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as shown in the figure below, the side length of square abcd is 8. on e…

Question

as shown in the figure below, the side length of square abcd is 8. on each side, segments ae = bf = cg = dh = 6 are cut. the area of quadrilateral efgh is __. a. 34 b. 36 c. 40 d. 100 in the figure, in trapezoid abcd, $overline{ad}paralleloverline{bc}$, $overline{bd}$ is the diagonal, and the median $overline{ef}$ intersects $overline{bd}$ at point o. if fo - eo = 3, then bc - ad is equal to __. a. 4 b. 6 c. 8 d. 10

Explanation:

Step1: Calculate area of square ABCD

The area of square \(ABCD\) with side - length \(a = 8\) is \(S_{ABCD}=a^{2}=8^{2}=64\).

Step2: Calculate area of four right - angled triangles

In right - angled triangle \(AEH\), \(AE = 6\), \(AH=8 - 6 = 2\).
The area of right - angled triangle \(S_{\triangle AEH}=\frac{1}{2}\times AE\times AH=\frac{1}{2}\times6\times2 = 6\).
Since \(\triangle AEH\cong\triangle BFE\cong\triangle CGF\cong\triangle DHG\) (by Side - Angle - Side congruence criterion as the angles of the square are right - angles and the corresponding side - lengths are equal), the total area of the four right - angled triangles is \(S_{total\ triangles}=4\times6 = 24\).

Step3: Calculate area of quadrilateral EFGH

The area of quadrilateral \(EFGH\) is \(S_{EFGH}=S_{ABCD}-S_{total\ triangles}\).
\(S_{EFGH}=64 - 24=40\).

for second question:

Step1: Recall the property of the median of a trapezoid

The median \(EF\) of trapezoid \(ABCD\) with \(AD\parallel BC\) has the property that \(EF=\frac{AD + BC}{2}\), and also, since \(AD\parallel EF\parallel BC\), in \(\triangle ABD\), \(E\) is the mid - point of \(AB\) and \(EO\parallel AD\), then \(EO=\frac{1}{2}AD\) (by the mid - point theorem in a triangle). Similarly, in \(\triangle BCD\), \(FO=\frac{1}{2}BC\).

Step2: Use the given condition \(FO - EO=3\)

Substitute \(FO=\frac{1}{2}BC\) and \(EO=\frac{1}{2}AD\) into \(FO - EO = 3\).
We get \(\frac{1}{2}BC-\frac{1}{2}AD = 3\).
Factor out \(\frac{1}{2}\): \(\frac{1}{2}(BC - AD)=3\).

Step3: Solve for \(BC - AD\)

Multiply both sides of the equation \(\frac{1}{2}(BC - AD)=3\) by 2, we get \(BC - AD=6\).

Answer:

C. 40