in a triangle pqr if s is any point on side qr show that pq+qr+rp is greater than 2ps
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A/c to yr question,
We know that the perimeter of pqr always be greater than the smaller pqs and psr.
So, by keeping these point in our mind we can solve it like that,
At first take pqr and pqs
Then, pqr > pqs (by perimeter)
So, pq+qr+rp>pq+qs+sp ----------(1)
Now similarly, pqr >triangle psr
So, pq+qr+rp>sp+rp+sr--------(2)
Now, add these (1)+(2)
Then, 2pq+2qr+2rp>pq+qs+sp+sp+rp+sr
2pq-pq+2qr-(qs+sr) +2rp-rp>2sp
Pq+qr+rp>2ps. ( Since qs+sr=qr)
Hence, proved that pq+qr+rp>2ps..
We know that the perimeter of pqr always be greater than the smaller pqs and psr.
So, by keeping these point in our mind we can solve it like that,
At first take pqr and pqs
Then, pqr > pqs (by perimeter)
So, pq+qr+rp>pq+qs+sp ----------(1)
Now similarly, pqr >triangle psr
So, pq+qr+rp>sp+rp+sr--------(2)
Now, add these (1)+(2)
Then, 2pq+2qr+2rp>pq+qs+sp+sp+rp+sr
2pq-pq+2qr-(qs+sr) +2rp-rp>2sp
Pq+qr+rp>2ps. ( Since qs+sr=qr)
Hence, proved that pq+qr+rp>2ps..
uditri:
thanks a lot this is what i wanted
Its ok
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