Bom dia! Good morning

@PreMath

15 күн бұрын

Hello dear 🌹 Thanks ❤️

@PreMath

15 күн бұрын

Hello dear🌹 Thanks ❤️

Realising ADC is half of ABC is the slick way of doing this easily. Brill.

@PreMath

15 күн бұрын

Excellent! Thanks for the feedback ❤️

Thanks Sir That’s nice and useful methods for solve . With glades

Nicely worked out. Both the methods are good

Easy to compute to get x=52, cos B=40/104=5/13, so h=40×sin B=40×12/13=480/13, therefore the area is 1/2×(104+52)×(480/13)=1/2×156×(480/13)=1/2×12×480=12×240= 2880.😊

@PreMath

15 күн бұрын

Thanks for the feedback ❤️

I calculated AB with Pythagorean Theorem, Then made a perpendicular segment AM on AB. afterwards I wrote this relation: Area of triangle CAB= (AC*CB)/2 = (CM*AB)/2. => 3840=104CM => CM=480/13. Afterwards area of the trapezoid is [(52+104)*480/13]/2 = 2880 square units.

The lesson to learn is when answer is not the unknown itself, solving the unknown may not be necessary. The first method uses the base side ratio hence area ratio 2x:x = 2:1 for equal height triangles without solving x. The second method treats hx as a single unknown in trapezoid area formula i.e. (3/2)hx and then find hx by triangle area formula i.e. (1/2)(2x)h = (1/2)(40)(96). By the way, hx can also be found by similar triangles side ratio when triangle ABC is divided into similar triangles by h but it is unnecessarily complicated. Solving x first seems to be a trap for students to waste time in exam.

Thank you!

1/ Area of the triangle ABC= 20.96= x.h--> h= (20.96)/x Area of the trapezoid= 3x/2 . (20.96/x)= 2880 sq units

@PreMath

15 күн бұрын

Thanks for sharing ❤️

AD intersect CB at E. In triangle EAB, DC parallel AB and DC=AB/2 so C is midpoint of EB. Then BE=2*CB=2*40=80. The area of triangle AEB=AC*BE/2=96*80/2=3840. Triangle EDC similar to triangle EAB with similar ratio is 1/2 so the area of ADC=1/4*area of EAB. Therefore the area of yellow tranpezoid=3/4*area of triangle EAB=3/4*3840=2880

Very easy way just bu observation Calcilate ratio of 96 and 40 and you will get 12/5 Recall 5-12-13 triplet of triangle and just find x by x =8×13/2 Now further calculate heights height by Pythagoras and calculate area

2x = 104 (Pythagoras) x = 52 104h = 96 * 40 h = 3840/104 = 480/13 m = 3x/2 = 78 A = mh = 78 * 480/13 = 2880 square units

Draw the median trangle ABC CF=AB/2=x, draw line DF=CB=40. Area trangle ADF=FCB=CDF=48*40/2=960. Area of theYellow Trapezoid ABCD = 3*960=2880.

Method 1: Triangle ∆BCA: BC² + CA² = AB² 40² + 96² = (2x)² 1600 + 9216 = 4x² 4x² = 10816 x² = 2704 x = √2704 = 52 Draw CE, where E is the point on AB where CE is perpendicular to AB. As ∠CEB = BCA = 90° and ∠EBC = ∠ABC, ,∆CEB and BCA are similar triangles. Triangle ∆CEB: CE/BC = CA/AB CE/40 = 96/104 = 12/13 CE = 40(12/13) = 480/13 Trapezoid ABCD: A = h(a+b)/2 A = (480/13)(52+104)/2 A = 480(4+8)/2 A = 240(12) = 2880 sq units 2nd method: Draw a circle so that ∆BCA is inscribed in said circle. As A, B, and C are all on rhe circumference and ∠BCA is 90°, then AB is the diameter of the circle, and the center is at the midpoint of AB, O. Draw OC. OB = OA = OC = r = x. Draw OE, where E is the midpoint of BC. As OB = OC, ∆COB is isosceles and thus OE bisects ∆COB, forming two congruent right triangles ∆BEO and ∆OEC. As OA = CD = x, AOCD is a parallelogram, and thus OD mutually bisects AC. As CA is also a chord of circle O, OD must also be perpendicular to CA if it bisects it. Let G be the point of intersection between OD and CA. Since OD and CB are both perpendicular to CA, OE and CA are both perpendicular to BC, OF bisects CA, and OE bisects BC, then OECG is a rectangle of side lengths 40/2 = 20 and 96/2 = 48. The area of trapezoid ABCD is the sum of the sreas of triangles ∆ABC and ∆CDA. Trapezoid ABCD: A = CA(BC)/2 + CA(GD)/2 A = 96(40)/2 + 96(20)/2 A = 48(40) + 48(20) A = 1920 + 960 = 2880 sq units

X=√(40²+96²)/2=52→ 2*52*h=40*96→ h=480/13→ Área ABCD =(3*52/2)*(480/13) =2880. Gracias y saludos.

@PreMath

15 күн бұрын

Excellent! Thanks for sharing ❤️

1) 96^2 + 40^2 = (2X)^2 ; 9.216 + 1.600 = 4X^2 ; 10.816 = 4X^2 ; X^2 = 10.816/4 ; X^2 = 2.704 ; X = sqrt(2.704) ; X = 52 2) Minor Basis = DC = X = 52 lin un 3) Major Basis = AB = 2X = 104 lin un 4) Calculating Height : 5) sin (BAC) = 5/12 6) sin (CAC') = 5/12 7) 40 / 104 = CC' / 96 ; 20 / 52 = CC' / 96 ; 10 / 26 = CC' / 96 ; 5 / 13 = CC' / 96 ; CC' = 480 / 13 8) So h = 480/13 9) AT = (52 + 104) * (h / 2) ; AT = 156 * (480/26) ; AT = 2.880 sq un 10) Answer : The Yellow Trapezoid Area is Equal to 2.880 Square Units.

@PreMath

15 күн бұрын

Excellent! Thanks for sharing ❤️

Let's find the area without finding x: . .. ... .... ..... The height of the trapezoid is also the height of the right triangle ABC according to its base AB. It can be calculated as follows: A(ABC) = (1/2)*AC*BC = (1/2)*AB*h(AB) ⇒ h(AB) = AC*BC/AB Now we can calculate the area of the trapezoid: A(ABCD) = (1/2)*(AB + CD)*h(AB) = (1/2)*(AB + CD)*(AC*BC/AB) = (1/2)*(2*x + x)*AC*BC/(2*x) = 3*x*AC*BC/(4*x) = 3*AC*BC/4 = 3*96*40/4 = 2880 Best regards from Germany

@PreMath

15 күн бұрын

Excellent! Thanks for sharing ❤️

I did it just like your second method.

You can even use similar triangles to find "h".

40^2+96^2=(2x)^2 1600+9216=4x^2 10816=4x^2 x^2=2704 x=52 2x=104 h^2=40^2-y^2 h^2=96^2-(104-y)^2 1600-y^2=9216-10816+208y-y^2 208y=3200 y=200/13 h^2=1600-40000/169 =270400/169-40000/169=230400/169 h=480/13 area of the yellow trapezoid : (52+104)＊480/13＊1/2=2880

@PreMath

15 күн бұрын

Excellent! Thanks for sharing ❤️

1st meathod is best

@ 1:39 the real dilemma is this "we" thing about sometimes calling a Trapezoid a Trapesium. I never call a Trapezoid a Trapesium. And not once did you. They can't be both at the same time. It's like Schrodinger's Cat. The difference is a Trapezoid is a geometric shape and a Trapesium is a neck muscle (this side of the pond). Nothing personal, Love too tease people and love etymology. So I'm cracking up! 🙂

@PreMath

15 күн бұрын

😀 Thanks for the feedback ❤️

Calcolo x col triangolo rettangolo 4x^2=96^2+40^2..x=52...96/sinα=40/sin(90-α).tgα=12/5...h=40sinα..h^2=40^2(tgα)^2/(1+(tgα)^2)=40^2(144/25)/(169/25)..h=40*12/13....Ay=((3*52)*40*12/13)/2=2880..α=CBA

@PreMath

15 күн бұрын

Excellent! Thanks for sharing ❤️

(96)^2+(40)^2=(2x)^2 x=52 Area of the yellow trapezoid=1/2(52+104)(480/13)=2880 square units.❤❤❤ Thanks sir.

@PreMath

15 күн бұрын

Excellent! You are very welcome! Thanks for sharing ❤️

Third method-hx = 1920. Area is 3/2 hx.

@PreMath

15 күн бұрын

Thanks for sharing ❤️

Can you please explain how the height of ADC is also = h?

@PreMath

15 күн бұрын

Dear Eyad, please watch the video again. Height h of the Triangle ADC is the vertical distance from the base to the very top! Hope I explained well. Thanks ❤️

@nickdsp8089

14 күн бұрын

For a visual representation you must extend CD from the point D to the left. Then from point A draw a perpendicular upwards to the extension. They intersect at a point. If you name this point F then AF is the height of triangle ADC which is the same as h as you can now see from the rectangle AFCE.

S = 96 × 40 × 0.75 = 2880 As simple as that...

My way of solution ▶ Area of ΔABC A(ΔABC)= 40*96/2 A(ΔABC)= 20*96 A(ΔABC)= 2x*h/2 ⇒ hx= 20*96 A(ΔACD)= h*x/2 A(ΔACD)= 20*96/2 A(ΔACD)= 10*96 A(ΔACD)= 960 square units A(ΔABC)= 20*96 A(ΔABC)= 1920 square units Ayellow= A(ABCD) A(ABCD)= A(ΔACD) + A(ΔABC) A(ABCD)= 960 + 1920 A(ABCD)= 2880 square units ✅

S=2880

@PreMath

15 күн бұрын

Thanks for sharing ❤️