Maximum Triangle Perimeter

Let the length of the first side is $4x$ and the length of the second side is $x,$ then by applying Triangle Inequality we may obtain:

$4x < x + 20 \Rightarrow x < 6\frac{2}{3}$

Hence, $x = 6$

The largest possible perimeter of the triangle is $4x + x + 20 = 50$

Usamos a desigualdade triangular que nos diz que a soma de dois lados tem que ser maior que um lado.

OK lets start let side be x, 4x , 20 using Triangle inequality - x+4x>=20 => x>=4 ...(1) - 20+x>=4x => 3x<=20 ..(2) - 20+4x>=x => 3x<=-20 ..(3) combining all three we know 'x' should be INTEGER so we have => x<=6.66 x=6 => so perimeter = (4*6) + 4 + 20 =50

In a triangle smaller side + smaller side > biggest side

so, If the 3 side lengths are x , 4x & 20

the range of x is $5<=x<=7$

putting the 3 value of x (5,6,7) we get

if x = $5$ , perimeter of the triangle = $45$

x = $6$ , perimeter of triangle = $50$

x = $7$ , perimeter of triangle = $45$

And get the largest value of the perimeter = $50$

Since, any two sides of a triangle must add up to greater than the other side we can see that 6,20,24 would be the largest possible combination, thus 6+20+24=50

Let $x$ be the shorter unknown side length. By the triangle inequality, we must have $x + 20 > 4x$ and $x+4x>20$ . This tells us that $4<x<\frac{20}{3}$ , and since $x$ must be an integer, we have a maximum perimeter when $x=6$ . $6+24+20 = 50$

Let me present a more rigorous solution (since the Triangle Inequality horse has been beaten WAY too death!) :

Suppose for $x \in \mathbb{N} \Rightarrow$ $x$ , $4x$ , and $20$ are the triangle's side lengths. Also, let $\theta \in (0, \pi)$ be the angle opposite the side of length $20$ . By the Law of Cosines, one obtains:

$20^2 = x^2 + (4x)^2 - 2(x)(4x)cos(\theta)$ ;

and solving for $cos(\theta)$ yields:

$cos(\theta) = \frac{17x^2 - 400}{8x^2}$ (i)

Since we require $-1 < cos(\theta) < 1$ for all $\theta \in (0, \pi)$ , we now substitute (i) into this inequality to produce:

$-1 < \frac{17x^2 - 400}{8x^2} < 1$ (ii)

and solving (ii) for $x$ yields $2\sqrt{2} < x < \frac{20}{3}$ . The maximum perimeter of the triangle is attained for $x = \lfloor \frac{20}{3} \rfloor = 6$ , or $6 + 24 + 20 = \boxed{50}.$

Suppose the length of the second side is x Hence, the length of the first side is 4x We all know that in a triangle the sum of two side lengths cann't be bigger than the third side length. So, x+20>4x =>20>3x =>x<20/3 =>x<6.667 =>x=6 ,as we must take an integer So the largest possible perimeter of the triangle is 4×6+6+20=50

Triangle Inequality: for all triangles

a + b > c where a,b,c are the sides of a triangle

1) x+4x > 20 or x > 4

2) x+20 > 4x or 20/3 > x

1. 4x+20 > x or x > -20/3

Thus -20/3 < 4 < x < 20/3 = 6+2/3

Since they are all integers the max of x = 6

Thus 4x = 24

And 6 + 24 + 20 = 50

Our three side lengths are (x,4x,20). We can use the triangle inequality to determine the maximum integer value for x. x+20>4x, so x<20/3. Since x must be an integer, its maximum value is 6. This gives us side lengths (6,24,20) and a perimeter of 50.

assume the lengths is x,4x,20

the inequality of triangle is a+b>c

c is the longest side of triangle , it just possible for 4x and 20

if 4x is the longest side

so x + 20 > 4x

$\frac{20}{3}$ > x

x < 6,66...

so the largest integer for x is 6

x , 4, 20 = 6, 24, 20

so it perimeter is 50

Let the side lengths of the triangle be $a$ , $4a$ , and $20$ . By the Triangle Inequality, we have $a+4a>20$ $a+20>4a$ $4a+20>a$ Note that the third inequality is essentially trivial because for any positive integer $a$ , $4a>a$ . The first inequality gives $a>4\Rightarrow a\ge5$ The second inequality gives $a<\dfrac{20}{3}\Rightarrow a\le6$ To maximize the perimeter which is $a+4a+20=5a+20$ , we need to maximize $a$ . The maximum $a$ can be is $6$ . We plug $6$ in $5a+20$ to get the answer of $\boxed{50}$ .

It can oly be even numbers until 50! So, 26!

Let $x$ be the fist side, $4x$ be the second, and $20$ be the third. Then, by the Triangle inequality theorem

$3x<20\ \\ 5x>20 \\$

The highest integral value of $x$ that satisfies both condition $(x>4 \quad and \quad x<6\frac{2}{3})$ is $x=6$ . Therefore, the maximum perimeter is 50 .

let one side be a. 4a+a>=20 i.e a>=5. a+20>+4a i.e. a<=6.7. 4a+20>=a i.e. a>=-6.7. drawing the number line, the only possible integer of a is 6 implying that the sides are 6, 20 and 24. therefore, the largest perimeter is 50.

We know that, the sum of two side lengths of a triangle is greater than the other side length...

Let's suppose that, the smaller side length is $x$ ... So, the larger side length is $4x$ ... And given the third side length is $20$ ...

Now let's work out the inequalities... Considering side lengths $x, 4x$ and $20$ ...

1st, $x+4x>20 \Longrightarrow x >5$

2nd, $x+20 > 4x \Longrightarrow x < \frac {20}{3}$

3rd, $4x+20>x \Longrightarrow x > - \frac {20}{3}$

Since, all of the side lengths are integers, we can rewrite the 2nd inequality, as...

$x≤6$

Now, to get the largest possible perimeter, we need to get the largest possible value of $x$ which is $6$ considering the inequalities...

So, the largest possible perimeter is...

$x+4x+20 = 5x+20 = (5 \times 6) +20 = 30+20 = 50$

Hence, the required value is $\fbox {50}$ ...

Let the sides be x, 4x, and 20. Since x+4x>20, therefore x>4. Then this inequality satisfy all integers greater than 4 for x. But x+20>4x, which gives x<20/3 and yields the largest integer x=6. Thus, the sides of the triangle are 6, 24 and 20 with P=50.

By triangle inequality,

4 a + a >20 and

20+ a > 4a

Thus,

5a >20 and

20> 3a

Therefore, the maximum value

a =6

For every triangle ABC with sides a, b and c it goes that the longest side is shorter than or equal than the sum of the other two. Let $a=20$ , $b=x$ and $c=4x$ . To get the largest perimeter we let c be the longest side of ABC. Then we have: $c < a + b <=> x < \frac{20}{3}$

Largest integer less than $\frac{20}{3}$ is $6$ so we get the perimeter of ABC:

$a + b + c = 20 + 6 + 4 \times 6 = 50$

Let the lengths are 20,a,4a with inequality 4a<20+a and 20<5a. Now, we get 4<a<6,67. So, the largest possible integer for a is 6 and we will get the parameter of the triangle is 6+24+20=50

The sum of two sides of a triangle is greater than the third side. Therefore:

(x + 20) > 4x

20 > 3x

Since x has to be an integer, x = 6 and the perimeter = x + 4x + 20 = 50

Let the sides be x , 4x and 20. By the triangle inequality we have x + 4x > 20 ,, x > 4 ,, 20 + x > 4x ,, 20 > 3x ,, 4 < x < |20/3| ,, Since we require x to be at its maximum, and that x is an integer, we obtain x = |20/3|= 6. Plugging in x to the equation 5x + 20 = 5(6) + 20 = 50

Let one side be x. Then other side is 4x. And the third side is already given as 20.

But we, already know that "sum of two sides of a triangle is always greater than the third side." - Triangle inequality

Hence we get the following three inequalities. But for solving the above question only one inequality is sufficient that is -

$20 + x > 4x \Rightarrow x < \frac{20}{3} \Rightarrow x < 6.666...$

Since the sides can only be integers, so x must also be an integer. Therefore $[x] = 6$ That is greatest integer that x can be is equal to 6.

Hence, we get the following sides of the triangle $6, 20, 24$ .

Therefore, the largest possible perimeter is $= 6+20+24 = 50$

Let the sides of the triangle be $x,4x$ and $20$ respectively. Then from triangle inequality, we can have a triangle with sides only in which the third side is less than the sum of the other two So, we have

$x+20>4x$

$\Rightarrow x<(20/3)=6.67...(1)$ And $5x>20$

$\Rightarrow x>4...(2)$

Also, $20+4x>x$

$\Rightarrow 20+3x>0...(3)$

From the above 3 inequalities, the maximum value of x we can have is 6.

So, the maximum perimeter = $x+4x+20 \Rightarrow 5(6)+20 \Rightarrow \boxed{50}$

4a > a + 20

3a = 20

for a an integer , a = 6

the perimeter : 6 + 24 + 20 = 50

The sides are x , 4x and 20 .

As, the sum of two sides of a triangel is greater than the third one .

so , x + 4x >20

$\Rightarrow$ x > 4 ...................... (1)

again , x + 20 > 4x

$\Rightarrow$ x < $\frac{20}{3}$

$\Rightarrow$ x $\leq$ 6 < $\frac{20}{3}$ ...................... (2)

again, 4x + 20 > x

$\Rightarrow$ x > $\frac{- 20}{3}$

but we know , x is a no-negative so, number x > 0 .................... (3)

The perimeter is maximum when x is maximum . From equation (2) we can see that the maximum integer value of x is 6 and (x = 6) also satisfy equation (1) and (3)

so, x = 6

thus, Preimeter = x + 4x + 20 = 5x + 20 = 30 + 20 = 50 .

If the first side of the triangle is $x$ then length of third side is $4x$ . It is given that the length of the third side is 20. According to the triangle inequality, we have $4x + x > 20$
$\Rightarrow$ $x$ > 4

Again using triangle inequality, $20 + x > 4x$
$\Rightarrow$ 20 > 3x According to which maximum possible value of $x$ = 6.

Therefore, perimeter = 20 + 6 + 4(6) = 26 + 24 = 50

a=4b c=20 we have ane condition: a+b>20 5b>20 b>4 a+20>b 4b+20>b 3b>-20 b>-20/3 b>0 b+20>a 3b<20 b<20/3 4<b<20/3 K = a+b+c = 5b+20 = 50

Consider triangle ABC, with AB = 20, BC = x, AC = 4x. From this explanation, I use assumption that AC is the side which have the longest measure, therefore we have inequality that $4x > 20 > x$ . Then, if angle ABC = $\alpha$ , so

$\cos \alpha = \frac{AB^{2}+BC^{2}-AC^{2}}{2.AB.BC}$

$\cos \alpha = \frac{400+x^{2}-16x^{2}}{2.20.x}$

$\cos \alpha = \frac{400-15x^{2}}{40x}$

We know the interval value of cosinus is $-1 \le \cos \alpha \le 1$ , But, in the triangle, $\alpha < 180^{0}$ , and therefore we got range $-1 < \cos \alpha \le 1$ , (since $cos 180^{0} = -1$

When it comes to triangles, there is a rule stating that the sum of any $2$ sides is always bigger than the third side.

Taking this into account, we get $2$ inequalities needed for the triangle to exist

$5x>20$ and $x+20>5x$ , assuming that if the second one is true $4x+20>x$ will always be true. Therefore we get the system

$\left\{ \begin{array}{l l} 5x>20 & \quad \\ x+20>4x & \quad \end{array} \right.$

From it, we can see that $4<x<\frac{20}{3}$ , which leaves us with only $2$ solutions for $x$ , those being $x=5$ and $x=6$ . Since we are looking for the largest possible perimeter , we will use $x=6$ .

Furthermore, the sides of the triangle are $6,24$ and $20$ , which means that the perimeter is

$L=24+6+20 \Rightarrow L=50$

Let the length of the "one side" stated in the problem be $x$ . Thus the triangle has sides: $x$ , $4x$ , and $20$ .

To satisfy the triangle equality and simultaneously generate the largest possible perimeter, $4x$ will have to be the longest side.

Thus, $$4x< x+20$$ $$3x<20$$ $$x< \frac{20}{3}$$ Thus the largest possible value of $x$ is $6$ and this generates a perimeter of: $$P=5x+20$$ $$P=30+20$$ $$P=\boxed{50}$$

Call the sides of the triangle 20, x, and 4x . Then, by Triangle Inequality, 4x + x > 20. So, we get that x > 4. Additionally, from the Triangle Inequality, x + 20 > 4x, giving us $\frac{20}{3}$ > x. Our final inequality is 6.666666 > x > 4. Clearly, the greatest integer x which satisfies this is 6. Thus, the side lengths of the triangle are 20, 24, and 6, so the perimeter is 50 .

The side lengths can be expressed as x,4x,20. Using the Triangle Inequality Theorem, 4x < x + 20 --> 3x<20 --> x< 20/3. Since the sides must be integers, x must be an integer, so x=6. Adding the sides, 6+24+20= 50 .

Let the sides of the triangle be x, 4x, and 20 units. So, the largest value of x which satisfies the Triangle Inequality ( Sum of two sides of a triangle must be greater than the third side ) is found to be 6 . Therefore, the sides of the triangle must be 6, 24, and 20 for its perimeter to be largest. Hence, Perimeter= 50

Let $a$ be the side length of the shorter unknown side. The longer unknown side has length $4a$ . By the triangular inequality and assuming that the triangle is non-degenerate, $a + 20 > 4a$ . Rearranging yields $a < \frac{20}{3}$ which implies that the maximum value of $a$ given the constraints is 6. The largest possible perimeter is $20 + 24 + 6 = 50$ .

Let second side be x and one side be y. A rule for triangles is that the sum of any two sides is greater than the third side. For the largest possible perimeter, we don't want the third side to be the largest, therefore it is one of the smaller two sides. We can then make the equation 20+x>4x, so 20>3x, and x<6.66666... Because the sides are all integers, we want x to be 6, so our longest side is 24, and our perimeter is 6+20+24=50

Let the first two sides of the triangle have lengths x and 4 x . From the triangle inequality theorem, we know that 4 x - x < 20 or 3 x < 20. The largest multiple of three that is less than 20 is 18, so the largest integer that x could be is 6. Thus the largest perimeter that the triangle could have is $5 \times 3$ + 20, or 50.

Due to the Triangle Inequality, the two shorter sides must add up to more than the longest side. In this case, it means that the length of side 2 + 20 must be more than that of side 1, which is 4 times the length of side 2. This means that 20 must be greater than 3 times the length of side 2. The maximum integer length of side 2 for which this is true is 6, making side 1 24. Adding, we get 6+20+24=50.

By the inequality of a triangle we know that any side of it must be lower that the sum of the other two sides. Since that, we get that 20 + x > 4x Solving this inequality we get 6 by the higher integer solution. Since the perimeter equals to the sum of all three sides we get: Perimeter = 20 + 6 + 6*4 = 50

Let the side length of the triangle be of the form a,4a and 20. By triangle Inequality,difference of any two sides<the third side i.e. 3a<20 or, a<6.67 or, a(max)=6 Hence, maximum perimeter=24+6+20=50

By triangle inequality we have $$a + b > c$$ $$b + c > a$$ $$c + a > b$$ Let $a = 4b$ and $c = 20$ . Then $$5b > 20$$ $$20 > 3b$$ $$20 + 3b > 0$$ If we maximize $b$ , then $a$ gets maximized and hence perimeter gets maximized, largest $b$ satisfying $20 > 3b$ is $b = 6$ , so $$(a,b,c) = (24,6,20)$$ We don't forget to check that all triangle inequalities are indeed satisfied and we get maximized perimeter $$24 + 6 + 20 = \boxed{50}$$

Call the length of one side $x$ and the other one $4x$ , notice the total perimeter equals $20 + 5x$ , so in order to obtain the maximum perimeter, we need to maximize $x$ . Through the triangle inequality we obtain $20 + x > 4x$ , then $\frac{20}{3} > x$ The maximum integer value of $x$ that satisfies this inequality is 6, and therefore the maximum possible perimeter is $20 + 5(6) = 50$ .

Let x be the length of side no 2. So the length of side no 1 is 4x. As we know in every triangle the sum of any two sides is greater then the third side. So $20+x>4x => 20>3x => 6 \frac{2}{3} >x$ . Every side's length is an integer so the biggest x is 6. The perimeter of the triangle is equal $20+6+4 \times 6=50$

4x+20>x x+20>4x By putting the values 1,2,3,4,5,6,7,8 then from 7 the value does not satisfied so x=6 6+24+20=50

By the triangle inequality,

$x+20 \geq 4x$

$x \leq \frac{20}{3}$

$x\leq 6.\dot{6}$

since x is an integer,

$x=6$

It is known that a triangle can be formed only when the sum of the lengths of its any two sides is greater than the length of the remaining side.

Now, let 'x' units be the side length of any one of the two unknown sides(say,side A).

Consequently, the other side(say,side B) would be '4x' units long.

We have, length of side C=20 units.

Applying the aforementioned statement of triangle formation,we get

side A + side B > side C => x + 4x > 20 => 5x > 20 => x > 4.

Similarly, side A + side C > side B => x + 20 > 4x => 20 > 3x => x < 6.67.

Since x is to be an integer, therefore, possible values for x = 5 and 6.

We know,perimeter of a triangle ABC = side A + side B + side C

Putting x = 5 : perimeter = 5 + 20 +20 = 45 units.

Putting x =6 : perimeter = 6 + 24 + 20 = 50 units.

Therefore, required perimeter of given triangle = 50 units.

By the triangle inequality, we get

$x + 4x > 20\\ 4x + 20 > x\\ 20 + x > 4x$

This gives us $4 < x \le 6$ , which means the maximum perimeter is $4x + x + 20 = 50$ .

Let x, 4x and 20 be the sides of the triangle, and p be the perimeter.

By triangle inequality,

x+4x>20; x>4

4x+20>x; x>-20/3

and 20+x>4x; x>20/3

Then, getting the intersection of the 3 inequalities, we have:

4<x<20/3. Therefore if x is an integer, then x=5 or 6. To get the maximum perimeter, we must have x=6. So the sides will be 6, 24 and 20. So p=6+24+20=50

The triangle has length 20, x and 4x. From triangle inequality, 20 + x < 4x or 4x + x < 20. From the first inequality, we get a maximum x of 6. Hence, the largest possible perimeter of the triangle is when x = 6, x + 4x + 20 =50

side 1= n

side 2= 4n

side 3= 20

side 1 + side 2 > side 3

side 2 + side 3 > side 1

side 1 + side 3 > side 2

5n > 20

n > 4

4n + 20 > n

n +20 > 4n n< 20/3

so, the largest n is 6 side 1 = 6 side 2 = 24 side 3 = 20 thus, the largest perimeter = 6+24+20 = 50

The length of three sides are x; 4x; and 20. Therefore, x+4x>20 and 4x-x<20 => x>4 and x<6.7 => x=5 or x=6. But we choose x=6 that will lead to the largest possible perimeter of the triangle, which is x+4x+20 = 5x+20 = 30+20 =50.

LET the length of the sides of the triangle be x,4x,20... as we know "THE SUM OF TWO SIDES OF A TRIANGLE IS ALWAYS GREATER THAN THE THIRD SIDE"...SO if the third side(length 20 ) is longest ,then x+4x>20 .. or 5x>20 . or x>4.....(1) OR,if the second side(length 4x) is longest, then x+20>4x.. or 3x<20...or x<6.667 now THE QUESTION asked the largest PERIMETER, then x=6 ONLY(as 4<x<6.667) therefore length of sides will be 6,24,20...hence perimeter...6+24+20=50...ANSWER..