Interesting one☺

@PreMath

11 ай бұрын

Glad you think so! Thank you! Cheers! 😀

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

@PreMath

11 ай бұрын

True! Thank you! Cheers! 😀

@soli9mana-soli4953

11 ай бұрын

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

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

11 ай бұрын

. Thank you! Cheers! 😀

@dalesmith7250

11 ай бұрын

This is exactly what I did

@NYlivinginTN

11 ай бұрын

Same here

@Mattdemon08

11 ай бұрын

Lovely! 👍🏾

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

11 ай бұрын

You are right! Thank you! Cheers! 😀

@xof-woodworkinghobbyist

11 ай бұрын

I did the same way.

@bb55555555

9 ай бұрын

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

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

11 ай бұрын

. Thank you! Cheers! 😀

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

11 ай бұрын

Thank you! Cheers! 😀

Interesting and easy at the same time

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

11 ай бұрын

Thank you! Cheers! 😀

@gassawi

11 ай бұрын

Where 30^2 is coming?

@anthonycheng1765

11 ай бұрын

@@gassawi sorry it is 32^2

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.

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

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

11 ай бұрын

Thank you! Cheers! 😀

Excellent!!! Thanks!

@PreMath

11 ай бұрын

Glad you liked it!

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

11 ай бұрын

So nice of you dear Thank you! Cheers! 😀

@allanflippin2453

11 ай бұрын

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

11 ай бұрын

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

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. 🤠

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

@PreMath

11 ай бұрын

Thank you too

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

11 ай бұрын

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

Thank you!

@PreMath

10 ай бұрын

You're welcome!

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.

Very good 👍

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

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.

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

11 ай бұрын

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.

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

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.

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

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

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

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

A little convoluted but worthwhile!🥂👍❤

@PreMath

11 ай бұрын

Thank you! Cheers! 😀

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

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

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

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

👍

@PreMath

11 ай бұрын

Thank you! Cheers! 😀

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

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

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

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

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

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

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.

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

Yeah nice, I did it the other way though, 345's everywhere 🙂

@PreMath

11 ай бұрын

Nice job!

@theoyanto

11 ай бұрын

@@PreMath cheers 🤓

S=300

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.

easy

@PreMath

11 ай бұрын

Thank you! Cheers! 😀

Pourquoi calculer l’aire du trapèze ? Une fois la base obtenu, la formule de l’aire d’un triangle suffit !

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

300

300 but I made a mistake first

A=300 square units

That was too long. BCD is 3-4-5 so we know BD = 40 Then find DA from small triangle then 1/2 b*h

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 √ )

Area triangle ABD = AB*h/2 = 25*24/2= 300 Area ABCD - area BCD -is redundant