College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 1 - Measurement and Problem Solving - Learning Path Questions and Exercises - Exercises - Page 32: 73

Answer

31.8 m .

Work Step by Step

The land is equilateral in shape. Height of the triangle $= \sqrt (200^{2}-100^{2})=100\sqrt 3$ So, area $=\frac{1}{2}\times 200 \times 100\sqrt 3=100^{2}\times \sqrt 3\, m^{2}$ Let the base of the upper smaller triangle be (2B). By comparison of 2 triangles, $\frac{100\sqrt 3-h}{100\sqrt 3}=\frac{2B}{200}$ $2B=\frac{2}{\sqrt 3}\times (100\sqrt 3-h)$ Since the area of the upper triangle is two-third of the whole area, $\frac{1}{2}\times (100\sqrt 3-h)\times \frac{2}{\sqrt 3}\times (100\sqrt 3-h)=\frac{2}{3}\times 100^{2} \times \sqrt 3$ or, $100\sqrt 3-h=100\sqrt 2$ or, $h=100\times (\sqrt 3-\sqrt 2)=31.8\,m$
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