Can you find area of the Green shaded square? | (Squares) |

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Can you find area of the Green shaded square? | (Squares) | #math #maths | #geometry
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[@GentlemanH]

I enjoyed that one and managed to solve it first time. Thanks. 🙂

[@PreMath]

I'm glad you liked it! 👍 Thanks for the feedback ❤️

[@allanflippin2453]

I found an intuitive approach, but I'm not sure the steps are valid: 1) Consider that BD cuts the larger square into two equal areas. 2) Isosceles triangles DEF and BFG each occupy half the area of the AEFG square. So each half of the larger square is 198 3) Isosceles triangles DHK and BML each occupy half the area of the green square 4) Add a point to complete the square formed by KLC. Then it becomes obvious that the KLC triangle occupies 1/4 the area of the green square 5) The green square = 198 / 2.25 or 88.

[@andrepiotrowski5668]

BC = b * sqrt(2) + b / sqrt(2) = b * (sqrt(2) + 1 / sqrt(2)) = 3 * b / sqrt(2) 4 * 99 = (BC)^2 = 9 * b² / 2 => (BC)² = 4 * 2 * 99 / 9 = 88

[@jamestalbott4499]

Thank you!

[@timmcguire2869]

The top left half of the square can easily be shown to be made of 4 congruent triangles. Two of those together equal 99, so 1/2 the area of the square is 2x99=198. The bottom right half of the square can easily be shown to be made of 9 congruent triangles, 4 of which make up the smaller green square. So 4/9 x 198 is 88. No Pythag needed :-)

[@RealQinnMalloryu4]

360°ABC/99=3.63ABC 1.2^3.1 (ABC ➖ 3ABC+2).

[@alexundre8745]

Bom dia Mestre Obrigado por mais uma aula Grato

[@MarcoPolo-xu9te]

I used other way. If you draw the line GF to the line DK, you will find out that DK = (4/3) a; therefore, b = DK sqrt ; b^2 = (8/9) a^2 = (8/9) . (3sqrt 11)^2 = 88. Fortunately, the same result, just nearly no geometry.

[@marcgriselhubert3915]

Be c the side length of the big ABCD square. The side length of the blue square is c/2. Then DB = sqrt(2).c, and the side lenght is the green square is DB/3 as DH = HM = MB (see right isosceles triangles DHK and DML), so the side length of gthe green square is (sqrt(2)/3).c The ratio (side length of the green square)/(side length of the blue square) is then (2.sqrt(2))/3), so the rartio of their areas is ((2.sqrt(2))/3)^2 = 8/9 As the area of the blue square is 99, we conclude that the area of the green square is 99.(8/9) = 88.

[@santiagoarosam430]

F es punto medio de DB y es el centro del cuadrado ---> ABCD =4*99. Si LMHK =a ---> CBD =a +a/2 +a/2 +a/4 =9a/4---> ABCD =18a/4 =4*99 ---> a=8*99/9 =88 cm². Gracias y saludos.

[@cyruschang1904]

white square (99 cm^2) is 1/4 of the big square area green square is 4/18 (= 2/9) of the big square area green square = (2/9)(4)(99 cm^2) = 88 cm^2

[@sorourhashemi3249]

thanks. So challenging.AG=9.95, .focus on ∆FGB, the side on front of angle 45° = √2/2 BF. So FB =14.11. DH=x, DB =2X-b. ∆LCK~BML, LC=y. And LK= √2y. So √2y/y=19.95-y/√2y. 3y^2-19.95y ====> y= 6.65. ===>b=6.65×√2=9.3765. b^=87.919

[@yalchingedikgedik8007]

Very nice and enjoyable Thanks Sir Thanks PreMath Good times ❤❤❤❤

[@lukeheatley4148]

i made it hard for myself by focusing on triangle KCL with side lengths (6.sqrt11 - sqrt2.b) and hypotenuse b. so b = sqrt2 . (6.sqrt11 - sqrt2.b) 3b = 6.sqrt2.sqrt11 b=2.sqrt2.sqrt11 b² = 88

[@himo3485]

99*2=198 Green Square area = 198*4/9 = 88cm²

[@tedn6855]

I did it a bit easier you can see all the angles are 45 therefore the diagnol is made up of 3 equal parts. So find the diagonol and divide by 3 then square. We know the top square has sides sqrt99 therefore sqrt 2 of that is that diagonol making it sqrt 198. Double that is sqrt 792. Now divide by 3 and get sqrt 88. Making the square 88.

[@marioalb9726]

A= ½d²=99cm² ---> d²=2A=198 2d = 3s s² = 4/9 d² = 88cm² ( Solved √ )

[@AmirgabYT2185]

S=88 cm²

[@rey-dq3nx]

AF=FC 3√ 22=3x/2 x=2√22 x²=88

[@phungpham1725]

1/ Connect CF. Because F is the midpoint of the diagonal DB -> CF perpendicular to DB-> C,F and A are collinear. -> the triangle CKL is a right isosceles -> CF=b+ b/2= 3/2 b=AF Because AF= asqrt2= sqrt 99x sqrt 2= sqrt 198 -> 3b/2 = sqrt 198 -> b=2/3 sqrt198 Area of the green square=4/9 x 198= 88 sq units😅😅😅

[@MrPaulc222]

This proved trickier than I first tbhought. I went wrong initially because I over complicated it by using irrationals instead of variables too early. But I got there in the end.

[@quigonkenny]

Let the side lengths of the large square ABCD, the 99cm² square AGFE, and the green square HMLK be a, b, and s respectively. Square AGFE: A = b² 99 = b² b = √99 = 3√11 As BD is the diagonal of large square ABCD and GF = FE = 3√11, then F must be the center point of the ABCD. Therefore, AGFE has half the side length of ABCD, so a = 2b = 2(3√11) = 6√11. Draw diagonal MK. As side HM of green square HMLK is collinear with diagonal BD, then as ∠CBD = ∠KMH = 45°, MK and BC are parallel. As angle pairs ∠LMK and ∠MLB and ∠LBM and ∠MKL are alternate interior angles and thus congruent, and as ∠MKL = ∠LMK = 45°, then all four angles equal 45°. As ML is common, ∆KLM and ∆BML are thus congruent isosceles right triangles and MK = BL. Draw diagonal LH let P be the point where MK and LH intersect, and as both lines are diagonals of HMLK, then P is the central point of HMLK and the midpoint of both MK and LH. As all internal angles of PLCK are 90° and KP and PL are congruent and adjacent, then PLCK is a square. BC = BL + LC 6√11 = (MK) + LC 6√11 = MK + (KP) 6√11 = MK + (MK/2) = 3MK/2 MK = (2/3)6√11 = 4√11 Green Triangle ∆KLM: KL² + LM² = MK² s² + s² = (4√11)² 2s² = 176 [ s² = 176/2 = 88 cm² ]

[@gelbkehlchen]

Solution: b = side of the black square = √(4*99) = 2*3*√11 = 6*√11, d = diagonal of the black square = √(b²+b²) = b*√2 = 6*√11*√2 = 6*√22. g = side of the green square = HM = ML = MB [because isosceles triangle MLB with angles 90°, 45° and 45°] = HK = DH [because isosceles triangle DKH with angles 90°, 45° and 45°] ⟹ g = 1/3*d = 1/3*6*√22 = 2*√22 area 0f the green square = g² = 4*22 = 88[cm²]

[@JSSTyger]

8(99)/9 = 88

[@ludosmets2018]

Draw perpendicular from F to DK, intersecting side b in point M. The small triangle FHM is isosceles (because angle FHM = 90° and angle HFM = 45°). FH = b/2 (because F is midpoint). DF= sqrt 198 (Pyth). Triangle DHK is also isoceles (because angle DHK = 90° and angle HDK = 45°). So, DH = HK or sqrt 198 - b/2 = b or 2 sqrt 198 - b = 2b or 2 sqrt 198 = 3b. So b= (2 sqrt 198) /3 and b^2= 88.

[@PrithwirajSen-nj6qq]

FH=b/2 ( ?) It is neither a given condition nor it has been proved .

[@ludosmets2018]

@@PrithwirajSen-nj6qq It is easy to prove that triangle EDF = triangle BFG and that triangle DHK = triangle BML and that consequently F is the midpoint of BD and of the side b.

[@PrithwirajSen-nj6qq]

@@ludosmets2018 Thanks a lot for the clear discussion u offered.

[@nandisaand5287]

A²=99 A=Sqrt(99)=3•Sqrt(11) Recognizing 🔺️ DHK is a special 45°-45°-90°🔺️, with sides B, B and B•Sqrt(2), DK=B•Sqrt(2). KC•Sqrt(2)=B KC=B/Sqrt(2) KC=B•Sqrt(2)/2 DC=2•A=DK+KC 2•3•Sqrt(11)=B•Sqrt(2)+B•Sqrt(2)/2 6•Sqrt(11)=3/2•B•Sqrt(2) B•Sqrt(2)=4•Sqrt(11) B=4•Sqrt(11)/Sqrt(2) B=4•Sqrt(22)/2 Area=B² =[4•Sqrt(22)/2]² =16•22/4 =4•22 =88 Put a box around it. [88cm²] How exciting.

[@zupitoxyt]

I jus forgot to focus on triangles besides the Green square 😢 well , good question

[@PrithwirajSen-nj6qq]

First I like to prove that the vertex of the 99 sq cms square(2a*2a=4a^2) is at the mid point(F) of the diagonal. 1)in triangle DEF Angle E= 90 degrees Any D =45 degrees Angle F= 45 degrees Hence DE= EF=2a Then DF=√(4a^2+ 4a^2)=a√8 2) we have proved above that DE = EF Hence E is the mid point of AD EF II AB Hence F is the mid point of BD. 3) in (1) we have proved DF= a√8 Hence BD = 2*DF=a 2√8 4) triangle DHK Ang H= 90 degrees Ang D = 45 degrees Hence ang K = 45 degrees Hence DH =HK = b--(1) 5) in triangle BML Ang M= 90 degrees Ang B =45 degrees Hence any L=45 degrees Hence BM =ML = b-- (2 ) 6) from (1) & (2) We may say BD =3b=a 2√8 b=a2√8/3 b^2=(a^2*4)*8/9=4a^ 2*8/9=99*8/9=88 sq cms Comment please

[@prossvay8744]

Green area=88cm^2

[@Birol731]

My way of solution ▶ The triangles ΔFBG and ΔDFE are congruent isosceles right triangles, and the triangles ΔMLB and ΔDKH are also congruent isosceles right triangles, so: [HK]= [KL]=[LM] =[MH]= x [MB]= [ML]= x [DH]=[HK]= x [HM]= x ⇒ [DB]= x+x+x [DB]= 3x [FG]= [GB] ⇒ [AB]= 2[AG] II) A(EFGA)= 99 cm² a= √99 a= 3√11 unit lengths ⇒ [AB]= 6√11 [DB]= √2*[AB] [DB]= √2*6√11 [DB]= 6√22 III) [DB]= 3x 3x= 6√22 x= 2√22 ⇒ x²= 2²*22 x²= 88 Agreen= 88 square units ✅

[@dustinhigh9035]

99

[@wasimahmad-t6c]

Saim 99

[@raffaeleguerrieri5482]

Su che base AEFG è un quadrato?

[@wasimahmad-t6c]

88 100% raite

[@LuisdeBritoCamacho]

STEEP-BY-STEP RESOLUTION PROPOSAL : 01) If Square [AEFG] = 99 sq cm ; Then : AE = AG = EF = FG = 3sqrt(11) cm 02) Square [ABCD] = (4 * 99) sq cm = 396 sq cm. Sides : AB = AD = CD = BC = sqrt(396) = 6sqrt(11) cm 03) Square Diagonal = BD = Side * sqrt(2) ; BD = 6sqrt(11) * sqrt(2) ; BD = 6sqr(22) cm 04) We can easy demonstrate by means of Algebra that : DH = HM = MB = X. I am not going to spare my time here in Routine Calculations. 05) BD / 3 = 2sqrt(22) cm 06) X = 2sqrt(22) cm 07) X^2 = (4 * 22) sq cm ; X^2 = 88 sq cm OUR BEST ANSWER : Green Square Area equal 88 Square Centimeters.

[@wackojacko3962]

I've got the Halloween blues and am trapped in a parallel universe far, far away trying too convince myself that ghosts don't exist. The only way back is too keep studying the concepts of Euclid and figure out how to work the useless Set Square and navigate back home. Wish me luck! 🙂

[@santiagoarosam430]

Que los vientos te sean favorables y que cuando llegues a casa, Argos te reconozca.

[@onix460]

a = sqrt(99); DE=sqrt(99); DF=sqrt(a²+DE²) = sqrt(198) DB=2sqrt(198); DH=HM=MB => HM= 2sqrt(198)/3 HM²(square HMLK) = 4*198/9 = 4*22 = 88cm² That's all