### What is the area of a quadrilateral with coordinates at 4 0 and 14 11 and 0 6 and -10 -5?

As the quadrilateral has no sides aligned with the x- and y- axes finding lengths requires used of Pythagoras and square roots. To make it easier to find lengths the quadrilateral could be divided up into smaller shapes, the area of each which could be added together to find the area of the quadrilateral, but to do so would require finding points within the quadrilateral.An easier method which works for ALL quadrilaterals in the place is to draw the smallest rectangle aligned with the x- and y-axes that contains the quadrilateral and then the area of the quadrilateral can be found as the area of the rectangle less the area between the quadrilateral and the rectangle. This is easier as the area between the rectangle and the quadrilateral can be divided up into triangles which have points already known (drawn from the corners of the rectangle to points of the quadrilateral) and have one side parallel to the x- or y- axis and its height is found as the difference between the x- or y-coordinates, and possibly some squares which have easy x- and y-coordinates to find.The coordinates of the rectangle are given by:A = (Xmin, Ymax), B = (Xmax, Ymax), C = (Xman, Ymin), D = (Xmin, Ymin)whereXmin = the minimum x-coordinate of all the pointsXmax = the maximum x-coordinate of all the pointsYmin = the minimum y-coordinate of all the pointsYmax = the maximum y-coordinate of all the pointsFrom this it can be seen that at least two of the points of the quadrilateral will be on the rectangle.In the case of the given coordinates, the quadrilateral is a convex quadrilateral PQRS with P = (-10, -5), Q = (0, 6), R = (14, 11), S = (4, 0).This gives: A = (-10, 11), B = (14, 11), C = (14, -5), D = (-10, -5)Points P & D and R & B coincide.Draw in the lines AQ and CS.The area of the quadrilateral PQRS is given by:(Note:Line lengths are difference in x-coordinates for horizontal lines and y-coordinates for vertical lines;vd XYZ is the vertical distance between horizontal line XY and point Z = difference in y-coordinates of X or Y and Z; andhd XYZ is the horizontal distance between vertical line XY and point z = difference in x-coordinates of X or Y and Z.)area PQRS = area ABCD - (area ABQ + area BCS + area CDS + area DAQ) = (AB × BC) - (½ × AB × vd ABQ + ½ × BC × hd BCS + ½ × CD × vd CDS + ½ × AD × hd ADQ)= (|14 - -10| × |11 - -5|) - ½ × (|14 - -10| × |11 - 6| + |11 - -5| × |14 - 4| + |14 - -10| × |-5 - 0| + |-5 - -11| × |-10 - 0|)= (|24| × |16|) - ½(|24| × |5| + |16| × |10| + |24| × |-5| + |-16| × |-10|)= (24 × 16) - ½(24 × 5 + 16 × 10 + 24 × 5 + 16 × 10)= 104 square units.(|n| means the absolute value of n, the value of n ignoring the sign.)--------------------------------------------The given coordinates when plotted on the Cartesian plane forms the shape of a 4 equal sided rhombus with diagonals that bisect each other at 90 degrees and by using the area formula of 0.5 times product of its diagonals the area of the rhombus is 104 square units.---------------------------------------------In mathematics (especially at degree level and beyond) it is not enough to say that something looks like something - it has to be proved. (Once proved, that statement can be used without having to reprove it).The second answer stated that when plotted the shape is a rhombus, but it is stated without proof. As such, there is no guarantee that the formula for the area of a rhombus (which has been proved and so can be used without proof here) will work. A rough sketch can help with formulating a proof - - not having any graph paper handy to plot the points I can only draw a sketch and it looks like it could be a parallelogram).Using P = (-10, -5), Q = (0, 6), R = (14, 11), S = (4, 0)Let T = (0, -5), U = (0, 11), V = (14, 0), W = (-10, 0)Now consider side PQ with point T to make triangle PQT, and side SR with point V to make triangle SRVPoint T is directly horizontally right of P (no change in y) and directly vertically below Q (no change in x); thus angle PTQ is a right angle as lines PT and QT are perpendicular.Point V is directly horizontally right of S (no change in y) and directly vertically below R (no change in x); thus angle SVR is a right angle as lines SV and QV are perpendicular.PT = 0 - -10 = 10 units long (can use the change in x as the length as there is no change in y)SV = 14 - 4 = 10 units longQT = 6 - -5 = 11 units long (can use the change in y as the length as there is no change in x)RV = 11 - 0 = 11 units longThus triangles PQT and SRV are congruent through Side (PT = SV) Angle (PTQ = SVR) Side (QT = RV)So PQ = SR and slope PQ = angle TPQ = slope SR = angle VSRTherefore PQ and SR are parallel and equal in length (= √(PT² + QT²) = √(10² + 11²) = √221 units).Now consider side QR with point U to make triangle QRU, and side PS with point W to make triangle PSWPoint U is directly horizontally left of R (no change in y) and directly vertically above Q (no change in x); thus angle RUQ is a right angle as lines RU and QU are perpendicular.Point V is directly horizontally left of S (no change in y) and directly vertically above P (no change in x); thus angle SWP is a right angle as lines SW and PW are perpendicular.UR = 14 - 0 = 14 units long (can use the change in x as the length as there is no change in y)WS = 4 - -10 = 14 units longQU = 11 - 6 = 5 units long (can use the change in y as the length as there is no change in x)PW = 0 - -5 = 5 units longThus triangles QRU and PSW are congruent through Side (QU = PW) Angle (RUQ = SWP) Side (UR = WP)So QR = PS and slope QR = angle URQ = slope PS = angle WSPTherefore QR and PS are parallel and equal in length (= √(QU² + UR²) = √(5² + 14²) = √221 units).Thus PQRS is at least a parallelogram as PQ and SR are parallel and QR and PS are parallel.However, since PQ = QR = RS = SP = √221 units it is a rhombus.Its area can now be calculated as:area rhombus = ½ × product diagonals= ½ × QS × PR= ½ × √((4 - 0)² + (0 - 6)²) × √((14 - -10)² + (11 - -5)²)= ½ × √(4² + 6²) × √(24² + 16²)= 104 square units