Question

what are the values of ab and de in parallelogram abcd? ab = (type an integer or a decimal) image of parallelogram abcd with sides labeled: ae=14, bc=20, cd=12, and a right triangle ced with right angle at e

Explanation:

Step1: Use the property of parallelogram

In a parallelogram \(ABCD\), \(AB = CD\).

Step2: Use the Pythagorean theorem in right - triangle \(CDE\)

In right - triangle \(CDE\), \(CD=\sqrt{CE^{2}+DE^{2}}\). Given \(CE = 12\), \(AE=14\), and \(BC = 20\) (since \(AD = BC\) in parallelogram \(ABCD\)), then \(DE=\sqrt{BC^{2}-CE^{2}}=\sqrt{20^{2}-12^{2}}=\sqrt{(20 + 12)(20-12)}=\sqrt{32\times8}=\sqrt{256}=16\).

Step3: Find the length of \(AB\)

Since \(AB = CD\) and \(CD=\sqrt{12^{2}+16^{2}}=\sqrt{144 + 256}=\sqrt{400}=20\) (using \(DE = 16\) from step 2 and \(CE=12\) in right - triangle \(CDE\) to find \(CD\) again, or using the property that in a parallelogram opposite sides are equal and \(AB = CD\), and also \(AD = BC = 20\), \(AE = 14\), so \(DE=AD - AE=20 - 14=6\) is wrong. Wait, no, correction:
In parallelogram \(ABCD\), \(AB = CD\). In right - triangle \(CDE\), by Pythagorean theorem \(CD=\sqrt{CE^{2}+DE^{2}}\). Since \(AD = BC = 20\) (opposite sides of parallelogram), \(AE = 14\), then \(DE=AD - AE=20 - 14 = 6\) (wrong, no. Wait, no, the correct way:
In parallelogram \(ABCD\), \(AB = CD\), \(AD = BC\). In right - triangle \(CDE\), \(CD=\sqrt{CE^{2}+DE^{2}}\). Since \(AD = BC = 20\) (opposite sides of parallelogram), and \(AE = 14\), but actually, in parallelogram \(AB = CD\), and in right - triangle \(CDE\) with \(CE = 12\) and \(BC = 20\) (so \(AD=BC = 20\)), using Pythagorean theorem \(DE=\sqrt{BC^{2}-CE^{2}}=\sqrt{20^{2}-12^{2}}=\sqrt{400 - 144}=\sqrt{256}=16\). Then \(AB = CD=\sqrt{12^{2}+16^{2}}=\sqrt{144 + 256}=\sqrt{400}=20\)

Answer:

\(AB = 20\)