How to Find the Area of a Right Triangle Using the Pythagorean Theorem | Geometry | https://epicentreconcerts.org

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Steps to Finding the Area of a Right Triangle Using the Pythagorean Theorem

Step 1: Because you must find all three side lengths of the triangle, begin by labeling those side lengths a, boron, and c. The longest side, hypotenuse, must be labeled vitamin c, and the stature of the triangle labeled a. The base of the triangulum will be labeled barn. You will be missing the lengths for either a or boron .
Step 2: Plug in the values of the variables into the Pythagorean Theorem equation and solve :
$ $ a^2 + b^2 = c^2 $ $
Since a= height and b= infrastructure, we can rewrite the equation as :
$ $ h^2 + b^2 = c^2 $ $
fireplug this count into the miss side distance in the diagram. You have now found the hypotenuse, the nucleotide, and the stature .
Step 3: Find the area of the triangle using the formula :
$ $ A = \frac { 1 } { 2 } bh $ $

Vocabulary and Equations for Finding the Area of a Right Triangle

Hypotenuse : The longest side of the triangle that connects the two legs
Variable : unknown share of an equation
Base : the bottom of a polygon
Height : the measure of how grandiloquent a figure is
Pythagorean Theorem :
$ $ a^2 + b^2 = c^2 $ $
This equality is used to find the lacking side lengths of a triangulum .
Area of a Right Triangle : The area of a right triangulum, or triangles in general, can be determined using the adopt recipe .
$ $ A = \frac { 1 } { 2 } bh $ $
The exemplar problems below will demonstrate how to find the area of a right triangle by using the Pythagorean Theorem. Example 1 will demonstrate how to find the area when given the lengths of the height and the hypotenuse. case 2 will demonstrate how to find the area when given the lengths of the al-qaeda and the hypotenuse.

Example Problem 1: Find the Area of the Triangle by Solving for the Base

Step 1: Because you must find all three side lengths of the triangle, begin by labeling those side lengths { equivalent } a, bacillus, { /eq } and { eq } c { /eq }. The longest side, hypotenuse, MUST be labeled { equivalent } c { /eq }. Label the miss side duration { equivalent } b { /eq }, and the early side length { equivalent } a { /eq } .

right triangle with lengths a and c

Step 2: Plug in the values of the variables into the Pythagorean Theorem equality and solve for the lacking side, { equivalent } b-complex vitamin { /eq }
$ $ a^2 + b^2 = c^2 $ $
Since { equivalent } a= \text { stature } { /eq } and { eq } b= \text { basis } { /eq }, we can rewrite the equation as
$ $ \begin { align* } h^2 + b^2 & = c^2\\ 3^2 + b^2 & = 5^2\\ 9 + b^2 & = 25\\ b^2 & = 16\\ \sqrt { b } & = \sqrt { 16 } \\ b & = 4\ \mathrm { in } \end { align* } $ $
spark plug this number into the miss side duration in the diagram. You have now found the hypotenuse, the base, and the stature. Label these in the triangle to help with finding the area .

Label the sides of the triangulum

Step 3: Find the area of the triangle, using the formula :
Remember to designate the stature as { equivalent } heat content { /eq } and the foundation as { equivalent } b-complex vitamin { /eq } .
$ $ A = \frac { 1 } { 2 } bh $ $
Plug your side lengths into the formula .
$ $ \begin { align* } A & = \frac { 1 } { 2 } ( 4 ) ( 3 ) \\ A & = \frac { 1 } { 2 } ( 12 ) \\ A & = 6\ \mathrm { in^2 } \end { align* } $ $
Therefore, the area of the right triangle is {eq}6\ \mathrm{in^2}

{/eq}.

Example Problem 2: Find the Area of the Triangle by Solving for the Height

Step 1: Because you must find all three side lengths of the triangle, begin by labeling those side lengths { equivalent } a, barn, { /eq } and { eq } c { /eq }. The longest slope, hypotenuse, MUST be labeled { equivalent } c { /eq }. Label the miss side length { equivalent } a { /eq } ( stature ), and the early side length { equivalent } barn { /eq } ( base ) .

triangulum with side lengths bacillus and degree centigrade

Step 2: Plug in the values of the variables into the Pythagorean Theorem equation as follows, using the miss side for { equivalent } a { /eq } .
$ $ \begin { align* } a^2 + b^2 & = c^2\\ a^2 + 10^2 & = 12^2\\ a^2 + 100 & = 144\\ a^2 & = 44\\ \sqrt { a } & = \sqrt { 44 } \\ a & = 6.63\ \mathrm { molarity } \end { align* } $ $
ballyhoo this total into the missing side length in the diagram. You have nowadays found the hypotenuse, the base, and the height. Label these in the triangulum to help with finding the area .

Step 3: Find the area using the rule :
$ $ A = \frac { 1 } { 2 } bh $ $
Plug your side lengths into the formula.

$ $ \begin { align* } A & = \frac { 1 } { 2 } ( 10 ) ( 6.63 ) \\ A & = \frac { 1 } { 2 } \cdot 66.3\\ A & = 33.15\ \mathrm { m^2 } \end { align* } $ $
Therefore, the area of the right triangle is {eq}33.15\ \mathrm{m^2}

{/eq}.

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