Question

suppose an architect draws a segment on a scale drawing with the end points (0,0) and (3/4,9/10). the same segment on the actual structure has the end points (0,0) and (30,36). what proportion could model this situation?

Answer 1

Answer with explanation:

Distance between two points (a,b) and (c,d) on two dimensional coordinate plane is given by

                  $=\sqrt{(a-c)^2+(b-d)^2}$

A=(0,0)

$B=(\frac{3}{4},\frac{9}{10})\\\\AB=\sqrt{(\frac{3}{4}-0)^2+(\frac{9}{10}-0)^2}\\\\=\sqrt{\frac{9}{16}+\frac{81}{100}}\\\\=\sqrt{\frac{2196}{1600}}$

⇒P=(0,0) and Q= (30, 36)

$PQ=\sqrt{(30-0)^2+(36-0)^2}\\\\PQ=\sqrt{900+1296}\\\\PQ=\sqrt{2196}\\\\\frac{PQ}{AB}=\frac{\sqrt{2196}}{\sqrt{\frac{2196}{1600}}}\\\\P Q=AB \times\sqrt{1600}\\\\PQ=40\times AB\\\\\frac{PQ}{AB}=40:1$

⇒Actual Structure of segment =40 × Length of Segment on Scale

Answer 2

Let the segment be represented by AB where A(0,0) = $A(x_{1}, y_{1})$ and B(3/4,9/10) = $B(x_{2}, y_{2})$.

The length of the segment drawn by architect can be calculated using distance formula:

AB =$\sqrt{}( x_{2}- x_{1})^ {2} + (y_{2}- y_{1})^ {2}$

$AB=\sqrt{(3/4-0)^{2}+(9/10-0)^{2}$

$AB=\sqrt{9/16+81/100}$

$AB = (6\sqrt{61})/40$

Similarly, Let the actual end points of segment be AC where A(0,0) = $A(x_{1}, y_{1})$ and C(30,36) = $C(x_{2}, y_{2})$.

The length of the original segment can be calculated using distance formula:

AC =$\sqrt{}( x_{2}- x_{1})^ {2} + (y_{2}- y_{1})^ {2}$

$AC=\sqrt{(30-0)^{2}+(36-0)^{2}$

$AC=\sqrt{900+1296}$

$AC = (6\sqrt{61})$.

Thus, the actual length is 40 times the length of the segment drawn by the architect.

Thus, the proportion of the model is 1:40