Can you find area of the Pink Quadrilateral? | (Nice Geometry problem) |

Learn how to find the area of the Pink Quadrilateral. Important Geometry and Algebra skills are also explained: area of a triangle formula; isosceles triangles; Pythagorean Theorem. Step-by-step tutorial by PreMath.com
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Can you find area of the Pink Quadrilateral? | (Nice Geometry problem) | #math #maths
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[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks ❤️

[@jimlocke9320]

At 3:45, BD has been constructed, dividing quadrilateral ABCD into 2 triangles, ΔABD and ΔBCD. We find that length BD is 13. First, let's find the area of ΔABD. In the video, a method is shown. An alternative method of finding the area of ΔABD is to construct 3 triangles congruent to it and assemble the 4 triangles into a square with sides of length 13. The area of the square is (13)² = 169. The area of each triangle is 1/4 of that square, or 169/4. Now to find the area of ΔBCD. Drop a perpendicular from D to BC, dividing ΔBCD into 2 15°-75°-90° triangles. The 15°-75°-90° triangle appears frequently enough in geometry problems that I feel all of us should know its properties. One of the properties is that the area is equal to the hypotenuse squared, divided by 8. So each triangle has area (13)²/8 = 169/8 and the combined area is twice that, or 169/4. The total area of the quadrilateral is 169/4 + 169/4 = 169/2, as PreMath also found. Of interest, the area of the quadrilateral is equal to the area of a square with sides equal to length AB or AD. If we construct a square ABED, working back from the solution, we know that the enclosed white triangular area must equal the shaded triangular area outside ABED. Once we find these 2 triangles are equal in area, as well as find the length of the square's diagonal, we have an alternative method of finding the area. However, this solution method is certainly more challenging than the actual method used!

[@PreMath]

Thanks for the feedback ❤️

[@jimlocke9320]

Solving it a straightforward but hard way: Construct a line parallel to AD and through point C. Drop a perpendicular to it from D and label the intersection as point E. Drop a perpendicular to it from B and label the intersection as point F. ADEF is a rectangle. The shaded area region is equal to the area of ADEF less the areas of ΔCDE and ΔBCF. We note that ΔCDE is a 15°-75°-90° right triangle with known hypotenuse. Applying the known ratio of sides of a 15°-75°-90° right triangle: (√3 - 1):(√3 + 1):(2√2), DE = (13)(√3 + 1)/(2√2) and CE = (13)(√3 - 1)/(2√2). Let AB = AD = x. From properties of a rectangle, EF = AD = x. EF = CE + CF, so CF = EF - CE = x - (13)(√3 - 1)/(2√2). ΔBCF is a 30°-60°-90° right triangle. The long side is √3 times as long as the short side. So, CF = (√3)BF and BF = CF/√3. We note that AF = AB + BF and AF = DE, so AB + BF = DE. x + (x - (13)(√3 - 1)/(2√2))/√3 = (13)(√3 + 1)/(2√2). I solved this equation for x and got x = 13/(√2) after a lot of mathematical manipulation. Then, I went back and got BF = (13)(√2)(√3 - 1)/4. The area of the rectangle is (AD)(AF) = (AD)(AD + BF) = (13/√2)((13/√2 + (13)(√2)(√3 - 1)/4). The area of a 15°-75°-90° right triangle, ΔCDE in this case, is hypotenuse squared divided by 8, or (13)²/8 The area of a 30°-60°-90° right triangle is (1/2)(short side)(long side), but the long side is √3 times as long as the short side, so for ΔBCF, (1/2)((13)(√2)(√3 - 1)/4)²(√3). If you write an equation for the area of the rectangle less the areas of the two triangles and do the mathematical manipulation to solve the equation, you should find that the shaded area is 169/2 = 84.5 square units, as I did.

[@CloudBushyMath]

Superb!👍

[@PreMath]

Thank you! Cheers!🌹

[@dhronegajjar1420]

Just remember if any side of quadrilateral is = it's diagonal then just do( side²/2) =13²/2

[@ChuzzleFriends]

Use the Corollary to the Polygon Interior Angle Sum Theorem (or the theorem itself). m∠A + m∠ABC + m∠C + m∠ADC = 360° 90° + 120° + m∠C + 75° = 360° m∠C + 285° = 360° m∠C = 75° Connect points B and D. This diagonal forms isosceles right △BAD. So m∠ABD = m∠ADB = 45°. Use the Angle Addition Postulate. m∠ABD + m∠CBD = m∠ABC 45° + m∠CBD = 120° m∠CBD = 75° m∠ADB + m∠BDC = m∠ADC 45° + m∠BDC = 75° m∠BDC = 30° So, △BDC is isosceles. BD = CD = 13 by the Base Angles Converse. △BAD is an isosceles right triangle. c = a√2 13 = (AB)√2 AB = 13/(√2) AB = AD = (13√2)/2 Pink Quadrilateral ABCD Area = △BAD Area + △BDC Area A = (bh)/2 = 1/2 * (13√2)/2 * (13√2)/2 = 42.25 A = 1/2 * a * b * sinC = 1/2 * 13 * 13 * sin(30°) = 42.25 ABCD Area = 42.25 + 42.25 = 84.5 So, the area of the pink quadrilateral is 84.5 square units.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@prossvay8744]

Connect B to C AB=AD LADB=LABD=45° and LBDC=30° ; LCBD=75° So LCBD=180°-30°-75°=75° BD=DC=13 Let AB=AD=x In ∆ABD x^2+x^2=BD^2 2x^2=13^2=169 x=13/√2 Area of the ABCD=area of ∆ABD + area of ∆DBC Area of ∆ABD=1/2(13/√2)^2=169/4 square units Area of ∆DBC=1/2(13)^2sin(30°)=169/4 square units So area of ABCD=2(169/4)=169/2=84.5 square units. Thanks Sir.❤❤❤

[@adhiraj073]

Connect B to D*

[@PreMath]

You are very welcome! Thanks for sharing ❤️

[@JangirBK]

The same one as above, but in short 🤙

[@sergeyvinns931]

Draw DB, angie ADB = angie ABD = 45 degrees, angie BDC = 30 degrees, angieDBC = angie DCB, DC = DB =13. Area trangie ABD = 13^2/4 = 42,25. Area trangie BCD = 13^2*sin 30/2 = 42,25. Area of the Pink Quadriter5ai = 84,5!

[@PreMath]

Excellent! Thanks for sharing ❤️

[@kennethstevenson976]

I found the area of the first triangle using 2x^2 = 13^2 ; x^2 = 169/2 ; x = 169/4 ; x = 42.25. The A = 1/2 (13)(13) sin(30) was needed for the area of the second triangle. Total area = 84.5 square units.

[@Ihsan403]

It is a woderful exercise❤

[@PreMath]

Thanks Ihsan dear ❤️🙏

[@phungpham1725]

1/ Connect DB. We have the triangle DAB is a right isosceles and the triangle BDC is a isosceles of which the each of base angles = 75 degrees and the angle BDC= 30 degrees. So we have BD= 13 and AD=AB=a=13/sqrt2 Notice that the height from A to BD = the height from C to DB= 13/2 ( from C just drop the height to BD, we have a special 30-90-60 triangle of which the height= DC/2 So the area of the two triangles DAB and BDC are equal ( same base, same height) Area of the quadrilateral = sq a= 169/2= 84.5 sq units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@santokhsidhuatla7045]

JoinBD, In triangle ABD, Angle BDA=45 Angle BDA=45 In triangle DBC, angleDBC=75, angleDCB=75 CD=BD=13 DA=AB=13/under root 2 In triangle CBD, perpendicular =6.5 Area=1/2x13x6.5=42.5 1/2x13/ur2x13/ur2=42.5 Total=84.5 square units

[@AmirgabYT2185]

S=84,5😊

[@PreMath]

Thanks for sharing ❤️

[@himo3485]

ABCD=ADB+CBD ADB=13＊6.5/2=42.25 CBD=13＊6.5/2=42.25 ABCD=42.25+42.25=84.5

[@PreMath]

Excellent! Thanks for sharing ❤️

[@alster724]

Very easy. Although there is no need for the oblique triangle area though for ∆BAD since it is already a rt∆. Instead, I use the basic triangle area formula (A= bh/2) since AD is the height and AB is the base. I only use the oblique area for ∆BCD since there is no height. Regardless, I got the same answer. Side note as well, I used the special properties of a 45-45-90 for ∆BAD which are c= b√2 or a√2 and the legs are equal.

[@TTTT-h5k]

👍

[@wackojacko3962]

No matter how one looks at this situation, the element that the sum of the interior angles of a quadrilateral is always 360 degrees must always be included. People have jobs...me I don't work , I just get paid. 🙂

[@PreMath]

Thanks for the feedback ❤️ Stay blessed!

[@justshubham8117]

Did it orally 🙌

[@santiagoarosam430]

∠ADB=45º=∠DBA→ ∠BDC=30º : ∠DBC=120º-45º=75º→ ∠DCB=180º-30º-75º=75º→ DC=13=DB. Transformamos el triángulo DBC en otro de superficie equivalente, desplazando el vértice C según una línea ascendente paralela a BD hasta alcanzar la horizontal por D y la vertical por B, obteniendo el punto "T"→ ABTD es un cuadrado y su diagonal es DB=13→ Áreas ABCD=ABTD=13²/2=169/2=84,50. Gracias y un saludo cordial.

[@Birol731]

My solution is ➡ if we draw a line between D and B we get 2 triangles: ΔABD is a isosceles right angle triangle: ∠DAB ∠ABD = ∠BDA ⇒ ∠ABD = ∠BDA = 45° the other triangle is ΔDBC ∠ABC = 120° ∠ABD= 45° ⇒ ∠DBC= 75° ∠CDA= 75° ∠BDA= 45° ⇒ ∠CDB= 30° ⇒ ∠BCD= 180°-(∠DBC+∠CDB) ∠BCD= 180°-(75°+30°) ∠BCD= 75° ⇒ ΔDBC is an isosceles triangle ! CD= DB= 13 according to the Pythagorean theorem for the ΔABD a²+a²= b² 2a²= 13² 2a²= 169 a²= 169/2 ⇒ A(ΔABD)= a*a/2 A(ΔABD)= a²/2 A(ΔABD)= (169/2)/2 A(ΔABD)= 169/4 square units. A(ΔDBC)= 1/2*sin(CDB)*CD*DB ∠CDB= 30° CD= 13 DB= 13 ⇒ A(ΔDBC)= 1/2*sin(30°)*13*13 A(ΔDBC)= 1/2*1/2*13*13 A(ΔDBC)= 169/4 square units Apink= A(ΔABCD) Apink= A(ΔABD) + A(ΔDBC) Apink= 169/4 + 169/4 Apink= 338/4 Apink= 84,5 square units ✅

[@PreMath]

Excellent! Thanks for sharing ❤️

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... First of all we divide the quadrilateral into two triangles: ABD and BCD. Since ABD is an isosceles right triangle, we can conclude: ∠ABD = ∠ADB = (180° − ∠BAD)/2 = (180° − 90°)/2 = 90°/2 = 45° ⇒ ∠CBD = ∠ABC − ∠ABD = 120° − 45° = 75° ∧ ∠BDC = ∠ADC − ∠ADB = 75° − 45° = 30° ⇒ ∠BCD = 180° − ∠CBD − ∠BDC = 180° − 75° − 30° = 75° Therefore BCD is also an isosceles triangle (∠CBD=∠BCD), so we can conclude: BD = CD = 13 Now we apply the Pythagorean theorem to calculate the area of the isosceles right triangle ABD: BD² = AD² + AB² = AB² + AB² = 2*AB² ⇒ AB² = BD²/2 ⇒ A(ABD) = (1/2)*AB*AD = (1/2)*AB² = (1/2)*BD²/2 = (1/2)*13²/2 = 169/4 The area of the isosceles triangle BCD can be calculated in the following way: A(BCD) = (1/2)*BD*CD*sin(∠BDC) = (1/2)*CD²*sin(∠BDC) = (1/2)*13²*(1/2) = 169/4 So the two triangles ABD and BCD have the same area and the total area turns out to be: A(ABCD) = A(ABD) + A(BCD) = 169/4 + 169/4 = 169/2 = 84.5 Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️

[@LuisdeBritoCamacho]

Let's do it!! 1) AB = AD 2) Draw a Line from Point B to Point D 3) Depict a Right Triangle with Sides AB ; AD ; BD 4) Observe that, if AB = AD we have a Right Triangle [ABD] with Internal Angles (45º ; 45º ; 90º) 5) Angle ABD = 45º, so : Angle DBC = 120º - 45º = 75º. And Angle BCD = Angle DBC = 75º 6) Angle ADB = 45º, so : Angle BDC = 75º - 45º = 30º 7) Now it's clear for all, that Triangle BCD is an Isosceles Triangle with Internal Angles (30º ; 75º ; 75º) 8) So : BD = CD = 13 9) Let consider that AB = AD = X 10) X^2 + X^2 = 169 ; 2X^2 = 169 ; X^2 = 169 / 2 ; X = 13 / sqrt(2) ; X = 13*sqrt(2) / 2 ~ 9,2 11) Area of Right Triangle [ABD] = 2A = (13*sqrt(2) / 2)^2 ; 2A = 169 / 2 ; A = 169 / 4 ; A ~ 42,25 12) Finding the Area of Isosceles Triangle [BCD] 13) 13 / sin(75º) = BC / sin(30º) ; BC = 13*sin(30º) / sin(75º) ; BC = 6,5 / sin(75º) ; BC ~ 6,73 14) Triangle [BCD] with Side Lengths (13 ; 13 ; 6,73) Using Heron's Formula (Online Calculator): 14) A = 42,25 15) Total Area = 42,25 + 42,25 = 84,50 16) My Best Answer is Area = 84,50 Square Units Working with Windows Calculator is an inferno!!

[@PreMath]

Excellent! Thanks for sharing ❤️

[@mouradbelkas598]

Another solution is, once AB and AD are known, Area = 1/2 (AB + 13) * AD

[@quigonkenny]

Draw BD. As DA = AB, ∆DAB is an isosceles triangle, and as ∠DAB = 90°, ∠ABD = ∠BDA = 45°. Let DA = AB = x. As BD divides ∠ABC and ∠CDA, ∠DBC = 120°-45° = 75°, and ∠CDB = 75°-45° = 30°. Thus ∠BCD = 180°-(30°+75°) = 75°, ∆CDB is an isosceles triangle, and CD = DB = 13. Triangle ∆DAB: DA² + AB² = BD² x² + x² = 13² 2x² = 169 x² = 169/2 x = √(169/2) = 13/√2 Quadrilateral ABCD: A = [DAB] + [CDB] A = bh/2 + absinC/2 A = (13/√2)(13/√2)/2 + (13)(13)sin(30°)/2 A = (169/2)/2 + 169(1/2)/2 A = 169/4 + 169/4 = 169/2 sq units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@abhikDebnath403]

How?

[@吃饭睡觉打豆豆-j6e]

i don’t know why the areaBCD=1/2*13*13*sin30? Someone can explain?

[@marcgriselhubert3915]

Fine.

[@PreMath]

Thanks for the feedback ❤️