SAT Question eb70d2d0: Area and Volume answer and explanation with Desmos solution

5$.I{t}^{\prime }sgiventhatatriangleisformedbyconnectingthethreepointsshown,whichare$\left( - 3 , 4 \right)$,$\left( 5 , 3 \right)$,and$\left( 4 , - 3 \right)$.LetthistrianglebetriangleA.TheareaoftriangleAcanbefoundbycalculatingtheareaoftherectanglethatcircumscribesitandsubtractingtheareasofthethreetrianglesthatareinsidetherectanglebutoutsidetriangleA.Therectangleformedbythepoints$\left( - 3 , 4 \right)$,$\left( 5 , 4 \right)$,$\left( 5 , - 3 \right)$,and$\left( - 3 , - 3 \right)$circumscribestriangleA.Thewidth,inunits,ofthisrectanglecanbefoundbycalculatingthedistancebetweenthepoints$\left( 5 , 4 \right)$and$\left( 5 , - 3 \right)$.Thisdistanceis＄4-\left(-3\right)$, or ＄7$.Thelength,inunits,ofthisrectanglecanbefoundbycalculatingthedistancebetweenthepoints$\left( 5 , 4 \right)$and$\left( - 3 , 4 \right)$.Thisdistanceis＄5-\left(-3\right)$, or ＄8$.Itfollowsthatthearea,insquareunits,oftherectangleis$\left( 7 \right) \left( 8 \right)$,or＄56$. One of the triangles that lies inside the rectangle but outside triangle A is formed by the points $\left(-3,4\right)$, $\left(5,4\right)$, and $\left(5,3\right)$. The length, in units, of a base of this triangle can be found by calculating the distance between the points $\left(5,4\right)$ and $\left(5,3\right)$. This distance is ＄4 - 3$,or＄1$. The corresponding height, in units, of this triangle can be found by calculating the distance between the points $\left(5,4\right)$ and $\left(-3,4\right)$. This distance is ＄5 - \left( - 3 \right)$,or＄8$. It follows that the area, in square units, of this triangle is $\frac{1}{2}\left(8\right)\left(1\right)$, or ＄4$.AsecondtrianglethatliesinsidetherectanglebutoutsidetriangleAisformedbythepoints$\left( 4 , - 3 \right)$,$\left( 5 , 3 \right)$,and$\left( 5 , - 3 \right)$.Thelength,inunits,ofabaseofthistrianglecanbefoundbycalculatingthedistancebetweenthepoints$\left( 5 , 3 \right)$and$\left( 5 , - 3 \right)$.Thisdistanceis＄3-\left(-3\right)$, or ＄6$.Thecorrespondingheight,inunits,ofthistrianglecanbefoundbycalculatingthedistancebetweenthepoints$\left( 5 , - 3 \right)$and$\left( 4 , - 3 \right)$.Thisdistanceis＄5-4$, or ＄1$.Itfollowsthatthearea,insquareunits,ofthistriangleis$\frac{1}{2} \left( 1 \right) \left( 6 \right)$,or＄3$. The third triangle that lies inside the rectangle but outside triangle A is formed by the points $\left(-3,4\right)$, $\left(-3,-3\right)$, and $\left(4,-3\right)$. The length, in units, of a base of this triangle can be found by calculating the distance between the points $\left(4,-3\right)$ and $\left(-3,-3\right)$. This distance is ＄4 - \left( - 3 \right)$,or＄7$. The corresponding height, in units, of this triangle can be found by calculating the distance between the points $\left(-3,4\right)$ and $\left(-3,-3\right)$. This distance is ＄4 - \left( - 3 \right)$,or＄7$. It follows that the area, in square units, of this triangle is $\frac{1}{2}\left(7\right)\left(7\right)$, or ＄24.5$.Thus,thearea,insquareunits,ofthetriangleformedbyconnectingthethreepointsshownis＄56-4-3-24.5$, or ＄24.5$. Note that 49/2, 24.5, and 24.50 are examples of ways to enter a correct answer.