in this video we're going to focus on finding the area of different shapes that you encounter in a typical geometry course the first shape that we need to talk about is the rectangle let's say the rectangle has a length of eight and a width of five what is the area of the rectangle the area is simply length time width so in this example it's just 8 * 5 which is 40 now you may need to keep in mind the units of the area let's say if this is 8T by 5T then the area is 40 square ft now the next shape that we need to talk about is the triangle let's start with a right triangle so let's say if the base of the triangle is 10 and the height is eight what is the area of the triangle the area is 1 12 base times height so we said 10 is the base and the height is eight so this is going to be 1 12 10 * 8 half of 10 is 5 and 5 * 8 is 40 so it's 40 square units now what if you have a triangle that looks like this let's say this is nine and this is five and this is six what's the area of the triangle so for any right triangle you could use the same formula it's 1 12 base time height in this example the height is 9 but the length of the base is 5 + 6 the base of the triangle is this entire length which is 11 so it's going to be 12 9 * 11 9 * 11 is 9 9 and half of 100 is 50 so half of 99 is 49.5 now what if we have an equilateral triangle let's say all sides are 10 what is the area of this triangle for an equilateral triangle the equation that you need is the < TK 3 4 * s^ 2 so it's < tk3 over 4 * 10 2 10 2 that's 100 and 100 / 4 is 25 so it's 253 that's how you could find the area of an equilateral triangle now let's say if we have a square and the length of the square is n what is the area of this figure now first Square all sides are the same so so the length and the width of the square is 9 the area of a square is simply s squ or side squ so it's going to be 9 s which is just 81 now let's talk about the area of a circle so let's say if you're given the diameter of the circle and let's say the diameter of the circle is 10 cm what is the area of the circle to find the area it's equal to p r 2 the radius R is 12 of the diameter so this is the diameter D and this portion is the radius of the circle so the radius is half of 10 which is five so it's 5 cm so the area is going to be piun * 5^ 2 or simply 25 Pi square cm now consider this problem let's say we have a circle and we want to find the area of a sector of the circle and let's say the angle here is 60 so we want to find the area of this shaded region and let's say the uh Circle has a a radius of 10 what is the area of the green region just the sector to find the area of a portion of a circle it's going to be the angle in degrees / 360 so that's the fraction of the circle times the area of the entire circle which isk r s so the angle Theta is 60 so it's going to be 60 / 36 mtip Pi * the radius squ which the radius is 10 Now 60 divided 360 what we can do is cancel a zero so we have 6 ID 36 and 10^ 2 is 100 now 36 is basically 6 * 6 so we can get rid of one of the six values so we're left with 100 Pi / 6 and we could reduce the fraction if we divide both numbers by two half of 100 is 50 half of 6 is 3 so it's 50 piun / 3 so that's the area of the sector and let's say if you have a semicircle and let's say the radius of that circle is 8 if you need to find the area of a semicircle is 12 pi r 2 because we have half a circle so it's 12 Pi * 8^2 and uh 8 squ that's 64 and half of 64 is 32 so in this example the area of the semicircle is 32 Pi square units next up we have the area of a parallelogram let's say if the base is 8 and the height is 12 what's the area of the parallelogram the area is very similar to a rectangle it's base times height so it's 8 * 12 and 8 * 12 is 96 so that's the area of this particular parallelogram now let's say if we have another one that looks like this let's say if you're given the slant height instead let's say the slant height is five and this section is nine but this part let's say is three and this part is six what is the area of this parallelogram in order to find the area you got to find the height first and notice that we have a right triangle the hypotenuse of the right triangle is five and one of the legs is three so we got to find the missing side so we could use a 2 + B2 is = to c^2 so a is 3 we're looking for the missing side B and hypotenuse C is 5 3 S is 9 5 SAR is 25 25 - 9 is 16 and the S < TK of 16 is four so the missing side or the height is four so now we could use the formula area as base time height we have a a base of N9 and a height of four so 9 * 4 is 36 next up we have the area of a trapezoid so let's say the first base has a length of 10 and the second one 20 and let's say the height of the triangle I mean the trapezoid rather is8 find the area of this trapezoid the area is 12 the sum of the two bases B1 + B2 times the height B1 is 10 and B2 the second base is 20 and the height is 8 so 10 + 20 is 30 and half of eight if we just multiply these two that's four so this becomes 4 * 30 which is 120 so that's the area of this particular trapezoid now let me give you another problem that's similar to this one so let's say say that these two sides are 10 and let's say this is 12 and this is 24 find the area of the trapezoid in order for us to do that we need to break it up into parts so we need to make two right triangles now keep in mind this entire length is 24 which means that this is 12 so this section has to be six and that has to be six so that it adds up to 24 so now we can find the height of the right triangle so this is 10 this is six we got to find the missing side so using the same equation a 2 + b^2 is equal c^2 let's say a is 6 B is the height that we're looking for and C is 10 6 squar is 36 10 squar is 100 and 100 - 36 is 64 and the square root of 64 is 8 so the height is 8 now that we have the height we can now find the area so the area is going to be 12 B1 which is 12 plus B2 which is 24 times the height of 8 so 12 + 24 that's 36 and half 36 is 18 so 18 * 8 let's break 18 into 10 and 8 and distribute 8 8 * 10 is 80 8 * 8 is 64 and 80 + 64 is 144 so that's the area of this particular trapezoid now the next shape that we need to talk about is the rhombus so let's say the length of the first diagonal is 10 and the length of the second diagonal is 12 what is the area of this rhombus by the way for a rhombus all sides are congruent like a square and the diagonals bisect each other at 90° so those are some things to keep in mind the area of a diagonal I mean not a diagonal but a rhombus is 12 D1 * D2 so it's the product of the two diagonals * 1 12 so it's 1 12 * 10 * 12 half of 10 is 5 5 5 * 12 is 16 so that's the area of this particular ramus now here's another problem so let's say the length of one of the sides is 13 and this side is five where five is only half of one of the diagonals find the area of this particular rhombus so let's focus on the right triangle this is 5 this is 13 it helps to know the special right triangles for example there's a 34 five triangle also if you multiply these numbers by two you can get the 6 8 10 triangle which we covered there is the 7 24 25 triangle the 8 15 17 right triangle and also the 5 12 13 triangle so we have two of these numbers 5 and 13 therefore the missing side must be 12 now that we know what the missing side is you can also use the Pagan theorem to get 12 if you want to you can use a s plus b s = c^2 to get the same answer but once we have the missing side we now have the length of the two diagonals so diagonal one is five and five these two sides are equivalent so diagonal 1 has a length of 10 diagonal 2 has a length of 12 + 12 which is 24 so a is equal to 12 D1 D2 so it's 12 of 10 * 12 I mean not 12 but 24 now half of 24 is 12 and 12 * 10 is 120 so the area of this particular rhombus is 120 square units now let's say if you have a a triangle where you're given two sides and the included angle what is the area of this particular triangle the area is 12 AB s c so let's call this angle a angle B angle C this is side C side a and side B so in this case a lowercase a is 12 and lowercase b or side B is 10 and the angle between them that's angle C that's 30 half of 12 is 6 and sin 30 is 1 12 half of 10 is 5 and 6 * 5 is 30 so that's the area of this particular triangle so if you have two sides of a triangle and the included angle you can use this formula so this works if you do not have a right triangle how can we find the area of a scalan triangle let's say this is 9 10 and this is 11 so this is not a right triangle and we don't have any angles so what can we do to find the area of a triangle if we're given all three sides the first thing we need to do is find S which is 12 of the perimeter of the triangle so it's going to be 9 + 10 + 11 ID 2 9 + 10 is 19 19 + 11 is 30 and 30 ID 2 is 15 so that's the value of s so now that we have the value of s we can use Heron's formula to find the area so it's going to be the square Ro T of s * s - A * s - B * s - c s is 15 and then it's going to be 15 - 9 * 15 - 10 * 15 - 11 15 - 9 is 6 15 - 10 is 5 15 - 11 is 4 now 15 well let's see if we could simplify this an easier way 15 can be broken into five and three and six is 3 * 2 I'm not going to change the five and four I'm going to leave it alone so basically these two can come out as a single five these two can come out out as a single three and the square root of four is two so that comes out as a two so we're left over with a two inside 5 * 3 is 30 I mean not 30 but 15 and 15 * 2 is 30 so the final answer is 30 < tk2 so that's the area of the triangle using Heron's formula let's say if we have aare square and we're given the length of the diagonal and let's say the length of that diagonal is 10 < tk2 what is the area of the square now we know that the area is basically side squared all sides of the square are congruent so therefore we could find the side by using the Pagan theorem if this is s s and that's 10un 2 we can solve it so using equation a 2 + b 2 is equal to c^ 2 A and B are both s in this example and C is 10 < tk2 S2 + s^2 is 2 s^2 10 s is 100 and the square < TK of two squar which is like the square < TK of 2 * theun of two that's the < TK of 4 which is two so we can divide both sides by two so therefore for s^2 is 100 and the square otk of 100 is 10 now the area is S2 and S2 is 100 so that is the area so that's the answer 100 now let's say if we have a circle and there's a right triangle in a circle let's call this a B and C and C is the center of the circle and you're given the length of one side of the triangle let's say it's eight your task is to find the area of the Shaded region to find the area of the Shaded region it's going to be the area of the large object the object on the outside which is the area of the circle minus the area of the object on the inside which is the area of the triangle so the area of the Shaded region is going to beunk r^ 2 minus 12 base * height now notice that 8 is the radius of the triangle the radius is the distance between the center of the circle and any point on a triangle so AC is also the radius of the circle which means this section is eight and that's the base and the height of the triangle so the radius which is eight is the same as the base of the triangle and is the same as the height of the triangle so this is going to be 64 Pi minus 8 * 8 is 64 and half 64 is 32 so the area of the Shaded region is 64 Pius 32