Expressing the Trapezoid Area: Using X to Account for Base Differences

Trapezoid Area with Variable Height Expressions

Given the trapezoid where the height is equal to the sum of the two bases.

It is known that the difference between the large base and the small base is 5

We will mark the small base with X

Express the area of the trapezoid using X

XXXX+5X+5X+5hhh

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Express the area of the trapezoid using X
00:03 We will use the formula for calculating the area of a trapezoid
00:07 (sum of bases) multiplied by height) divided by 2
00:25 The height equals the sum of bases according to the given data
00:41 We will use the shortened multiplication formulas to expand the brackets
00:58 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

Given the trapezoid where the height is equal to the sum of the two bases.

It is known that the difference between the large base and the small base is 5

We will mark the small base with X

Express the area of the trapezoid using X

XXXX+5X+5X+5hhh

2

Step-by-step solution

To solve this problem, we will find the area of the trapezoid using the given expressions for the bases and height.

Step 1: Determine the height of the trapezoid.

  • The height, h h , is given as the sum of the two bases: h=X+(X+5)=2X+5 h = X + (X + 5) = 2X + 5

Step 2: Apply the formula for the area of a trapezoid.

  • The formula for the area of a trapezoid is: Area=12×(small base+large base)×height \text{Area} = \frac{1}{2} \times (\text{small base} + \text{large base}) \times \text{height}
  • Substitute the expressions for the bases and height: Area=12×(X+(X+5))×(2X+5) \text{Area} = \frac{1}{2} \times (X + (X + 5)) \times (2X + 5)
  • Further simplifying: Area=12×(2X+5)×(2X+5) \text{Area} = \frac{1}{2} \times (2X + 5) \times (2X + 5)
  • This becomes: Area=12×(2X+5)2 \text{Area} = \frac{1}{2} \times (2X + 5)^2
  • Expand (2X+5)2 (2X + 5)^2 using the square of a binomial formula: (2X+5)2=4X2+20X+25 (2X + 5)^2 = 4X^2 + 20X + 25
  • Thus the area simplifies to: Area=12(4X2+20X+25) \text{Area} = \frac{1}{2} (4X^2 + 20X + 25)

Therefore, the expression for the area of the trapezoid in terms of X X is 12(4X2+20X+25) \frac{1}{2}(4X^2 + 20X + 25) .

3

Final Answer

12[4x2+20x+25] \frac{1}{2}\lbrack4x^2+20x+25\rbrack

Key Points to Remember

Essential concepts to master this topic
  • Formula: Area = 12×(b1+b2)×h \frac{1}{2} \times (b_1 + b_2) \times h for trapezoids
  • Technique: Height equals sum of bases: h=X+(X+5)=2X+5 h = X + (X + 5) = 2X + 5
  • Check: Expand (2X+5)2=4X2+20X+25 (2X + 5)^2 = 4X^2 + 20X + 25 carefully ✓

Common Mistakes

Avoid these frequent errors
  • Forgetting the 1/2 factor in trapezoid area formula
    Don't calculate just (small base + large base) × height = wrong answer that's double the correct area! Students often remember triangle area needs 1/2 but forget trapezoids also need it. Always use the complete formula: Area = 1/2 × (sum of bases) × height.

Practice Quiz

Test your knowledge with interactive questions

Choose the expression that has the same value as the following:

\( (x+y)^2 \)

FAQ

Everything you need to know about this question

Why is the height equal to the sum of the bases?

+

This is a special condition given in the problem! In most trapezoids, the height is independent of the bases. Here, we're told that h=X+(X+5)=2X+5 h = X + (X + 5) = 2X + 5 as a constraint.

How do I expand (2X + 5)² correctly?

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Use the pattern (a + b)² = a² + 2ab + b². So (2X+5)2=(2X)2+2(2X)(5)+52=4X2+20X+25 (2X + 5)^2 = (2X)^2 + 2(2X)(5) + 5^2 = 4X^2 + 20X + 25 .

Can I factor out the 1/2 at the end instead?

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Yes! You can write the answer as 12(4X2+20X+25) \frac{1}{2}(4X^2 + 20X + 25) or as 2X2+10X+12.5 2X^2 + 10X + 12.5 , but the factored form is usually preferred in algebra.

What if X is negative?

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Since X represents a length (the small base), it must be positive. Also, we need X + 5 > X, so the large base is indeed larger than the small base.

Why don't I just substitute numbers for X?

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The problem asks for an algebraic expression in terms of X. This gives a general formula that works for any valid value of X, which is more powerful than a single numerical answer.

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