Find Area of the Blue Shaded Region in the right triangle | Important Geometry skills explained

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[francismoles]

The triangle BDE is a reduction of ABC with a 3/4 rapport, area of BDE is equal of 3/4 at square, 9/16, we can Say that the area of the trapezoid is equal at 7/16 of the area of the triangle ABC

[JMan]

Perfect problem to practice similar triangles! You can also use the scale factor from the 3-4-5 special triangle to find out the sides of the smaller triangle. For the big triangle, it's 8. For the smaller triangle, it's 6. Love the multi-step nature of the problem. Kids get so overwhelmed at the sight of a multi-step problem like this but that creates a perfect opportunity to teach patience in math. All will be well by breaking down the problem!

[MrPaulc]

I constructed an imaginary trangle in the top left which has side lengths of one-quarter the larger triangle (6-8-10).This has an area of 24. To this I added the rectangle part which is 8 by 18. 24+144=168.

[txokomates]

Very happy to have found your youtube channel and very grateful for sharing your knowledge of teaching. Thank you so much.
I don't know English and I watch your videos without sound and they are perfect, it is perfectly understood.
I write to you with a translator.
I would like to know what program you use to make these videos so that it looks like you are writing on a whiteboard.
Greetings and good day

[Mediterranean]

Area of ABC = 1/2*24*32= 384
by the pythagorean theorem
BC=40 or by scaling up 3;4;5 triangle (8×)
so BE=40-10=30
we notice that BED and ABC are similar triangle since they share a common angle and both have a right angle
so their ratio is k=30/40=3/4
Area of BED= Area of ABC *k^2
=384*9/16
=216
Blue area = ABC area - BED area
= 384-216
=168

blue area = area of ABC- area of BDE
= 24

[mahatmapodge]

Similar triangles again, once you recognise that its a 3-4-5 triangle its just a triangle and rectangle, no need for complexity,

draw a line from E perpendicular to AC, to point F

Area == rectangle AFED + triangle FEC

Area == (AF)(FE) + (FC)(FE)/2


FEC is similar to ABC so, if CE == 10

then FE == 8 and FC == 6

AF == AC - FC == 24 - 6 == 18

Area == (AF)(FE) + (FC)(FE)/2

Area == (18)(8) + (6)(8)/2 == 168 units

[richardslater]

For me, it’s more straightforward using trig. The angle at Cis easy to calc. Tan-1 32/24. Draw a horizontal line at E. it’s now easy to calculate the dimensions of that small blue triangle, knowing one side length and the angle at C. The area of the small blue triangle plus the blue rectangle gives 168.

[unknownidentity]

I also started with the calculation of the lengths x = DB and y = DE. From the lengths of AB and DB I calculated the length of AD as the next step. Because AC and DE are parallel, ACDE is a trapezoid and its area can be calculated by:
Area(ACDE) = (1/2) * (AC + DE) * AE

Same result, of course. Best regards from Germany.

[triangles]

Find angle ABC: tanx = 0.75.. approximately 37 degrees. Consider C as origin of a cartesian plane. The clockwise angle of vector CE subtends is 37 degrees. In this circumstance: X component is cos 37 x 10 (approx 8). Y component is sin 37 x 10 (approx 6). Calc area in 2 parts: top part is 1/2.b.h = 1/2x8x6 = 24. Rectangle below is l.b (24-6)x8 = 18x8 = 144. Final answer: 168 square units.

[JSSTyger]

I tried it all in my head and got 168. It helped realizing the 3-4-5 triangles.

[Kame]

∆ABC = 3x4x5 triangle with an 8 multiplier
BC = 5 * 8 = 40
BE = BC - EC = 40 - 10 = 30
Multiplier of ∆DBE = 30 ÷ 5 = 6
∆DBE = 3x4x5 triangle with a 6 multiplier
DB = 4 * 6 = 24
DE = 3 * 6 = 18

Area of shade = Area of ∆ABC - Area of ∆DBE.
Area of triangle = ½(base)(height)
S = ½(32)(24) - ½(24)(18) = 384 - 216 = 168 units squared

[santiagoarosam]

8X3=24 y 》Azul =(3×4/2)×(8^2) - 6(6^2)=168
Gracias y saludos.

[saltydog]

As EDB is a smaller version of the triangle CBA, once we use Pythagoras' T to find the length of CB we can see that EB is 3/4 of CB and so DB will be 3/4 of AB - 32*3/4 = 24 making AD 8 and the distance from A to a point on the line AC level with E will be 3/4 of 24 = 18. Therefore the rectangle in the shaded area is 18*8 = 144 plus the area of the triangle at the top of the shaded area which is 1/2*8*6=24 and 144+24=168

[alster]

An alternate way for the Pythagorean for ∆ABC

(24)²+(32)²= (10+c)²
576+1024 = 100+20c+c²

1600= c²+20c+100

c²+20c-1500= 0
(c+50)(c-30)

c= -50, 30


Since -50 is rejected for geometric figures, so c= 30


Then the rest is easy

[sameerqureshi-khcc]

Sir that blue area is a trapezoid. You can also use the following formula for calculating area of trapezoid:

1/2 × height × sum of parallel sides

[XR]

I like this channel. 🥰
My shortest version ;
If ABC triangle is 3-4-5 --> CB =40
CEF = 1/4 ABC dimensions 3-4-5 triangle.
CE = 10, EF = 8 and CF = 6
A(CEF ) = 1/2 ( CF . EF ) = 24
A (ADEF) = (AC-AF) . EF = ( 24-6 ) . 8 = 144
A ( ACDE ) = A (CEF) + A(ADEF ) = 24 + 144 = 168 //

[quigonkenny]

As AC/AB = 24/32 = 3/4 and 32/4 = 8, ∆CAB is a 8:1 ratio (3, 4, 5) Pythagorean Triple triangle and BC = 5(8) = 40.

BE = 40 - 10 = 30

By observation, ∆EDB and ∆CAB are similar, therefore since 30/5 = 6, ∆EDB is a 6:1 ratio (3, 4, 5) Pythagorean Triple triangle and DE = 3(6) = 18 and DB = 4(6) = 24

ADEC = ∆CAB - ∆EDB
A = (32)24/2 - 24(18)/2 = 384 - 216 = 168

[khatharrmalkavian]

By Pythagoras, triangle ABC has hypotenuse 40. Thus the smaller triangle has hyp 30. The base of BDE will be 3/4 of ABC's since it projects perfectly to the hypotenuse, so 32*3/4=24 base and hyp 30. By Pythagoras the line segment ED will be 18. 18 * 24/2 = 216 area for BDE and 24*32/2 = 384 area for ABC, so 384-216=168 area for the blue region.

[thefunpolice]

Solved this one two different ways both of which are mentioned in the comments section. I enjoy these videos very much. Thanks @PreMath

[alexniklas]

The large triangle is Egyptian, hypotenuse = 40. Hypotenuse of the smaller triangle = 40-10 = 30. The smaller triangle is also Egyptian, legs 18 and 24.
Trapezium area:
(24+18)8/2= 168
Thanks sir