QUESTION IMAGE
Question
- ac = _ bd = _ be = _ ab = _ bc = _
Step1: Properties of rectangle diagonals
In a rectangle, diagonals are equal and bisect each other. Given \(AE = 15\), then \(AC=BD = 2\times AE\).
\(AC = BD=2\times15 = 30\)
Step2: Find \(BE\)
Since \(E\) is the mid - point of \(BD\) (diagonals of a rectangle bisect each other), \(BE=\frac{BD}{2}\).
\(BE=\frac{30}{2}=15\)
Step3: Use Pythagorean theorem to find \(AB\)
In right - triangle \(ABC\), \(AC = 30\), \(BC\) is adjacent side, \(AB\) is opposite side. Let's first note that in rectangle \(ABCD\), \(DC = AB\) (opposite sides of a rectangle are equal). Wait, no, using right - triangle \(ADC\) (since \(\angle ADC = 90^{\circ}\) in rectangle \(ABCD\)), \(AD\) and \(DC\) are sides and \(AC\) is the hypotenuse. Wait, no, using right - triangle \(ABC\) (wrong, should use right - triangle \(ADC\)). In rectangle \(ABCD\), \(\angle ADC=90^{\circ}\), \(AC = 30\), \(DC = 27\) (given \(DC = 27\), and \(AB = DC\) (opposite sides of a rectangle are equal)). By Pythagorean theorem \(AD=\sqrt{AC^{2}-DC^{2}}\). But \(AB\) and \(BC\): Wait, in rectangle \(ABCD\), \(AB\parallel DC\), \(AD\parallel BC\), \(AB = DC\), \(AD = BC\). Using right - triangle \(ADC\) (\(\angle ADC = 90^{\circ}\)), \(AC^{2}=AD^{2}+DC^{2}\). Wait, no, we want \(AB\): Since \(AB = DC\) (opposite sides of a rectangle are equal) is wrong. Wait, no, in rectangle \(ABCD\), \(AB\) and \(DC\) are opposite sides, so \(AB = DC\) (but \(DC = 27\) is given? No, wait the length given as \(DC = 27\). Wait, no, in the problem, the figure shows \(DC = 27\). In rectangle \(ABCD\), \(AB = DC\) (opposite sides of a rectangle are equal). So \(AB = 27\). Then for \(BC\), using right - triangle \(ABC\) (\(\angle ABC=90^{\circ}\)), \(AC = 30\), \(AB = 27\). By Pythagorean theorem \(BC=\sqrt{AC^{2}-AB^{2}}\)
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\(AC = 30\)
\(BD = 30\)
\(BE = 15\)
\(AB = 27\)
\(BC=\sqrt{30^{2}-27^{2}}=\sqrt{900 - 729}=\sqrt{171}=3\sqrt{19}\approx13.08\)