Isosceles Right Triangle Dimensions: Solving for Side Lengths with Area 32

Triangle Area with Algebraic Side Expressions

In front of you is an isosceles right triangle.

The expressions listed next to the sides describe their length.

( x>8 x>-8 length measurements in cm).

Since the area of the triangle is 32.

Find the lengths of the sides of the triangle.

323232x+8x+8x+8

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find the lengths of the triangle's sides
00:03 We'll use the formula for calculating triangle area
00:07 (height times side) divided by 2
00:12 Sides are equal according to the given data
00:15 We'll substitute the area value according to the given data
00:25 We'll multiply by the denominator to eliminate the fraction
00:32 Taking the square root, remember we get 2 solutions
00:35 One positive and one negative
00:40 We'll isolate the unknown and find the solution for each possibility
00:46 This is one solution, but it doesn't fit the domain
00:53 This is the second solution, we'll substitute to find the sides
00:59 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

In front of you is an isosceles right triangle.

The expressions listed next to the sides describe their length.

( x>8 x>-8 length measurements in cm).

Since the area of the triangle is 32.

Find the lengths of the sides of the triangle.

323232x+8x+8x+8

2

Step-by-step solution

To solve this problem, we'll utilize the properties of an isosceles right triangle and the area formula:

  • Step 1: Identify the expression for the legs of the triangle as x+8 x + 8 .
  • Step 2: Use the area formula for a right triangle, 12×(leg)2=32 \frac{1}{2} \times (\text{leg})^2 = 32 .
  • Step 3: Given that the triangle is isosceles, solve for x x by substituting and expanding the terms.
  • Step 4: Derive the hypotenuse using the known leg value and calculate leg×2 \text{leg} \times \sqrt{2} .

Now, let's work through each step:

Step 1: Recognize both legs of the isosceles triangle are x+8 x + 8 .

Step 2: Apply the area formula for right triangles:
We know 12×(leg)×(leg)=32 \frac{1}{2} \times (\text{leg}) \times (\text{leg}) = 32 . Therefore, the equation is:

12×(x+8)2=32 \frac{1}{2} \times (x + 8)^2 = 32

Step 3: Simplify and solve for x x .

(x+8)2=64(x + 8)^2 = 64 x+8=64x + 8 = \sqrt{64} x+8=8orx+8=8x + 8 = 8 \quad \text{or} \quad x + 8 = -8

Given the constraint x>8x > -8, we discard x+8=8x + 8 = -8 since it violates the condition. Therefore,

x+8=8x + 8 = 8 x=0x = 0

Recalculate x+8 x + 8 , which states the leg is 8.

Step 4: Determine the hypotenuse.

hypotenuse=82hypotenuse = 8\sqrt{2}

Therefore, the side lengths of the triangle are 8,8,82 8, 8, 8\sqrt{2} . Match this to the choices given, which is option 3.

The lengths of the sides of the triangle are 8,8,82 8, 8, 8\sqrt{2} .

3

Final Answer

8,8,82 8,8,8\sqrt{2}

Key Points to Remember

Essential concepts to master this topic
  • Isosceles Right Triangle: Both legs are equal, hypotenuse equals leg × √2
  • Area Formula: Area = ½ × leg² = ½ × (x + 8)² = 32
  • Verification: Check legs are 8 cm and area is ½ × 8² = 32 ✓

Common Mistakes

Avoid these frequent errors
  • Using wrong area formula for isosceles right triangle
    Don't use Area = ½ × base × height with different values = wrong equation! Students often forget both legs are equal in isosceles triangles. Always use Area = ½ × leg² since both legs have the same length.

Practice Quiz

Test your knowledge with interactive questions

Find the value of the parameter x.

\( 2x^2-7x+5=0 \)

FAQ

Everything you need to know about this question

Why do both legs have the same expression x + 8?

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In an isosceles right triangle, the two legs are always equal in length. The diagram shows both legs labeled as x + 8, confirming this is an isosceles triangle.

How do I find the hypotenuse once I know the legs?

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For any right triangle, use the Pythagorean theorem: c2=a2+b2 c^2 = a^2 + b^2 . But for isosceles right triangles, there's a shortcut: hypotenuse = leg × √2.

What does the constraint x > -8 mean?

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This constraint ensures the side length x+8 x + 8 is positive. Since lengths cannot be negative, we need x+8>0 x + 8 > 0 , which means x>8 x > -8 .

Why do I get two solutions when solving (x + 8)² = 64?

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Taking the square root gives x+8=±8 x + 8 = ±8 , so x=0 x = 0 or x=16 x = -16 . But x=16 x = -16 violates the constraint x>8 x > -8 , so we only use x=0 x = 0 .

How can I double-check my final answer?

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Substitute back into the area formula: with legs of 8 cm each, Area=12×8×8=32 \text{Area} = \frac{1}{2} \times 8 \times 8 = 32 ✓. The hypotenuse should be 82 8\sqrt{2} cm.

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