Question

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Answer 1

Answer:

One base is $\large \boxed{\sf 9}$  inches for one of the bases and $\large \boxed{\sf 15}$ inches for the other base.

$\large \boxed{\sf 24}=\sf(b_1+b_2)$.  Use guess and check to find two numbers that add to $\large \boxed{\sf 24}$ with one number 6 more than the other to get

$\large \boxed{\sf 9}$  inches for one of the bases and $\large \boxed{\sf 15}$ inches for the other base.

Step-by-step explanation:

$\textsf{Area of a trapezoid}= \sf \dfrac{1}{2}(b_1+b_2)h \quad \textsf{(where b are the bases and h is the height)}$

Given:

• Area = 48 in²
• h = 4 in

Substitute given values into the formula to find (b₁ + b₂) :

$\implies \sf 48=\dfrac{1}{2}(b_1+b_2)4$

$\implies \sf \dfrac{48}{4}=\dfrac{1}{2}(b_1+b_2)$

$\implies \sf 12=\dfrac{1}{2}(b_1+b_2)$

$\implies \sf 12 \cdot 2=(b_1+b_2)$

$\implies \sf 24=(b_1+b_2)$

Therefore:

$\large \boxed{\sf 24}=\sf(b_1+b_2)$

Let:

• Base b₁ = x in
• Base b₂ = (x + 6) in

$\implies \sf (b_1+b_2)=24$

$\implies \sf x+x+6=24$

$\implies \sf 2x+6=24$

$\implies \sf 2x=18$

$\implies \sf x=9$

$\sf If\:b_1=9\:in,\:then\:b_2=9+6=15\:in$

Therefore,

• Base b₁ = 9 in
• Base b₂ = 15 in

Answer 2

Answer:

See below ~

Step-by-step explanation:

Let the bases be x and x + 6.

Substituting the known values into the formula :

• A = 1/2 x (a + b) x h
• 48 = 1/2 x (x + x + 6) x 4
• 2x + 6 = 24
• 2x = 18
• x = 9

The base lengths are :

1. x = 9 inches
2. x + 6 = 9 + 6 = 15 inches