \(a = 10\), \(b=10 + 6\) is wrong. Wait, no, the correct way is:
The figure can be seen as a rectangle of \(10\times10\) and a triangle of base \(6\) and height \(10\). The area of the rectangle is \(10\times10=100\), area of the triangle is \(\frac{1}{2}\times6\times10 = 30\). The total area is \(100+30 = 130\). But if we assume that in the original problem, the area was calculated with some irregularity (like the jagged part). If the new shape is a "filled - in" shape (rectangle + triangle), and the original was 130 (maybe with the jagged part having an area that was considered as part of the 130). But if we use the formula for the new shape (a more "regular" shape):
The new area is more than if the original had a concave shape (jagged part). Since the jagged part in the original might have made the area less (if we consider the "hole" due to jaggedness). So the new area (rectangle + triangle) is more than the original 130 (assuming the jagged part was a depression).
A good estimate:
The area of the rectangle is \(10\times10 = 100\). The area of the triangle is \(\frac{1}{2}\times6\times10=30\). So the total area is \(100 + 30=130\). But if we assume that the original 130 was an underestimate (because of the jagged part), the new area (a more "solid" shape) is more. A good estimate for the new area (if we consider standard geometric shapes) is \(10\times(10 + 6)-\frac{1}{2}\times6\times10\) (no, wrong). Wait, correct formula for the new shape (a trapezoid - like, but it's a rectangle and a triangle). The area is \(10\times10+\frac{1}{2}\times6\times10=100 + 30=130\). But if we assume that the original 130 was for a shape with a "dent" (jagged part), the new area (with the "dent" filled as a triangle) is more. A good estimate: \(10\times10+\frac{1}{2}\times6\times10=130\). But if we consider that maybe the height of the triangle is a bit less (but the problem gives height as 10). So, if we go by the calculation:
The new area is more than 130 (assuming original 130 was for a shape with a concave part). A good estimate: \(10\times10+\frac{1}{2}\times6\times10 = 130\), but if we consider a more "generous" estimate (if there was an error in measuring the height of the triangle in the original), but based on the given numbers:

Answer:

The new area of the park would be more than 130 square yards. A good estimate for the new area of the park would be about 130 square yards.