Question

Find the value of x in the triangle shown below NEED HELP ASAP

Answer 1

To find the value of x we can apply Pythagoras theorem in this right angled triangle.

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So let's apply!

`{ \boxed{ \sf{ \mapsto {Hypotenuse}^(2) ={Base }^(2) +{Perpendicular }^(2) }}}`

`\sf{ : \implies {x}^(2) = {9}^(2) + {3}^(2) }`

`\sf{ : \implies {x}^(2) = 81 + 9 }`

`\sf{ : \implies {x}^(2) = 90 }`

`\sf{ : \implies {x} = √(2 * 3 * 3 * 5) }`

`\sf{ : \implies {x} = 3 √(10) }`

Hence this is the value of x.

Answer 2

x = 3√10

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Equality Properties

• Multiplication Property of Equality
• Division Property of Equality
• Addition Property of Equality
• Subtract Property of Equality

Trigonometry

[Right Triangles Only] Pythagorean Theorem: a² + b² = c²

• a is a leg
• b is another leg
• c is the hypotenuse

Explanation:

Step 1: Identify Variables

Leg a = 9

Leg b = 3

Hypotenuse c = x

Step 2: Solve for x

1. Substitute [PT]: 9² + 3² = x²
2. Rearrange: x² = 9² + 3²
3. Exponents: x² = 81 + 9
4. Add: x² = 90
5. Isolate x: x = 3√10