How to find the perimeter of an acute / obtuse triangle

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Geometry › How to find the perimeter of an acute / obtuse triangle

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An acute isosceles triangle has an area of square units and a base with length . Find the perimeter of this triangle.

$2\sqrt{73} +6$

$\sqrt{73}+6$

$2\sqrt{73}+8$

$\sqrt{73}+8$

Explanation

To solve this problem, first work backwards using the formula:

$Area=\frac{b\cdot h}{2}$

Plugging in the given information and solving for height.

$24=\frac{6h}{2}$

$24\times2=6h$

$h=\frac{48}{6}=8$

Now that you've found the height of the triangle, use the Pythagorean Theorem to find the length of one of the equivalent sides of the triangle.

(Note, when applying the formula divide the measurement of the base in half--so the triangle will have sides of and .

Thus, the solution is:

$c=\sqrt{73}$

In this problem, the isosceles triangle will have two sides with the length of $\sqrt{73}$ and one side length of . Therefore, the perimeter is: $2\sqrt{73}+6$

For the triangle below, the perimeter is . Find the value of .

Explanation

The dashes on the two sides of the triangle indicate that those two sides are congruent. Thus, the length of the missing side is also.

Now, use the information given about the perimeter to set up an equation to solve for . The sum of all the sides will give the perimeter.

A triangle is defined by the following points on a coordinate plane:

What is the perimeter of the triangle?

Explanation

In order to find the perimeter of the triangle, we will first need to find the length of each side of the triangle by using the distance formula.

Recall the distance formula for a line:

$\text{distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The first side of the triangle is the line segment made with $(1, 5)\text{ and }(2, 12)$ as its endpoints.

$\text{side 1}=\sqrt{(2-1)^2+(12-5)^2}=\sqrt{50}=5\sqrt2$

The second side of the triangle is the line segment that has $(2, 12)\text{ and }(-3, 4)$ as its endpoints.

$\text{side 2}=\sqrt{(2-(-3))^2+(12-4)^2}=\sqrt{89}$

The third side of the triangle is the line segment that has $(-3, 4)\text{ and }(1, 5)$ as its endpoints.

$\text{side 3}=\sqrt{(1-(-3))^2+(5-4)^2}=\sqrt{17}$

Now, add up these three sides with a calculator to find the perimeter of the triangle.

$\text{Perimeter}=5\sqrt{2} +\sqrt{89}+\sqrt{17}=20.63$

Make sure to round to places after the decimal.

Inter_geo_tri_series_

Find the perimeter of the triangle shown above.

$7+\sqrt{26.5}in$

$\sqrt{26.5}in$

$\sqrt{26.5+7}in$

$4.5+\sqrt{26.5}in$

Explanation

Since the two side lengths provided in the question are not the hypotenuse, first use the Pythagorean Theorem to find the length of the third side.

$c=\sqrt{26.5}$

Perimeter is equal to .

$2.5+4.5+\sqrt{26.5}=7+ \sqrt{26.5}$

A triangle is defined by the following points in a coordinate plane: .

What is the perimeter of the triangle?

Explanation

In order to find the perimeter of the triangle, we will first need to find the length of each side of the triangle by using the distance formula.

Recall the distance formula for a line:

$\text{distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The first side of the triangle is the line segment made with $(5, 6)\text{ and }(7, 2)$ as its endpoints.

$\text{side 1}=\sqrt{(7-5)^2+(2-6)^2}=\sqrt{20}=2\sqrt5$

The second side of the triangle is the line segment that has $(7, 2)\text{ and }(9, 11)$ as its endpoints.

$\text{side 2}=\sqrt{(9-7)^2+(11-2)^2}=\sqrt{85}$

The third side of the triangle is the line segment that has $(5, 6)\text{ and }(9, 11)$ as its endpoints.

$\text{side 3}=\sqrt{(9-5)^2+(11-6)^2}=\sqrt{41}$

Now, add up these three sides with a calculator to find the perimeter of the triangle.

$\text{Perimeter}=2\sqrt{5} +\sqrt{85}+\sqrt{41}=20.09$

Make sure to round to places after the decimal.

Find the perimeter of the triangle below. Round to the nearest tenths place.

Explanation

Draw in the height to create a right triangle.

Now, using the relationship between the lengths of sides in a triangle, where the long leg is the length of the short leg times $\sqrt3$ and the hypotenuse is two times the length of the short side. We can find out that the height of the triangle is since it is the short leg and the hypotenuse is .

The dashes on two sides of the triangle indicate that these two sides are congruent. The three side lengths of the triangle are $6, 6, \text{ and }6\sqrt3$ .

Now, add up these side lengths to find the perimeter.

$\text{Perimeter}=6+6+6\sqrt3=22.4$

Find the perimeter of the triangle below. Round to the nearest tenths place.

Explanation

Draw in the height to create a right triangle.

Now, using the relationship between the lengths of sides in a triangle, where the long leg is the short leg times $\sqrt 3$ and the hypotenuse is two times the short leg, we can find out that the height of the triangle is $\frac{3}{2}$ and the hypotenuse is .

The dashes on two sides of the triangle indicate that these two sides are congruent. The three side lengths of the triangle are $3, 3, \text{and }3\sqrt3$ .

Now, add up these side lengths to find the perimeter.

$\text{Perimeter}=3+3+3\sqrt3=11.2$

An obtuse Isosceles triangle has an area of square units and a base with length . Find the perimeter of this triangle.

$8+8\sqrt{5}$

$8+2\sqrt{5}$

$4\sqrt{5}$

$4+\sqrt{5}$

Explanation

To solve this solution, first work backwards using the formula: $Area=\frac{b\cdot h}{2}$

$32=\frac{8h}{2}$

Now that you've found the height of the triangle, use the Pythagorean Theorem to find the length of one of the equivalent sides of the triangle.

(Note, when applying the formula divide the measurement of the base in half--so the triangle will have sides of and .

Thus, the solution is:

$c=\sqrt{80}=4\sqrt{5}$

In this problem, the isosceles triangle will have two sides with the length of $4\sqrt{5}$ and one side length of . Therefore, the perimeter is: $8+8\sqrt{5}$ .

A triangle is defined by the following points on a coordinate plane: .

What is the perimeter of the triangle?

Explanation

In order to find the perimeter of the triangle, we will first need to find the length of each side of the triangle by using the distance formula.

Recall the distance formula for a line:

$\text{distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The first side of the triangle is the line segment made with $(-5, 1)\text{ and }(6, -2)$ as its endpoints.

$\text{side 1}=\sqrt{(6-(-5))^2+(-2-1)^2}=\sqrt{130}$

The second side of the triangle is the line segment that has $(6, -2)\text{ and }(7, 0)$ as its endpoints.

$\text{side 2}=\sqrt{(7-6)^2+(0-(-2))^2}=\sqrt{5}$

The third side of the triangle is the line segment that has $(-5, 1)\text{ and }(7, 0)$ as its endpoints.

$\text{side 3}=\sqrt{(7-(-5))^2+(0-1)^2}=\sqrt{145}$

Now, add up these three sides with a calculator to find the perimeter of the triangle.

$\text{Perimeter}=\sqrt{130} +\sqrt{5}+\sqrt{145}=25.68$

Make sure to round to places after the decimal.

Find the perimeter of the triangle below. Round to the nearest tenths place.

Explanation

Draw in the height to create a right triangle.

Now, using the relationship between the lengths of sides in a triangle, where the long leg is the short leg times $\sqrt3$ and the hypotenuse is twice the length of the short leg, we can find out that the height of the triangle is and the hypotenuse is .

The dashes on two sides of the triangle indicate that these two sides are congruent. The three side lengths of the triangle are $8, 8, \text{ and }8\sqrt3$ .

Now, add up these side lengths to find the perimeter.

$\text{Perimeter}=8+8+8\sqrt3=29.9$

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