Can you find area of the Blue shaded Quadrilateral?

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Can you find area of the Blue shaded Quadrilateral? | Square and right triangles |

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Жыл бұрын

Learn how to find the area of the Blue shaded Quadrilateral. Square has an area 20. Important Geometry and algebra skills are also explained; area of the Triangle formula; area of the Square formula; Similar triangles; Pythagorean Theorem. Step-by-step tutorial by PreMath.com
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Пікірлер: 48

@bigm383 Жыл бұрын

Thanks for another great problem, Professor!❤🥂👍🍺

@PreMath Жыл бұрын

You are very welcome! Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

@colinedwards-qd5jn Жыл бұрын

😊Aaaàaa

@AllmondISP Жыл бұрын

There were really no need to calculate FE or using the Pythagorean theorem on triangle FCD. The minute we know FCD and BCE are a similar triangles you can find DF and you're basically done.

@davidellis1929 Жыл бұрын

Here's a cool shortcut once we have used the Pythagorean Theorem to determine EC=2. Triangles CDF and BCE are similar, since their larger acute angles are both complementary to angle DCF. We know CE is half of BE, so DF is half of DC, or sqrt(5). This gives us the legs of both right triangles, which yields their areas as 5 and 4.

@cyruschang1904 11 ай бұрын

DC = BC = ✓20 ✓[(✓20)^2 - 4^2] = ✓4 = 2 area of the triangle on the right = 4 x 2v÷ 2 = 4 area of the triangle on top = 4 x [✓20) ÷ (4)]^2 = 4 x 20 ÷ 16 = 5 Area sought = area of the square - sum of the area of the two triangles = 20 - 4 - 5 = 11

@danieldennis9831 Жыл бұрын

⏢ABEF=11 square units BTW, the characters I use below are from a list I compiled so I could do things like this (I love math videos and doing the work). The characters are (can be copied from my comment) ≈≡√|∠△□⏢±°¹²³⌠±≥≤µ!⌠¯•⋅÷¼⅓½⅔¾αβπδΩ Okay... I will show my work (Math teachers love seeing the work) A□ABCD=20 20=s² |EB=4 |EC=x |BC=s (side of □ABCD) △ BCE has height side of 4. x²+16=s² x²+16=20 x²=4 x=2 this is 1/2 |EB (opposite side of right triangle) A△BCE=4(2)(½)=4 square units s=√20=√4√5=2√5 ∠BCE is ∠α ∠CBE is ∠β With ∠BEC a right angle, ∠α+∠β=90. ∠FCD≡∠CBE ∠CFD≡∠BCE tan∠α=4/2=2 tan∠α=2=2√5/(½•2√5)=2√5/√5 |FD=√5 A△CDF=(½)2√5•√5=5 A□ABCD=A⏢ABEF+A△CDF+A△BCE=20 ⏢ABEF+4+5=20 ⏢ABEF=11 square units

@PreMath Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@JLvatron Жыл бұрын

Very good! I got it!

@TurquoizeGoldscraper Жыл бұрын

When I got to the complementary triangles step, I used it to determine DF directly: CE/BE = DF/CD 2 / 4 = DF / 20^0.5 1/2 * 20^0.5 = DF 5^0.5 = DF

@unknownidentity2846 Жыл бұрын

After calculating the length CE, you can get z=DF directly from the similarity of BCE and CDF: CE / BE = DF / CD 2 / 4 = DF / 2√5 ⇒ DF = √5 Best regards from Germany

@PreMath Жыл бұрын

Great! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@theoyanto Жыл бұрын

Yeah, nice problem thanks 👍🏻

@PreMath Жыл бұрын

You are very welcome! Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

@KAvi_YA666 Жыл бұрын

Thanks for video.Good luck sir!!!!!!!!!

@batavuskoga Жыл бұрын

Nice exercise.

@misterenter-iz7rz Жыл бұрын

It seems to be quite cumblesome 😮. EC^2=20-16=4, so EC=2. As similar triangles, DE=root 20/2, therefore the answer is 20-4x2/2-root 20xroot 20/4=20-4-5=11.😊

@KenFullman Жыл бұрын

Exactly what I was thinking. There was no need to calculate EF because we could just get the full length of CF directly by using the similar triangles method. It's nice to get one up on him now and again eh? 😁

@PreMath Жыл бұрын

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

@marioalb9726 Жыл бұрын

1) Square: A₁ = 20 cm² Side = √20 cm Cos α = 4/√20 α = 26,565° tan α = x / √20 x = 2,236 cm 2) Area upper right triangle A = ½ b.h = ½ √20 . 2,236 A = 5 cm² 3) Area lower right triangle: b = 4 h = x cos α = 2,236 cos 26,565° h = 2 cm Area = ½ b.h = ½ 4.2 Area = 4 cm² 4) Blue shaded area: Area = A1 - A2 - A3 Area = 20 - 5 - 4 Area = 11 cm² ( Solved √ )

@murdock5537 10 ай бұрын

Nice! ∆BCD = 2 - 4 - 2√5 → area ∆BCD = 4 ∆CDF = √5 - 2√5 - 5 → area ∆CDF = 5 → blue area = 20 - 9 = 11 sin⁡(φ) = √5/5 = CE/BC = DF/CF

@rolandhutapea6390 Жыл бұрын

The area of the blue region ABEF can also be calculated by drawing a line from B to F, to form right ∆ BAF and ∆ BEF. The area of ABEF is the sum of the area of ∆ BAF (½ x 2√5 x √5 = 5 sq. units) and ∆ BEF (½ x 3 x 4 = 6 sq. units) = 11 sq. units. Tx.

@santiagoarosam430 Жыл бұрын

Lado del cuadrado =√20=2√5 → EC²=(2√5)²-4² =4 → EC=√4=2 → Los triángulos rectángulos CEB y FDC son semejantes y los catetos de cada triángulo están en relación 2/4=1/2 → DF/DC=DF/2√5=1/2 → DF=√5 →→ Área azul =20 -(√5*2√5 /2) - (2*4/2) =20 -5 -4 =11 Gracias y un saludo cordial.

@amitsinghbhadoriya6318 Жыл бұрын

Thanks 🙏

@PreMath Жыл бұрын

You’re welcome 😊 Thank you! Cheers! 😀 You are awesome. Keep it up 👍

@svmartins1 Жыл бұрын

I did solve it 😀. Kind regards from Portugal!

@PreMath Жыл бұрын

Super! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@AnonimityAssured Жыл бұрын

I think I've cracked this one. Spoiler alert: s² = 20 = 4 ∙ 5 s = √20 = √4 ∙ √5 = 2√5. By Pythagoras, a² = s² − 4². a² = 20 − 16 = 4. a = √4 = 2, so triangle BEC has orthogonal sides in ratio 2:1. Area of triangle BEC = 2 ∙ 4 / 2 = 4. Triangle CDF must also have orthogonal sides in ratio 2:1, so line segment DF has length √5. Area of triangle CDF = √5 ∙ 2√5 / 2 = 10 / 2 = 5. Area of blue quadrilateral = 20 − 4 − 5 = 20 − 9 = 11 square units.

@PreMath Жыл бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@fco.antoniosanchezmunoz5399 Жыл бұрын

When used similarity of triangles, knowing that "z" was what we need, why did we focus on "y" instead, if at that step effort was the same? That meant an additional calculation

@gelbkehlchen Жыл бұрын

Solution: Square = 20 square units ⟹ square side = √20 ⟹ EC = √(20-4²) = 2 Triangle FCD is similar to triangle EBC (3 angles). ⟹ DF/DC = EC/EB ⟹ DF = EC/EB*DC = 2/4*√20 = √20/2 ⟹ area of blue region = area of the square - area of triangle FCD - area of triangle EBC = 20-1/2*√20/2*√20-1/2*2*4 = 20-5-4 = 11

@raya.pawley3563 Жыл бұрын

Thank you

@wackojacko3962 Жыл бұрын

Inciteful No! Unless you're trying too start a riot. Insightful YES, VERY! 🙂

@PreMath Жыл бұрын

😀 Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

@gelbkehlchen Жыл бұрын

It is easier to make a relation from leg to leg according to the 2 similar triangles.

@Copernicusfreud Жыл бұрын

Yay! I solved it.

@kennethstevenson976 11 ай бұрын

Got the answer by using triangle theory and proportion of similar triangles without looking.

@williamwingo4740 Жыл бұрын

Got it without peeking. Once you know triangles DFC and CEB are similar, you can get CF by the proportion 4/2 sqrt(5) = 2 sqrt(5)/x so 4x = (4)(5) = 20 and x = 5. DF and EC follow by Pythagoras. Otherwise it works out pretty much the same. 🤠

@xsilata Жыл бұрын

Why Y, straight Z

@FreestyleViewer Жыл бұрын

Solving a Problem somehow is not a right way of teaching the young enquiring minds. Firstly a problem should be solved by use of a Natural-Sequencing, and then only should we aspire for the more fun coming from inventing the Fast-Trick methods of the solution. Let’s take the problem at hand as an example Natural-Sequencing Method: Given the area is 20, so the square is fixed with side √20. Now let CF is drawn at some angle with CD, say θ. So, angle CBE is also θ. Hence, we can find θ easily, but we’ll do it later as per our requirement. Now the sum of the areas of two triangles CDF and CBE is (1/2)(√20)(√20tanθ) + (1/2)(4)(√20sinθ) From ∆CBE, cosθ = 4/√20. Therefore, sinθ = √4/√20 = 2/√20, and tanθ = (√4)/4 = 1/2. Putting these values we get the sum of the two Areas as 10/2 + 4 = 9. Therefore, the Green Area is 20-9 = 11 units. Fast-Trick Method: Comparing the Angles in the two Triangles, we see they are similar such that EC/DF = EB/DC = CB/FC. Putting the values of the known lengths we get 2/DF = 4/√20 = √20/FC Hence, we can get DF as (√20)/2. Now the sum of the Areas of the two triangles is (1/2)(4)(2) + (1/2)((√20)/2)(√20) = 4 + 5 = 9. Therefore the required Green Area is 11 units.