Calculate area of the Blue Shaded Trapezoid | Trapezoid | (Trapezium) |

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Learn how to find the area of the Blue Shaded Trapezoid. Important Geometry and algebra skills are also explained: Trapezoid; Trapezium; Trapezoid area formula; Heron's formula. Step-by-step tutorial by PreMath.com
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Комментарии : 101

@yalchingedikgedik8007 Месяц назад

Thanks Sir That’s very useful and enjoyable Very thanks ❤❤❤❤❤

@PreMath Месяц назад

So nice of you dear🌹 Glad to hear that! Thanks for the feedback ❤️

@jimlocke9320 Месяц назад

Alternative to using Heron's Formula at 3:15: Drop perpendicular from C to AB and label the intersection as point F. Let CF = h and EF = x. BF = BE - CF = 15 - x. So, we have right ΔCEF with hypotenuse CE = 13, CF = h and EF = x, applying the Pythagorean theorem, x² + h² = (13)² = 169. We also have right ΔCBF with hypotenuse CB = 14, CF = h and BF = 15 - x, applying the Pythagorean theorem, (15- x)² + h² = (14)² = 196. Expanding, (15)² - 30x + x² + h² = 196. Replace (15)² by 225 and x² + h² by 169: 225 - 30x + 169 = 196, 30x = 198 and x = 6.6. Substituting 6.6 for x in x² + h² = 169, we get h = 11.2 or, equivalently, 56/5. Skip ahead to 7:05 and complete the computation of the area of the trapezoid.

@phungpham1725 Месяц назад

Thank you! I have just tried that way

@georgebliss964 Месяц назад

👍I also solved it that way.

@allanflippin2453 Месяц назад

Yes, I also solved it that way. That's because I'm having trouble remembering Heron's Formula :D This approach works fine.

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@monkey6114 Месяц назад

The general formula for any trapezoid is 2s(s-x)(s-y)(s-B+b)÷(B²-2abB+b²)*B Where s=(x+y+B-b)÷2 and x,y are the 2 nonparalel lines and B is the biger base and b is the smaller one

@himo3485 Месяц назад

h^2=13^2-x^2 h^2=14^2-(15-x)^2 169-x^2=-29+30x-x^2 30x=198 x=33/5 169-1089/25=4225/25-1089/25=3136/25 h=56/5 Trapezoid area : (12+27)＊56/5＊1/2=39＊28/5=1092/5

@joansoldevilacaba4138 Месяц назад

I have done the same way as yours

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@mintprathomkrumint4499 24 дня назад

Pythagoras theorem

@marcgriselhubert3915 Месяц назад

Something different: Let's use an orthonormal center A and first axis (AB). D is on the circle center A and radius 13, so D(13.cos(t); 13.sin(t)) with 0< t < 90°. Then C(12 + 13.cos(t); 13.sin(t)), B(27; 0), VectorBC(-15 +13.cos(t); 13.sin(t)), then BC^2 = 225 - 390.cos(t) +169.(cos(t)^2) + 169.(sin(t)^2) = 225 -390.cos(t) + 169 = -390.cos(t) +394. Knowing that BC^2 = 14^2 = 196, we have 390.cos(t) = 394 -196 = 198 and cos(t) = 198/390 = 33/65. Then cos(t)^2 = 1089/4225 and sin(t)^2 = 1 - (1089/4225)= 3136/4225 and sin(t) = sqrt(3136/4225) =56/65. Finally the height of the trapezoïd is the ordinate of D, so it is 13.sin(t) = 56/5. The area of the trapezoïd is ((AB + CD)/2).height = ((12 +27)/2).(56/5) = 1092/5.

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@PrithwirajSen-nj6qq Месяц назад

Drop perpendicular from C and D on AB so that CP and DQ cut AB at P and Q respectively. Now 🔺 CBP CP^2 +BP ^2=14^2=196 (1) In 🔺 ADQ DQ^2 +AQ^2 =169 (2) By 1 - 2 BP^2 - AQ ^2= 27 >(BP + AQ)(BP - AQ)=27. (3) And from figure BP + AQ =27- 12=15--(4) BP-AQ =27/15=9/5 BP = [15 + 9/5]/2= 84/10 CP=√(196-8.4^2) =11.2 Area =1/2*11.2*(12+27) =1/2*11.2*39= 218.4 so unit Comment please

@phungpham1725 Месяц назад

Thank you so much for your alternative solution!😊

@PrithwirajSen-nj6qq Месяц назад

@@phungpham1725 Thanks , u read.

@artstocker60 Месяц назад

I also did it this way.

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@Copernicusfreud Месяц назад

That is how I solved it too.

@RAG981 Месяц назад

I found angle ABC using cosine rule. I.e. cosB =(14^2 +15^2 - 13^2)/2.14.15 = 3/5, so sinB = 4/5 and h = 14sinB=56/5 so area =39/2 x 56/5 = 2184/10.

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@Arjun07. Месяц назад

वाह... आपका गणित कौशल अविश्वसनीय है! आपने मुझे अपना टेस्ट पेपर पास करने में मदद की!!! आपका बहुत-बहुत धन्यवाद!

@PreMath Месяц назад

प्रिय आपका स्वागत हैं। देखते रहें। आप सर्वश्रेष्ठ हैं🌹

@mostafalak-jv5qq Месяц назад

Very excited

@giuseppemalaguti435 Месяц назад

13^2-(15-a)^2=14^2-a^2...a=42/5...h^2=14^2-(42/5)^2...h=56/5...A=(27+12)*56/5/2=39*28/5=1092/5

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@ConvivialWorldTravel Месяц назад

Thanks. It can be solved another way. Area of Trapezoid = area of triangle + area of Parallelogram.

@santiagoarosam430 Месяц назад

En el triángulo de altura "h", base 27-12=15=a+(15-a) , y lados 13 y 14 : 13²-a²=h²=14²-(15-a)²→ a=33/5=6,6 → h=√[13²-(33/5)²]=56/5 → Área ABCD =[(27+12)/2]*(56/5)=1092/5=218,40 ud². Gracias y un saludo cordial.

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@phungcanhngo Месяц назад

Thank you professor for smart solution.

@toninhorosa4849 Месяц назад

I solve like this: From point "D" we trace the height "h" until we reach the straight line AB at point "P". We name the line AP = y. From point "C" we draw the height "h" until we reach the straight line AB at point Q. We name the straight line QB = x. y = 27 - 12 - x y = 15 - x (1st equation) we apply Pythagoras to the triangles: CQB and DAP. 196 = h^2 + x^2 169 = h^2 + (15 - x)^2 subtracting the 2nd equation from the 3rd 27 = x^2 - (15 - x)^2 27 = x^2 - ( 225 - 30x + x^2) 27 = x^2 - x^2 + 30x - 225 30x = 225 + 27 30x = 252 x = 252/30 x = 8,40 h^2 = 14^2 - (8,4)^2 h^2 = 196 - 70,56 h^2 = 125,44 h = 11,2 Trapezium area = (1/2)*(12+27)*11,2 Trapezium area = 218,4.

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@MrPaulc222 Месяц назад

27-12=15, so plit that 15 between the two triangles. On the left: 13^2 - (15/2)^2 = h^2 169 - (225/4) = h^2, so h^2 = 169 - 56.25 = 112.75 or 451/4 On the right: 14^2 - (15/2)^2 = 196 - ... (well, it is 27 more) so 563/4. Average them out for h^2 = 507/4 Square root it for h = sqrt(507)/2 approximates to 11.258... (39/2) * 11.258... = 219.54 sq units (rounded). Okay, I was a bit out, and possibly by more than would be normal just for rounding, so maybe averaging those two solutionsd for h aren't a good way. Pretty close though: 1.14 out in a solution that is 218.4.

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@JobBouwman Месяц назад

The enclosing triangle has sides 27 , 13/15*27 and 14/15*27. Use Heron to calculate the area, and multiply with factor (1 - (12/27)^2) to remove the top part.

@quigonkenny Месяц назад

Draw EC, so that E is the point on AB so that EC is parallel to DA. As CD is also parallel to AE (as CD is parallel to AB since ABCD is a trapezoid), then CDAE is a parallelogram, AE = 12, and EC = 13. In triangle ∆BCE, EB = 27-12 = 15. Let ∠BCE = α. cos(α) = (13²+14²-15²)/2(13)(14) cos(α) = (169+196-225)/26(14) cos(α) = 140/364 = 5/13 sin(α) = √{13²-5²)/13 sin(α) = √(169-25)/13 sin(α) = √144/13 = 12/13 Drop a perpendicular from C to EB at F. Note that there are two formulas to calculate the area of ∆BCE: bh/2 = absinC/2 EB(CF)/2 = CE(BC)sin(α)/2 15CF = 13(14)(12/13) 15CF = 168 CF = 168/15 = 56/5 Trapezoid ABCD: A = h(a+b)/2 A = (56/5)(12+27)/2 A = (28/5)(39) = 1092/5 = 218.4 sq units

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@artstocker60 Месяц назад

Solved it different. I enjoy seeing how others correctly solved the same problem by alternative solutions.

@PreMath Месяц назад

Excellent! Thanks for the feedback ❤️

@luigipirandello5919 Месяц назад

Nice solution. Thank you.

@PreMath Месяц назад

Glad to hear that! You are very welcome! Thanks for the feedback ❤️

@jorgeenriquefreirepena8482 24 дня назад

Felicitaciones, excelente explicación.

@phungpham1725 Месяц назад

Honestly, my method is like yours!😅 A= 218.4 sq units😊 Thank you so much

@PreMath Месяц назад

Excellent!🌹 Thanks for sharing ❤️

@marcelowanderleycorreia8876 Месяц назад

Very good!! 👍

@PreMath Месяц назад

Glad to hear that! Thanks for the feedback ❤️

@murdock5537 Месяц назад

Nice! φ = 30°; AB = 27 = AE + BE; BC = 14; AD = 13 = CE; CD = 12; AE = AT + ET = a + (12 - a) → sin⁡(ATD) = 1 BE = AS + BS = a + (15 - a) = a + b; DAB = CEB = α; ABC = β; DT = CS = h (14)^2 =169 + 225 - 2(13)(15)cos⁡(α) → cos⁡(α) = 33/65 = a/13 → a = 33/5 → b = 15 - a = 42/5 → h = 56/5 → ∆ ATD = pyth. triple = (1/5)(33 - 56 - 65) → ∆ BCS = pyth. triple = (14/5)(3 - 4 - 5) → area ABCD = (1/2)(27 + 12)h = 1092/5

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@hervechampagne195 Месяц назад

Isn't it easier, when you have the area of the triangle, to calculate the area of the parallelogram which sides are well known (13 and 12) ?

@AmirgabYT2185 Месяц назад

S=218,4 sq un

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@sergeyvinns931 Месяц назад

Area Trapezoid=area triangleA(CD)B+areaof a rectagle12*h/ Area triangle = \/p*(p-13)*(p-14)*(p-15), p=P/2=42/2=21, area =84, h=84*2/15=11,2, area rectangle = 12*11,2=134,4. Area of the Blue Trapezoid = 134,4+84=218,4!

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@cosmolbfu67 Месяц назад

169-k^2 = h^2 196-(15-k)^2 = h^2 k=33/5 , h=56/5 A= (12+27)(56/5)/2 =2184/10 = 218.4***

@raya.pawley3563 Месяц назад

Thank you

@PreMath Месяц назад

You are very welcome! Thanks ❤️

@user-ps2lw3rg2x 24 дня назад

Благодарю. Красивая задача

@johnbrennan3372 Месяц назад

Did it another way. Dropped a perp. from A and B onto DC extended both ways. This gave me the height = 56/5 etc

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@Waldlaeufer70 Месяц назад

27 - 12 = 15 => 2 triangles: i) h² + x² = 13² ii) h² + (15 - x)² = 14² h² = h² 13² - x² = 14² - (15 - x)² 169 - x² = 196 - 225 + 30x - x² 30x = 169 + (225 - 196) = 198 x = 198/30 = 99/15 = 33/5 = 6.6 h² = 169 - x² = 169 - 6.6² = 169 - 43.56 = 125.44 h = 11.2 A = (27 + 12)/2 * 11.2 = 19.5 * 11.2 = 218.4 square units

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@amitavadasgupta6985 Месяц назад

ar.=12×11.2+84=134.4+84=218.4 sq unit

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@wackojacko3962 Месяц назад

You know, the Nagal!

@user-ud6ui7zt3r Месяц назад

Do we assume that DC is parallel to AB ?

@user-lo7hg8zy4x 17 дней назад

Drop two perpendicular lines from the top . Let perpendicular height= C A+B =15 B = 15-A A^2 + C^2 = 169..........(1) (15-A)^2 + C^2 = 196 225-30A+A^2+ C^2 =196.............(2) (2)-(1) 225-30A = 27 A= 6.6 B=15-6.6 = 8.4 C= √169-(6.6)^2 = 11.2 Area= (12+27)/2 x 11.2 = 218.4

@devondevon4366 Месяц назад

218.4 Answer Let the height of the trapezoid = h Draw two perpendicular line from AB to isolate the 12 by h triangle forming two triangles ADP and BCR (P and R are new points). label AP= n, hence BR = 15-n (27 - 12= 15) Hence, for ADP using Pythogrean Theorem h^2 = 13^2 - n^2 and for BCr h^2 = 14^2- (n-15)^2 Hence 13^2 - n^2 = 14^2 - (n^2 + 225 - 30n) 169 - n^2 = 196 - n^2 - 225 + 30n 198 = 30n 6.6 =n hence 15-n = 8.4 Using Pythaogrean height= 11.2 Hence area of Trapezoid = (27 + 12)* 11.2 *1/2

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@Tmwyl 2 дня назад

I got this one using a different method!

@calvinmasters6159 28 дней назад

Dropped 2 verticals to make a rectangle and 2 Pythagoreans. Solve 2 triangle bases as 6.6 & 8.4 Vertical is 11.2 Take it home

@unknownidentity2846 Месяц назад

Let's find the area: . .. ... .... ..... Let's add the points E and F on AB such that CDEF is a rectangle. In this case the triangles ADE and BCF are right triangles, so we can apply the Pythagorean theorem. With h being the height of the trapezoid we obtain: AD² = AE² + DE² = AE² + h² BC² = BF² + CF² = BF² + h² BC² − AD² = BF² − AE² = (AB − AF)² − AE² = (AB − AE − EF)² − AE² = (AB − AE − CD)² − AE² 14² − 13² = (27 − AE − 12)² − AE² 196 − 169 = (15 − AE)² − AE² 27 = 225 − 30*AE + AE² − AE² 30*AE = 198 ⇒ AE = 6.6 h² = AD² − AE² = 13² − 6.6² = 125.44 ⇒ h = 11.2 Let's check this result: BF = AB − AE − CD = 27 − 6.6 − 12 = 8.4 h² = BC² − BF² = 14² − 8.4² = 125.44 ✓ Now we are able to calculate the area of the trapezoid: A(ABCD) = (1/2)*(AB + CD)*h = (1/2)*(27 + 12)*11.2 = 218.4 Best regards from Germany

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@newjoisey9424 Месяц назад

Once the area of triangle EBC is found, wouldn't it just be easier to add the area of the remaining parallelogram ADCE (12 x 13)?

@PreMath Месяц назад

Hello dear, area of the parallelogram=base*height Thanks for the feedback ❤️

@claytonhough3044 25 дней назад

@@PreMath I did it this way but 84 + (13x12=156) = 240 exactly??? not 218.4 WHY???

@kalavenkataraman4445 Месяц назад

218.4 Sq.units

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@LuisdeBritoCamacho Месяц назад

STEP-BY-STEP RESOLUTION : 1) (B - b) = (27 - 12) = 15 2) AD' = X 3) BC' = Y 4) Y > X and : X + Y = 15 5) h^2 = 169 - X^2 6) h^2 = 196 - Y^2 7) h^2 = h^2, so : 8) 196 - Y^2 = 169 - X^2 9) 196 - 169 = Y^2 - X^2 10) Y^2 - X^2 = 27 11) (Y + X) * (Y - X) = 27 12) As : (Y + X) = 15, 13) 15 * (Y - X) = 27 14) (Y - X) = 27 / 15 15) (X - Y) = 9 / 5 16) System of Two Equations with Two Unknowns : a) X + Y = 15 b) Y - X = 9 / 5 17) Solutions : a) X = 33/5 b) Y = 42/5 18) Height = h^2 = 196 - (42/5)^2 or h^2 = 169 - (33/5)^2 19) h = sqrt(3.136/25) 20) h = 56/5 21) Trapezoid Area = Z 22) Z = (27 + 12) * (56/10) 23) Z = 39 * 5,6 ; Z = (1.092 / 5) ; Z = 218,4 24) ANSWER : The Trapezoid Area is equal to 218,4 Square Units.

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️