Take a square w/ sides 6. Cut that square diagonally twice to create quarters. One of the quarters is the triangle we are searching. -> 6*6=36, 36:4=9 qed
Easiest way: The triangle is a special right triangle 45-45-90. Proof: Solve for x, then each angle, and should be able to see 2x is 90 degrees. From there: AC = BC = AB/sqrt(2) = 6/sqrt(2) because of the special isosceles right triangle. Area of this right triangle = 0.5(AC)(BC) = 0.5(18) = 9 Please comment about this solve
Dear, Minh, you are absolutely correct. However, we must prove that this is indeed a 45-45-90 triangle by using the "perpendicular Bisector Theorem!" Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Dear Viyush, there is another theorem called "Angle Bisector Theorem!" We could have used that one as well. However, that was unnecessary. Thanks for your input. You are awesome 👍 Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
I have seen a lot of your videos on finding areas and other properties of triangles. I have also seen ones by others. One thing I have never seen is one which uses Heron’s Formula. Have you ever done one? Do mathematicians ever use it? Just wander why?
If two angles are equal The triangle is isosceles as wel as right angle so if we calculate side a by pithogorous theorem We get a=h=sqrt18 Then A=(1/2.)bh So. A=(1/2)18= So.....A=9sq units I tried by this method sir I caluculated in my mind and got Answer.... I hope your method is also good
No need of drawing perpendicularcular bisector. As A=B=45 degree and C=90 degree, ABC is a right angled isosceles triangle. We can find the lengths of the equal sides AC and BC by using Pythagoras theorem and then area is half*AC*BC
Right in fact he correctly stated the perpendicular bisector theorem but he didn't prove that the perpendicular bisector of AB goes through c (though that is not a difficult proof). So your method is simple and more correct.
x + 2x + 45= 180 (as the interior angles are , x , 2x and 45) Answer 9 8:28 3x =135 x =45 Triangle is isoscles 45 45 90 Since one side is 6, then the other two sides = 2 a^2 = 36 a^2 =18 a = sqrt 18 So the sides are sqrt 18, sqrt 18 and 6 Area hence = (sqrt 18)(sqrt18)/2 =18/2 Answer =9
Hello One more solution. ∆ABC is a right angled triangle. Angle CAB = Angle CBA BC = AC Hence, ∆ABC is an isosceles traingle. AB becomes the hypotenuse. Hypotenuse² = (Side 1)² + (Side 2)² 6² = AC² + BC² 36 = AC² + AC² 36 = 2AC² 18 = AC² AC = √18 AC = BC = 3√2 Area of a ∆ = ½ × Base × Altitude = ½ × AC × BC = ½ × (3√2)² = ½ × 18 = 9 units. Comment if correct.
So nice of you Govinda! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃 Greetings from USA!
@@PreMath I am bengali. and poor communicator in English. I follow your vedio to understand math to teach my son. who read in class nine. From India. 🇮🇳🇮🇳🇮🇳
@@govindashit6524 My dear Bengali friend. Thank you so much for your continued love and support. Your English is very good 😀 Don't underestimate yourself. Please keep sharing premath channel with your family and friends. Take care dear and all the best😃
Sir, In this mathematical problem, after calculating the value of AC=BC=3√2 unit [ by using pythagorean theorem], can we use this formula : √[s(s-AB)(s-BC)(s-CA)] to find the area of Triangle 🔺️ ABC ?? since s = 1/2 × (AB+BC+CA)
After finding Angle BCA is 90deg and Angle CAB is 45deg, Triangle ABC is a right-angled Isosceles Triangle with AB as hypotenuse, and let BC = CA = x By Pythagorus Theorem, AB^2 = BC^2 + CA^2 6^2 = x^2 + x^2 ==> 2(x^2) = 36 x^2 = 18 x = 18^(1/2) Hence, BC = CA = 18^(1/2) With Angle BCA = 90deg, Area of Triangle ABC = (1/2) x BC x CA = (1/2) x 18^(1/2) x 18^(1/2) = (1/2) x 18 = 9
The interior angles are X, 2X, and 45. 3x+45=180. X=45. This is an isosceles right triangle. Hypotenuse has length 6. Sides have lengths 3*sqrt(2). Area = 3*sqrt(2)*3*sqrt(2)/2 = 9.
Sir So Easy problem first we find this triangle is an isoscesles right angled triangle from given information and I find equal sides 3 root under 2 and apply area of triangle formula we get area 9
3rd Method: Using Pythagoras twice. First , Triangle ABC is an isosceles triangle, with angle 2X being 90 deg. This from 2X +X +45 = 180, X = 45 deg. Then solve for the sides AC & BC = 3root 2 Then draw bisector CD and solve for this (height) using Pythagoras again =3 The ½ Base X Height = 9 A bit drawn out, but I am not that familiar with Trig.
Just to mention that there is a formula to calculate area based on one side and 3 angles. You generate this formula from S=0,5absin(gama) by replacing b by a times sin(beta) divide by sin(alpha) (which comes from the sin theorm).
Once you realize, that the triangle is 45-90-45 (or one half of a square cut diagonally) and that the smaller triangles are similar, you realize that the two half-sized triangles together form a square with a side length of three (half the base). Thus the area is 3 times 3, which is nine. Fullstop.
Work smarter not harder. Angle ACB is ninety degrees. Side AB is the hypotenuse. It’s a 45 45 90 so we know ac and ab are both 6/sqrt(2) one of them is the base the other the height A=(6/sqrt(2))*(6/sqrt(2))/2 A=(36/2)/2=18/2=9
Easier method. Once you figure out its a right triangle (2x + x + 45 = 180, x = 45) w/ hypoteneuse of 6, turn it on its side. Sides (base and height) are equal and are 6*sin(45). 1/2 * (6*sin(45))^2 = 9. You're welcome.
i dont like the perpendicular bisector usage here, i dont see how it is useful in triangles in general, and it only works here cuz gamma at C it 90° and the p. bisector HAPPENS to go through it cuz of that...or i misunderstood its definition. pythagoras and isosceles is quicker here.
Another method.. 45-45-90 triangle whose sides are: AB= 6 AC= 3√2 (solved by using the special property of a 45-45-90 triangle which is AB= AC√2). BC= 3√2 USING HERON'S FORMULA, IT YIELDS THE SAME RESULT.😊😝
Take a square w/ sides 6. Cut that square diagonally twice to create quarters. One of the quarters is the triangle we are searching. -> 6*6=36, 36:4=9 qed
Easiest way: The triangle is a special right triangle 45-45-90. Proof: Solve for x, then each angle, and should be able to see 2x is 90 degrees.
From there: AC = BC = AB/sqrt(2) = 6/sqrt(2) because of the special isosceles right triangle.
Area of this right triangle = 0.5(AC)(BC) = 0.5(18) = 9
Please comment about this solve
Dear, Minh, you are absolutely correct. However, we must prove that this is indeed a 45-45-90 triangle by using the "perpendicular Bisector Theorem!"
Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Yeah. Dunno why he Cares about Projecting D from C whereas it's a triangle rectangle.
Trigonometry method:
x+2x+45°=180°
x=45°
Area of triangle=(1/4 )X sin(2x)h²
=(1/4) X (sin90°)6²
=9 units ²
I'm happy to know that most of the viewers take keen interest in the problems .Moreover the teacher is
competent enough I love his method
very much.
So nice of you Jasbir dear
Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Love and prayers from the USA!
Congratulations Master
A Hug From Brazil 🇧🇷
Grateful
Wow...cool 👍👍👍👍👍
Wow so good teacher I will teach my students the same to you
Because your skill is very nice
How can you say perpendicular bisected intersect angle C
Dear Viyush, there is another theorem called "Angle Bisector Theorem!" We could have used that one as well. However, that was unnecessary. Thanks for your input. You are awesome 👍 Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
@@PreMath thanks for your time Sir
You are doing a great job Sir
Wish you all the best sir
@1:23, its an equilateral triangle,with sum of the smaller sides bring 36 , or twice 3 times 6. Area is 3×6=18
Thanks Christopher for the feedback. I really appreciate that my friend. You are awesome 👍 Take care dear and stay blessed😃
I have seen a lot of your videos on finding areas and other properties of triangles. I have also seen ones by others. One thing I have never seen is one which uses Heron’s Formula. Have you ever done one? Do mathematicians ever use it? Just wander why?
angels:
2x + x + 45 = 180 --> 3x = 180 - 45 --> 3x = 135 --> x = 45°
A = B = 45
C = 90
lengths of order sides
[AC] = [BC] = [AB] / √2 = 6 / √2 = 3√2
Area
heigth = [AC] / √2 = [BC] / √2 = 3√2 / √2 = 3 = [AB] / 2
S = 6 * 3 / 2 = 9
Thanks Gregory for the feedback. You are awesome 👍 Take care dear and stay blessed😃
If two angles are equal
The triangle is isosceles as wel as right angle so if we calculate side a by pithogorous theorem
We get a=h=sqrt18
Then A=(1/2.)bh
So. A=(1/2)18=
So.....A=9sq units
I tried by this method sir
I caluculated in my mind and got Answer.... I hope your method is also good
Bravo Rangaswamy! Very smart👍 Great feedback.
Take care dear and stay blessed😃
Thanks sir for this video 😊😊
No need of drawing perpendicularcular bisector. As A=B=45 degree and C=90 degree, ABC is a right angled isosceles triangle. We can find the lengths of the equal sides AC and BC by using Pythagoras theorem and then area is half*AC*BC
Right in fact he correctly stated the perpendicular bisector theorem but he didn't prove that the perpendicular bisector of AB goes through c (though that is not a difficult proof). So your method is simple and more correct.
Had this in 5 minutes. I like
x + 2x + 45= 180 (as the interior angles are , x , 2x and 45) Answer 9 8:28
3x =135
x =45
Triangle is isoscles 45 45 90
Since one side is 6, then the other two sides =
2 a^2 = 36
a^2 =18
a = sqrt 18
So the sides are sqrt 18, sqrt 18 and 6
Area hence = (sqrt 18)(sqrt18)/2
=18/2
Answer =9
Hello
One more solution.
∆ABC is a right angled triangle.
Angle CAB = Angle CBA
BC = AC
Hence, ∆ABC is an isosceles traingle.
AB becomes the hypotenuse.
Hypotenuse² = (Side 1)² + (Side 2)²
6² = AC² + BC²
36 = AC² + AC²
36 = 2AC²
18 = AC²
AC = √18
AC = BC = 3√2
Area of a ∆ = ½ × Base × Altitude
= ½ × AC × BC
= ½ × (3√2)²
= ½ × 18
= 9 units.
Comment if correct.
This problem seems tricky, and you did an excellent job explaning it with two different methods. I liked the one involving trig. Awesome job, sir!
S=9
super,,
❤love you Sir ❤
So nice of you Govinda! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Greetings from USA!
@@PreMath
I am bengali. and poor communicator in English. I follow your vedio to understand math to teach my son. who read in class nine.
From India. 🇮🇳🇮🇳🇮🇳
@@govindashit6524 My dear Bengali friend. Thank you so much for your continued love and support. Your English is very good 😀 Don't underestimate yourself.
Please keep sharing premath channel with your family and friends. Take care dear and all the best😃
Sir, In this mathematical problem,
after calculating the value of AC=BC=3√2 unit [ by using pythagorean theorem], can we use this formula : √[s(s-AB)(s-BC)(s-CA)] to find the area of Triangle 🔺️ ABC ?? since s = 1/2 × (AB+BC+CA)
Yes i also use this method to solve and the answer is approximately 9
@@divyapadariya4755 Thanks 😊
After finding Angle BCA is 90deg and Angle CAB is 45deg, Triangle ABC is a right-angled Isosceles Triangle with AB as hypotenuse,
and let BC = CA = x
By Pythagorus Theorem,
AB^2 = BC^2 + CA^2
6^2 = x^2 + x^2
==> 2(x^2) = 36
x^2 = 18
x = 18^(1/2)
Hence, BC = CA = 18^(1/2)
With Angle BCA = 90deg,
Area of Triangle ABC
= (1/2) x BC x CA
= (1/2) x 18^(1/2) x 18^(1/2)
= (1/2) x 18
= 9
Great
Thanks Keerthi for the feedback. You are awesome 👍 Take care dear and stay blessed😃
The interior angles are X, 2X, and 45. 3x+45=180. X=45. This is an isosceles right triangle. Hypotenuse has length 6. Sides have lengths 3*sqrt(2). Area = 3*sqrt(2)*3*sqrt(2)/2 = 9.
Nice puzzle. Though I think the proof that the perpendicular bisector of AB includes C is missing?
Sir So Easy problem first we find this triangle is an isoscesles right angled triangle from given information and I find equal sides 3 root under 2 and apply area of triangle formula we get area 9
Thanks Raj for the feedback. You are awesome 👍 Take care dear and stay blessed😃
3rd Method: Using Pythagoras twice.
First , Triangle ABC is an isosceles triangle, with angle 2X being 90 deg.
This from 2X +X +45 = 180, X = 45 deg.
Then solve for the sides AC & BC = 3root 2
Then draw bisector CD and solve for this (height) using Pythagoras again =3
The ½ Base X Height = 9
A bit drawn out, but I am not that familiar with Trig.
triangle ABC: angles 45, 45, 90. Therefore, AC ^ Sqrt (2) = AB. Hence AC = AB / Sqrt (2). The area is AC ^ 2/2. (6 / Sqrt (2)) ^ 2/2 = 9
i watched and liked the video
Got it but used the right angle triangle ABC. Sin 45=6BC. So BC=4.2426 Area = 0.5x4.2426x4-2426= 8.9999. Love the challenge friend.
Thanks my dear friend for the feedback. You are awesome 👍 Take care dear and stay blessed😃
When your calculator gives you an answer like 8.9999, it's worth going back to see if there's a rounding issue.
@@wwoods66 did not use a calculator bro
Более быстрое решение. Внешние углы треугольника равны соответствующим внутренним. 3x÷45=180, x=45, 2x=90. Гипотенуза равнобедренного прямоугольного треугольника равна 6, катеты 6/(корень квадратный из 2). Считая один катет основанием, второй -высотой получаем искомую площадь.
3rd method:
Let us consider right triangle ACB:
let a:=|AC|=|CB|;
|AB|^2=|AC|^2 + |CB|^2;
|AB|^2=2*a^2;
a^2= |AB|^2 /2 = 6^2/2= 18;
A= a^2 /2= 18/2=9
Observe 2x = 90
AC = 6sin(45) = 6/surd(6)
ABC = 1/2 x 6/surd(2) x 6 x sin(45) = 9
Thank you sir. May I write (6 × 3)/2 =9
Thank you sir
So nice of you dear! You are awesome 👍 I'm glad you liked it! Take care dear and stay blessed😃
Just to mention that there is a formula to calculate area based on one side and 3 angles. You generate this formula from S=0,5absin(gama) by replacing b by a times sin(beta) divide by sin(alpha) (which comes from the sin theorm).
Once you realize, that the triangle is 45-90-45 (or one half of a square cut diagonally) and that the smaller triangles are similar, you realize that the two half-sized triangles together form a square with a side length of three (half the base). Thus the area is 3 times 3, which is nine. Fullstop.
You're right. Why do we always make things so complicated?
Work smarter not harder. Angle ACB is ninety degrees. Side AB is the hypotenuse. It’s a 45 45 90 so we know ac and ab are both 6/sqrt(2) one of them is the base the other the height A=(6/sqrt(2))*(6/sqrt(2))/2 A=(36/2)/2=18/2=9
3x + 45 =180, So x = 45. So Angle ACB = 90. So AC + BC.
ACsq + BCsq = 6*6 = 36. 2ACsq = 36. ACsq = 18
Areaof ABC= 0.5*AC*BC = 0.5*AC*AC = 0.5*18 = 9
|AC|^2 +|CB|^2 = 36. Therefore |AC|=|CB|=square root of 18. Area of triangle = 1/2 by sq. root of 18 by sq. root of 18 = 9
Ans : Area of ABC = 9 square Units
#RightTriangle
Easier method. Once you figure out its a right triangle (2x + x + 45 = 180, x = 45) w/ hypoteneuse of 6, turn it on its side. Sides (base and height) are equal and are 6*sin(45).
1/2 * (6*sin(45))^2 = 9. You're welcome.
Right triangle ADC
Right triangle ABC
3x = 135"
x = 135÷3 = 45°
#Trigonometry
i dont like the perpendicular bisector usage here, i dont see how it is useful in triangles in general, and it only works here cuz gamma at C it 90° and the p. bisector HAPPENS to go through it cuz of that...or i misunderstood its definition.
pythagoras and isosceles is quicker here.
Valde valde facilis Area= 9 sq. Unit. Responsi.
S = (a² • sinB • sinC) / (2 • sinA)
#tangent
Right triangle BDC
A = 6h ÷ 2
tan (45°) = 1 = CD÷3 = 3÷3
Another method..
45-45-90 triangle whose sides are:
AB= 6
AC= 3√2 (solved by using the special property of a 45-45-90 triangle which is AB= AC√2).
BC= 3√2
USING HERON'S FORMULA, IT YIELDS THE SAME RESULT.😊😝
Since AC is perpendicular to BC, Heron’s formula is not not needed. Area = 1/2AC*BC.
2x = 90°
Why? So convolute. The original is right isosceles as well, no need for perpendicular bisector.
9 in secs..😎
Great job! Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
9?
Yes it is!
Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
CD = 3
x = 45°
x variable
Base 6 con 2angoli da 45 gradi è un quadrato quindi 6 x6 =36 36/4=9 area triangolo
I think we have easy solution
THE AREA = 0.5(3sqr2)^2=9
9 sq. units
Thanks Mumtaz for the feedback. I really appreciate that dear. You are awesome 👍 Take care dear and stay blessed😃
h = 3
A = 9
....a bit too simple for my taste but maybe ok for children
Thanks Robert for the feedback. You are awesome and smart 👍 Take care dear and stay blessed😃
Hello
Hello cool guy, thanks for this.
You are awesome 👍 Take care dear and stay blessed😃
Trop long et maladroit,