Question

solve this problem...​

Answer 1

$\underline{\mathfrak{\huge{Your\:Answer:}}}$

$\sf{Given:}$

ABCD and ABEF are on the same base

ABCD is a rectangle

ABEF is a parallelogram

$\sf{To\:Prove:}$

Perimeter of ABEF > Perimeter of ABCD

$\sf{Proof:}$

Consider only two triangles, $\triangle FDA\:and\:\triangle ECB$.

In $\tt{\triangle FDA}$

DA > FA ...(1)

In $\tt{\triangle ECB}$

CB > EB ...(2)

Now, the since both the figures stand on the same and equal base, we can say that :-

FE = AB = DC

Perimeter of ABCD = AB + BC + CD + DA

Perimeter of ABEF = AB + BE + EF + FA

On comparing, we get [ if we see (1) and (2) ] :-

AB + BC + CD + DA > AB + BE + EF + FA ( Since DA + CB > FA + EB )

Hence Proved !