In ABC we have 6.x = 180°, so x = 30°. Now AB/sin(30°) = BC/sin(60°) = AC/sin(90°) = 32, so AB = 16 and BC = 16.sqrt(3) and the perimeter is 16..(3 + sqrt(3)). The area is (1/2).AB.BC as the triangle is a right triangle, so the area is (1/2).(16).(16.sqrt(3) = 128.sqrt(3)). (No difficulty.)
Let's face this challenge: . .. ... .... ..... From the interior angle sum of the triangle ABC we can conclude: 180° = ∠BAC + ∠ABC + ∠ACB = 2x + 3x + x = 6x ⇒ x = 180°/6 = 30° ⇒ ∠BAC = 2x = 60° ∧ ∠ABC = 3x = 90° ∧ ∠ACB = x = 30° Therefore ABC is a 30°-60°-90° triangle and we obtain: AB = AC/2 = 32/2 = 16 BC = AB*√3 = 16√3 Now we are able to calculate the area A and the perimeter P of the triangle: A = (1/2)*AB*BC = (1/2)*16*16√3 = 128√3 P = AB + AC + BC = 16 + 16√3 + 32 = 48 + 16√3 Best regards from Germany
Total angle (sum of angles) in a triangle =180 degrees. P = Perimeter = AB+BH+AH. Here, it is x+2x+3x = 6x. Hence x = (180/6) degrees = 30 degrees. Name the vertices at angles x as A, that at 2x as B & that at 3x as H (sides designated AB, BH & AH). The angles are x = 30 degrees, 2x = 60 degrees, 3x = 90 degrees and the "side 32" lies between x & 2x facing 3x = 90 degrees. Its a right-angled triangle. And hypotenuse = AB = 32. Construction: The triangle is flipped over at H (angle 3x = 90 deg.), around A (angle x) & joined to the original triangle. Call the flipped vertex as C. Then in the composite triangle, at vertex A, the angles get added to 90+90 degrees = 2(3x) = 180 degrees. Hence the continuation is a straight line BH+HC = 2 BH The new triangle is an equilateral triangle with angles each 2x = 60 degrees. Hence it has equal sides AC = 2BH = AB = 32. BH = 32/2 = 16 = CH. In the Pythagoras (right-angled) triangle ABH, hypotenuse^2 = altitude^2+base^2 hypotenuse^2 - altitude^2 = base^2; AH is the base AH^2 = AB^2 - BH^2 = 32^2 - (32/2)^2 = (16 x 2)^ - 16^2 = 16^2(2^2 - 1) = 16^2(3) AH = √[(16^2)3] = 16√3. P = AB+BH+AH = 32+16+16√3 = 16(3+√3). Area of (the first) triangle, ABH = ½[altitude x base] = ½[BH x AH] = ½[16 x 16√3] = ½.16^2[√3] = 128√3
It is very simple. First name the corner where you have the angle 2x as point A. Then the point in which you have the angle x as B. And the third one which has 3x as Point C. From A draw a perpendicular to meet BC at a point extended byCB as D. Let us assume AC as P. And BC as Q. In the triangle ABC the sides are proportional to their opposite angle. So AB which is in length 32/3x=P/x=Q/2x. The sum of three angles is 6 •x=180.So the angle 3•x=90 being a Rt Angle. Now two Rt.angled triangles are there only two unknown involving P one side and another Q. So you can solve easily for Pand Q. Once you know Pand Q and the other side given as 32 you get perimeter and half of it semi perimeter. Use the Area of a triangle using the three sides formula. √s(s-a) ( s-b)(s-c) /2. You get the area and Perimeter.
Extend CB to D and angle ABD =2x +x =3x Now we see AB is on CBD and two adjacent angles ABD and ABC are 3x Hence AB is perpendicular to BC. Hence angle ABC=3x =90 degrees Now angle ACB =x =3x/3=90/3=30 degrees Angle BAC =2x =30*2=60 degrees . ABC is a 30-60-90 triangle Side opposite to 90 is 32 Hence side opposite to 30 is 32/2=16 side opposite to 60 is 16√3 Area =1/2*16*16√3 =128√3 sq units Perimeter =32+16+16√3 =48+16√3=16(3+√3) units
x = 30deg so it's a 30,60,90 drawn totally out of scale (well, you did warn us!). 32 is the hypotenuse as it's opposite the 90deg. Therefore, the two shorter sides are 16 and 16*sqrt(3) Perimeter is 48 + 16*sqrt(3) which approximate to 75.71 Area is 8*16*sqrt(3) = 128*sqrt(3) which approximates to 221.7 un^2. The key to this is figuring out the angles first: i.e. 6x = 180deg
Solution: The sum of the interior angles of a triangle is always 180°, therefore: x + 2x + 3x = 180° 6x = 180° x = 30° Thus, we are dealing with a special triangle 30° - 60° - 90° The segment opposite 90° is 32 and it is "2x" The segment opposite 30° is "x" And the segment opposite 60° is "x√3" Therefore 2x = 32 x = 16 The sides of the triangle ABC is 32, 16 and 16√3 Perimeter = 32 + 16 + 16√3 Perimeter = 48 + 16√3 Units ✅ Perimeter ≈ 75.7128 Units ✅ Area = ½ base height Area = ½ 16√3 . 16 Area = 128√3 Square Units ✅ Area ≈ 221.7025 Square Units ✅
Solution: x+2x+3x = 6x = 180° |/6 ⟹ x = 30° ⟹ the triangle is the famous 30°-60°-90° triangle. This means that AB = 32/2 = 16 and BC = √(32²-16²) = √786 = √(16*16*3) = 16*√3. Perimeter = 32+16+16*√3 = 16*2+16+16*√3 = 16*(3+√3) ≈ 75.7128 Area = AB*BC/2 = 16*16*√3/2 = 128*√3 ≈ 221.7025
Καλησπέρα σας. Διαφωνώ πλήρως με το παραπλανητικό σχήμα που δόθηκε. Άσχετα αν ένας μαθηματικός μπορεί εύκολα να καταλάβει τι συμβαίνει, φανταστείτε να δοθεί έτσι σε ένα διαγώνισμα. Ο μέσος μαθητής θα αμφιβάλλει για αυτό που βλέπει και θα χάσει ώρα ψάχνοντας τι πάει στραβά. Κατά τη γνώμη μου, θα μπορούσε να δοθεί η απλή αυτή άσκηση περιγραφικά, χωρίς σχήμα, ως εξής. "Δίνεται τρίγωνο ΑΒC με ΑC=32 και γωνίες: C=x, A=2x και B=3x. Να βρείτε το εμβαδόν και την περίμετρο του τριγώνου ABC".
𝗩𝗲𝗿𝘆 𝗶𝗻𝘁𝗲𝗿𝗲𝘀𝘁𝗶𝗻𝗴 𝘀𝘂𝗺
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waooo very helpful vedio thanks for sharing ❤❤❤
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In ABC we have 6.x = 180°, so x = 30°. Now AB/sin(30°) = BC/sin(60°) = AC/sin(90°) = 32, so AB = 16 and BC = 16.sqrt(3) and the perimeter is
16..(3 + sqrt(3)). The area is (1/2).AB.BC as the triangle is a right triangle, so the area is (1/2).(16).(16.sqrt(3) = 128.sqrt(3)). (No difficulty.)
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Ohh god! Was it so easy!? I used trigonometry and then got the right answer.. This way is much easier! Thank you sir!
Let's face this challenge:
.
..
...
....
.....
From the interior angle sum of the triangle ABC we can conclude:
180° = ∠BAC + ∠ABC + ∠ACB = 2x + 3x + x = 6x ⇒ x = 180°/6 = 30°
⇒ ∠BAC = 2x = 60°
∧ ∠ABC = 3x = 90°
∧ ∠ACB = x = 30°
Therefore ABC is a 30°-60°-90° triangle and we obtain:
AB = AC/2 = 32/2 = 16
BC = AB*√3 = 16√3
Now we are able to calculate the area A and the perimeter P of the triangle:
A = (1/2)*AB*BC = (1/2)*16*16√3 = 128√3
P = AB + AC + BC = 16 + 16√3 + 32 = 48 + 16√3
Best regards from Germany
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Area 245.76
Perimeter 76.8
For those complaining about graphics. Graphics are and should never be drawn to scale. Trust your math and not your eyes.
True dat. But this was egregiously misleading.
Total angle (sum of angles) in a triangle =180 degrees.
P = Perimeter = AB+BH+AH.
Here, it is x+2x+3x = 6x.
Hence x = (180/6) degrees = 30 degrees.
Name the vertices at angles x as A, that at 2x as B & that at 3x as H (sides designated AB, BH & AH).
The angles are x = 30 degrees, 2x = 60 degrees, 3x = 90 degrees and the "side 32" lies between x & 2x facing 3x = 90 degrees.
Its a right-angled triangle.
And hypotenuse = AB = 32.
Construction: The triangle is flipped over at H (angle 3x = 90 deg.), around A (angle x) & joined to the original triangle. Call the flipped vertex as C. Then in the composite triangle, at vertex A, the angles get added to 90+90 degrees = 2(3x) = 180 degrees.
Hence the continuation is a straight line BH+HC = 2 BH
The new triangle is an equilateral triangle with angles each 2x = 60 degrees. Hence it has equal sides
AC = 2BH = AB = 32.
BH = 32/2 = 16 = CH.
In the Pythagoras (right-angled) triangle ABH,
hypotenuse^2 = altitude^2+base^2
hypotenuse^2 - altitude^2 = base^2; AH is the base
AH^2 = AB^2 - BH^2 = 32^2 - (32/2)^2 = (16 x 2)^ - 16^2 = 16^2(2^2 - 1) = 16^2(3)
AH = √[(16^2)3] = 16√3.
P = AB+BH+AH = 32+16+16√3 = 16(3+√3).
Area of (the first) triangle, ABH = ½[altitude x base] = ½[BH x AH] = ½[16 x 16√3] = ½.16^2[√3] = 128√3
Easy.
After figuring out it was a 30-60-90 rt∆, I solved this myself and skipped to the end to double check if my answers match and they did.
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S=128√3≈221,696≈221,7
P=48+16√3=16(3+√3)≈75,712≈75,71
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Peculiarly drawn 90' triangle.
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It is very simple. First name the corner where you have the angle 2x as point A. Then the point in which you have the angle x as B. And the third one which has 3x as Point C. From A draw a perpendicular to meet BC at a point extended byCB as D. Let us assume AC as P. And BC as Q. In the triangle ABC the sides are proportional to their opposite angle. So AB which is in length 32/3x=P/x=Q/2x.
The sum of three angles is 6 •x=180.So the angle 3•x=90 being a Rt Angle. Now two Rt.angled triangles are there only two unknown involving P one side and another Q. So you can solve easily for Pand Q. Once you know Pand Q and the other side given as 32 you get perimeter and half of it semi perimeter. Use the Area of a triangle using the three sides formula. √s(s-a) ( s-b)(s-c) /2. You get the area and Perimeter.
Extend CB to D and angle ABD =2x +x =3x
Now we see AB is on CBD and two adjacent angles ABD and ABC are 3x
Hence AB is perpendicular to BC.
Hence angle ABC=3x =90 degrees
Now angle ACB =x =3x/3=90/3=30 degrees
Angle BAC =2x =30*2=60 degrees .
ABC is a 30-60-90 triangle
Side opposite to 90 is 32
Hence side opposite to 30 is 32/2=16
side opposite to 60 is 16√3
Area =1/2*16*16√3
=128√3 sq units
Perimeter =32+16+16√3
=48+16√3=16(3+√3) units
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Bom dia Mestre
Obrigado pela aula
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Got it using trigonometric principles.
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x = 30deg so it's a 30,60,90 drawn totally out of scale (well, you did warn us!).
32 is the hypotenuse as it's opposite the 90deg.
Therefore, the two shorter sides are 16 and 16*sqrt(3)
Perimeter is 48 + 16*sqrt(3) which approximate to 75.71
Area is 8*16*sqrt(3) = 128*sqrt(3) which approximates to 221.7 un^2.
The key to this is figuring out the angles first: i.e. 6x = 180deg
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x + 2x + 3x = 180°
It is a 30° - 60° - 90° triangle
The three sides are 16 - 16√3 - 32
Perimeter = 48 + 16√3
1969 I was 20 klicks up the Mekong and had to secure the perimeter . So similarly as @ 2:16 , I was stared at like whoose in charge here? 😊
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Solution:
The sum of the interior angles of a triangle is always 180°, therefore:
x + 2x + 3x = 180°
6x = 180°
x = 30°
Thus, we are dealing with a special triangle 30° - 60° - 90°
The segment opposite 90° is 32 and it is "2x"
The segment opposite 30° is "x"
And the segment opposite 60° is "x√3"
Therefore 2x = 32
x = 16
The sides of the triangle ABC is 32, 16 and 16√3
Perimeter = 32 + 16 + 16√3
Perimeter = 48 + 16√3 Units ✅
Perimeter ≈ 75.7128 Units ✅
Area = ½ base height
Area = ½ 16√3 . 16
Area = 128√3 Square Units ✅
Area ≈ 221.7025 Square Units ✅
Solution:
x+2x+3x = 6x = 180° |/6 ⟹
x = 30° ⟹ the triangle is the famous 30°-60°-90° triangle. This means that AB = 32/2 = 16 and BC = √(32²-16²) = √786 = √(16*16*3) = 16*√3.
Perimeter = 32+16+16*√3 = 16*2+16+16*√3 = 16*(3+√3) ≈ 75.7128
Area = AB*BC/2 = 16*16*√3/2 = 128*√3 ≈ 221.7025
My way of solution ▶
For the given triangle ΔABC, the angles are:
∠CAB= 2x
∠ABC= 3x
∠BCA= x
⇒
2x+3x+x= 180°
x= 30°
⇒
∠CAB= 60°
∠ABC= 90°
∠BCA= 30°
ΔABC is a right triangle.
sin(30°)= [AB]/[CA]
[CA]= 32
⇒
1/2= [AB]/32
[AB]= 16
cos(30°)= [BC]/[CA]
[CA]= 32
⇒
√3/2= [BC]/32
[BC]= 16√3
P(ΔABC)= [AB]+ [BC] + [CA]
P(ΔABC)= 16+ 16√3 + 32
P(ΔABC)= 48+ 16√3
P(ΔABC)= 16(3+√3)
P(ΔABC)≈ 75,713 length units
A(ΔABC)= [AB]*[BC]/2
A(ΔABC)= 16*16√3/2
A(ΔABC)= 128√3
A(ΔABC)≈ 221,70 square units
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180÷6=30×2=60÷30=2×2=4+1×1=5squroth=2.236+3=5.236)(32÷2.236×5.236=74.9338
angle 180=nx?
Area 245.76
Perimeter. 76.8
STEP-BY-STEP RESOLUTION PROPOSAL :
01) X + 2X + 3X = 180º ; 6X = 180º ; X = 30º
02) Triangle (ABC) Angles = (30º ; 60º ; 90º)
03) sin(30º) = AB / 32 ; AB = 32 * sin(30º) ; AB = 16
04) sin(60º) = BC / 32 ; BC = 32 * sin(60º) ; BC = (32 * sqrt(3)) / 2 ; BC = 16sqrt(3)
05) A = (16sqrt(3) * 16) / 2
06) A = 128sqrt(3) ; A ~ 221,70 sq un
07) P = 32 + 16 + 16sqrt(3) ; P = 48 + 16sqrt(3) ; P ~ 75,8 lin un
Therefore,
Area ~ 221,70 Square Units
Perimeter ~ 75,8 Linear Units
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Area =128.*3(*= read as square root )
Perimeter =16(3+*3).......May be
Καλησπέρα σας. Διαφωνώ πλήρως με το παραπλανητικό σχήμα που δόθηκε. Άσχετα αν ένας μαθηματικός μπορεί εύκολα να καταλάβει τι συμβαίνει, φανταστείτε να δοθεί έτσι σε ένα διαγώνισμα. Ο μέσος μαθητής θα αμφιβάλλει για αυτό που βλέπει και θα χάσει ώρα ψάχνοντας τι πάει στραβά. Κατά τη γνώμη μου, θα μπορούσε να δοθεί η απλή αυτή άσκηση περιγραφικά, χωρίς σχήμα, ως εξής. "Δίνεται τρίγωνο ΑΒC με ΑC=32 και γωνίες: C=x, A=2x και B=3x. Να βρείτε το εμβαδόν και την περίμετρο του τριγώνου ABC".
X=30..3x=90???
Yes. That's why he redrew it as a right triangle.
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Very beautiful video stay connected Sir ❤❤🎉🎉🎉
took me 10 seconds
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Sorry, but it's not a good problem at solving at all....