Find the area of the Rectangle | A Very Nice Geometry Problem

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[@davidellis1929]

There's a much simpler solution, without resorting to Heron's formula or trig. If you draw a perpendicular from A to BE, you can see that the isosceles triangle ABE consists of two 5-12-13 right triangles joined along their legs of length 12. Each 5-12-13 triangle has area 30, so the area of ABE is 60. But ABE has half the area of ABCD, since they share the base AB and equal altitudes to AB, so the area of ABCD is 120.

[@Alleswirdgut0501]

Great. I love that solution.

[@murdock5537]

Nice one!

[@skwest]

This one's in the head: Heron's formula for the 13-13-10 triangle yields an area of 60, then either backtrack to find the altitude of the triangle (i.e. the short side of the rectangle), or just recognize that the triangle represents ½ the area of the rectangle... either way, the final calc for area of the rectangle is 120. Now to watch and see if that's correct.

[@michaeldoerr5810]

Super agree!!!

[@marioalb9726]

ABE is an isosceles triangle : Pytagorean theorem h₁² = 13² - (10/2)² -> h₁= 12cm Area of rectangle : A = b.h = (h₁/cosα)(10.cosα) A = 12*10 = 120 cm² ( Solved √ )

[@kateknowles8055]

Triangle ABE has a base BE and if we draw the corresponding height, it is 12 because this triangle is isosceles and 169-25= 144. That area ABE is 60. If we cut off triangle BEC and draw it again with BC imposed onto AD then DF is the extension of ED and equals EC. The area of the rectangle ABCD equals the area of the new shape. It is a parallelogram. The triangle ABE is half of it. The area of ABCD was twice 60 . ABCD's area was : 120 (but we just cut it up)

[@WahranRai]

Simply apply the Pythagorean theorem Let x = AD et y= DE ➔ 13^2 = x^2 + y^2 et 10^2 = x^2 + (13- y)^2 ➔ 13^ 2 - 10^2 = x^2 + y^2 - ( x^2 + (13-y)^2) = y^2 - (13-y)^2 = (y + 13-y)*(y - 13+y) ➔ 69 = 13*(2*y-13) ---> y = 119/13 ---> x^2 = 13^2 - (119/13)^2 = 14400/169 ➔ x= 120/13 ➔ Area of rectangle ABCD = 13*120/13 = 120

[@andersandersson1561]

I did the same.

[@shadrana1]

At 2:15, Let AB=13,AD=BC=y,EC=x, DE=13-x Consider triangle BEC, EB^2=BC^2+EC^2 (Pythagoras) 10^2=y^2+x^2,y^2=100-x^2.....................(1) Consider triangle ADE, AE^2=AD^2+DE^2 13^2=y^2+(13-x)^2 Pythagoras) This breaks down to; y^2=26x-x^2............................................(2) Equate (1) and (2), 100-x^2=26x-x^2 26x=100 x=100/26=50/13 units. Substitute x back into (1), 100=y^2-(50/12)^2 16900=169y^2-2500 y^2=14400/169 y=120/13 units. IABCDI=AB*BC=13*y=13*120/13 =120 square units. Thanks for the Puzzle professor.

[@marioalb9726]

ABE is an isosceles triangle : Pytagorean theorem h² = 13² - (10/2)² -> h = 12cm Area of isosceles triangle: A₁ = ½ b.h = ½*10*12 = 60 cm² Area of rectangle = 2 times area of isosceles triangle : A = 2A₁ = 120 cm² ( Solved √ )

[@Bhogichand]

This is much simpler. You tuber has made it complicated.

[@bakrantz]

Draw a perpendicular bisector from point A to EB. Find the height of the triangle as 12 determine area 0.5*12*10=6. That triangle is half the area of the rectangle. So the rectangle area 2*60 or 120 sq. units.

[@santiagoarosam430]

h=√(13²-5²)=12---> ABCD/2 =5*12=60---> ABCD =2*60 =120. Gracias y saludos.

[@engr.amirshehzad6026]

Using law of cosine (13*13+13*13-10*10) /2*13*13 = 0.70 Then cos^-1(0.70) = 45Degree The sin45=hyp/13 Hyp= 9.25 units Then rectangle area = 13*9.25=119.85 Units approximately

[@aromaclinic4112]

5:12:13 BC=120/13

[@ludmilaivanova1603]

if we drop a perpendicular from E we will see two rectangulars with thier diagonals AE adn BE. Diagonal devides a rectangular into two equal triangles. So it is enough to find the area of AEB which is an isosceles triangle with the base 10 and sides-13. The hight is sqrt 169-25=144 (12). 10x12/2x2=120.

[@sorourhashemi3249]

Thanks. Easy.

[@ناصريناصر-س4ب]

We calculate the area of triangle ABE using Iron's method. We find s=√p(p-13)²(p-10)=60. From this, the height of triangle ABE related to side AB is equal to H=2S/13=120/13=AD. From this, the area of the rectangle is (120/13)*13=120.

[@imishan8981]

Herons method ig

[@michaeldoerr5810]

This is one of the best summations of this video

[@murdock5537]

BAE = δ → 100 = 2(169)(1 - cos⁡(δ)) → cos⁡(δ) = 119/169 → sin⁡(δ) = 120/169 → area ∎ABCD = 169sin⁡(δ) = 120

[@quigonkenny]

Let BC = AD = y and ED = x. As ABCD is a rectangle and thus AB = CD = 13, then CE = 13-x. Triangle ∆EDA: ED² + DA² = AE² x² + y² = 13² x² + y² = 169 Triangle ∆BCE: BC² + CE² = EB² y² + (13-x)² = 10² y² + 169 - 26x + x² = 100 x² + y² - 26x + 69 = 0 169 - 26x + 69 = 0 26x = 238 x = 238/26 = 119/13 x² + y² = 169 (119/13)² + y² = 169 14161/169 + y² = 169 y² = (28561-14161)/169 y² = 14400/169 y = √(14400/169) = 120/13 Rectangle ABCD: A = hw = (120/13)13 = 120 sq units

[@ritwikgupta3655]

ABCD is given to be a rectangle. Hence [ABCD] = 2*∆ABE. Area of ABE can be calculated using many methods. No need to find the height.

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال BAE = BEC cosBAE = 5/13 sinBAE =12/13 DC =120/13 S(ABCD) =120

[@prossvay8744]

Area of the rectangle=(13)(120/13)=120.❤❤❤

[@imetroangola17]

*Solução:* Traçando a altura h do vértice a A até o ponto M do lado BE, temos que AM=h é também mediana e bissetriz, pois o triângulo ∆ABE é isósceles. Assim, ME=MB=5, por Pitágoras no ∆AME: 13² = h² + 5² → h² = 169 - 25 = 144 h = √144 → *h = 12.* Seja a área [ABE]=S. Assim, S= h × BE/2 = 12 × 10/2 → S=60. Por outro lado, observe que: S = BC × AB/2 →BC × AB = 2S [ABCD] = 2× 60 *[ABCD] = 120 unidades quadradas*

[@adgf1x]

ar. of triangle=12×5=60 dq.unit.now sr. of rectangle=60×2=120 sq unit(as area ot rectangle=ar o g traiangle ×2 as eith same base and between same parallels).

[@marioalb9726]

This exercise was already solved in minute 5:14 of video. A = b.h = AB. EF = 120 cm² All the following is totally unnecessary !!

[@gravitheist5431]

169 - 25 = 144 SQR 144 = 12 12 x 5 =60 60/6.5 = 9.23 9.23 x 13 = 120

[@cyruschang1904]

the area of the 13 - 13 - 10 triangle is half the area of the rectangle. the height of the 13 - 13 - 10 triangle is ✓[(13^2) - (5^2)] = ✓(169 - 25) = 12 area of the rectangle = (12 x 10) ÷ 2 x 2 = 120

[@jkhazraji7358]

AD=BC, EC=DC-DE=13-DE, (AD)²+ (DE)² =13² Eq....(1), (EC)²+ (BC)² =10^2 => (EC)²+(AD)²=10² Eq...(2), By solving the 2 equations b=120/13, So area=13x120/13=120

[@Bhogichand]

What mistake initially done? Did anyone notice ? Initially triangle was drawn and denoted as ABC instead of ABE. In next step it was written correctly as ABE.

[@Mathematician3.14]

S∆=AB*AD/2=60 ---> S=AB*AD=60*2=120. Easy

[@himo3485]

AF⊥EB EF=FB=5 AF=√[13^2-5^2]=12 triangle AEB = 10*12*1/2 = 60 area of rectangle ABCD = 60*2 = 120

[@jkhazraji7358]

[√18*(18-13)(18-13)(18-10)]*2=120

[@hans7831]

I've never heard of Heron's theorem. Although it is quite practical and quick, it is not absolutely necessary. Heron may forgive me.

[@wasimahmad-t6c]

9.1546+3.8453=13×9.23=120

[@SGuerra]

A questão é muito boa. Parabéns. Brasil Novembro de 2024.

[@nenetstree914]

120

[@jairoeveliogordillomarin5780]

13²= (13-X)²+X² 10²= (X²+Y²) Despejo Y de ambas ecuaciones y las igualo. Entonces hallo el valor de X=3.85 u. Luego hallo el valor de Y= 9.23 u. Área del rectángulo= 13×9.23= 120 u².

[@marioalb9726]

Cosine rule : c² = 2b²(1-cosα) cos α = 1 - ½c²/b² = 1 - ½10²/13² α = 45,24° Height of rectangle: h = b sin α = 9,23 cm Área of rectangle: A = b.h = 13*9,23 = 120cm² ( Solved √ )

[@AmirgabYT2185]

S=120

[@RealQinnMalloryu4]

(13)^2 (13)^2 (10)^2={169+169+100}=338{60°A60°B60°C60°DE}=360°ABCDE/338=1.22ABCDE 1.2^11 1.2^11^1 1.2^1^1 2^1(ABCDE ➖ 2ABCDE+1).

[@xz1891]

5 sec, 120

[@ManojkantSamal]

^=read as to the power *read as square root AD^2=13^2 -(13-x)^2 =169-169-x^2+26x =26x-x^2......eqn1 Again BC^2=10^2-x^2 =100-x^2....eqn2 AD=BC So, 26x-x^2=100-x^2 26x=100 X=100/26=50/13 BC^2 =100-x^2 =100-(50/13)^2 =100-(2500/169) =(16900-2500)/169 =14400/169 BC=*(14400/169)=120/13 Area of the rectangle =ABxAD =13×(120/13)=120.... May be

[@bpark10001]

Way too complicated! Bisect BC at point G. Draw GA. AGB is right 5/12/13△. △BCE is similar with BE = 10 so scaling factor is 10/13. BC = (10/13)(12) = 120/13. Area rectangle = (13)(120/13) = 120. (There is no need to calculate any area other then the rectangle directly from its sides.)

[@CracksMeUp-182]

It is clear that area of🔺is half of the rectangle. So, when you get area of the 🔺= 60, just double it. Area the rectangle = 2 x 60 = 120

[@alexniklas8777]

120