Can you find the area of the right triangle? | (Algebra) |
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- Learn how to find the area of the triangle. Important Geometry and Algebra skills are also explained: Pythagorean Theorem; area of a triangle formula. Step-by-step tutorial by PreMath.com
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Very nice and enjoyable
I like these exercises
Thanks Sir for your efforts
With my respects ❤❤❤
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Funny how what was once tedious is now satisfying.
Sir, there is no need to calculate "a" and "b" seperately.
a^2+b^2=148^2
(a+b)^2=188^2
2ab= (188+148) (188-148)
2ab = 336×40
1/2 ab= 3360 units
I use this method
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That's what I did!
a*a + 2ab+ b*b = 188*188; a*a+b*b=148*148 (right triangle rule) so we get 2ab=188*188-148*148 area = ab/2 = (188*188-148*148)/4
Very clear there is no need to calculate a and b
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a²+b²=21904
a+b=188
ab/2=?
(a+b)²=35344
a²+2ab+b²=35344
21904+2ab=35344
2ab=13440
ab/2=3360
S=3360 square units
That's how I did it. That way you completely bypass having to find a and b.
@@johnbutler4631 Yes
Exactly wat I did and it what easier than resolving a quadratic equation.
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Quite so. It is shorter in steps to get at the answer rather than calculating individual values of a and b.
(a+b)^2 = a^2 + b^2 + 2ab => (148)^2 + 2ab = (188)^2 => 2ab = (188)^2 - (148)^2 => ab = (((188+148)(188-148)))/2 => ab = 336×20 => area of the triangle=> 336×10 = 3360 u^2
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Wow very beautiful sharing thank you for sharing 💘💘💘
Thanks for visiting
3,360
The perimeter of the triangle = (a+ b) or 188 + 148 = 336
The factors of 148 = 2, 37, 74
The Pythagorean triplet, 12, 35, 37 and a sum of 84
84 * 4 = 366
The triangle is a 12-35-37 right triangle scaled up by 4.
Hence, the bases are 48 and 140
Hence, area = 24 * 140 = 3,360
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(a+b)^2= a^2 +b^2+2ab
a^2+b^2=148^2
By solving these 2 equations we get (ab) as 6720
Then for area we can do ab/2 or 6720/2 = 3360.
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a+b=188--->a=188-b---> (188-b)²+b²=148²---> b=48 ---> a=140---> a*b/2=140*48/2= 3360 ud².
Gracias y saludos.
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a+b=188 (1)
a^2+b^2=(148)^2=21904
{1) (a+b)^2=(188)^2
a^2+2ab+b^2=35344
2ab=35344-21904
ab=6720 (2)
(1) b=188-a
(2) a(188-a)=6720
a^2-188a+6720=0
a=140 ; a=48
b=48 ; b=140
Area of the triangle=1/2(140)(48)=3360 square units.❤❤❤
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△ACB is a right triangle.
a² + b² = c²
a² + b² = 148²
= 21904
It is given that a + b = 188.
(a + b)² = a² + 2ab + b²
188² = c² + 2ab
35344 = 21904 + 2ab
2ab = 13440
(2ab)/4 = 13440/4
(ab)/2 = 3360 (This is the area formula in terms of a & b)
So, the area of the triangle is 3360 square units.
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Fun.
More please.
Keep watching...
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area = ½(a.b)
a²+b² = 148²
(a+b)² - 2ab = 148²
188² - 148² = 2ab
ab = (188² - 148²)/2
area = ½(188² - 148²)/2 = 3360
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a^2+b^2=(148)^2
a^2+b^2+2ab =(188)^2
ab/2=(188-144)*(188+144)/4 . a and b aren't needed
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a^2+b^2=148^2;
(a+b)^2=188^2;
2ab= 188^2-(a^2+b^2);
2ab= 188^2-148^2= 13440;
S= 2ab/4=13440/4=3360
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148^2 = 21,904
Try a^2 + b^2 100 and 88 due to 10,000 + 7,744 respectively = 17,744.
As a^2 + b^2 needs to be higher, the numbers need to be farther apart, and they can't both be odd. As one needs to be even, both must be because they add up to 188.
As c^2 ends in a 4, and even square numbers must end in 0, 4, or 6, that eliminates many of the possibilities.
Try 140 and 48
19,600 + 2,304
That seems to be the solution to side lengths as 140 + 48 = 188 and 19,600 + 2,304 = 21,904.
Area is (140*48)/2
7*48=336, so 3,360 un^2
Probably not the type of solution you seek, as a fair chunk of it (which I didn't show) was using maths logic to eliminate possible solutions, but I suppose my way is valid too.
Thank you.
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You can also solve it more quickly with no need to calculate a and b :
(a+b)² = 188² = a² + b² + 2ab = c² + 2ab
⇒ 188² = 148² + 2ab ⇒ 2ab = 188² - 148²
⇒ ab/2 = (188² - 148²)/4 = 3360 = Area
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a²+b²= 21904
(188)²= a²+2ab+b²
35344= 21904+2ab
13440= 2ab
Since A∆= ab/2, divide both sides by 4
13440/4= 2ab/4
A∆= 3360 units²
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Let's find the area:
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Since the triangle is a right triangle, we can apply the Pythagorean theorem:
AC² + BC² = AB²
AC² + (188 − AC)² = 148²
AC² + 35344 − 376*AC + AC² = 21904
2*AC² − 376*AC + 13440 = 0
AC² − 188*AC + 6720 = 0
AC = 94 ± √(94² − 6720) = 94 ± √(8836 − 6720) = 94 ± √2116 = 94 ± 46
So we have two possible solutions:
AC = 94 + 46 = 140 ⇒ BC = 188 − AC = 188 − 140 = 48
AC = 94 − 46 = 48 ⇒ BC = 188 − AC = 188 − 48 = 140
In both cases the area turns out to be:
A = (1/2)*AC*BC = (1/2)*140*48 = 3360
Best regards from Germany
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148=4×37
188=4×47
37=36+1÷6^2^+1^2
One side is 2×6=12
37^2-12^2=49×25
3rd side is 35.
Area =16×1/2×(12×35)=16×6×35
=16×210=3320 sq units
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a²+b² = a²+2ab+b² = 148²
(a+b)² = 188²
2ab + 188² = 148²
ab = ½(188² - 148²) = 6720
Area ∆ = ½ab = 3360 u²
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Area = (1/2).a.b = (1/4).[(a + b)^2 - (a^2 + b^2)]
= (1/4).[188^2 - 148^2)] = (1/4).[35344 - 21904)]
= (1/4).13340 = 3360
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We can just use heron's formula, because you need to add up all 3 sides to get the semiperimeter, which in this case they've already provided
Thanks for the feedback ❤️
Hi. You have the semi perimeter pretty easy: 168 but don’t you need to know all 3 side
Lengths individually for the formula? Or is this one of the alternate formulas you used?
Actually I see the only unknown in the equation is ab which you can solve for ok although it’s a bit messy
a^2 + b^2 = 148^2 ***1
(a+b)^2 = 188^2 ***2
2-1
2ab = (40)(336)
ab/2 = 3360 = A
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a^2 + b^2 = c^2 => (a+b)^2 - 2ab = c^2 => 188^2 - 4S = 148^2 => 4S = (188-148)(188+148) => S = 3360
Площадь- полупериметр на радиус вписаной окружности: S=p×r.
p= (188+148)/2=168
r=(a+b-c)/2=(188-148)/2=20
S=168×20=3360
3360
another method
area of the triangle = 2ab/4
a^2 + b^2 = c^2
c^2 = 148^2
(a+b)^2 = 188^2
a^2 + b^2 + 2ab = 188^2
c^2 = 188^2 - 2ab
2ab = 188^2 - 148^2
2ab = 13,440
2ab/4 = 3,360 Answer
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Thank you!
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Triangle abc ( a² + b².) = c² = 148²
( a + b ) ² = ( a² + b² ) + 4 ab
( a + b )² - ( a² + b² ) = 4ab
((( a + b )² - ( c²)/4) = ab
( 188² - 148² ) / 4 = ab
( 35344 - 21904 ) ÷ 4 = 3360
Several viewers have pointed out that we don't need the lengths of a and b to calculate the area. Actually, if a circle is constructed with the hypotenuse as the diameter, the vertex of the 90° angle will lie on the circle. The vertex can be moved on the circle and the triangle remains a right triangle. The minimum perimeter is just slightly more than twice the length of the hypotenuse. The maximum perimeter is reached when ΔABC becomes an isosceles right triangle and, if the hypotenuse has length c, the maximum perimeter is c(1 +√2). For a triangle with hypotenuse 148, the perimeter can range continuously from just over 296 to a little more than 357. The given triangle has a perimeter of 148 + 188 = 336 and is the Pythagorean triple 12 - 35 - 37 multiplied by 4. However, there is no need for a and b to be integers even if the hypotenuse, perimeter and area are integers. Example: hypotenuse 148, a + b = 172, so perimeter = 320. (a + b)² = (172)², so a² + 2ab + b² = 29584. Hypotenuse c = 148, c² = 21904, so c² = a² + b² = 21904, therefore 21904 + 2ab = 29584, 2ab = 7680, area = ab/2 = 7680/4 = 1920.
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a² + b² = c²
a² + b² = 148²
A = ½(ab)
(a+b)² = 188²
a² + b² + 2ab = 188²
188² - (a² + b²) = 2ab
¼(188² - 148²) = ¼(2ab)
¼(188+148)(188-148) = ½(ab)
¼(336)(40) = A
336*10 = A
3360 = A 9:38
I dug deep into my unconscious psyche and used an Ouija Board for a positive transformation from question Area=? ...to problem solved! 🙂
😀
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Sir,
Starting with application of Pythagorean theorem followed by 2ab to be added both sides for completing the square with intervention of (a*a)-(b*b) to avoid long calculating process,will lead us to find ab/2 =3360 units square directly without approaching the values of a and b separately. 🙏with honour
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I got this one!
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Il y a plus simple: Rac (a2+b2) = 148 ; a+b = 188 ; surface =ab/2
Rac ((a+b)2- 2ab) = 148 ; Rac (188*188 - 2av )= 148 ; en élévant au carré: 188*188 - 2ab = 148*148 ; soit 2ab= (188*188) -(148*148) = 35344 - 21904 = 13540
surface ab/2 = 13540/4 = 3360
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My way of solution ▶
a+b= 188
c= 148
A(ΔABC)= ?
⇒
(a+b)²= a²+2ab+b²
a+b= 188
a²+b²= c²
c= 148
⇒
188²= 148²+2ab
2ab= 188²-148²
2ab= (188-148)*(188+148)
2ab= 40*336
ab= 20*336
ab= 6720
A(ΔABC)= ab/2
A(ΔABC)= 6720/2
A(ΔABC)= 3360 square units
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3360...
Heh, finally said the answer before anyone else for the first time
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(188^2-148^2)/4
The area is 3360 square units. This is kind of similar to that other problem that I needed to correct myself on. That "given the perimeter find the area" problem. No lie.
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Area can be calculated without finding the other sides also.
Thanks for the feedback ❤️
STEP-BY-STEP RESOLUTION PROPOSAL :
01) a^2 + b^2 = 21.904 (= 148^2)
02) (a + b)^2 = 35.344 (= 188^2)
03) a^2 + 2ab + b^2 = 35.344
04) (a^2 + b^2) + 2ab = 35.344
05) 21.904 + 2ab = 35.344
06) 2ab = 35.344 - 21.904
07) 2ab = 13.440
08) ab = 6.720
09) A = ab / 2
10) A = 6.720 / 2
11) A = 3.360
OUR ANSWER :
Area of Right Triangle equal 3.360 Square Units.
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ar.=3360 sq unit
Чисто алгебра,не интересно
Certains vont me trouver cruel, mais dans ce problème on ne pose que la question facile, en effet il suffit de connaître le théorème de Pythagore et de maîtriser le calcul élémentaire. il aurait fallu et c'est semble-t-il un défaut courant que je constate chez mes jeunes collègues qui font des vidéos, il ne se pose pas le problème de la construction de la figure et encore moins de son éventuelle existence ou non existence . Ici il faut dire que l'on ne peut pas faire de construction géométrique classique de la figure c'est à dire à la grecque ancienne "au compas et à la règle".Il faut calculer un angle dont les lignes trigonométriques sont hélas irrationnelles non algèbriques , donc on doit se contenter d'une valeur approchée certes très précise à la calculatrice ....
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