Question

given trapezoid (abcd) with midsegment (ef), (ef parallel ad), (ef parallel bc)
prove: (ef = \frac{1}{2}(ad + bc))
(ad = sqrt{(8 - 0)^2 + (0 - 0)^2} =)
(bc = sqrt{(6 - 2)^2 + (6 - 6)^2} =)
(ef = sqrt{(7 - 1)^2 + (3 - 3)^2} =)
(\frac{1}{2}(ad + bc) = \frac{1}{2}()
does (ef = \frac{1}{2}(ad + bc))?

Explanation:

Step1: Calculate length of AD

Using distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) for \(A(0,0)\) and \(D(8,0)\):
\(AD=\sqrt{(8 - 0)^2+(0 - 0)^2}=\sqrt{64 + 0}=\sqrt{64}=8\)

Step2: Calculate length of BC

Using distance formula for \(B(2,6)\) and \(C(6,6)\):
\(BC=\sqrt{(6 - 2)^2+(6 - 6)^2}=\sqrt{16+0}=\sqrt{16}=4\)

Step3: Calculate length of EF

Using distance formula for \(E(1,3)\) and \(F(7,3)\):
\(EF=\sqrt{(7 - 1)^2+(3 - 3)^2}=\sqrt{36+0}=\sqrt{36}=6\)

Step4: Calculate \(\frac{1}{2}(AD + BC)\)

Substitute \(AD = 8\) and \(BC = 4\):
\(\frac{1}{2}(AD + BC)=\frac{1}{2}(8 + 4)=\frac{1}{2}\times12 = 6\)

Step5: Compare EF and \(\frac{1}{2}(AD + BC)\)

Since \(EF = 6\) and \(\frac{1}{2}(AD + BC)=6\), we have \(EF=\frac{1}{2}(AD + BC)\)

Answer:

• \(AD = 8\)
• \(BC = 4\)
• \(EF = 6\)
• \(\frac{1}{2}(AD + BC)=6\)
• Yes, \(EF=\frac{1}{2}(AD + BC)\)