To find the area of a figure with sides of different lengths, we first need to determine the shape of the figure. In this case, the sides are 5m, 3m, 4m, and 6m, which could form a quadrilateral or a triangle depending on the configuration. If it is a quadrilateral, we would need more information such as the angles between the sides to calculate the area. If it is a triangle, we could use Heron's formula to find the area. More details or a diagram would be necessary to provide an accurate calculation.
It is not possible to answer the question.
The fact that there are four lengths given in the question suggests that the shape is a quadrilateral. Unfortunately, the lengths of a quadrilateral's sides does not determine its area. One way to see this is that a square can be flexed into a rhombus and the top and bottom sides of the rhombus brought closer and closer together until its area is almost zero.
It is not possible to answer the question.
The fact that there are four lengths given in the question suggests that the shape is a quadrilateral. Unfortunately, the lengths of a quadrilateral's sides does not determine its area. One way to see this is that a square can be flexed into a rhombus and the top and bottom sides of the rhombus brought closer and closer together until its area is almost zero.
It is not possible to answer the question.
The fact that there are four lengths given in the question suggests that the shape is a quadrilateral. Unfortunately, the lengths of a quadrilateral's sides does not determine its area. One way to see this is that a square can be flexed into a rhombus and the top and bottom sides of the rhombus brought closer and closer together until its area is almost zero.
It is not possible to answer the question.
The fact that there are four lengths given in the question suggests that the shape is a quadrilateral. Unfortunately, the lengths of a quadrilateral's sides does not determine its area. One way to see this is that a square can be flexed into a rhombus and the top and bottom sides of the rhombus brought closer and closer together until its area is almost zero.
It is not possible to answer the question.
The fact that there are four lengths given in the question suggests that the shape is a quadrilateral. Unfortunately, the lengths of a quadrilateral's sides does not determine its area. One way to see this is that a square can be flexed into a rhombus and the top and bottom sides of the rhombus brought closer and closer together until its area is almost zero.
To find the area of a quadrilateral with sides of 7m, 4m, 5m, and 3m, you can use Brahmagupta's formula for the area of a cyclic quadrilateral: Area = √(s-a)(s-b)(s-c)(s-d), where s is the semiperimeter (s = (a + b + c + d) / 2) and a, b, c, and d are the lengths of the sides. Plug in the values of the sides into the formula to calculate the area.
-15m but -15 metres doesnt exist
5m+3=3m+9 5m+3-3=3m+9-3 5m=3m+6 5m-3m=3m+6-3m 2m=6 (2m)/2=6/2 m=3
-3m = 5m+8 -3m-5m = 8 -8m = 8 m = -1
4m+9+5m-12=42 9m=45 m=5
5m,4m,3m respectively
5*4*3 60?
To find the area of a quadrilateral with sides of 7m, 4m, 5m, and 3m, you can use Brahmagupta's formula for the area of a cyclic quadrilateral: Area = √(s-a)(s-b)(s-c)(s-d), where s is the semiperimeter (s = (a + b + c + d) / 2) and a, b, c, and d are the lengths of the sides. Plug in the values of the sides into the formula to calculate the area.
-15m but -15 metres doesnt exist
The area of rectangle is : 20.0
-7m - 3m + 5m = -5m
5m+3=3m+9 5m+3-3=3m+9-3 5m=3m+6 5m-3m=3m+6-3m 2m=6 (2m)/2=6/2 m=3
Every one of them could be a side of a rectangle. It is not possible to give an answer in respect of an unspecified rectangular object. however, i feel the question is incorrect as it should be sides of the triangle and according to it 3rd is the answer as 3+5<9 :)
The area is 15 square metres.
-3m = 5m+8 -3m-5m = 8 -8m = 8 m = -1
The pressure exerted by the body is calculated by dividing the force by the area. In this case, the pressure would be 450N / (5m x 4m) = 22.5 Pascal.
Rectangles are really simple. The perimeter is twice the length plus twice the width (in this case 4m x 2 = 8m, 5m x 2 = 10m, and 8m + 10m = 18m). The area is simply the length times the width (in this case, 4m x 5m = 20m2).