How to think outside the Box without Calculators? | Can you find the area? |

Learn how to find the area of the isosceles triangle. Important Geometry and Algebra skills are also explained: area of a triangle formula. Step-by-step tutorial by PreMath.com
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[@jimlocke9320]

The 15°-75°-90° right triangle appears often enough in geometry problems that I recommend that all of us become familiar with its properties. In this problem, after determining that AC = BC by virtue of ΔABC being isosceles, drop a perpendicular from C to AB, dividing ΔABC into two 15°-75°-90° right triangles. Each has a hypotenuse of length 100. One formula for the area of a 15°-75°-90° right triangle is hypotenuse squared divided by 8. So, the area of each triangle is (100)²/8 = 10000/8 = 1250. There are 2 15°-75°-90° right triangles, so area ΔABC = 2(1250) = 2500 square units, as PreMath also found.

[@phungpham1725]

Thank you! ❤

[@marcelowanderleycorreia8876]

Very cool the first method!! Very good sir!!!

[@marcgriselhubert3915]

Let's be more general and calculate the area of ABC if angleA = angleB = t and if CA = CB = L. This problem happens rather frequently. Be H the orthogonal projection of C on (AB). We have CH = CB.sin(t) = L.sin(t), and BH = CB.cos(t) = L.cos(t) The height of ABC is then L.sin(t) and its basis is AB = 2.BH = 2.L.cos(t), so its area is (1/2).(L.sin(t)).(2.L.cos(t)) = (L^2).sin(t).cos(t) = (1/2).(L^2).sin(2.t). Now in the case of L = 100 and t = 15°, the area of ABC is (1/2).(100^2).sin(30°) = (1/2).(10000).(1/2) = 2500.

[@LudwigSpiegel]

Nice!

[@jamestalbott4499]

Thank you!

[@scottdort7197]

I chopped it in half and put it back together. This gave me an isosceles triangle with two 100 long legs and an angle of 30°. Then I use the sin formula for triangle area and that gave me 2500.

[@jimlocke9320]

Actually, your method is a variation on PreMath's second method. He recognized that the original triangle had two sides of length 100 with a 150° angle in between. He used the sine formula area = (1/2)a b sin(c) to get (1/2)(100)(100)sin(150°). After your rearrangement, the sine formula becomes (1/2)(100)(100)sin(30°). However, sin(150°) = sin(30°) = 1/2, so the end result is the same. After you have chopped the triangle in half, you can construct 3 congruent copies of one of the triangular pieces and assemble them into a square with the hypotenuses of length 100 as sides. A smaller square opening is left in the middle with sides equal to the difference between the side lengths of the original triangular piece. The original triangular piece is a 15°-75°-90° right triangle, with side ratios short:long:hypotenuse of (√3 - 1):(√3 + 1):2√2. So the difference of sides is (√3 + 1)/(2√2)(100) - (√3 - 1)/(2√2)(100) = (100)/(√2). The combined area of the 4 triangles equals the area equals the larger square's area, (100)² = 10000, less the smaller square's area, ((100)/(√2))² = 10000/2 = 5000, area = 5000. The original triangle's area is equal to two of those four triangles. So, its area = (2/4)(5000) = 2500.

[@alster724]

The oblique triangle area (Method 2) made it easy

[@CloudBushyMath]

Looking Beyond Boundaries💡

[@PrithwirajSen-nj6qq]

Use projection of cosine rule AB=100cos15deg+100cos15deg=2*100cos15deg =200*cos15 degree Then three sides of the given triangle is known. Using Heron Theorem we may derive the area of the given triangle

[@bobbyheffley4955]

By applying the difference formulas for sine and cosine and letting 15°=45°-30°, I determined that the height is 50[sqrt (6)-sqrt (2)] and that the base is 25 [sqrt (6)+sqrt (2)]. By multiplying base and height and dividing by 2, I got 2500 square units as the area.

[@hongningsuen1348]

30, 45, 60 are lovely angles for trigonometric ratios as they give side ratios. Angles that can be derived from them are also lovely such as 15 = 30/2, 22.5=45/2, 75 = 30 + 45, 105 = 45+60. (also 120 = 0 + 90, 135 = 45+90, 150 = 60+90). When these other angles appear in triangles of the problem, they may guide possible construction. In this problem the exterior angle at C = 15 + 15 = 30 gives an obvious hint for the construction used.

[@devondevon4366]

2500 Draw another triangle symmetric to the above and attach it to the other to form a quadrilateral and thus, two isosceles triangles with degrees 75, 75, and 30 Draw a perpendicular from C to a new point P. CP is twice the height of the ACB On the right (or left), draw a perpendicular line through the isosceles triangle to create a 60, 90, 30 triangle and a 15, 75, 90 degree triangle Hence, the length of the perpendicular line is 50 (since its hypotenuse is 100), its other base is 50 sqrt 3 Let's focus on the right triangle on the left (15, 75, and 90) Since one of its bases is 50, the other base is 100 - 50 sqrt 3 = 50 (2- sqrt 3) or 13.3974596 Hence CP^2 = 50^2 + 13.3974596^2' = 2500 +179.491942 =2679.4919 CP = 51.7638, note that CP is twice the height of ACB hence, CP/2 = 25.8819 is the height of triangle ACB Since its hypotenuse = 100, then it other base= sqrt (100^2 - 25.8819^2) =96.59258 its area hence = 96.59258 x 25.8819 = 2500 Answer

[@davoodzamani4254]

Hello . This problem is easily solved with geometry. Project triangle ABC on side AB to form a rhombus (ACBD). Now make a perpendicular from the point C to the side AD and call it h. The area of the triangle ACD is equal to( h×Ad)/2 h is the opposite side at an angle of 30 degrees, and therefore its length is half of the chord, i.e. 50 units. (100×50)/2 = 2500 squ And according to symmetry, the area of triangle ACD is equal to the area of triangle ABC cheers😉

[@himo3485]

30° 60° 90° → 1 : √3 : 2 area of the Triangle : 100＊50/2=2500

[@misterenter-iz7rz]

Isosceles triangle, AC=BC=100, 1/2×100^2×sin150=1/2×100^2×1/2=10000/4=2500.😊 Alternatively, area =base×height/2, 1/2×100×(100×sin 30)=2500.😊

[@prossvay8744]

LBAC=ABC=15° So CA=CB=100 LABC=150° Area of triangle=1/2(100)(100)sin(150°) sin(150°)=sin(90°+60°)=cos(60°)=1/2 So area =(100)(100)/4=2500 square units.❤❤❤ Thanks sir.

[@engralsaffar]

The way I did it is the following: Sin30=2sin15cos15 1/2=2*h/100*1/2*b/100 A=1/2*b*h=10000/4=2500 square units

[@santiagoarosam430]

Tomando AB como eje, hacemos la simetría de la figura y obtenemos un trapecio de lado 100. La distancia ortogonal de C al simétrico de AC es 100/2=50. Área del trapecio obtenido =2*100*50/2=5000. Área ABC =5000/2=2500. Gracias y saludos.

[@santiagoarosam430]

No es un trapecio, es un rombo. Disculpas por el error.

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال S(ABC) =1/2 .100.100 sin150 =2500

[@sergeyvinns931]

Area = 100^2*sin150/2= 10000/4= 2500! sin150= sin(180-150)= sin30=1/2!

[@LuisdeBritoCamacho]

The only way I found to solve this Problem was the following: Let the Vertical Line passing through C and intercepting Line AB, be C' 1) Let CC' = Height = b 2) Let AC' = Base = a 3) CC'^2 + AC'^2 = 10.000 ; a^2 + b^2 = 10.000 4) b = tan(15º)*a 5) As tan(15º) = (2 - sqrt(3)), we have: 6) b = [2 - sqrt(3)] * a 7) Now I have a System of two Equations; one Linear and another Non Linear. 8) Solutions: a ~ 96,5926 lin un ; b ~ 25,8819 lin un 9) Triangle Area = a * b = 96,5926 * 25,8819 = 2.500 sq un. 10) My answer is . The Area of the given Triangle is equal to 2.500 Square Units. Note : I am gonna see the Video!! Thank you and Greetings from the Great Iberia!!

[@PrithwirajSen-nj6qq]

We extend

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... Since ∠ABC=∠BAC, the triangle is an isosceles triangle (AC=BC). Therefore with D being the midpoint of AB the two triangles ACD and BCD are congruent. So we can add the point E such that CE=2*CD and BE=BC to obtain the triangle BCE which has the same area than the triangle ABC because the triangle BDE is congruent to ACD and BCD: A(ABC) = A(BCE) = (1/2)*BC*BE*sin(∠CBE) = (1/2)*BC²*sin(2*∠ABC) = (1/2)*100²*sin(2*15°) = (1/2)*10000*sin(30°) = (1/2)*10000*(1/2) = 2500 Best regards from Germany

[@jameswarren3925]

But 50 is the height of triangle ADB, not of ABC. Or am I missing something?

[@JamesDavy2009]

Obtuse triangles when asked to find their area are shown with their perpendicular heights as dotted auxiliary lines. The first method simply used one of the congruent sides as the base of the triangle.

[@srirajan1933]

But it is not clear that AD is a congruent side, and does not relate to triangle ABC’s height.

[@Shawn_RHVAC]

D can’t be a 90 degree angle??

[@stevehart175]

No doubt the ans is 2500^2 units…but not following how he’s determining those base & height values for ABC? How can 50 be the height, or 100 be base for that matter? Happy to be guided😊

[@devondevon4366]

Because any side of the three sides of a triangle can be used for the base, and he uses 100 as that side, then extend it to get the height as one can extend the base to form the height, which is a perpendicular segment from the base to the vertex opposite it. Doing this will form a 30, 60, 90 right triangle in which 30 = x 60 = x sqrt 3 and 90 = 2x Keep in mind that AC = BC since both face 15 degrees, So, since BC =100 then AC =100, but since AC faces 90 degrees, then its other two bases (30 and 60) then 30 will = 50 hence, area = base * height *1/2 = 100 * 50*1/2 = 5,000 *1/2 =2,500

[@stevehart175]

@@devondevon4366 thanks (the extending base to form height was the blind spot in my understanding 🙈👍)

[@devondevon4366]

@@stevehart175 Sure, no problem

[@AmirgabYT2185]

S=2500

[@JLvatron]

I used the triangle formula a*b/2 * Sin x. Here, x=120 which is the same as Sin 30, which is a known value of 1/2. So that all reduces to 50*50= 2500. [Sorry, i meant Sin 150 is same as Sin 30]

[@user-tf1ti2dw7g]

Sin120 is not same as sin 30

[@JLvatron]

@@user-tf1ti2dw7g Sorry, I meant Sin 150 is same as Sin 30

[@comdo777]

asnwer=55 os isit

[@akhtarraja7823]

I am getting 2500/(2-✓3)😢

[@somapatra5560]

3:48 excuse u u r still usin sine law and in the display u r showin without law of sines😤😤😠😠 In fact, the entire video is based on sine law very very dissatisfied with the video

[@scottl.1568]

Whut