Find the blue shaded area | Geometry Problem | Important Geometry and Algebra Skills Explained

Find the blue shaded area | Geometry Problem | Important Geometry and Algebra Skills Explained

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[florianbuerzle]

Even easier: the right angle altitude theorem. The altitude h of a right trinangle creates two line segments on the hypotenuse H, say p and q. The theorem states: pq = h^2. Let p = 8 and h = 4, we get q = 16/8 = 2. So H = 10 and area(BCD) = 10*4/2 = 20.

[randyrogers]

Ratio of these similar triangles is (root 80)/8 so the area ratio would be that number squared or 1.25. Multiply this times area of white triangle to get 20. Takes a fraction of the time in this example.

[misterenter-izrz]

BD=4sqrt(5), BC=2sqrt(5) by similar triangles, therefore the area is 4sqrt(5)*2sqrt(5)/2=20.😊

[Andyyen-fqjh]

Two triangles are similar.
So if we know the length of BD, we can solve it easier.
BD:AD=4√5:8=√5:2, hence the area of triangle BCD is [(√5)^2/2^2]×(4×8)÷2=20

[douglaspinheiro]

There is another way to do. the angle BDC is complimentary to the angle ADB and we know that ABD + ADB = 90, thus BDC = ABD = â. so tgâ = 4/8 = 0.5 and tgâ= sqrt(80) / BC, we find BC = 0.5*sqrt(80). finaly the area is BD*BC/2= sqrt(80) * 0.5(sqrt(80)/2 = 80/4=20

[santiagoarosam]

Punto E es la proyección ortogonal de B sobre DC →La razón de semejanza entre los triángulos BEC y DBC es s=4/8=1/2 → EC=s(AD)=4/2=2 → DC=8+2=10 →Azul =10*4/2 =20
Gracias y saludos

[murdock]

Nice! AB = DR = 8; AD = 4; CD = 8 + k = DR + CR → sin⁡(BRD) = 1
tan⁡(φ) = AD/AB = 1/2 = k/4 → k = 2 → area ∆BCD = 4 + 16 = 20
or: 16 = 8k → k = 2

[TheDerlick]

ADB° = BCD° (This can easily be proven with a quick calculation, even)

Therefore, just as AD (adjacent) is half of AB (opposite), BC (adjacent) is half of BD (opposite).

Since the pythagorean theorem tells us BD is 4root(5), BC is half of that, or 2root(5).

area of a triangle is 1/2bh
b = BC = 2root(5)
h = BD = 4root(5)

1/2 • 2root(5) • 4root(5) =
root(5) • 4root(5) =
(root(5))² • 4 =
5 • 4 = 20

[albertofernandez]

Why do you calc the hypotenuse, if it is not neccesary?.

[bozotheclown]

This was the long version. But a nice small challenge indeed. Thanks for that...

[bpark]

There is much simpler way. Once the hypotenuse is BD calculated, 4 sqrt(5), because the 2 triangles are similar, the leg BC is half that, 2 sqrt(5). Blue area is half the product of the 2 legs.

[drnandkishorbagul]

Simply can use Geometric Mean theorem.

[kaliprasadguru]

Sir, no need to find the length of BD to calculate the area of the triangle BDC .

[marcelomatos]

The two right triangles are similar. The side BC=(√︎80)/2

[sonamchoi]

ABD~DBC
2=BA/AD=DB/BC
피타고라스 정리 DB=4Root5
BC=8Root5

[marioalb]

Hypotenuse of white right triangle, is a leg of blue right triangle

H² = 4² + 8²
H= 8, 944 cm

B/H = 4/8
B = 4, 472 cm

Area = ½ B. H
Area = 20 cm² ( Solved √ )

[honestadministrator]

Height of this trapezium is AD
∆ BAD & ∆DBC be compared
angle BAD =angle DBC are both 90°
angle ABD =alternate interior angle
CAB
∆ BAD is similar to ∆DBC
DB / DC = AB / BD
DC = BD^2 /AB = (AB^2 + AD^2)/AB
= AB + AD^2 /AB
= 8 + 4^2 /8 unit = 10 unit
Hereby | ∆ BCD | = AD x CD /2
= 4 x 10 /2 unit
= 20 sqr unit

[leslieable]

In this video WE "ARE" or "HAVE BEEN" (not 'have') GIVEN...

We "are/have been" given... "means someone/body gave (to be exact 'have given') something to us"...

We "have given"... means "we 'gave something to' someone/body else"...

[harikatragadda]

CB=DB*AD/AB = ½√80
Blue Area = ½DB*CB = 20

[shashankkatiha]

No need to go this long. Calculate DB and put BC=x. In triangle BDC area= 1/2*h*DC = 1/2*DB*BC. DB is sqrt(80), h=4 and DC=sqrt(80+x^2). Solve for x and put the value in any of the two formulae to get the area.