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Geometry concepts - Similar triangle problems

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Geometry concepts - Similar triangle problems

by sumanr84 » Tue Jul 27, 2010 11:30 pm
1. The perimeters of two similar triangles is in the ratio 3 : 4. The sum of their areas is 75 cm2. Find the area of each triangle.

2. The areas of two similar triangles are 45 cm2 and 80 cm2. The sum of their perimeters is 35 cm. Find the perimeter of each triangle.

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by Rahul@gurome » Tue Jul 27, 2010 11:44 pm
Solution to first question.
Let the sides of one triangle be a, b and c and the ratio of the sides of two triangles be 1:k.
So the sides of other triangle are ka, kb and kc.
Perimeter of first triangle is a+b+c.
Perimeter of second triangle is k(a+b+c).
So (a+b+c):k(a+b+c) = 3:4.
Or k is 4/3.
So the sides are also in the ratio 3:4.
So areas are in the ratio 3^2 : 4^2 which is 9:16.
Or the areas are (9/25)*75 = 27 and (16/25)*75 = 48.

So areas are 27 and 48.
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by Rahul@gurome » Tue Jul 27, 2010 11:53 pm
Solution to second question:
Ratio of areas is 45:80 = 9:16.
Or sides are in the ratio sqrt(9) : sqrt(16) which is 3:4.
Or perimeters are also in the ratio 3:4.
The sum of perimeters is 35.
Or individual perimeters are (3/7)*35 = 15 and (4/7)*35 = 20.

So perimeters are 15 and 20.
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