1st side is = x-8
2nd side is = x
3rd side is = 4(x-8)
Equations - x-8+x+4(X-8)=28
x-8+x+4x-32=28
- Mr williams -
Side # 1 - 40Side # 2 - 20Side # 3 - 16
No. In the first place, the word is "multiply", not "times", and in the second place, to get the width you divide the perimeter by two and then subtract the length (there are alternative methods, but none of them is even close to multiplying the length by the perimeter).
Let's denote the perimeter of the first triangle as P. Since the triangles are congruent, the perimeter of the second triangle is also P. The sum of their perimeters is then 2P. According to the given statement, this sum is three times the perimeter of the first triangle. So we have the equation 2P = 3P. Simplifying, we find that P = 0, which is not a valid solution. Therefore, there is no triangle for which the sum of the perimeters of two congruent triangles is three times the perimeter of the first triangle.
Let's get at the idea by working backward. Suppose we know the scale factor; what will the ratio of perimeters be? For instance, suppose we have two triangles; one has sides of 3, 4, and 5 inches; the other has sides of 33, 44, and 55 inches. The scalefactor is 11: you multiply each side length of the first triangle to get the corresponding side length of the second triangle. Now look at the perimeters. The perimeter of the first triangle is 3+4+5 = 12 inches. The perimeter of the second triangle is 33+44+55 = 132 inches. The ratio of perimeters is 132/12 = 11. Do you notice that it's the same as the scale factor? This will always be true! Here is why. We can write the sides of the second triangle as 3*11, 4*11, and 5*11. Then the perimeter is 3*11 + 4*11 + 5*11 = (3 + 4 + 5)*11 using the distributive property. To find the ratio of perimeters, divide this by the perimeter of the first triangle: (3+4+5)*11 ---------- = 11 3+4+5 Let's continue and think about the ratio of areas. The triangles I chose happen to be right triangles (do you know how to show this?) so the area is half the product of the two shorter sides. Thus the area of the first triangle is (3*4)/2 = 6 square inches. The area of the second triangle is (33*44)/6 = 726 square inches. The ratio of areas is 726/6 = 121. This ratio happens to be 11 squared. It will always be true that the ratio of areas is the square of the scale factor. Again, we can see why this is true. Writing the sides of the second triangle as 3*11 and 4*11, the area is 3*11 * 4*11 = (3*4)*(11*11) Divide this by the area of the first triangle to find the ratio of areas: (3*4)*(11*11) ------------- = 11*11 = 11^2 3*4 Do you see how it works now? What is the answer to your problem?
Example 1:Find the perimeter of a triangle with sides measuring 5 centimeters, 9 centimeters and 11 centimeters.Solution:P = 5 cm + 9 cm + 11 cm = 25 cmExample 2:A rectangle has a length of 8 centimeters and a width of 3 centimeters. Find the perimeter.Solution 1:P = 8 cm + 8cm + 3 cm + 3 cm = 22 cmSolution 2:P = 2(8 cm) + 2(3 cm) = 16 cm + 6 cm = 22 cmIn Example 2, the second solution is more commonly used. In fact, in mathematics, we commonly use the following formula for perimeter of a rectangle:, where is the perimeter, is the length and is the width.In the next few examples, we will find the perimeter of other polygons.Example 3:Find the perimeter of a square with each side measuring 2 inches.Solution:= 2 in + 2 in + 2 in + 2 in = 8 inExample 4:Find the perimeter of an equilateral triangle with each side measuring 4 centimeters.Solution:= 4 cm + 4 cm + 4 cm = 12 cmA square and an equilateral triangle are both examples ofregular polygons. Another method for finding the perimeter of a regular polygon is to multiply the number of sides by the length of one side. Let's revisit Examples 3 and 4 using this second method.Example 3:Find the perimeter of a square with each side measuring 2 inches.Solution:This regular polygon has 4 sides, each with a length of 2 inches. Thus we get:= 4(2 in) = 8 inExample 4:Find the perimeter of an equilateral triangle with each side measuring 4 centimeters.Solution:This regular polygon has 3 sides, each with a length of 4 centimeters. Thus we get:= 3(4 cm) = 12 cmExample 5:Find the perimeter of a regular pentagon with each side measuring 3 inches.Solution:= 5(3 in) = 15 inExample 6:The perimeter of a regular hexagon is 18 centimeters. How long is one side?Solution:= 18 cmLet represent the length of one side. A regular hexagon has 6 sides, so we can divide the perimeter by 6 to get the length of one side ().= 18 cm ÷ 6= 3 cmSummary:To find the perimeter of a polygon, take the sum of the length of each side. The formula for perimeter of a rectangle is: . To find the perimeter of a regular polygon, multiply the number of sides by the length of one side.
Perimeter of a triangle = (length of the first side) plus (length of the second side) plus (length of the third side)
Perimeter of a triangle = (length of the first side) plus (length of the second side) plus (length of the third side)
the perimeter of a triangle is 86 inches. the largest side is four inches less than twice the smallest side. the third side is 10 inches longer than the smallest side. what is the length of each side?
Hz
== == The corresponding angle is 60 degrees.
5,12,13
13, 12,5
Side # 1 - 40Side # 2 - 20Side # 3 - 16
No. In the first place, the word is "multiply", not "times", and in the second place, to get the width you divide the perimeter by two and then subtract the length (there are alternative methods, but none of them is even close to multiplying the length by the perimeter).
Let's denote the perimeter of the first triangle as P. Since the triangles are congruent, the perimeter of the second triangle is also P. The sum of their perimeters is then 2P. According to the given statement, this sum is three times the perimeter of the first triangle. So we have the equation 2P = 3P. Simplifying, we find that P = 0, which is not a valid solution. Therefore, there is no triangle for which the sum of the perimeters of two congruent triangles is three times the perimeter of the first triangle.
It depends on two things. First, one length, by itself, does not define a triangle. And second, it depends on what the question about the triangle is!
No. The first, second and fifth lines are of similar length whilst the third and fourth are of a similar, shorter length.