How do you find the hypotenuse when given the perimeter and area?

With the hypotenuse and perimeter question you have a+b+h=P, which is insufficient to determine the values of a and b. If you have area and perimeter, you have 2ab=4A,a+b+h=P,a+b=P−h,a2+b2=h2.

How many right triangles with whole number leg lengths are there such that the area and the perimeter are equal?

4 Answers. These are the Equable triangles. There are only five: (5,12,13),(6,8,10),(6,25,29),(7,15,20), and (9,10,17).

How many Pythagorean triangles are there with the property that the area of the triangle is the same as the perimeter?

There exist exactly 2 Pythagorean triples which define a Pythagorean triangle whose area equals its perimeter: (1):(6,8,10), leading to an area and perimeter of 24. (2):(5,12,13), leading to an area and perimeter of 30.

Is it possible for a triangle to have the same area and perimeter?

Any triangles with the same perimeter and area and with one side the same are congruent. This is no surprise. The area of a triangle is half of the product of the base and the perpendicular height, so given the base and the area, the height is fixed.

Do 2 congruent triangles have the same perimeter?

If two triangles have same perimeter, then they are congruent.

How do you find the perimeter and area of a right angled triangle?

It is calculated with the help of the formula: Area = ½ × base × height. The perimeter of a right-angled triangle is the total length of its boundary or the sum of the lengths of all three sides, which includes the hypotenuse, the height, and the base.

How do you find the sides of a right triangle given the hypotenuse?

If you have the hypotenuse, multiply it by sin(θ) to get the length of the side opposite to the angle. Alternatively, multiply the hypotenuse by cos(θ) to get the side adjacent to the angle.

How do you find the area and perimeter of a triangle?

If we equate the expressions for area and perimeter for a right triangle with legs a and b, then 1 2 a b = a + b + a 2 + b 2. Solving that equation for b yields b = 4 a − 2 a − 4.

How do you find the integer value of a right triangle?

Such a right-angled triangle will have sides a, b, and a 2 + b 2. So we have the relation 1 2 a b = a + b + a 2 + b 2. a = 4 ( b − 2) b − 4. Hence a is an integer only when c = − 8, − 4, − 2, − 1, 1, 2, 4 or 8.

What is the value of 1+2+4 times the perimeter?

Adding those up gives 1+2+4 = 7. Similar results can be found for other small perimeters. It is also possible to derive a general formula which has the curious property that $a(2n-3) = a(2n)$, or in other words, starting with a triangle with an odd perimeter, we can find a related triangle with a perimeter 3 more just by adding 1 to each side.

How do you find the middle side of a 15-meter perimeter?

For example with a perimeter of 15, the longest side must be 5, 6 or 7. If it is 5 then the middle side can be 5; if it is 6 then the middle side can be 5 or 6; if it is 7 then the middle side can be 4, 5, 6 or 7. Adding those up gives 1+2+4 = 7. Similar results can be found for other small perimeters.