# I can't figure out how to do these problems. Any help is much appreciated. Thanks 1. The areas of two circles are in the ratio of 16:49. What is the ratio of their diameters? 2. The perimeters of...

I can't figure out how to do these problems. Any help is much appreciated. Thanks

1. The areas of two circles are in the ratio of 16:49. What is the ratio of their diameters?

2. The perimeters of two similar triangles are in the ratio 2:5. What is the ratio of their areas?

gsenviro | Certified Educator

1) The area of a circle is given as: area = pi/4 x d^2

where, d is the diameter of the circle.

In the given case, lets assume that areas of circles are A1 and A2, and the corresponding diameters are d1 and d2.

Thus, A1/A2 = [pi/4  x (d1)^2]/[pi/4  x (d2)^2] = (d1)^2/(d2)^2 = (d1/d2)^2  = 16/49 (given)

or, d1/d2 = (16/49)^(1/2) = 4/7

2) For similar triangles, the length of sides is in proportion to each other, i.e.

a/A = b/B = c/C  = x               (where x is some fixed ratio called as "scale factor")

for two similar triangles with side lengths: (a,b,c) and (A,B,C)

The perimeter of a triangle is given as sum of sides. Thus in our case

(a+b+c)/(A+B+C) = 2/5

It can be seen that the fixed ratio, x, mentioned in previous equation is also equal to 2/5.

(it can be easily found out by substituting the values of a, b and c from the ratio equation).

Using the properties of similar triangle: If the scale factor is x, then the ratio of areas of the triangles is x^2.

thus, in current case, Area1/area2 = x^2 = (2/5)^2 = 4/25

iamkaori | Student

1. areas of a circle is: r*r*pi

16:49
= 16pi : 49pi

so we can say that r1 : r2 is 4:7