Dear sir your action fields pathways are really awesome and unique so I have transparent and true love and support forever till my last breat ❤️❤️❤️🙏🙏🙏🙏my great love and greetings from India ... 🙏🙏I will try to meet with you one day ❤️❤️❤️😀😀😀😀😀😀😀😀😀❤️👋🙏🙏👋please call me for USA Visit 😀❤️❤️🙏🙏🙏🙏
135 is 90+45 so you have a rectange plus a 45 deg isocèle square triangle, partially cut off with one side of 9 units and the other long enough to make the width base of 15.
The measures in this puzzle are faulty. The triangle EDC is an isosceles triangle with a hypotenuse of 15; therefore DC cannot be 9 but must be 15/sqrt2. Moreover the auxiliary lines are not needed at all, this figure can be divided into a rectangle 15x9 and the isosceles triangle with a base of 15 and e heigth of 7.5
I really liked this method, which made calculation simpler than completing the rectangle, which is what I did: First, like you, I figured out that ∠BCD = 135°. Then I proceeded as follows: Draw line through D that is parallel to AB. Extend AE and BC until they intercept this line at points E and F respectively. We now have rectangle ABGF, with FG = AB = 15, and AF = BG = ? We also have 2 isosceles right triangles: △DEF and △CDG. In △CDG, hypotenuse CD = 9, so CG = DG = 9/√2 Since FG = AB = 15 and DG = 9/√2, then DF = 15−9/√2 In △CDG, EF = DF = 15−9/√2, and hypotenuse DE = 15√2−9 In rectangle ABGF side AF = AE + EF = 6 + (15−9/√2) = 21−9/√2 side BG = 21−9/√2 = BC + CG = BC + 9/√2 → BC = 21−9√2 Now if we draw line AC, we have: isosceles △CDE with legs CD = 9 and DE = 15√2−9 trapezoid ABCE with bases AE = 6, BC = 21−9√2 and height AB = 15 Area(ABCDE) = 1/2 × 9 × (15√2−9) + 1/2 × 15 × (6 + 21−9√2) = 162
I also solved it by completing the rectangle but slightly different. Once you have the two isosceles right triangles DEF and CDG and know the length of their sides (EF=DF=15-9/√2 and CG=DG=9/√2) you also know the height of rectangle ABGF: AF=EF+AE=15-9/√2 + 6=21-9/√2. The area of the rectangle ABGF then is 21-9/√2 x 15. The areas of the two isosceles triangles are DEF=1/2 x 15-9/√2 x 15-9/√2 and CDG=1/2 x 9/√2 x 9/√2. The area of ABCDE then is (21-9/√2 x 15) - (1/2 x 15-9/√2 x 15-9/√2) - (1/2 x 9/√2 x 9/√2) = 162.
Very nice solution! Thinking inside the box also provides a solution. Mine ends up with two rectangles and two isosceles right triangles. Of course, adding the four gives the same total as your subtraction.
I did enjoyed how you solved the problem. There is something strange w/ the problem : 1. draw a line E to C, then it form the CDE Right triangle 2. Because the angle of : a. AED is 135deg and BAE is 90deg, then CED is 45deg b. BCD is 135deg and ABC is 90deg then ECD is 45 3. There for triangle ECD with angles : 90, 45 and 45 deg, then CD must = DE Using phitagorian theory AB^2 = CD^2 + DE^2 So the 9 and 15 are not match. 2.
I fell for that as well, but the line EC does not split the angle AED into 90 and 45. It might be 95 and 40 and the split of angle BCD will have components that are complementary to those, 85 and 50 perhaps.
You made an assumption that I also made in the beginning. That is, you assumed that AE = BC, and therefore ABCE is a rectangle (and CDE is an isosceles right triangle). But there is nothing in the diagram that would lead to this assumption. In fact, it turns out that BC > AE and ABCE is actually a trapezoid
@@talgoam This is impossible since this figure has to be comletely symmetrical: the angles A, B, D are 90° and the angles C and E are both 135°. Therefore CDE has to be an isosceles triangle which sits on an 15 by 9 rectangle ABCE.
Yay! I solved it. I broke the pentagon into two rectangles and two isosceles right triangle and added all the areas. My answer was also 162. My largest rectangle was 15 * 6. My other rectangle was {9*sq rt(2)} * {15 - 9*sq rt(2)}. My largest isosceles triangle had the legs of 9. My smallest isosceles triangle had the legs of {15 - 9* sq rt(2)}. Adding all the areas gives 90 + 40.5 + 135*sq rt(2) -162 + 193.5 - 135*sq rt (2) = 162.
I made it in a similar way, but my smallest rectangle was (9/√2) * (15 - 18/√2). My smallest isosceles right triangle had legs 9/√2 and my other one had legs 15 - 9/√2. For areas: the largest rectangle was 90, the smallest rectangle was 135√2 - 81, the largest triangle was 132,75 - 135/√2 and the smallest triangle was 20,25. Total: 162.
I did the similar but introduced an X=9/sqrt(2). If substituted into the full Area equation and finally simplified step by step, sqrt(2) eliminates, no fractions involved and the final step is 90 + 72 = 162
My doubt. Can such a pentagon exist? Since angles A, B & D are 90⁰ each and angles C & D are 135⁰ each, wouldn't ABCDE be a pentagon with equal CD & DE sides? In that case, triangle CDE ought to be an isosceles right triangle. The area of the pentagon then would be area of rectangle ABCE (15 * 6 = 90) + area of triangle CDE (½ * 9 * 9 = 22.5) =112.5 Sq units. (If CDE were an isosceles right triangle with hypotenuse of 15, the sides ought to be 10.6 and not 9) Please correct me if the concept is wrong.
construct EC to form a rectangle EC=AB=15 lets focus on CDE it's a scaled up 3;4;5 triangle so (3;4;5)*3=(9;12;15) ED=12 we don't need the angle area = Rectangle area (6*15=90)+ triangle area (12*9/2=54) =144
At 6:30, you do not need ΔAEF. Drop a perpendicular from point E to BP and call the intersection point G. ABGE is a rectangle, sides 6 and 15, area 90. ΔEGP is a special 45°-45°-90° right triangle, sides of length 15, area 112.5. We add these areas to get the area of quadrilateral ABPE = 202.5. Now, we need to deduct the area of ΔPDC, which is a special 45°-45°-90° right triangle, sides of length 9, area 40.5. That leaves area = 162 for polygon ABCDE, as PreMath also found. Note that C and G are not the same point. Length PC = 9√2 and length PG = 15. ΔPDC does not overlap rectangle ABGE but the calculation method would be the same if it did.
Very nice problem with a really clever solution. For my solution an additional point F is required. F has the same x-coordinate as D and the same y-coordinate as E. The coordinates of all points are: A ( 0 ; 0 ) B ( 15 ; 0 ) C ( 15 ; yC ) D ( xD ; yD ) E ( 0 ; 6 ) F ( xD ; 0 ) By subtracting 90° from the angle AED we get 45°. Therefore the slope of the line ED must be tan(45°) = 1. The product of the slopes of two perpendicular lines is −1, so the slope of the line DC is −1. From the slope of ED we can conclude: yD − yE = xD − xE yD − 6 = xD From the slope of DC we can conclude: yD − yC = −(xD − xC) yD − yC = xC − xD yD − yC = 15 − xD In addition, we know the distance DC: (xC − xD)² + (yD − yC)² = 9² 2(xC − xD)² = 9² xC − xD = 9/√2 xC − 9/√2 = xD 15 − 9/√2 = xD yD = xD + 6 = 21 − 9/√2 yC = xD + yD − 15 = 21 − 18/√2 Finally, the pentagon is split into two trapezoids: Area(ABCDE) = Area(AFDE) + Area(FBCD) = (1/2)*(yE + yD)*(xD − xE) + (1/2)*(yD + yC)*(xC − xD) = (1/2)*(6 + 21 − 9/√2)*(15 − 9/√2) + (1/2)*(21 − 9/√2 + 21 − 18/√2)*[15 − (15 − 9/√2)] = (1/2)*(27 − 9/√2)*(15 − 9/√2) + (1/2)*(42 − 27/√2)*(9/√2) = (1/2)*(405 − 378/√2 + 81/2 + 378/√2 − 243/2) = (1/2)*(405 − 162/2) = (1/2)*(405 − 81) = (1/2)*324 = 162🙂 Best regards from Germany.
At last I thought I had cracked one of these problems! I drew a line joining points E & C. So, now a rectangle ABCE. Area = 15x 6 = 90 sq units. Line EC = 15 & is the hypotenuse of triangle CDE. Angle D = 90 degrees. So, side DE = Square Root of (ECsquared - CDsquared). = 15squared - 9squared. = 225 - 81 = 144 Sq Root of 144 = 12. Area of triangle CDE - 1/2 x 9 x 12 = 54. 90 + 54 = 144!. Thank God my schooldays are over!!
At a quick glance, for an n sided shape total interior angles = 180* (n-2). For the pentagon 180 * 3 = 540 . Angle C = 540-3*90 -135 =135. Marking a point , F , forming a right angled triangle DFC . interior angles = 180-135 = 45 degrees , isosceles triangle. then 9^2= 2 * DF^2 = . DF = 9/sqrt(2)= CF. Marking a point , G , forming a right angled triangle EGD . interior angles = 180-135 = 45 degrees , isosceles triangle. then (15- DF)^2 = (15-9/sqrt(2))^2 = DG = 8.636 = EG . Hence total area = rectangle AGFB = 15 * (6 +8.636) =219.4. Area of DFC = half base * height =0.5 *(9/sqrt(2) ^*2=0.5* 81/2= 20.25 and area of triangle EGD = 0.5 *8.636^2 = 37.3.Then area of blue shaded polygon = 219.4-20.25-37.3=.162. Area o fblue shaded polygon = 162
162 Angle BCD=135 degrees since it is a pentagon Draw a line MP parallel to point D; it's = 15 Extend line AE and line BC to MP, forming two triangles: DEM and CDP; both triangles are 45-45-90 Since CD is 9, then DP and CP= 9 sqrt 2/2 = 4.5 sqrt 2 or 6.364 Draw a line from point D which is perpendicular to line AD; label the point R; hence RD= 6.364 and AR = 8.634 [15 - 6.364]; hence line MD =8.634 . Hence ME= 8.634 since all two new triangles are 45-45-90 Area of DEM = 8.634^2/2 = 37.29 Area of CDP = 6.364^2/2 = 20.25 The total area of both = 57.54 Since line ME = 8.634, the length of the new rectangle is 14.634 (8,634 + 6] But since triangle DEM and CDP created a new rectangle 15 * 14.634 or 219.54, the area of the shaded region is 219.54 - 57.54 = 162. Answer.
The internal angles of a pentagon are (5-2)180 = 540°. ∠C + 90 + 90 + 135 + 90 = 540 ∠C = 540 - 405 = 135° As ∠E and ∠C are both 135°, either ∆CDE is an isosceles right triangle or BC ≠ EA. 2(9²) ≠ 15², so it's the latter. Extend lines left and right, parallel to AB, from D, and up from C and E. The lines from D and C will intersect at P and the lines from D and E will intersect at Q. Note that ABPQ is a rectangle, with all four corners being right angles. As ∠BCD and ∠DEA are each 135°, angles ∠DCP and ∠QED are 180-135 = 45°, as BP and QA are straight lines. As ∠CPD, ∠CDE, and ∠DQE are all right angles, ∠PDC and ∠EDQ are 45° as well, making ∆CPD and ∆DQE isosceles right triangles. Let m be the side leg length of ∆CPD and n be the side leg length of ∆DQE. Triangle ∆CPD: CP² + PD² = 9² m² + m² = 81 m² = 81/2 m = √(81/2) = 9/√2 = 9√2/2 Triangle ∆DQE: n = 15 - m = 15 - 9√2/2 n = (30-9√2)/2 DE² = DQ² + QE² DE² = n² + n² = 2n² DE = √(2n²) = n√2 DE = √2(30-9√2)/2 = 15√2 - 9 BC + m = 6 + n BC = 6 + (30-9√2)/2 - 9√2/2 BC = 6 + (30-18√2)/2 BC = 21 - 9√2 By observation, the area of the pentagon will be the area of rectangle ABPQ (X) minus the areas of triangles ∆CPD (Y) and ∆DQE (Z). Pentagon ABCDE: Z = (30-9√2)/2)²/2 = (15-9/√2)²/2 Z = 225/2 + 81/4 - (2(15)9/√2)/2 Z = 531/4 - 135√2/2 Y = (9/√2)²/2 = 81/4 X = (6 + n)15 = (6+(30-9√2)/2)15 X = (21-9√2/2)15 = 315 - 135√2/2 A = (315-135√2/2) - (81/4) - (531/4 - 135√2/2) A = 315 - 81/4 - 531/4 A = (1260-612)/4 A = 648/4 = 162
Sir, you can extend the a line between points E and C. AECD becomes a rectangle with sides 15 and 6 units. The side EB in the right triangle EBC is calculated using the Pythagorean theorem to get it's length. Then, the area of the irregular pentagon is the sum of the areas of the rect AECD and the area of right triangle EBC. However, the answer differs from yours. Please let me know if I have followed a wrong method.
See my solution. BC is greater than AE, so CE is not parallel to BC, and the 15 * 6 rectangle is EAB to an additional point below C. As most of the triangles are 45 degrees, we can just multiply or divide by the square root of two. and avoid using Pythagoras.
I have done the area is the sum of area of triangle + rectangle. rectangle area is 6 x 15= 90. while triangle area is 6 x the height that can be calculat with teorema of pitagora. The sum is 162
@premath in fact the height of triangle is 12. the weight is 9 , but is a irregular so we must leave out another triangle(144 is area but :2 is 72) 72 +90 are 162
Why LONG method? Your answer is wrong. Divide the figure to a triangle and rectangle. Area of tri. is 90sq. units. Area of rectan is 54sq. unit. Hence the area of the figure is 90 + 54 = 144 sq.units. FINAL !!!!?
Figure has errors. BCD has to be 135 to satisfy the sum of internal angles of a pentagon. The figure then has to be symmetrical. That means it could be a 15 x 6 rectangle with an upper 45/45/90 triangle. Rectangle's area is 90. Triangle's area is 15sq/4, which is 225/4 = 56 1/4. Total area should then be 146.25. Please correct if this in error.
I think that the problem is tricky. The most important clue is to find out that the line ED make a 45 degrees angle with the two vertical line, which leads to discover 2 right isosceles. The rest is easy.
Mine solution is little bit complicated but can be done with the help of calculator. First we have to increase the Length of AE and BC to take a rectangular. Then we will get two triangle in both sides having angle of 45° 45° and 90° . Then we have to calculate the Area of the right triangle and Then Left triangle . Then Have to calculate the area of the rectangular 15 × ( 6 + 8.63603897) . Then substracting The total area of two triangles from the Area of the rectangle. we will get 219.540585 - ( 37.2905846 + 20.25 ) = 162
Am I missing something? Question seems to be wrong at the very outset. Lets draw an imaginary line EC, dividing the polygon into a triangle DEC and a quadrilateral ABEC. Now as per Internal Angles total formula, it has been derived that angle DCB is 135° which is equal to angle DEA. So quadrilateral ABEC is basically a rectangle as angles EAB and ABC are both 90° and if it isn't a rectangle then it means sides EA and BC are of unequal length which would mean that angle EDC can't be 90°. Also it has been derived that angle DEC and DCE are both equal i e. 45° so basically traingle ECD is an isosceles triangle. Now if we apply Pythagoras theorem to right angle triangle DEC, then DE²=EC²-DC², which means DE = 12 as EC = AB = 15, which in turn doesn't make traingle DEC an isosceles triangle. Hence the question is wrong as per me. Pls correct me If I am wrong.
ABCE is a rectangle only if you assume that BC = AE, which it obviously is not, because that leads to the inconsistency that you mention. Since BC and AE are not equal, then ABCE is actually a trapezoid, and triangle CDE is not isosceles, because EC is not parallel to AB, so angles DEC and DCE are not 45°
You made the erroneous assumption that BC = AE, and therefore ABCE is a 6×15 rectangle, and that △CDE is isosceles with DE = DC = 9. But if this were so, then hypotenuse CE = 15, which is not possible when legs = 9.
It is nice for you to teach people MATH, however, please make sure the question is correct. Unfortunately this question is COMPLETELY WRONG. That is why many people below (including me) came up with different answer. What is wrong is not your calculation. What is very WRONG is your polygon. Your polygon DOES NOT exist. Please look at your diagram. The triangle AEF, PDC, BFP and even CDE are all isosceles triangle. I try to actually draw a polygon like that. It is IMPOSSIBLE because it does not exist. Since this polygon does not exist, you will come out with different answer depended which isosceles triangle you used to calculate the area. I gave your an example, I will just take the rectangle ABCE + the isosceles triangle CDE. The area of ABCE is 6 x 15 = 90. The area of triangle CDE is 9 x 9 divided by 2 = 40.5. So the area of the "POLYGON" is 90+40.5 = 130.5. One more, you said the triangle BFP is isosceles, so the length BP is the same as BF which is 21, but did you see since the triangle PDC is isosceles, you can also use the pythagorean formula (a2 x b2 = c2) to calculate the length of CP and it will come out with 12.727. So BF is not 21, but 6 +12.727 = 18.727. Please correct your mistake.
If angle at E and C is 135, EA=CB, and ED =CD, and ABCD is a rectangle with AB=CD. Area of rectangle ABCD + Area of Isc Dr angled Triangle CDE = 90+40.5 = 130.5
Wrong. Impossible figure. Draw line EC. That gives a 9x12x15 triangle on top. The angle !35 degrees is impossible. The top triangle would have to be isosceles with 9x9xsqrt(2x81) which is NOT 15. So, figure is impossible.
Figure is only impossible because you assumed that BC = AE, making ABCE a rectangle, which would make CE = 15. But nothing in diagram suggest that BC = AE. Turns out that ABCE is actually a trapezoid, with BC > AE and EC > 15
Given the information at the start, C MUST be 135 to keep the angles total to 540, so the polygon is symmetrical. So, ED MUST equal DC and EA MUST equal BC (because of the right angles). If they were not the same then angle C could not be 135. However, all this being true, DC can't be 9 while EC is also 15. Nonsense.
Please learn geometry before teaching it : this polygon is not possible. The DEC triangle is isocel (see angles DCB=DEA=135° and DEC=DCE=45°) therefore DC=ED... and ABCE is a rectangle (same angles to check)... therefore EC² = AB² = DC²+ED²... How funny : 225=162 !
Area (bottom rectangle) = bh = 15(6) = 90, Area top triangle = ½bh = ½(9)(12) = 6(9) = 54, (the top triangle is a 9, 12, 15 right triangle, which is a multiple of a 3, 4, 5 right triangle, and the legs are the base and height), the total area of the pentagon is, thus, 90 + 54 = 144 square units... But that's wrong... 🤔
По вашему построению получается дерьмо собачье, два разных угла равны по 45 градусов? Значит два разных по размкру катета, равны по 9 метрических единиц? Садитесь, 2! Нужно из точки Е провести параллельную линию АВ в сторону вертикали СВ, потом к этой горизонтали продлить сторону DC, а от точки С, провести горизонталь влево до пересечения со стороной ED, которая отсечёт На ней отрезок равный 9! После этих построений вычислим площадь прямоугольного треугольника с катетами 9, добавим к ним площадь прямоугольника со сторонами 6 и 15 т добавим площадь маленького прямоугольного треугольника с катетами равными разнице между большей стороной прямоугольника и гипотенузой треугольника с катетами 9, и получим площадь пятиугольника, равную 243 квадратных метрических единицы!
At a quick glance, for an n sided shape total interior angles = 180* (n-2). For the pentagon 180 * 3 = 540 . Angle C = 540-3*90 135 =135. Marking a point , F , forming a right angled triangle DFC . interior angles = 180-135 = 45 degrees , isosceles triangle. then 9^2= 2 * DF^2 = . DF = 9/sqrt(2)= CF. Marking a point , G , forming a right angled triangle EGD . interior angles = 180-135 = 45 degrees , isosceles triangle. then (15-9/sqrt(2))^2 = DG = 8.636 = EG . Hence total area = rectangle AGFB = 15 * (6 +8.636) =219.4. Area of DFC = half base * height =0.5 (9/sqrt(2) ^2=0.5 * 81/2= 20.25 and area of triangle EGD = 0.5 *8.636^2 = 37.3.Then area of blue shaded polygon = 219.4-20.25-37.3=.162. Area o fblue shaded polygon = 162
Very nice 😊and beautiful presentation ❤❤❤ Sir
So nice of you
Thanks for your continued love and support!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Dear sir your action fields pathways are really awesome and unique so I have transparent and true love and support forever till my last breat ❤️❤️❤️🙏🙏🙏🙏my great love and greetings from India ... 🙏🙏I will try to meet with you one day ❤️❤️❤️😀😀😀😀😀😀😀😀😀❤️👋🙏🙏👋please call me for USA Visit 😀❤️❤️🙏🙏🙏🙏
135 is 90+45 so you have a rectange plus a 45 deg isocèle square triangle, partially cut off with one side of 9 units and the other long enough to make the width base of 15.
The measures in this puzzle are faulty. The triangle EDC is an isosceles triangle with a hypotenuse of 15; therefore DC cannot be 9 but must be 15/sqrt2.
Moreover the auxiliary lines are not needed at all, this figure can be divided into a rectangle 15x9 and the isosceles triangle with a base of 15 and e heigth of 7.5
Never saw that coming... Thanks for broadening my view, excellent as usual 👍🏻
I really liked this method, which made calculation simpler than completing the rectangle, which is what I did:
First, like you, I figured out that ∠BCD = 135°. Then I proceeded as follows:
Draw line through D that is parallel to AB.
Extend AE and BC until they intercept this line at points E and F respectively.
We now have rectangle ABGF, with FG = AB = 15, and AF = BG = ?
We also have 2 isosceles right triangles: △DEF and △CDG.
In △CDG, hypotenuse CD = 9, so CG = DG = 9/√2
Since FG = AB = 15 and DG = 9/√2, then DF = 15−9/√2
In △CDG, EF = DF = 15−9/√2, and hypotenuse DE = 15√2−9
In rectangle ABGF
side AF = AE + EF = 6 + (15−9/√2) = 21−9/√2
side BG = 21−9/√2 = BC + CG = BC + 9/√2 → BC = 21−9√2
Now if we draw line AC, we have:
isosceles △CDE with legs CD = 9 and DE = 15√2−9
trapezoid ABCE with bases AE = 6, BC = 21−9√2 and height AB = 15
Area(ABCDE) = 1/2 × 9 × (15√2−9) + 1/2 × 15 × (6 + 21−9√2) = 162
I also solved it by completing the rectangle but slightly different. Once you have the two isosceles right triangles DEF and CDG and know the length of their sides (EF=DF=15-9/√2 and CG=DG=9/√2) you also know the height of rectangle ABGF: AF=EF+AE=15-9/√2 + 6=21-9/√2. The area of the rectangle ABGF then is 21-9/√2 x 15. The areas of the two isosceles triangles are DEF=1/2 x 15-9/√2 x 15-9/√2 and CDG=1/2 x 9/√2 x 9/√2. The area of ABCDE then is (21-9/√2 x 15) - (1/2 x 15-9/√2 x 15-9/√2) - (1/2 x 9/√2 x 9/√2) = 162.
Enhorabuena. ¡Que sencillez¡ Lo bueno, si breve, dos veces bueno.
Very nice solution! Thinking inside the box also provides a solution.
Mine ends up with two rectangles and two isosceles right triangles.
Of course, adding the four gives the same total as your subtraction.
I did enjoyed how you solved the problem.
There is something strange w/ the problem :
1. draw a line E to C, then it form the CDE Right triangle
2. Because the angle of :
a. AED is 135deg and BAE is 90deg, then CED is 45deg
b. BCD is 135deg and ABC is 90deg then ECD is 45
3. There for triangle ECD with angles : 90, 45 and 45 deg, then CD must = DE
Using phitagorian theory
AB^2 = CD^2 + DE^2
So the 9 and 15 are not match.
2.
I fell for that as well, but the line EC does not split the angle AED into 90 and 45. It might be 95 and 40 and the split of angle BCD will have components that are complementary to those, 85 and 50 perhaps.
You made an assumption that I also made in the beginning. That is, you assumed that AE = BC, and therefore ABCE is a rectangle (and CDE is an isosceles right triangle). But there is nothing in the diagram that would lead to this assumption. In fact, it turns out that BC > AE and ABCE is actually a trapezoid
ABCE is not a rectangle
Hi there , I think if the angle D is equal to 90 degrees then the line DC can not be 9 unit. There is a problem here. Would you check it ?
DC can be 9 units. I think you made the same error as I did, in assuming line BC to be equal to AE. It's not. =)
DC can be 9 unit because BC > AE. We can prove that BC is approx = 8.274. If we joint E,C together then we can see EC is not parrel to AB.
@@talgoam This is impossible since this figure has to be comletely symmetrical: the angles A, B, D are 90° and the angles C and E are both 135°. Therefore CDE has to be an isosceles triangle which sits on an 15 by 9 rectangle ABCE.
Yay! I solved it. I broke the pentagon into two rectangles and two isosceles right triangle and added all the areas. My answer was also 162. My largest rectangle was 15 * 6. My other rectangle was {9*sq rt(2)} * {15 - 9*sq rt(2)}. My largest isosceles triangle had the legs of 9. My smallest isosceles triangle had the legs of {15 - 9* sq rt(2)}. Adding all the areas gives 90 + 40.5 + 135*sq rt(2) -162 + 193.5 - 135*sq rt (2) = 162.
I made it in a similar way, but my smallest rectangle was (9/√2) * (15 - 18/√2). My smallest isosceles right triangle had legs 9/√2 and my other one had legs 15 - 9/√2.
For areas: the largest rectangle was 90, the smallest rectangle was 135√2 - 81, the largest triangle was 132,75 - 135/√2 and the smallest triangle was 20,25. Total: 162.
I did the similar but introduced an X=9/sqrt(2). If substituted into the full Area equation and finally simplified step by step, sqrt(2) eliminates, no fractions involved and the final step is 90 + 72 = 162
My doubt. Can such a pentagon exist? Since angles A, B & D are 90⁰ each and angles C & D are 135⁰ each, wouldn't ABCDE be a pentagon with equal CD & DE sides? In that case, triangle CDE ought to be an isosceles right triangle. The area of the pentagon then would be area of rectangle ABCE (15 * 6 = 90) + area of triangle CDE (½ * 9 * 9 = 22.5) =112.5 Sq units. (If CDE were an isosceles right triangle with hypotenuse of 15, the sides ought to be 10.6 and not 9) Please correct me if the concept is wrong.
When I did maths 0.5*9*9 equalled 40.5
@@mmyers8817 Yes. My mistake. Apologies for that. It is 40.5 and not 22.5. Still, it comes to only 130.5.
yeah it should be a bilateral polygon right? (or isosceles polygon)
You did this in a way that I absolutely was not expecting. Good to open the mind up to new concepts.
Certainly, completing the Triangle was by far better than completing the Rectangle.
Thanks!
That's a really cool problem! Thanks for sharing! 😎
construct EC to form a rectangle
EC=AB=15
lets focus on CDE
it's a scaled up 3;4;5 triangle
so (3;4;5)*3=(9;12;15)
ED=12
we don't need the angle
area = Rectangle area (6*15=90)+ triangle area (12*9/2=54)
=144
Thanks for video.Good luck sir!!!!!!!!!
At 6:30, you do not need ΔAEF. Drop a perpendicular from point E to BP and call the intersection point G. ABGE is a rectangle, sides 6 and 15, area 90. ΔEGP is a special 45°-45°-90° right triangle, sides of length 15, area 112.5. We add these areas to get the area of quadrilateral ABPE = 202.5. Now, we need to deduct the area of ΔPDC, which is a special 45°-45°-90° right triangle, sides of length 9, area 40.5. That leaves area = 162 for polygon ABCDE, as PreMath also found.
Note that C and G are not the same point. Length PC = 9√2 and length PG = 15. ΔPDC does not overlap rectangle ABGE but the calculation method would be the same if it did.
Beautiful solution
Love outside the box thinking!
Very nice problem with a really clever solution. For my solution an additional point F is required. F has the same x-coordinate as D and the same y-coordinate as E. The coordinates of all points are:
A ( 0 ; 0 )
B ( 15 ; 0 )
C ( 15 ; yC )
D ( xD ; yD )
E ( 0 ; 6 )
F ( xD ; 0 )
By subtracting 90° from the angle AED we get 45°. Therefore the slope of the line ED must be tan(45°) = 1. The product of the slopes of two perpendicular lines is −1, so the slope of the line DC is −1. From the slope of ED we can conclude:
yD − yE = xD − xE
yD − 6 = xD
From the slope of DC we can conclude:
yD − yC = −(xD − xC)
yD − yC = xC − xD
yD − yC = 15 − xD
In addition, we know the distance DC:
(xC − xD)² + (yD − yC)² = 9²
2(xC − xD)² = 9²
xC − xD = 9/√2
xC − 9/√2 = xD
15 − 9/√2 = xD
yD = xD + 6 = 21 − 9/√2
yC = xD + yD − 15 = 21 − 18/√2
Finally, the pentagon is split into two trapezoids:
Area(ABCDE)
= Area(AFDE) + Area(FBCD)
= (1/2)*(yE + yD)*(xD − xE) + (1/2)*(yD + yC)*(xC − xD)
= (1/2)*(6 + 21 − 9/√2)*(15 − 9/√2) + (1/2)*(21 − 9/√2 + 21 − 18/√2)*[15 − (15 − 9/√2)]
= (1/2)*(27 − 9/√2)*(15 − 9/√2) + (1/2)*(42 − 27/√2)*(9/√2)
= (1/2)*(405 − 378/√2 + 81/2 + 378/√2 − 243/2)
= (1/2)*(405 − 162/2)
= (1/2)*(405 − 81)
= (1/2)*324
= 162🙂
Best regards from Germany.
Very nice,perfect,out of the box🥳🥳🥳🥳🥳🥳🥳🥳🥳
Great explanetion👍
Thanks for sharing😊
Excellent work, my approach was different
Boa tarde Professor
Obrigado pelas Aulas
Grato
Good one!
At last I thought I had cracked one of these problems! I drew a line joining points E & C. So, now a rectangle ABCE. Area = 15x 6 = 90 sq units. Line EC = 15 & is the hypotenuse of triangle CDE. Angle D = 90 degrees. So, side DE = Square Root of (ECsquared - CDsquared). = 15squared - 9squared. = 225 - 81 = 144 Sq Root of 144 = 12. Area of triangle CDE - 1/2 x 9 x 12 = 54. 90 + 54 = 144!. Thank God my schooldays are over!!
Thank you sir for making us think out of the box
Well done.
excellent!
Hi, great video PreMath! Can someone post the area of a pentagon formula? TIA
excellent question
Nice one🎉🎉🎉
Glad to hear that!
Thanks for your continued love and support!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Thank you again
I thought the answer is 144 if BC=AE= 6
At a quick glance, for an n sided shape total interior angles = 180* (n-2). For the pentagon 180 * 3 = 540 . Angle C = 540-3*90 -135 =135. Marking a point , F , forming a right angled triangle DFC . interior angles = 180-135 = 45 degrees , isosceles triangle. then 9^2= 2 * DF^2 = . DF = 9/sqrt(2)= CF. Marking a point , G , forming a right angled triangle EGD . interior angles = 180-135 = 45 degrees , isosceles triangle. then (15- DF)^2 = (15-9/sqrt(2))^2 = DG = 8.636 = EG . Hence total area = rectangle AGFB = 15 * (6 +8.636) =219.4. Area of DFC = half base * height =0.5 *(9/sqrt(2) ^*2=0.5* 81/2= 20.25 and area of triangle EGD = 0.5 *8.636^2 = 37.3.Then area of blue shaded polygon = 219.4-20.25-37.3=.162. Area o fblue shaded polygon = 162
162
Angle BCD=135 degrees since it is a pentagon
Draw a line MP parallel to point D; it's = 15
Extend line AE and line BC to MP, forming two triangles: DEM and CDP; both triangles are 45-45-90
Since CD is 9, then DP and CP= 9 sqrt 2/2 = 4.5 sqrt 2 or 6.364
Draw a line from point D which is perpendicular to line AD; label the point R; hence RD= 6.364
and AR = 8.634 [15 - 6.364]; hence line MD =8.634 . Hence ME= 8.634 since all two new triangles are 45-45-90
Area of DEM = 8.634^2/2 = 37.29
Area of CDP = 6.364^2/2 = 20.25
The total area of both = 57.54
Since line ME = 8.634, the length of the new rectangle is 14.634 (8,634 + 6]
But since triangle DEM and CDP created a new rectangle 15 * 14.634 or 219.54, the area of
the shaded region is 219.54 - 57.54 = 162. Answer.
Was there a need for 135° please educate me. Thanks
The internal angles of a pentagon are (5-2)180 = 540°.
∠C + 90 + 90 + 135 + 90 = 540
∠C = 540 - 405 = 135°
As ∠E and ∠C are both 135°, either ∆CDE is an isosceles right triangle or BC ≠ EA. 2(9²) ≠ 15², so it's the latter.
Extend lines left and right, parallel to AB, from D, and up from C and E. The lines from D and C will intersect at P and the lines from D and E will intersect at Q. Note that ABPQ is a rectangle, with all four corners being right angles.
As ∠BCD and ∠DEA are each 135°, angles ∠DCP and ∠QED are 180-135 = 45°, as BP and QA are straight lines. As ∠CPD, ∠CDE, and ∠DQE are all right angles, ∠PDC and ∠EDQ are 45° as well, making ∆CPD and ∆DQE isosceles right triangles. Let m be the side leg length of ∆CPD and n be the side leg length of ∆DQE.
Triangle ∆CPD:
CP² + PD² = 9²
m² + m² = 81
m² = 81/2
m = √(81/2) = 9/√2 = 9√2/2
Triangle ∆DQE:
n = 15 - m = 15 - 9√2/2
n = (30-9√2)/2
DE² = DQ² + QE²
DE² = n² + n² = 2n²
DE = √(2n²) = n√2
DE = √2(30-9√2)/2 = 15√2 - 9
BC + m = 6 + n
BC = 6 + (30-9√2)/2 - 9√2/2
BC = 6 + (30-18√2)/2
BC = 21 - 9√2
By observation, the area of the pentagon will be the area of rectangle ABPQ (X) minus the areas of triangles ∆CPD (Y) and ∆DQE (Z).
Pentagon ABCDE:
Z = (30-9√2)/2)²/2 = (15-9/√2)²/2
Z = 225/2 + 81/4 - (2(15)9/√2)/2
Z = 531/4 - 135√2/2
Y = (9/√2)²/2 = 81/4
X = (6 + n)15 = (6+(30-9√2)/2)15
X = (21-9√2/2)15 = 315 - 135√2/2
A = (315-135√2/2) - (81/4) - (531/4 - 135√2/2)
A = 315 - 81/4 - 531/4
A = (1260-612)/4
A = 648/4 = 162
Nice👍🏽👍🏾
Glad you think so!
Thanks for your continued love and support!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Sir, you can extend the a line between points E and C. AECD becomes a rectangle with sides 15 and 6 units. The side EB in the right triangle EBC is calculated using the Pythagorean theorem to get it's length. Then, the area of the irregular pentagon is the sum of the areas of the rect AECD and the area of right triangle EBC. However, the answer differs from yours. Please let me know if I have followed a wrong method.
That's what I was thinking. I don't understand why it comes out differently.
See my solution. BC is greater than AE, so CE is not parallel to BC, and the 15 * 6 rectangle is EAB to an additional point below C. As most of the triangles are 45 degrees, we can just multiply or divide by the square root of two. and avoid using Pythagoras.
the pentagon can be divided by a triangle+a trapeze+rectangle;
I have done the area is the sum of area of triangle + rectangle. rectangle area is 6 x 15= 90. while triangle area is 6 x the height that can be calculat with teorema of pitagora. The sum is 162
@premath in fact the height of triangle is 12. the weight is 9 , but is a irregular so we must leave out another triangle(144 is area but :2 is 72) 72 +90 are 162
Why LONG method? Your answer is wrong. Divide the figure to a triangle and rectangle. Area of tri. is 90sq. units. Area of rectan is 54sq. unit. Hence the area of the figure is 90 + 54 = 144 sq.units. FINAL !!!!?
Figure has errors. BCD has to be 135 to satisfy the sum of internal angles of a pentagon.
The figure then has to be symmetrical. That means it could be a 15 x 6 rectangle with an upper 45/45/90 triangle. Rectangle's area is 90. Triangle's area is 15sq/4, which is 225/4 = 56 1/4.
Total area should then be
146.25. Please correct if this in error.
شكرا
You are very welcome!
Thanks for your continued love and support!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
It's impossible.
if AÊD is 135° , DÊC must be 45°
make length of ED is 9 unit also
and EC must be 9•root(2) but it isn't.
I think that the problem is tricky. The most important clue is to find out that the line ED make a 45 degrees angle with the two vertical line, which leads to discover 2 right isosceles.
The rest is easy.
The Question is wrong. Your polygon does not exist. 🤨
Mine solution is little bit complicated but can be done with the help of calculator. First we have to increase the Length of AE and BC to take a rectangular. Then we will get two triangle in both sides having angle of 45° 45° and 90° . Then we have to calculate the Area of the right triangle and Then Left triangle . Then Have to calculate the area of the rectangular 15 × ( 6 + 8.63603897) . Then substracting The total area of two triangles from the Area of the rectangle. we will get 219.540585 - ( 37.2905846 + 20.25 ) = 162
Am I missing something? Question seems to be wrong at the very outset. Lets draw an imaginary line EC, dividing the polygon into a triangle DEC and a quadrilateral ABEC.
Now as per Internal Angles total formula, it has been derived that angle DCB is 135° which is equal to angle DEA. So quadrilateral ABEC is basically a rectangle as angles EAB and ABC are both 90° and if it isn't a rectangle then it means sides EA and BC are of unequal length which would mean that angle EDC can't be 90°. Also it has been derived that angle DEC and DCE are both equal i e. 45° so basically traingle ECD is an isosceles triangle.
Now if we apply Pythagoras theorem to right angle triangle DEC, then DE²=EC²-DC², which means DE = 12 as EC = AB = 15, which in turn doesn't make traingle DEC an isosceles triangle. Hence the question is wrong as per me. Pls correct me If I am wrong.
ABCE is a rectangle only if you assume that BC = AE, which it obviously is not, because that leads to the inconsistency that you mention. Since BC and AE are not equal, then ABCE is actually a trapezoid, and triangle CDE is not isosceles, because EC is not parallel to AB, so angles DEC and DCE are not 45°
El salir de la caja ayuda mucho.
The correct answer is 130.5
I thought so too, but if you make CDE a right angled isosceles triangle, it doesn't work with a hypotenuse of 15, (9² + 9² ≠ 15²). Therefore BC ≠ 6.
You made the erroneous assumption that BC = AE, and therefore ABCE is a 6×15 rectangle, and that △CDE is isosceles with DE = DC = 9. But if this were so, then hypotenuse CE = 15, which is not possible when legs = 9.
Please correct me. I joined CE = 15. Then I found DE by using Pythagoras. Then I found area of the triangle.......
Area is 130.5 sq units
It is nice for you to teach people MATH, however, please make sure the question is correct. Unfortunately this question is COMPLETELY WRONG. That is why many people below (including me) came up with different answer.
What is wrong is not your calculation. What is very WRONG is your polygon. Your polygon DOES NOT exist.
Please look at your diagram. The triangle AEF, PDC, BFP and even CDE are all isosceles triangle. I try to actually draw a polygon like that. It is IMPOSSIBLE because it does not exist.
Since this polygon does not exist, you will come out with different answer depended which isosceles triangle you used to calculate the area.
I gave your an example, I will just take the rectangle ABCE + the isosceles triangle CDE. The area of ABCE is 6 x 15 = 90. The area of triangle CDE is 9 x 9 divided by 2 = 40.5. So the area of the "POLYGON" is 90+40.5 = 130.5.
One more, you said the triangle BFP is isosceles, so the length BP is the same as BF which is 21, but did you see since the triangle PDC is isosceles, you can also use the pythagorean formula (a2 x b2 = c2) to calculate the length of CP and it will come out with 12.727. So BF is not 21, but 6 +12.727 = 18.727.
Please correct your mistake.
If angle at E and C is 135, EA=CB, and ED =CD, and ABCD is a rectangle with AB=CD.
Area of rectangle ABCD + Area of Isc Dr angled Triangle CDE = 90+40.5 = 130.5
Wrong. Impossible figure.
Draw line EC. That gives a 9x12x15 triangle on top.
The angle !35 degrees is impossible. The top triangle would have to be isosceles with 9x9xsqrt(2x81) which is NOT 15.
So, figure is impossible.
Figure is only impossible because you assumed that BC = AE, making ABCE a rectangle, which would make CE = 15. But nothing in diagram suggest that BC = AE. Turns out that ABCE is actually a trapezoid, with BC > AE and EC > 15
@@MarieAnne. Good point.
It is often hard to see the assumptions we make.
Given the information at the start, C MUST be 135 to keep the angles total to 540, so the polygon is symmetrical. So, ED MUST equal DC and EA MUST equal BC (because of the right angles). If they were not the same then angle C could not be 135. However, all this being true, DC can't be 9 while EC is also 15. Nonsense.
the angle is wrong
the sum or the answer of 135-180
Is that not a triangle 3(3 4 5). So 12 will be the roof. 12*9/2=54. 54 + (15*6) = 144 What is wrong in my calculation?
Length of BC is not 6, therefore point C is higher than point E and EC is not 15
@@mohamadtaufik5770 thanks, I see. So it is not a rectangle, but a trapezoid...
I like to learn of your strategies, but I don't have the patience to follow you go over the most basic arithmetic.
👍🥂❤
Сижу, пытаюсь доказать, что 9 = неизвестной стороне)). Точно ж равно!))
162
130.5
A=144sqare unit
Area=21*21/2 -6*6/2-9*9/2=162
I found the video extremely slow, nice solution, but I skipped most of it.
And overcomplicated. No need for any auxiliary lines.
u can basically just connect e and c and find the area why so complicated😂
just use law of sines why so complicated
Please learn geometry before teaching it : this polygon is not possible. The DEC triangle is isocel (see angles DCB=DEA=135° and DEC=DCE=45°) therefore DC=ED... and ABCE is a rectangle (same angles to check)... therefore EC² = AB² = DC²+ED²... How funny : 225=162 !
Oh, okay... The bottom is not a rectangle...
Area (bottom rectangle) = bh = 15(6) = 90, Area top triangle = ½bh = ½(9)(12) = 6(9) = 54, (the top triangle is a 9, 12, 15 right triangle, which is a multiple of a 3, 4, 5 right triangle, and the legs are the base and height), the total area of the pentagon is, thus, 90 + 54 = 144 square units... But that's wrong... 🤔
I had a top triangle of about 54, + 17 + 90. BC is greater than AE, so there is an extra 17 triangle.
По вашему построению получается дерьмо собачье, два разных угла равны по 45 градусов? Значит два разных по размкру катета, равны по 9 метрических единиц? Садитесь, 2! Нужно из точки Е провести параллельную линию АВ в сторону вертикали СВ, потом к этой горизонтали продлить сторону DC, а от точки С, провести горизонталь влево до пересечения со стороной ED, которая отсечёт На ней отрезок равный 9! После этих построений вычислим площадь прямоугольного треугольника с катетами 9, добавим к ним площадь прямоугольника со сторонами 6 и 15 т добавим площадь маленького прямоугольного треугольника с катетами равными разнице между большей стороной прямоугольника и гипотенузой треугольника с катетами 9, и получим площадь пятиугольника, равную 243 квадратных метрических единицы!
At a quick glance, for an n sided shape total interior angles = 180* (n-2). For the pentagon 180 * 3 = 540 . Angle C = 540-3*90 135 =135. Marking a point , F , forming a right angled triangle DFC . interior angles = 180-135 = 45 degrees , isosceles triangle. then 9^2= 2 * DF^2 = . DF = 9/sqrt(2)= CF. Marking a point , G , forming a right angled triangle EGD . interior angles = 180-135 = 45 degrees , isosceles triangle. then (15-9/sqrt(2))^2 = DG = 8.636 = EG . Hence total area = rectangle AGFB = 15 * (6 +8.636) =219.4. Area of DFC = half base * height =0.5 (9/sqrt(2) ^2=0.5 * 81/2= 20.25 and area of triangle EGD = 0.5 *8.636^2 = 37.3.Then area of blue shaded polygon = 219.4-20.25-37.3=.162. Area o fblue shaded polygon = 162
162