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Learn how to find the area of Blue shaded triangle in the Trapezoid. Important Geometry and algebra skills are also explained. Step-by-step tutorial by PreMath.com
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Комментарии : 85

@walcholjacob4259 Год назад

Interesting one☺

@PreMath Год назад

Glad you think so! Thank you! Cheers! 😀

@burlfromlondon Год назад

After you know the value of AB then you can directly calculate the area just using half base times height

@PreMath Год назад

True! Thank you! Cheers! 😀

@soli9mana-soli4953 Год назад

And you can call not EA but AB = X, then X² = 24² + (32 - X)² it's a little faster

@pwmiles56 Год назад

By Pythagoras DB is 40. Drop a perpendicular from A to P on DB. As ABD is isosceles P is the midpoint of DB, DP=PB=20. Triangle APB is similar to triangle BCD (as angle ABP is complementary to CBD and PAB is complementary to that). So AP = (24/32)x20 = 15 Area of ABD = 40(15/2) = 300

@PreMath Год назад

. Thank you! Cheers! 😀

@dalesmith7250 Год назад

This is exactly what I did

@NYlivinginTN Год назад

Same here

@Mattdemon08 Год назад

Lovely! 👍🏾

@juuusturull1250 Год назад

Great video! You didn't even need to find the area of the trapezoid nor the white triangle. If the base of the blue triangle is 25 and height is 24, you can find the area of the blue triangle easily.

@PreMath Год назад

You are right! Thank you! Cheers! 😀

@xof-woodworkinghobbyist Год назад

I did the same way.

@bb55555555 11 месяцев назад

My thought exactly. As soon as you got the 25 for the base you already had the height of the triangle.

@anthonycheng1765 Год назад

angle BDC = atan(3/4). angle ABD = atan(3/4) (alt angles, DC // AB), AB = AD (given), angle ADB = angle ABD (base angles isos. triangles). Draw AE perpendicular to DB with BE = ED. AE = sqrt(24^2 + 30^2) = 40 (Pyth. Thm) BE = ED = 20. AE = 20*tan(atan(3/4)). Area = 1/2*AE*BD = 1/2*40*20*tan(atan(3/4)) = 300

@PreMath Год назад

Thank you! Cheers! 😀

@gassawi Год назад

Where 30^2 is coming?

@anthonycheng1765 Год назад

@@gassawi sorry it is 32^2

@misterenter-iz7rz Год назад

Clearly BD=40, it remains to find the height h of the triangle, in view of similar triangles, h/20=24/32=3/4, thus h=15, therefore the answer is 40x15/2=300.😃

@PreMath Год назад

. Thank you! Cheers! 😀

@himo3485 Год назад

AB=AD=x (32-x)^2+24^2=x^2 1024-64x+x^2+576=x^2 64x=1600 x=25 area of the Blue Triangle : 25＊24/2=300

@PreMath Год назад

Thank you! Cheers! 😀

@ybodoN Год назад

Draw a perpendicular from A to E on DB. As DC ∥ AB ⇒ ∠CDB ≅ ∠DBA ⇒ ◺BCD ~ ◺BEA. DB = √(24² + 32²) = 40 ⇒ EB = 20. AB = 40 ⋅ 20 / 32 = 25. Area △ABD = ½ 25 ⋅ 24 = 300 u².

@harikatragadda Год назад

Draw a perpendicular from A to DB at E. AE divides triangle ABD into two right triangles with base 20. Triangle AEB is Similar to Triangle DCB, AE = BE*CB/DC= 15 Blue area = 2*½*AE*BE = 300

@PreMath Год назад

Thank you! Cheers! 😀

@sumanmukherjee100 Год назад

It was a pretty good question sir.. I first drew a perpendicular from point A to line DC , which is AP then I took AB =x then i got DP=32-x as AP is perpendicular it was 24as well the by Pythagorean theorem I got AD is 25 . And the the area of the traingle is half times base times hight and the I got area =25×24/2 = 300units .. thank you sir love from India..

@PreMath Год назад

So nice of you dear Thank you! Cheers! 😀

@allanflippin2453 Год назад

Yes, I used this approach as well. It saves a couple steps. We no longer need the area of the trapezoid, just calculate the triangle area directly.

@sumanmukherjee100 Год назад

@@allanflippin2453 yeah , you are correct..I think this is one of the easiest approach..

@williamwingo4740 Год назад

A slightly different method: Let x = AB = AD, and draw triangle ADE as you did. Then by Pythagoras, x^2 = 24^2 + (32 - x)^2; multiply out (32 - x)^2: x^2 = 24^2 + 32^2 - 64 x + x^2; subtract out x^2: 0 = 24^2 + 32^2 - 64x; rearrange: 64x = 24^2 + 32^2; solve for x: x = (24^2 + 32^2)/64 = (576 + 1024)/64 = 1600/64 = 200/8 = 25. And the area of the blue triangle is (1/2)(24)(25) = (12)(25) = 300. It might also be worth noting that triangle ADE is integer Pythagorean, 7-24-25. Cheers. 🤠

@anahimaceda7070 Год назад

Excellent!!! Thanks!

@PreMath Год назад

Glad you liked it!

@TimBoulette Год назад

Extend line AB to the left, and drop an auxiliary line from point D to point P such that it completes rectangle BCDP. Line DP is equal to line CD, which is 24. Let x = line PA. Then, AB = 32-x, as does line AD. Pythagorean theorem: x^2 + 24^2 = (32-x)^2. Solve for x and you get x = 7. Therefore, AB= 25. Area of ABD = (1/2)*24*25 = 300.

@bahijsarhan8002 Год назад

Bd=40 حسب فيثاغورث نرسم الارتفاعam علىbd فهو متوسط الزاويةbdc=abd تبادل داخلي نحسب cosالزاويتان 20/x=32/40 نجدx=25 S=25×24/2=300

@misterenter-iz7rz Год назад

More generally, let BC=a, CD=b, then BD=sqrt(a^2+b^2), let AE=h be the height of the triangle, so 2h/sqrt(a^2+b^2)=a/b, thus h=a sqrt(a^2+b^2)/(2b), and then the area is sqrt(a^2+b^2)a sqrt(a^2+b^2)/(4b)=(a^2+b^2)a/(4b). It is 300, as a=24, b=32.

@ybodoN Год назад

Otherwise, DB = √(a² + b²) ⇒ EB = ½ √(a² + b²) ⇒ AB = ½ (a² + b²) / b ⇒ area △ABD = ¼ (a² + b²) a/b which, of course, is just an alternate form of the same general formula.

@Mattdemon08 Год назад

Good question. It was a good brain workout for me. I noted from your video and the comments how i could have skipped some of my steps. Thank you.

@PreMath Год назад

Great job! So nice of you. Thank you! Cheers! 😀

@engralsaffar Год назад

In 3,4,5 right triangle DB=40 We can connect A to point M in the middle of DB, to get a 2 right triangles Triangles DBC and AMB are similar triangles 24/32=AM/20 AM=15 A=15*20=300 square units

@KAvi_YA666 Год назад

Thanks for video.Good luck sir!!!!!!!!!

@PreMath Год назад

Thank you too

@mohanramachandran4550 Год назад

at. 6 : 38 we find AB = 25 Then area of the blue triangle is ½bh = ½ *25*24 = 300 square units

@jhouck1969 Год назад

A simpler solution: Triangle BCD is a 3-4-5 triangle, so length BD = 40. Because sides AB and CD are parallel, angle CDB = angle DBA. Bisect angle DAB creating a perpendicular bisector of side BD, making two identical 3-4-5 triangles with sides 15-20-25. This means the height of triangle ABD is 15 and the base is 40. Thus the area of triangle ABD is (1/2)(15)(40) = 300.

@juanalfaro7522 9 месяцев назад

It is good to find alternate ways of solving problems, but I would have used the direct way: I find the length AB by finding the angle ABD, which is the complement of angle CBD, then determine the area either as (1) AB x BD x sin (ABD / 2, or determining the H from A to the midpoint of BD and then calculating Area = BD x H /2. BD is easily determined by the 3-4-5 rule as 40 and angle ABD is the complement of 53.13, which is 36.87. So cos (36.87) = 0.6, sin (36.87) = 0.8, AB = 25, and H = 15. Clearly the area is 300.

@raya.pawley3563 Год назад

Thank you!

@PreMath Год назад

You're welcome!

@quigonkenny 7 месяцев назад

Let AB = DA = x. Let E be a point on CD such that EA is parallel to BC. By observation, ED = 32-x. Since the point of this video seems to be trapezoids, by the title, we'll determine the area by finding the difference in area between the trapezoid ABCD and triangle ∆BCD, though it would be simpler just to determine the area of ∆DAB directly once we have x. Triangle ∆AED: a² + b² = c² (32-x)² + 24² = x² 1024 - 64x + x² + 576 = x² 64x = 1600 x = 25 Triangle ∆BCD: A = bh/2 = 32(24)/2 = 384 Trapezoid ABCD: A = h(a+b)/2 = 24(25+32)/2 A = 12(57) = 684 Triangle ∆ABD: A = 684 - 384 = 300 Double check: A = bh/2 = 25(24)/2 = 600/2 = 300

@bearantarctic5843 Год назад

Or alternatively: Find angle CBD using trigonometry Find all of the angles in the blue triangles Find that DB is 40 units long using the Pythagorean theorem Find that AB is 25 units long using the Law of Cosines Calculate the area of the triangle using Heron's formula A little bit of trig didn't hurt anyone

@bobbyheffley4955 Год назад

Let the legs of the isosceles triangle be 25 units in length and the longer side be 40 units in length. 25+25+40=90 units. Half of that is 45. 45-25=20 45-40=5 By Heron's formula, the area is sqrt [45(20)(20)(5)]=sqrt (90,000)=300

@MrPaulc222 11 месяцев назад

i never found AB or AD. Instead, I found the diagonal to be 40 due to 8(3) + 8(4) = 8(5). Then

@MarieAnne. Год назад

To calculate area of triangle ABD, we need only use the formula Area = 1/2 * base * height where base = AB (which you calculated between 2:45 and 6:43) and height = 24 (given). So once we've found length of AB = 25, we get: Area = 1/2 * 25 * 24 = 300

@alster724 Год назад

Interesting and easy at the same time

@brianrsnes7875 Год назад

I too made the DEA triangle and found x to be 7. But I subtracted triangle DEA from triangle DEB. Thus half time heigt on both is 12. (12*32)-(12*7)=12*25=300

@sie_khoentjoeng4886 Год назад

Alternative way, we can calculate the blue area after calculate AE, here AE = 7: ABD = EBD - EAD ABD = (32*24)/2- (24*7) /2 ABD = 768/2- 168/2 ABD = 384 - 84 ABD = 300

@Abby-hi4sf Год назад

I love outside the box, because it expands our vision to any problem! Thank you!

@vidyadharjoshi5714 Год назад

DB = 40. Angles ABD & ADB = 36.87 each. Cos 36.87 = 20/AB = 0.8. So AB = 25, So Blue Area = 0.5*24*25 = 300

@cleiberrocha1449 8 месяцев назад

Once we find the value of AB = 25, just calculate the area of the triangle ABD = (1/2) 25×24 =25 ×12= 300

@giuseppemalaguti435 Год назад

A(blue) =bh/2=20/cos(90-arcsin32/40)*24/2=25*24/2=300

@mohamadtaufik5770 Год назад

DB^2=DC^2+CB^2 DB^2=32^2+24^2 DB=40 tan BDC=24/32=3/4 24^2+a^2=b^2 576+a^2=b^2 b=V576+a^2 h^2+20^2=576+a^2 h^2-a^2=176 a+b=32 --> b=32-a 576+a^2=(32-a)^2 576+a^2=1024-64a+a^2 64a=1024-576=448 a=7 --> h^2=176+7^2=225 h=15 Area of blue triangle=0.5*40*15=300 square units

@santiagoarosam430 Год назад

24²+32²=DB² → DB=40 → Si “M” es el punto medio de DB→DM=MB=40/2=20 → El punto A equidista de B y D → El punto A pertenece a la mediatriz de DB → La razón de semejanza “s” entre los triángulos rectángulos AMB y BCD es s=20/32=5/8 → AM=24*5/8 =15 → Área azul =20*15=300 Gracias y saludos.

@user-lk4ky6ch9s Год назад

△ABD=x AB=x/12=AD (32-AB)^2+24^2=(x/12)^2 x=300

@bigm383 Год назад

A little convoluted but worthwhile!🥂👍❤

@PreMath Год назад

Thank you! Cheers! 😀

@yamunaravi7132 Год назад

Very good 👍

@user-eo2ed7lr2n Год назад

We don't need calculate trapezoid's area. If we know AB and AD - we can find triangle's area. AB is a base, BC is a height. AB*BC/2 is a triangle's area

@chanpangchin9744 9 месяцев назад

Mentally guessed 7,24,25. Area of blue region=1/2 x 25x24=300

@BilalSozkesen-dr4im 11 месяцев назад

at the end why are u calculate area of abcd just do that height of abd=24 ab=25 area of abd=25*24/2 =300

@ybodoN Год назад

Let's play with trigonometry! But first we need to divide the isosceles triangle ABD into two congruent right triangles one side of which is half of DB🧐 Since DC ∥ AB ⇒ ∠CDB ≅ ∠DBA (alternate angles). Let's call this angle θ. Since tan θ = 24/32 = ¾ ⇒ sin θ = ⅗ ⇒ cos θ = ⅘ (trigonometric identities). Then DB = 32 / cos θ = 32 / (⅘) = 40 and AB = ½ DB / cos θ = 20 / (⅘) = 25. So, the area of the blue △ABD is ½ 40 ⋅ 25 ⋅ sin θ = ½ 40 ⋅ 25 ⋅ ⅗ = 300 u².

@marioalb9726 Год назад

Taking the white right triangle: c²=a²+b² c²=32²+24² c= 40 cm tan α = 32/24 α = 53,13° β = 90° - α β = 36.87° Taking half blue triangle, which is a right triangle: tan β = h / (c/2) h = c/2. tan β h = 40/2 . tan 36,87° h = 20 . 0,75 h = 15 cm Area of blue triangle: A = c. h / 2 = A = 40 . 15 / 2 A = 300 cm² ( Solved √ )