Similar Triangles , Area and Perimeter

If two triangles ABC and PQR are similar (ΔABC ∼ ΔPQR) , then their corresponding sides are proportional and corresponding angles are equal i.e. PQ/AB = QR/BC = RP/CA. 

The ratio of areas of similar triangles is the square of the ratio of their sides i.e.   Area PQR / Area ABC = (PQ/AB)² 

The ratio of perimeters of similar triangles is the ratio of their sides i.e. 
 Perimeter PQR / Perimeter ABC = PQ/AB


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Poles and Strings

A pair of telephone poles d metres apart is supported by two cables which run from the top of each pole to the bottom of the other. The poles are 4 m and 6 m tall. Determine the height above the ground of the point, T , where the two cables intersect. What happens to this height as d increases? This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com
Let the height of the poles QP = a and RS = b.
Let the distance PO = c. Let h be the height above ground of the point T.

Since ∆QPR and ∆TOR are similar , so a : d = h : (d – c) , or (d – c ) = dh / a.
Similarly , ∆SRP and ∆TOP are similar , so b : d = h : c . So , c = dh / b.

Eliminating c from the above two equations we get d = dh (1/a + 1/b) , which gives h = ab / (a+b)
If a = 4 m and b = 6 m then we get h = 12 / 5 m. Thus height of point T is 12/5 m , independent of d.This can be seen by dragging points P or Q and see the path taken by point T.

Construction of Similar Triangle

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Pythagoras Theorem by Similar Triangles

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Basic Proportionality Theorem (Thales Theorem)

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