I solved like this: I have : A big triangle ABC where: AB = AC = 5 BC = 6 From point "A" I draw de line AP perpendicular to line BC. This will be the height of triangle ABC. I have now : BP = 3 and PC = 3 To find AP I apply Pythagoras on triangle APC. AC^2 = PC^2 + AP^2 5^2 = 3^2 + AP^2 AP^2 = 5^2 - 3^2 AP^2 = 25 - 9 = 16 AP = 4 ( h = height) I have a square and his side a will call of "a" DF=DG=EF=GF = a The big triagle ABC and the small triangle ADE are similar ( three angles equals) In ∆ADE the height will be AH = (4 - a) Then: h/Base => 4/6 = (4-a)/a 4a = 6(4 - a) 4a = 24 - 6a 4a + 6a = 24 10a = 24 a = 24/10 a = 2,4 Square area = a^2 = (2,4)^2 Square area = 5,76 units^2
The altitude on BC is 4 by the Pythagorean theorem. Be x the side length of the square, we have by similarity that (x/2) / (4 - x) = 3/4. So x = 12/5 and x² = 144/25. Easy one 😊
At 7:25, Math Booster has found that AQ = 4 and BQ = 3. Consider ΔABQ. The ratio of short side BQ to long side AQ is 3/4. Now consider similar ΔDBG. The short side BG is therefore 3/4 of the length of long side DG, which has length 2a, so BG has length 3a/2. However, Math Booster has found BG = 3 - a. So, 3a/2 = 3 - a, 5a/2 = 3, 5a = 6 and a = 6/5. The side of the square has length 2a, or (2)(6/5) = 12/5. The area of the square is (12/5)² = 144/25, as Math Booster also found. The area can also be written in decimal form as 5.76.
imo it will be easier to use 3-4-5 phytagorean number so I use 12x as side of the square GD is 12x, BG will be 9x and BD will be 15x DP is 6x, PA will be 8x and DA will be 10x BD + DA = 5 15x + 10x = 5 25x = 5 => x= 1/5 side is 12x = 12/5 area is side^2= (12/5)^2 = 144/25 unit square
we can solve this problem without additional constructions by using the ratio of a base and side leg of similar triangles: ADE and ABC (6/5) and name a side of the square "a", for example. Based on that, we express FC using Pythagorean theorem and solve the quadratic equation. we got a=2.4 and Area is 5.76.
The answer is (12/5)^2. I must admit that I was really surprised that I overlooked this lesson: that similar triangles by the right angle leg justify both sides of the square to be 2a. I hope that it is a sufficient takeaway. Because that justifies both 3-a as well as how the triangles subdivides the square. Personally I would just use the Pythagorean Theorem, indent the right angles and thetas, show where the a gets cancelled out, and set the square sides to 2a. I might be repeating the same procedure. I hope that this means that I got this. Just like the last two problems which I have tested myself and timed myself on!!!
Answer 5.76 Draw a perpendicular line inside triangle ABC from BC to A to form two congruent triangles, APB and APC, with sides 3, 4, and 5 Triangles BDG and CEF are similar to triangle ABC Label the square 'n,' hence DE = n, DG=n, and EF=n If DG =n (and BDG is similar to ABC), then BG= 3/4n, If EF=n ( and CEF is similar to ABC), then FC = 3/4n Notice that line BC (on ABC) = BG + DE + FC = 6 Hence, 3/4n + n + 3/4 n = 6 (recall n =DE see above) 3/4n + 4/4n + 3/4 n = 6 10/4 n = 6/1 n = 6/1 * 4/10 n= 24/10 n = 2.4 So the length of the square = 2.4 Hence its area = 2.4^2 or 5.76 Answer
∎DEFG → DE = EF = FG = DG = a; BC = BO + CO ↔ BO = CO = 3; AB = BC = 5 sin(BOA) = 1 → AO = 4 ; ABO = δ → tan(δ) = 2a/(6 - a) = 2(4 - a)/a → a^2 = 144/25 = area ∎DEFG
I solved like this:
I have :
A big triangle ABC where:
AB = AC = 5
BC = 6
From point "A" I draw de line AP perpendicular to line BC. This will be the height of triangle ABC.
I have now :
BP = 3 and PC = 3
To find AP I apply Pythagoras on triangle APC.
AC^2 = PC^2 + AP^2
5^2 = 3^2 + AP^2
AP^2 = 5^2 - 3^2
AP^2 = 25 - 9 = 16
AP = 4 ( h = height)
I have a square and his side a will call of "a"
DF=DG=EF=GF = a
The big triagle ABC and the small triangle ADE are similar ( three angles equals)
In ∆ADE the height will be AH = (4 - a)
Then:
h/Base => 4/6 = (4-a)/a
4a = 6(4 - a)
4a = 24 - 6a
4a + 6a = 24
10a = 24
a = 24/10
a = 2,4
Square area = a^2 = (2,4)^2
Square area = 5,76 units^2
The altitude on BC is 4 by the Pythagorean theorem. Be x the side length of the square, we have by similarity that (x/2) / (4 - x) = 3/4. So x = 12/5 and x² = 144/25. Easy one 😊
Keep up working hard. Thank you
Parabéns pelos vídeos! 🎉
At 7:25, Math Booster has found that AQ = 4 and BQ = 3. Consider ΔABQ. The ratio of short side BQ to long side AQ is 3/4. Now consider similar ΔDBG. The short side BG is therefore 3/4 of the length of long side DG, which has length 2a, so BG has length 3a/2. However, Math Booster has found BG = 3 - a. So, 3a/2 = 3 - a, 5a/2 = 3, 5a = 6 and a = 6/5. The side of the square has length 2a, or (2)(6/5) = 12/5. The area of the square is (12/5)² = 144/25, as Math Booster also found. The area can also be written in decimal form as 5.76.
imo it will be easier to use 3-4-5 phytagorean number so I use 12x as side of the square
GD is 12x, BG will be 9x and BD will be 15x
DP is 6x, PA will be 8x and DA will be 10x
BD + DA = 5
15x + 10x = 5
25x = 5 => x= 1/5
side is 12x = 12/5
area is side^2= (12/5)^2 = 144/25 unit square
I called side of square x. AP = 4-x. DP = .5x and APD is similar to AQB, so (4-x)/.5x = 4/3 giving 12 - 3x = 2x, so x = 12/5.
we can solve this problem without additional constructions by using the ratio of a base and side leg of similar triangles: ADE and ABC (6/5) and name a side of the square "a", for example. Based on that, we express FC using Pythagorean theorem and solve the quadratic equation. we got a=2.4 and Area is 5.76.
The answer is (12/5)^2. I must admit that I was really surprised that I overlooked this lesson: that similar triangles by the right angle leg justify both sides of the square to be 2a. I hope that it is a sufficient takeaway. Because that justifies both 3-a as well as how the triangles subdivides the square. Personally I would just use the Pythagorean Theorem, indent the right angles and thetas, show where the a gets cancelled out, and set the square sides to 2a. I might be repeating the same procedure. I hope that this means that I got this. Just like the last two problems which I have tested myself and timed myself on!!!
Answer 5.76
Draw a perpendicular line inside triangle ABC from BC to A to form two congruent triangles, APB and APC, with sides 3, 4, and 5
Triangles BDG and CEF are similar to triangle ABC
Label the square 'n,' hence DE = n, DG=n, and EF=n
If DG =n (and BDG is similar to ABC), then BG= 3/4n,
If EF=n ( and CEF is similar to ABC), then FC = 3/4n
Notice that line BC (on ABC) = BG + DE + FC = 6
Hence, 3/4n + n + 3/4 n = 6 (recall n =DE see above)
3/4n + 4/4n + 3/4 n = 6
10/4 n = 6/1
n = 6/1 * 4/10
n= 24/10
n = 2.4 So the length of the square = 2.4
Hence its area = 2.4^2 or 5.76 Answer
let the side of the square =2x, makes it simpler. similar triangles 4 mins max for solution
6/2=3 √[5²-3²]=4
BG=FC=3x GF=DE=DG=EF=4x 3x+4x+3x=6 10x=6 x=3/5
area of square DEFG = 4x*4x = 12/5 * 12/5 = 144/25
4 : 3 = k : 1/2 x
k = 2x/3
x = 4 - 2x/3
x = 12/5
x^2'= 144/25= 5.76
Love 🎉❤it
∎DEFG → DE = EF = FG = DG = a; BC = BO + CO ↔ BO = CO = 3; AB = BC = 5
sin(BOA) = 1 → AO = 4 ; ABO = δ → tan(δ) = 2a/(6 - a) = 2(4 - a)/a → a^2 = 144/25 = area ∎DEFG
DE=2a ; M es punto medio de GF y N lo es de DE→ (4-2a)/a=2a/(3-a)→ a=6/5→ Área DEFG =(2a)²=2*6/5)²=144/25 ud².
Gracias y un saludo cordial.
Aclaración: Si BA=5 y BM=6/2=3 ---> MA=4. --->. AN/ND=DG/GB---> (4-2a)/a=(2a)/(3-a)
ACB=α . Briggs,.cos(α/2)=√(8*3/5*6)=√(4/5)..cosα=2*4/5-1=3/5...5sinα:3=l:(3-l/2)...l=12/5
3:4=a:(4-2a)
1/6+1/4=1/X
X=12/5
Area=144/25
I got this one!
5.76
Area =5.76 square units
Area =3^2
6
12
Essa eu acertei !
it is too long. There is simple way.
4-x:4=x/2:3
12-3x=2x
12=5x
x=12/5
area=144/25