Alternatively, to find the area of the blue region, we can construct DH to divide ADGH into 2 triangles, ΔDGH and ΔDAH, and sum their areas. Length DG = side of square DEFG = √(725/4) = √((25)(29)/4) = (5/2)√(29), call that the base of ΔDGH. By corresponding sides of similar triangles, GH is 3/5 as long = (3/5)(5/2)√(29) = (3/2)√(29), call that the height. Area ΔDGH = (1/2)(5/2)(√(29))(3/2)(√(29)) = 435/8. For ΔDAH, length AD = length CD = 25/2, call that the base, and length AH = length AB - length BH = 25/2 - 3 = 19/2, call that the height, so area ΔDAH = (1/2)(25/2)(19/2) = 475/8. Add the triangle areas: 435/8 + 475/8 = 910/8 = 455/4. Deduct from area of square DEFG = 725/4 - 455/4 = 270/4 = 67.5 square units, as PreMath also found.
1/ Let a and b be the sides of the small and big squares respectively. The triangles HBG and GCD are similar so HB/BG=GC/CD ---> 3/(a-5)=5/a---> a=25/2 2/ The angles EDA and GDC are equal so the triangles are congruent. We have EA=5 and the area of the triangle EDA= 1/2 x5x12.5= 31.25 (1) 3/The points E, A, H, and B are collinear. EH=EA+AH= 5+ (12.5-3)= 14.5 Label HF= h Because the triangle EFH and DCG are similar too, so ---> h/EH= CG/DG=5/b---> h= (14.5 x5)/b Area of EFH= 1/2 hxb= 1/2 x b x(14.5x5)/b =(14.5x5)/2= 36.25 (2) Area of the green area= (1) +(2) = 67.5 sq units
Llamaremos K a la proyección ortogonal de F sobre AB. Los triángulos GCD, EAD y FKE son congruentes; CG=EA=FK=5 ; EK=5+AK=AB ; KB=5 ; KH=5-3=2. Los triángulos HKF y HBG son semejantes: KF/KH=GB/3 ; GB=15/2 : CB=DA=EK=(15/2)+5=25/2. Área verde =EDAHF =EAD+FKE+HKF =2*EAD+HKF =(5*25/2)+(2*5/2)=135/2 =67.5 Gracias y un saludo cordial.
At 0:10, Let S1=DC=CB=BA=AD Let S2=DG=GF=FE=ED Triangle DCG and Triangle GBH are similar since all the angles are equal. S1/5=GB/3, S1/5=(S1-5)/3 3S1=5S1-25 S1=25/2 Consider triangle DCG, S2=sqrt(S1^2+CG^2) Pythagoras S2=sqrt(625/4+100/4)=sqrt(725/4)=sqrt((25*29)/4) =(5*sqrt29)/2 IDGHAI=IABCDI-IDCGI-IGBHI=625/4-125/4-45/4=455/4 IDGFEI=(S2)^2=((5*sqrt29)/2)^2=(25*29)/4 Igreen area EFHADI=(S2)^2-IDGHAI=(25*29)/4-455/4= 725/4-455/4 Igreen area EFHADI=(725-445)/4 =270/4 = 67.5 square units as required. Thanks for the puzzle Professor.
Start by labeling the side length of square ABCD as x. Therefore, CD = x & BG = BC - CG = x - 5. Let α and β be the measures of complementary angles. Let m∠CDG = α. Then, because ∠C is a right angle by definition of squares, m∠CGD = β. So, because ∠DGF is similarly a right angle, m∠BGF = α. And since ∠B is also a right angle by definition of squares, m∠BHG = β. So, △DCG ~ △GBH by AA Similarity. A proportion will take place. CD/CG = BG/BH x/5 = (x - 5)/3 3x/5 = x - 5 3x = 5x - 25 2x = 25 x = 12.5 So, the side length of square ABCD is 12.5 u. CD = 12.5, BG = 7.5, & AH = 9.5. Apply the Pythagorean Theorem on △DCG to find the area of square DEFG. a² + b² = c² (12.5)² + 5² = (DG)² (DG)² = 156.25 + 25 = 181.25 A = s² = (DG)² = 181.25 So, the area of square DEFG is 181.25 u². Draw a segment AG to separate the overlapping region (quadrilateral ADGH) into two triangles and find its area. Quadrilateral ADGH Area = △ADG Area + △AGH Area A = 1/2 * b * h △ADG Area = 0.5 * 12.5 * 12.5 = 78.125 △AGH Area = 0.5 * 9.5 * 7.5 = 35.625 ADGH Area = 78.125 + 35.625 = 113.75 Green region area = Square DEFG Area - ADGH Area = 181.25 - 113.75 = 67.5 So, the area of the green region is 135/2 square units, or 67.5 square units.
Drop a perpendicular from F to J on AB. Now ⊿CDG ≅ ⊿ADE ≅ ⊿JEF so CG = AE = JF = 5. Let DC = x so EH = x + 5 − 3. Then the area of ⊿ADE = ½ 5 x and the area of ⊿EHF = ½ 5 (x + 2). So the area of the green shaded region is 5 (x + 1). Use your favorite method to find that x = 12.5 and the final answer will be 5 (12.5 + 1) = 5 (13.5) = 67.5 square units.
Let AB = BC = CD = DA = x. Then ⅗ x = x − 5 ⇒ x = 25/2 and the area of ⊿CDG ≅ ⊿ADE is ½ · 25/2 · 5 = 31.25 sq. u. DE = EF = FG = GD = √((25/2)² + 5²) = ½ 5√29. Then ⊿ADE ~ ⊿FEH ⇒ FH = √29 and the area of ⊿FEH = 36.25 sq. u. Since the green shaded region is ⊿ADE + ⊿FEH, its area is 31.25 + 36.25 = 67.5 square units.
Reading comments is enjoyable as they show so many paths to climb a mountain. Some of the paths are very very difficult reserved for expert climbers, definitely not me. I would like to share how I find my easy path for this problem. First I notice that DC is the key to solution as it gives solution to areas of large square, small square and the two triangles. Second I notice the similar right-angled triangles by complementary angles such that DC/BG = 5/3 by corresponding sides proportionality equation. Third I notice an equation linking DC and BG such that I can substitute BG by DC. Reflection on how to get a solution is more rewarding that getting a correct answer. Solution follows that path: 1. ∆DCG and ∆GBH are similar triangles. (AAA by complementary angles) Hence DC/GB = CG/BH = 5/3 (corresponding sides proportionality equation) 2. GB = BC - CG = DC - 5 (property of square BC = DC and CG given) 3. DC/GB = DC/(DC - 5) = 5/3 (corresponding sides proportionality equation from above) Hence DC = 25/2 4. Area of green = Area DEFG - Area ABCD + Area DCG + Area GBH = (DG2 - DC2) + (5 x DC)/2 + (3 x GB)/2 = GC2 + (5 x 25/2)/2 + [3 x (25/2 - 5)]/2 (Pythagoras theorem for ∆DCG) = 25 + 125/4 + 45/4 = 25 + 170/4 = 270/4 = 67.5
To solve this problem, I used the Geometrical Proof of Pythagorean Theorem. I create a Square with Side Length = AB + 5 li un; in wich the hypotenuses are DG; DE; EF and FG. Let's call the Side Length of the Small Square [ABCD] "X". And angle HGB and angle GDC "alpha"; as they are congruent angles. So: If : tan (alpha) = 5 / X If : tan (alpha) = 3 / (X -5) Then : 5 / X = 3 / (X - 5) 5 * (X - 5) = 3X ; 5X - 25 = 3X ; 5X - 3X = 25 ; 2X = 25 ; X = 25/ 2 ; X = 12,5 linear units So the Area of Square [ABCD] = 12,5^2 = 156,25 sq un Now, I know all 3 Squares Areas : 156,25 sq un ; 5^2 + 12,5^2 = DG^2 = 181,25 sq un ; (12,5 + 5)^2 = 17,5^2 = 306.25 sq un. Now it's time to find the Green Area! We have 2 triangles : [ADE] and [EFH]; and Green Area is equal to Area [ADE] + Area [EFH] Area [ADE] = (5 * 12,5) / 2 = 62,25 / 2 = 31,25 sq un Area [EFH] = ? First we have to calculate Length FH!! EH^2 - EF^2 = FH^2 EH = X - 3 + 5 = X + 2 = 12,5 + 2 = 14,5 li un EF = sqrt(181,25) ~ 13,5 li un 14,5^2 - 13,5^2 = 210,25 - 181,25 = 29 FH = sqrt(29) Area [EFH] = (FH * FE) / 2 = (sqrt(29) * sqrt(181,25)) / 2 = 72,5 / 2 = 36,25 Answer: The Area of the Green Shaded Region is equal to 36,25 sq un + 31,25 sq un = 67,5 square units.
Oh Gawd, that looks complicated. Unsure which way to go so will assimilate info. Start with the sides of the smaller square which I will call x.: GCD and HBG are similar, so (x-5)/3 = x/5. (5x-25) = 3x. 2x = 25, so x=25/2 or 12.5, whichever is easier to work with. Call the larger square's sides y. (25/2)^2 + 5^2 = y^2. 625/4 + 25 = y^2. 725/4 = y^2. Two squares of areas 625/4 and 725/4 square units. Area ADE is (5*12.5)/2 = 31.25 because AE=5 (due to CG (given)). The relevant sides of HFE should be 5 and (5+12.5-3), so 5 and 14.5. 72.5/2=36.25. 31.25 and 36.25=67.5 sq units. Your way was cleaner and I caused myself to make an extra calculation or two. Thank you.
Just superb. Masterful solution presented masterfully.
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Alternatively, to find the area of the blue region, we can construct DH to divide ADGH into 2 triangles, ΔDGH and ΔDAH, and sum their areas. Length DG = side of square DEFG = √(725/4) = √((25)(29)/4) = (5/2)√(29), call that the base of ΔDGH. By corresponding sides of similar triangles, GH is 3/5 as long = (3/5)(5/2)√(29) = (3/2)√(29), call that the height. Area ΔDGH = (1/2)(5/2)(√(29))(3/2)(√(29)) = 435/8. For ΔDAH, length AD = length CD = 25/2, call that the base, and length AH = length AB - length BH = 25/2 - 3 = 19/2, call that the height, so area ΔDAH = (1/2)(25/2)(19/2) = 475/8. Add the triangle areas: 435/8 + 475/8 = 910/8 = 455/4. Deduct from area of square DEFG = 725/4 - 455/4 = 270/4 = 67.5 square units, as PreMath also found.
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Great teacher! I wish all those who visit your page shows their gratitude by thumbs up! Thank you for making Math fun!
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You are a rare gem teacher❤. We do not many people explaining tricks in free❤
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1/ Let a and b be the sides of the small and big squares respectively.
The triangles HBG and GCD are similar so HB/BG=GC/CD ---> 3/(a-5)=5/a---> a=25/2
2/ The angles EDA and GDC are equal so the triangles are congruent.
We have EA=5 and the area of the triangle EDA= 1/2 x5x12.5= 31.25 (1)
3/The points E, A, H, and B are collinear. EH=EA+AH= 5+ (12.5-3)= 14.5
Label HF= h
Because the triangle EFH and DCG are similar too, so ---> h/EH= CG/DG=5/b---> h= (14.5 x5)/b
Area of EFH= 1/2 hxb= 1/2 x b x(14.5x5)/b =(14.5x5)/2= 36.25 (2)
Area of the green area= (1) +(2) = 67.5 sq units
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Ohh! What a nice trick it is ❤ it
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Greatly done!
For the similar triangles, I used 5 / DC = 3 / (DC - 5).
Then 5 ( DC - 5 ) = 3 DC.
5DC - 25 = 3 DC.
2 DC = 25.
DC = 12.5.
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I also solved it this way. For some reason, introducing an unrelated constant such as K adds confusion to something which seems straightforward.
Just brilliant!
Thanks Sir
Very simplified explain
Excellent Sir
With respects
Brilliant!
Very good professor!!
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Respect button for premath❤
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Sir Can' t we use GHB as isoceles right angle triangle and solve further
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Veri nice sar❤❤❤
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Love it!!!!!!!!!!!!!!!
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to find CD , BG -->
Let angle CDG = t
--> angle BGH = t
Let CD=X --> BG=X-5 =Y
tan(t) = 5/X = 3/(X-5)
5(X-5) = 3X
2X=25 --> X= 25/2
Y= 25/2 - 5 = 15/2
.
Llamaremos K a la proyección ortogonal de F sobre AB.
Los triángulos GCD, EAD y FKE son congruentes; CG=EA=FK=5 ; EK=5+AK=AB ; KB=5 ; KH=5-3=2.
Los triángulos HKF y HBG son semejantes: KF/KH=GB/3 ; GB=15/2 : CB=DA=EK=(15/2)+5=25/2.
Área verde =EDAHF =EAD+FKE+HKF =2*EAD+HKF =(5*25/2)+(2*5/2)=135/2 =67.5
Gracias y un saludo cordial.
∎ABCD → AB = a = AH + BH = (a - 3) + 3 = BG + CG = (a - 5) + 5 = CD = AD = a
∎DEFG → DE = EF = FG = DG = k; tan(δ) = 5/a = 3/(a - 3) → a = 25/2 → k = 5√29/2 →
green shaded area = k^2 - 455/4 = 135/2; sin(δ) = 2√29/29
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@ 4:15 .....absolutely beautiful! 🙂
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At 0:10,
Let S1=DC=CB=BA=AD
Let S2=DG=GF=FE=ED
Triangle DCG and Triangle GBH are similar since all the angles are equal.
S1/5=GB/3,
S1/5=(S1-5)/3
3S1=5S1-25
S1=25/2
Consider triangle DCG,
S2=sqrt(S1^2+CG^2) Pythagoras
S2=sqrt(625/4+100/4)=sqrt(725/4)=sqrt((25*29)/4) =(5*sqrt29)/2
IDGHAI=IABCDI-IDCGI-IGBHI=625/4-125/4-45/4=455/4
IDGFEI=(S2)^2=((5*sqrt29)/2)^2=(25*29)/4
Igreen area EFHADI=(S2)^2-IDGHAI=(25*29)/4-455/4= 725/4-455/4
Igreen area EFHADI=(725-445)/4 =270/4 = 67.5 square units as required.
Thanks for the puzzle Professor.
l:5=(l-5):3..l=12,5..Ag=12,5^2+5^2-(12,5^-5*12,5/2-7,5*3/2)=12,5^2+5^2-(12,5^2-85/2)=25+42,5=67,5
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Start by labeling the side length of square ABCD as x. Therefore, CD = x & BG = BC - CG = x - 5.
Let α and β be the measures of complementary angles.
Let m∠CDG = α. Then, because ∠C is a right angle by definition of squares, m∠CGD = β.
So, because ∠DGF is similarly a right angle, m∠BGF = α.
And since ∠B is also a right angle by definition of squares, m∠BHG = β.
So, △DCG ~ △GBH by AA Similarity. A proportion will take place.
CD/CG = BG/BH
x/5 = (x - 5)/3
3x/5 = x - 5
3x = 5x - 25
2x = 25
x = 12.5
So, the side length of square ABCD is 12.5 u. CD = 12.5, BG = 7.5, & AH = 9.5.
Apply the Pythagorean Theorem on △DCG to find the area of square DEFG.
a² + b² = c²
(12.5)² + 5² = (DG)²
(DG)² = 156.25 + 25
= 181.25
A = s²
= (DG)²
= 181.25
So, the area of square DEFG is 181.25 u².
Draw a segment AG to separate the overlapping region (quadrilateral ADGH) into two triangles and find its area.
Quadrilateral ADGH Area = △ADG Area + △AGH Area
A = 1/2 * b * h
△ADG Area = 0.5 * 12.5 * 12.5
= 78.125
△AGH Area = 0.5 * 9.5 * 7.5
= 35.625
ADGH Area = 78.125 + 35.625
= 113.75
Green region area = Square DEFG Area - ADGH Area
= 181.25 - 113.75
= 67.5
So, the area of the green region is 135/2 square units, or 67.5 square units.
Excellent!
Thanks ❤️
Drop a perpendicular from F to J on AB. Now ⊿CDG ≅ ⊿ADE ≅ ⊿JEF so CG = AE = JF = 5. Let DC = x so EH = x + 5 − 3.
Then the area of ⊿ADE = ½ 5 x and the area of ⊿EHF = ½ 5 (x + 2). So the area of the green shaded region is 5 (x + 1).
Use your favorite method to find that x = 12.5 and the final answer will be 5 (12.5 + 1) = 5 (13.5) = 67.5 square units.
Extremely beautiful ❤😊😊😊
Thank you! Cheers!❤️🌹
شكرا لكم على المجهودات
يمكن وضع z=Cd
tanCDG =5/z=3/(z-5)
z=25/2
......
S=135/2
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Area of green region=(5√29/2)^2-113.75=67,5 square units. ❤❤❤ Thanks sir.
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Let AB = BC = CD = DA = x. Then ⅗ x = x − 5 ⇒ x = 25/2 and the area of ⊿CDG ≅ ⊿ADE is ½ · 25/2 · 5 = 31.25 sq. u.
DE = EF = FG = GD = √((25/2)² + 5²) = ½ 5√29. Then ⊿ADE ~ ⊿FEH ⇒ FH = √29 and the area of ⊿FEH = 36.25 sq. u.
Since the green shaded region is ⊿ADE + ⊿FEH, its area is 31.25 + 36.25 = 67.5 square units.
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I did it the same way.
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Reading comments is enjoyable as they show so many paths to climb a mountain. Some of the paths are very very difficult reserved for expert climbers, definitely not me. I would like to share how I find my easy path for this problem.
First I notice that DC is the key to solution as it gives solution to areas of large square, small square and the two triangles. Second I notice the similar right-angled triangles by complementary angles such that DC/BG = 5/3 by corresponding sides proportionality equation. Third I notice an equation linking DC and BG such that I can substitute BG by DC. Reflection on how to get a solution is more rewarding that getting a correct answer.
Solution follows that path:
1. ∆DCG and ∆GBH are similar triangles. (AAA by complementary angles)
Hence DC/GB = CG/BH = 5/3 (corresponding sides proportionality equation)
2. GB = BC - CG
= DC - 5 (property of square BC = DC and CG given)
3. DC/GB = DC/(DC - 5) = 5/3 (corresponding sides proportionality equation from above)
Hence DC = 25/2
4. Area of green = Area DEFG - Area ABCD + Area DCG + Area GBH
= (DG2 - DC2) + (5 x DC)/2 + (3 x GB)/2
= GC2 + (5 x 25/2)/2 + [3 x (25/2 - 5)]/2 (Pythagoras theorem for ∆DCG)
= 25 + 125/4 + 45/4
= 25 + 170/4
= 270/4
= 67.5
DC=5a GB=3a
5a=3a+5 2a=5 a=5/2
DG^2=(25/2)^2+5^2=625/4+100/4=725/4
DG=5√29/2 GH=3√29/2
DAH=25/2＊19/2＊1/2=475/8
DGH=5√29/2＊3√29/2＊1/2=435/8
DEFG=5√29/2＊5√29/2=725/4
area of Green shaded region :
725/4 - 475/8 - 435/8 = 1450/8 - 910/8
= 540/8 = 135/2 = 67.5
Thanks ❤️
To solve this problem, I used the Geometrical Proof of Pythagorean Theorem.
I create a Square with Side Length = AB + 5 li un; in wich the hypotenuses are DG; DE; EF and FG.
Let's call the Side Length of the Small Square [ABCD] "X". And angle HGB and angle GDC "alpha"; as they are congruent angles.
So:
If : tan (alpha) = 5 / X
If : tan (alpha) = 3 / (X -5)
Then : 5 / X = 3 / (X - 5)
5 * (X - 5) = 3X ; 5X - 25 = 3X ; 5X - 3X = 25 ; 2X = 25 ; X = 25/ 2 ; X = 12,5 linear units
So the Area of Square [ABCD] = 12,5^2 = 156,25 sq un
Now, I know all 3 Squares Areas : 156,25 sq un ; 5^2 + 12,5^2 = DG^2 = 181,25 sq un ; (12,5 + 5)^2 = 17,5^2 = 306.25 sq un.
Now it's time to find the Green Area!
We have 2 triangles : [ADE] and [EFH]; and Green Area is equal to Area [ADE] + Area [EFH]
Area [ADE] = (5 * 12,5) / 2 = 62,25 / 2 = 31,25 sq un
Area [EFH] = ?
First we have to calculate Length FH!!
EH^2 - EF^2 = FH^2
EH = X - 3 + 5 = X + 2 = 12,5 + 2 = 14,5 li un
EF = sqrt(181,25) ~ 13,5 li un
14,5^2 - 13,5^2 = 210,25 - 181,25 = 29
FH = sqrt(29)
Area [EFH] = (FH * FE) / 2 = (sqrt(29) * sqrt(181,25)) / 2 = 72,5 / 2 = 36,25
Answer:
The Area of the Green Shaded Region is equal to 36,25 sq un + 31,25 sq un = 67,5 square units.
Oh Gawd, that looks complicated.
Unsure which way to go so will assimilate info.
Start with the sides of the smaller square which I will call x.:
GCD and HBG are similar, so (x-5)/3 = x/5. (5x-25) = 3x. 2x = 25, so x=25/2 or 12.5, whichever is easier to work with.
Call the larger square's sides y. (25/2)^2 + 5^2 = y^2. 625/4 + 25 = y^2. 725/4 = y^2.
Two squares of areas 625/4 and 725/4 square units.
Area ADE is (5*12.5)/2 = 31.25 because AE=5 (due to CG (given)).
The relevant sides of HFE should be 5 and (5+12.5-3), so 5 and 14.5. 72.5/2=36.25.
31.25 and 36.25=67.5 sq units.
Your way was cleaner and I caused myself to make an extra calculation or two.
Thank you.
Thanks ❤️
Ill say A = 67.5