Mathematics
Grade9
Easy

Question

The perimeters of two similar triangles is in the ratio 3 : 4. The sum of their areas is 75 square cm. Find the area of each triangle.

  1. 27 and 48
  2. 15 and 60
  3. 25 and 50
  4. 30 and 45

hintHint:

Finding the area and adding them

The correct answer is: 27 and 48


    Let 3x = a side in increment subscript 1
    and 4x = the corresponding side in increment subscript 2.
    Then, fraction numerator area space triangle subscript 1 over denominator area space triangle subscript 2 end fraction equals open parentheses fraction numerator 3 straight x over denominator 4 straight x end fraction close parentheses squared
fraction numerator area space triangle subscript 1 over denominator area space triangle subscript 2 end fraction equals fraction numerator 9 straight x squared over denominator 16 straight x squared end fraction
area space triangle subscript 1 space plus space area space triangle subscript 2 equals 9 straight x squared plus 16 straight x squared
75 space equals space 25 straight x squared
3 space equals space straight x squared
area space triangle subscript 1 space equals space 9 straight x squared space equals space 9 left parenthesis 3 right parenthesis space equals space 27 space cm squared
area space triangle subscript 2 space equals space 16 straight x squared space equals space 16 left parenthesis 3 right parenthesis space equals space 48 space cm squared

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