UPSC 2014 CDS I Math Paper

Perimeter of ΔABC = 3 units
Now,DE=

1

2

BC=

1

2

unit
DF=

1

2

‌AC=

1

2

unit,
EF=

1

2

‌AB=

1

2

(Using mid-point theorem)
Therefor of perimeter ΔDEF
=3×

1

2

=

3

2

unit
PQ=

1

2

‌DE=

1

2

×

1

2

=(

1

2

)2 unit
RQ=

1

2

DF=(

1

2

)2unit,
PR=

1

2

EF=(

1

2

)2unit
(Using mid-point theorem)
Therefore, perimeter of ΔPQR
=3×(

1

2

)2units
So, the perimeters of triangles formed are 3 units,

3

2

units,
Perimeters of the triangle are in GP with first term, a = 3 and common ratio r=

1

2

Therefore, sum of perimeters of all triangles