A Very Nice Geometry Problem | You should be able to solve this!
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1. Calculate the area of △DAB, which is 4•8/2 =16.
2. Drop an altitude of △DAB from A perpendicular to its hypotenuse, BD, intersecting at E.
3. Calculate the length of hypotenuse BD via Pythagoras, or √(8² + 4²). That length is 4√5.
4. Knowing the area of △DAB (16), and the length of its hypotenuse (4√5), work backwards to calculate the length of the constructed altitude (AE), which will be 16 = (4√5)(AE)/2, so that the length of AE =(8/5)√5.
5. The altitude AE forms two new triangles, △BEA and △AED. All three triangles, △BEA, △AED, and △BAD are similar, so their relative sides are in ratios to each other. This allows us to calculate a segment length we need (ED). So, BA:AD::AE:ED, or 8:4::(8/5)√5:ED. Therefore, the length of ED = (4/5)√5.
6. Knowing BD from #3 and now ED from #5, lets us calculate BE. BE = BD - ED. Or, 4√5 - (4/5)√5 = (16/5)√5.
7. Observe that △ABC is isosceles so that altitude AE bisects BC. Therefore, BE = EC, so EC = (16/5)√5.
8. Our last necessary segment is DC. But DC = EC - ED. Or, DC = (16/5)√5 - (4/5)√5. Therefore, DC = (12/5)√5.
9. To figure the shaded area (△ACD) we need a base (DC from #8) and an altitude (AE from #4). So the area will be (DC)•(AE)/2, or (12/5)√5 • (8/5)√5 / 2 = (96/5)/2 = 48/5 or 9⅗.
So, 9⅗ is the answer.
Now to watch the video to see if I got it right.
Cheers!
- s.west
AB=AC --> ABC=ACB. Now ABC = arctan (4/8) = 26.565 = ACB. Thus BAC = 180 - 2 * 26.565 = 126.87 = DAC + BAD = DAC + 90 --> DAC = 36.87. Then:
[ADC] = AD*AC/2 * sin (DAC) = 4*8/2 * 3/5 = 16*3/5 = 48/5 = 9.6 sq. units
Thats also my way
Let angle ABC=angle ACB=b, angle ADB=d ,then angle DAC=d-d and b+d=90
sin b=cos d=4/(8^2+4^2)=1/sqrt(5)
cos b=sin d=8/(8^2+4^2)=2/sqrt(5)
sin(d-b)=sin d*cos b-cos d*sin b=3/5
Area ACD=1/2*AD*AC*sin(d-b)=1/2*4*8*3/5=48/5
The solution is simpler without trigonometry. By Pythagoras in ABD, BD=4*sqrt(5). Since the area of ABD is 16, the length of its altitude AE is 8/sqrt(5), which gives us the altitude of the shaded triangle. By Pythagoras in ACE, CE=16/sqrt(5). By Pythagoras in AED, ED=4/sqrt(5). Then the base of the shaded triangle is DC=EC-ED=12/sqrt(5), so the shaded area is easy to calculate as 48/5.
Los triángulos AEB y AEC son congruentes y la razón de semejanza entre DEA y AEB es s=4/8=1/2--->s²=1/4---> Si AEB=4a---> DEA=a---> CDA=3a. ---> DAB=4a+a=5a=4*8/2=16---> a=16/5---> 3a=48/5.
Gracias y saludos.
Let AE⊥BC
ΔABD => BD²=AB²+AD²=64+16=80 => BD=4√5
(ABD)=(AB·AD)/2=8.4/2=16 (1) and
(ABD)=(BD⋅AE)/2=(4√5⋅AE)/2=2√5⋅AE (2)
(1),(2) => 2√5⋅AE=16 => *AE=8/√5*
Right triangles AED,ABE are similar with ratio of similarity 1:2.
So *ED=4/√5* and *BE=16/√5*
DC=EC - ED=16/√5 - 4/√5=12/√5
Now area (ADC)=(DC·AE)/2=……..=48/5 unit square
ACD=α...teorema dei seni..8/sin(180-arctg8/4)=4/sinα..sinα=sinarctg2/2=1/√5...Ared=(1/2)4*8sin(180-(180-arctg2+arcsin(1/√5))=16sin(arctg2-arcsin1/√5)=16(4/5-1/5)=48/5
*Solução:*
O ∆ABC é isósceles. Logo,
∠ABC=∠ACD=θ. Pelo teorema do ângulo externo no ∆ACD, temos:
∠ADC=90° + θ →∠CAD=90°-2θ
Seja S=[ADC]. Daí,
S=8×4×sin(90°-2θ)/2=16×sin(90°-2θ)
Ora, sin(90°-2θ)=cos2θ, por sua vez,
cos2θ=2(cos θ)² -1, como cosθ=2/√5 então
cos2θ=8/5 -1=3/5. Portanto,
S=16×3/5 → *S=48/5.*
I did not use trigonometry, just Pythagoras and formula for the Area. Nice problem!
We can, alternatively, compute the area of ΔACD by using CD as its base and AE as its height. At 7:00, we have computed length AE as 8/(√5) and have BC = BE + CE = 16/(√5) + 16/(√5) = 32/(√5). We can compute length BD by applying the Pythagorean theorem to ΔABD and find that BD = √(8² + 4²) = √(64 + 16) = √(80) = 4√5. So, CD = BC - BD = 32/(√5) - 4√5 = (32√5)/5 - (20√5)/5 = (12√5)/5. Area = (1/2)bh = (1/2)(CD)(AE) = (1/2)((12√5)/5)(8√5) = 48/5, as Math Booster also found.
*Solução 2 -3 - 4:*
Por Pitágoras no ∆ABD, temos BD=4√5. Além disso, seja
∠ABC=θ, assim, cosθ=2/√5.
Usando a lei dos cossenos no ∆ABC:
8²= 8²+BC² -2×8×BCcosθ
0= BC² -16×BC×2/√5 ÷(BC)
0= BC -32/√5 → BC=32/√5
*BC=32√5/5.*
*_Solução 2:_*
DC=BC-BD→ *DC=12√5/5.*
Altura h do ∆ABC é dada por:
h= AB×AD/BD=32/4√5=8/√5.
Seja S=[ADC]. então
S=h×DC/2 = (8/√5× 12√5/5)/2
*S=48/5.*
*Solução 3:*
[ABC]= (AB×BC× sinθ)/2
como sinθ=1/√5, então
[ABC]=(8×32√5/5×1/√5)/2
[ABC]=128/5, por outro lado,
[ABD]=8×4/2=16. Logo,
[ADC]=[ABC] -[ABD]
S=128/5 -16=(124-80)/5
*S=48/5.*
*Solução 4:*
As medidas dos lados do ∆ACD, são: 8, 4 e 12√5/5.
Usando a fórmula de Heron:
P= (8+4+ 12√5/5)/2= 6+6√5/5
Seja S=[ADC]. então
S=[p(p-8)(p-4)(p-12√5/5)]½
S=[(6+6√5/5)(2+6√5/5)(-2+6√5/5)(6-6√5/5)]½
S=[(6+6√5/5)(6-6√5/5)(6√5/5+2)(6√5/5-2)]½
Use: a²-b²=(a+b)(a-b)
S=[(36-36/5)(36/5 - 4)]½
S=[36(1-1/5)(36/5 - 4)]½
S=[36(4/5)16/5]½
S=[36×4×16/25]½
S= 6×2×4/5
*S=48/5*
Love it!!!!!!!!!!
Triangle ∆DAB:
DA² + AB² = BD²
4² + 8² = BD²
BD² = 16 + 64 = 80
BD = √80 = 4√5
Drop a perpendicular from A to E on BD. As ∠DAB = ∠AED = 90° and ∠EDA is common, ∆AED and ∆DAB are similar triangles.
Triangle ∆AED:
AE/DA = AB/BD
h/4 = 8/4√5 = 2/√5
h = 4(2/√5) = 8/√5
ED/DA = DA/BD
ED/4 = 4/4√5 = 1/√5
ED = 4/√5
As CA = AB, ∆CAB is an isosceles triangle, so perpendicular AE bisects ∆CAB and forms two congruent right triangles, ∆BEA and ∆AEC.
DC = EC - ED = BE - ED
DC = (BD-ED) - ED = BD - 2ED
DC = 4√5 - 2(4/√5)
DC = 20/√5 - 8/√5 = 12/√5
Triangle ∆ADC:
Aᴛ = bh/2 = DC(AE)/2
Aᴛ = (12/√5)(8/√5)/2 = 96/10 = 48/5 = 9.6 sq units
This is something that I have noticed: Do you think that you could have scaled it down to just one-fourth? The reason why I ask this is because this triangle is scaled up by 4. I kind of did the first solution in the back of my mind using the Pythagorean Theorem for the right-angled triangle one. I think that if all of the sides were scaled down and simplified, that the end result would easily be multiplied by 16 and have given you the same result. I have just tried it and I have to say that I am in the wrong. Looks like I kind of understand it and I need to practice that special way of simplifying at the very end!!! I hope that this shows that I should be able to solve this!!!
I must concede defeat. Looks like I cannot simplify this problem down to one-fourth. And given that I have understood AND am familiar with your way of simplifying, I will practice it again!!!
This is basically triangle similarity being used to justify an equilateral triangle contained so that a subtraction of an area can occur. Please correct me if I am wrong. I have just practiced that!!!
(8)^2=64H/A/Sino°.(4)^2=16A/0/Coso°.(8)^2=64O/A/ Tano° {64H/A/ASino°+16A/O/BCoso°+64O/A/CCoso°}=144H/AA/O O/A/ASino°BCoso°CTano° { 180°H/A A/O O/A144/ASino°BCoso°CCoso°}= 1.36H/A A/O O/A//ASino°BCoso°CTano°= 1^1.6^6 3^2^3^2 3^1^1^2 3^2 (H/A A/O O/A /ASino°BCoso°CCoso° ➖ 3 H/A A/O O/A /ASino°BCoso°CTano°+2) .