then you can use law of cosines to find a diagonal and you have two triangles.
(if you just have the 4 lengths and no angle, the shape is not uniquely determined.)
(if you just have the 4 lengths and no angle, the shape is not uniquely determined.)
Area of a right angled triangle is easy since the two short sides serve as the base and height. You can probably figure the rest out from here.
In this case, we have the side lengths AB = 5 cm, BC = 13 cm, CD = 9 cm, and DA = 15 cm, and the diagonal AC = 12 cm.
To calculate the area of quadrilateral ABCD, we can use the formula:
Area = 1/2 * (AC * BD)
First, we need to find the length of the diagonal BD. We can use the Law of Cosines to find the length of BD:
BD^2 = AB^2 + AD^2 - 2 * AB * AD * cos(angle BDA)
In this case, angle BDA is an opposite angle to angle BCD, which can be found using the Law of Cosines as well:
cos(angle BDA) = (AB^2 + AD^2 - BD^2) / (2 * AB * AD)
Substituting the given values:
(AB^2 + AD^2 - BD^2) / (2 * AB * AD) = (5^2 + 15^2 - BD^2) / (2 * 5 * 15)
Simplifying:
25 + 225 - BD^2 = 150
BD^2 = 250
BD = √250 = 5√10 cm
Now that we have the lengths of the diagonals AC and BD, we can calculate the area of quadrilateral ABCD:
Area = 1/2 * (AC * BD)
Substituting the values:
Area = 0.5 * (12 cm * 5√10 cm)
Area = 6 cm * 5√10 cm
Area = 30√10 cm²
Therefore, the area of quadrilateral ABCD is 30√10 square centimeters.