The area of trapezoid ABCD*ABCD* (AB∣∣CD*AB*∣∣*CD*) is 140 cm2.140 cm2. Through vertex D*D* we have constructed segment DN*DN* (with N*N* on side AB*AB*) such that DN∣∣BC.*DN*∣∣*BC*. If AB=20 cm*AB*=20 cm and CD=8 cm*CD*=8 cm, find the area of △AND.△*AND*.

Given that the area of trapezoid \( ABCD \) is \( 140 \, \text{cm}^2 \), \( AB = 20 \, \text{cm} \), and \( CD = 8 \, \text{cm} \), we are asked to find the area of triangle \( \triangle AND \) where \( DN \parallel BC \).

### Step 1: Area of the Trapezoid
The formula for the area of a trapezoid is:
\[
\text{Area of } ABCD = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
\]
Here, \( \text{Base}_1 = AB = 20 \, \text{cm} \) and \( \text{Base}_2 = CD = 8 \, \text{cm} \).

Let the height of the trapezoid be \( h \). Then,
\[
140 = \frac{1}{2} \times (20 + 8) \times h
\]
\[
140 = \frac{1}{2} \times 28 \times h
\]
\[
140 = 14h
\]
\[
h = \frac{140}{14} = 10 \, \text{cm}
\]
So, the height of the trapezoid \( ABCD \) is \( 10 \, \text{cm} \).

### Step 2: Understanding Triangle \( \triangle AND \)
Since \( DN \parallel BC \), triangle \( \triangle AND \) and trapezoid \( ABCD \) share the same height from vertex \( D \) to line \( AB \). The height of triangle \( \triangle AND \) is thus also \( 10 \, \text{cm} \).

Let the length of segment \( AN = x \, \text{cm} \).

### Step 3: Ratio of Similarity
Since \( DN \parallel BC \), triangles \( \triangle AND \) and \( \triangle BCD \) are similar. The ratio of similarity is given by:
\[
\text{Ratio} = \frac{AN}{AB} = \frac{x}{20}
\]
The length of \( DN \) corresponds to the length of \( CD = 8 \, \text{cm} \).

### Step 4: Area of Triangle \( \triangle AND \)
The area of triangle \( \triangle AND \) is given by:
\[
\text{Area of } \triangle AND = \frac{1}{2} \times \text{Base} \times \text{Height}
\]
\[
\text{Area of } \triangle AND = \frac{1}{2} \times x \times 10 = 5x \, \text{cm}^2
\]

### Step 5: Relationship Between Areas
The area of the trapezoid \( ABCD \) is composed of the area of \( \triangle AND \) and the area of \( \triangle BCD \). Because \( \triangle AND \) and \( \triangle BCD \) are similar, the areas of these triangles are proportional to the squares of the corresponding sides:
\[
\frac{\text{Area of } \triangle AND}{\text{Area of Trapezoid } ABCD} = \left(\frac{x}{20}\right)^2
\]
Let \( k = \frac{x}{20} \), then:
\[
\text{Area of } \triangle AND = k^2 \times 140
\]

Since \( k^2 \) represents the proportion of the total area that belongs to \( \triangle AND \) and also equals the ratio \( \frac{8}{20} \), we can calculate:
\[
k = \frac{8}{20} = 0.4
\]
Thus:
\[
\text{Area of } \triangle AND = (0.4) \times 140 \, \text{cm}^2 = 56 \, \text{cm}^2
\]

The correct area calculation is thus:
\(\boxed{56}\) cm².

Learn how to calculate the area of triangle AND within trapezoid ABCD with a detailed mathematical solution. This problem involves using the properties of trapezoids and similar triangles, specifically focusing on area calculations. Suitable for students in Grades 7-9.

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