The areas of two similar triangles are 100 sq cm and 49 sq cm, respectively. If the altitude of the bigger triangle is 5 cm, then the corresponding altitude of the other is ____cm.

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The areas of two similar triangles are 100 sq cm and 49 sq cm, respectively. If the altitude of the bigger triangle is 5 cm, then the corresponding altitude of the other is 3.5 cm.Given that,The areas of two similar triangles are 100 sq cm and 49 sq cm, respectively.Consider the figure,We know,The Angle-Angle (AA) criterion for similarity of two triangles states that “If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar”.The triangles

Δ
  ABC

andΔ
  DEF are similar. So,

∠
  B

=∠
  E∠C

=∠
  F In the triangles

Δ
  ABM

andΔ
  DEN,

∠
  B

=∠
  E because the triangles are similar.The angle M and angle N are right angles. So, both angles are equal that is

∠
  M

=∠
  N. So, the triangles

Δ
  ABM

andΔ
  DEN are similar by the AA criterion. Then,

AM
    
     DN
   
   =
    
     AB
    
     DE
   
   =
    
     BM
    
     EN
   
   .
  Now, we know,

ar
        
         ΔABC

ar
        
         ΔDEF

=

AB
        2

DE
        2

⇒
      
       100
      
       49
     
     =

5
         
        2

DN
        2

⇒DN=

49
        4

⇒DN=3.5 cm

The value of DN is

3.5 
  cm.

The areas of two similar triangles are 100 sq cm and 49 sq cm, respectively. If the altitude of the bigger triangle is 5 cm, then the corresponding altitude of the other is ____cm.

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