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# CA Geometry: Pythagorean theorem, area

## Video transcript

we're on problem 36 and it says what is the area in square units of the trapezoid shown below so when you just look at this you're like okay trapezoid do I know the the formula for the area of a trapezoid and you might get confused and all of that you say well trapezoid I can break it up into a rectangle and a triangle if I were to draw a line right here I draw a line right there then I've broken up the traps into a rectangle and a triangle and then if I know the dimensions of each of those I know the dimensions I know the area of each of them and then I know the area of the entire thing let's see what's this height right here or this width I should say well we're going from zero to what x is equal to X is equal to eight here all right I just went straight down from X is equal to eight y is equal to five so this dimension is eight and then then when we go from X is equal to eight to X is equal to 12 how far is that well that's going to be four all right so this is four and this is eight fair enough and then how high is this rectangle we're going from Y is equal to zero to y is equal to five so that's five and of course this is five as well so we're done we're ready to figure out the area the area of the rectangle part is eight times five that's 40 the area of this triangle is five times four times one-half right if we didn't put that one-half we'd be figuring out the area of this rectangle right there so it's five times four is 20 times one half is ten so the area of both of these combined is 10 plus forty fifty 37 the figure below is a square with four congruent parallelograms inside this looks interesting what is the area in square units of the shaded portion so the shaded portion is the whole square minus the area of the parallelogram so the whole square that's easy it's 12 and the height is 12 but since we know it's a square we know the width also has to be 12 right so the area the whole square is 144 and now we have to figure out the area if we know the area of one of the parallelograms we know the area of all of the parallelograms because they are congruent so let's see we can figure out the area of one of the parallelogram so I'll there is actually a formula for the area of a parallelogram it's actually just the base times the height and they actually give us that but let me show you that they give us that because it might not be obvious to you so if I let me try to draw it so I want to use my line tool nope that's not the line tool so one side it goes straight like that and I come down like that good enough okay now if I look at just this parallelogram they tell us that the height here is three right this height is three and I know it's the height because they told me this is a 90-degree angle and they tell us that the base is five and I'm telling you that the area of a parallelogram is just the base times the height is equal to 15 but you shouldn't just take my word for it you that should make intuitive sense to you and the way to think about it intuitively is imagine if we were to take if we were to take this part of the parallelogram if we just cut it right here and if we were to move it over here if we were to cut that off and move it over here then the parallelogram would look something like this you'd have the part that we didn't cut off right and then you move the cut off part over here and now the dimensions this base would be five and then this height would be three in the area of this rectangle is 15 and there's no reason why the area of this should be any different than that we just rearranged its parts so that's why the area of a parallelogram is just the base times the height so the area of each of these parallelograms are 15 so the area of all of them combined is 15 times 4 which is 60 so 144 minus 60 is what ad 484 and that's choice B problem 38 what is the area in square meters of the trapezoid shown below so to figure out the area we could break it up into these rectangles and triangles to figure out the area of this rectangle we need to know we need to know its height we know this height and actually we need that we'll need that to figure out the area of the trying as well so what's this height right there let's see we know that this this distance is going to be six it's a rectangle that distance is going to be six if that distance is six and both of these are five both of these triangles here are going to be congruent right they because this length is equal to this length this length is equal to this length and we also what we can also make you know this angle is equal to that angle but anyway so what is are these is let me do it in another color what's the length of these two green sides let's call it X all right well we know that when you add X plus 6 plus X it has to equal 12 that whole top part so you get X plus X is 2x plus 6 is equal to 12 2x is equal to 6 X is equal to 3 and you might have been able to solve that in your head that that's 6 and these are the same that both of these are going to be 3 and now we can use that information to figure out this height right there right because if we just draw this triangle right there that's 3 that's 5 this is some unknown side a and you might already recognize we're gonna use a Pythagorean theorem but this is a very typical type of right triangle so you might already be able to guess what a is but we'll solve for it so we can know that a squared plus 3 squared is equal to the hypotenuse squared the side opposite the 90-degree angle squared so that is equal to 25 5 squared is 25 a squared plus 9 is equal to 25 a squared is equal to 16 a is equal to 4ei is equal to 4 and now we're ready to figure out the area what's the area of the rectangle 6 times 4 is 24 what's the area of each of these triangles 3 times 4 times 1/2 3 times 4 is 12 times 1/2 is 6 so the area of that triangle 6 the air of this triangle is 6 so 6 plus 24 plus 6 is 36 B problem 39 what is the area and square inches of the triangle below interesting okay so this is a equilateral triangle all the sides are equal and so we could actually say that this since these two triangles are in symmetric we can say that's equal to that and you know this comes to a general formula for the area of an equilateral triangle but let's just figure it all out so this is so this side is going to be five and this side is going to be five if this is five and that's 10 what is this side right here let's call it X Pythagorean theorem this is the hypotenuse so x squared plus five squared plus 25 is going to be equal to the hypotenuse squared is equal to 100 x squared is equal to 100 minus 25 75 X is equal to the square root of 75 75 is 25 times 3 right so that's equal to the square root of 25 times 3 which is equal to the square root of 25 times the square root of 3 which is equal to 5 roots of 3 so X is equal to 5 roots of 3 and now what's the area of just this right triangle right here this one on the right side well it's base is 5 its height is 5 roots of 3 so it's going to be 1/2 times the base 5 times this height 5 roots of 3 and that's what 1/2 times 5 times 5 so it's 25 root 3 over 2 and that's just this triangle right there well this triangle is going to have the same is we're going to have the exact same area right there congruent triangles so the area of the figure is this times 2 so 2 times that is equal to just the 25 root 3 and that's choice B next problem problem 40 the perimeter of 2 squares are in a ratio of 4 to 9 what is the ratio between the areas of the two squares so let me draw let me draw straw so let's draw two squares so that's one square and let me draw another square that's another square and let's say that this the sides of this rx right and the sides of this one or Y so they're saying the perimeters of the two squares are in a ratio of four to nine so the perimeter of the first square is for X right X plus X plus X plus X so the perimeter of the first square is 4x the perimeter of the second square is 4y so that's the ratio of the perimeter of the first squared of the perimeter of the second Square and then that is equal to four to nine that is equal to four to nine and they say what is the ratio between the areas of the two squares so when they want us to figure out the area of the first square is x squared and the ratio right X base times height x times X and the area of the second square is y times y so they want us to figure out what that is equal to right well this is x squared over Y squared this is the same thing as x over Y squared so if we can figure out what x over Y is equal to we can just square it and we'll get x squared over Y squared so let's try to do that so they give us this let's see if we divide well this is simplifies write four over for no reason x over Y is equal to four over nine right X over nine is equal to four over nine so let's substitute that here so x squared over Y squared is equal to x over Y squared which is equal to 4/9 squared which is equal to 16 over 81 or the ratio of the areas of the two squares is 16 to 80 one choice D maybe we could fit one more problem in there actually no I'm over ten minutes I'll stop right there see you in the next video