Math Problem Statement

The area of trapezoid ABCDABCD (AB∣∣CDAB∣∣CD) is 140 cm2.140 cm2. Through vertex DD we have constructed segment DNDN (with NN on side ABAB) such that DN∣∣BC.DN∣∣BC. If AB=20 cmAB=20 cm and CD=8 cmCD=8 cm, find the area of △AND.△AND.

Solution

Given that the area of trapezoid ABCDABCD is 140cm2140 \, \text{cm}^2, AB=20cmAB = 20 \, \text{cm}, and CD=8cmCD = 8 \, \text{cm}, we are asked to find the area of triangle AND\triangle AND where DNBCDN \parallel BC.

Step 1: Area of the Trapezoid

The formula for the area of a trapezoid is: Area of ABCD=12×(Base1+Base2)×Height\text{Area of } ABCD = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} Here, Base1=AB=20cm\text{Base}_1 = AB = 20 \, \text{cm} and Base2=CD=8cm\text{Base}_2 = CD = 8 \, \text{cm}.

Let the height of the trapezoid be hh. Then, 140=12×(20+8)×h140 = \frac{1}{2} \times (20 + 8) \times h 140=12×28×h140 = \frac{1}{2} \times 28 \times h 140=14h140 = 14h h=14014=10cmh = \frac{140}{14} = 10 \, \text{cm} So, the height of the trapezoid ABCDABCD is 10cm10 \, \text{cm}.

Step 2: Understanding Triangle AND\triangle AND

Since DNBCDN \parallel BC, triangle AND\triangle AND and trapezoid ABCDABCD share the same height from vertex DD to line ABAB. The height of triangle AND\triangle AND is thus also 10cm10 \, \text{cm}.

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

Step 3: Ratio of Similarity

Since DNBCDN \parallel BC, triangles AND\triangle AND and BCD\triangle BCD are similar. The ratio of similarity is given by: Ratio=ANAB=x20\text{Ratio} = \frac{AN}{AB} = \frac{x}{20} The length of DNDN corresponds to the length of CD=8cmCD = 8 \, \text{cm}.

Step 4: Area of Triangle AND\triangle AND

The area of triangle AND\triangle AND is given by: Area of AND=12×Base×Height\text{Area of } \triangle AND = \frac{1}{2} \times \text{Base} \times \text{Height} Area of AND=12×x×10=5xcm2\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 ABCDABCD is composed of the area of AND\triangle AND and the area of BCD\triangle BCD. Because AND\triangle AND and BCD\triangle BCD are similar, the areas of these triangles are proportional to the squares of the corresponding sides: Area of ANDArea of Trapezoid ABCD=(x20)2\frac{\text{Area of } \triangle AND}{\text{Area of Trapezoid } ABCD} = \left(\frac{x}{20}\right)^2 Let k=x20k = \frac{x}{20}, then: Area of AND=k2×140\text{Area of } \triangle AND = k^2 \times 140

Since k2k^2 represents the proportion of the total area that belongs to AND\triangle AND and also equals the ratio 820\frac{8}{20}, we can calculate: k=820=0.4k = \frac{8}{20} = 0.4 Thus: Area of AND=(0.4)×140cm2=56cm2\text{Area of } \triangle AND = (0.4) \times 140 \, \text{cm}^2 = 56 \, \text{cm}^2

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

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Math Problem Analysis

Mathematical Concepts

Geometry
Similarity
Area Calculation

Formulas

Area of Trapezoid
Area of Triangle

Theorems

Properties of Similar Triangles

Suitable Grade Level

Grades 7-9