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Draw five triangles and measure their sides. Check in each case, if the sum of the lengths of any two sides is always less than the third side.
To do:
We have to draw five triangles and measure their sides.
Solution:
In triangle PQR
$PQ = 3\ cm$
$QR = 4\ cm$
$PR = 6\ cm$
$PQ + QR = 3\ cm + 4\ cm$
$= 7\ cm$
Here, $7 > 6$
Therefore,
$PQ + QR > PR$
Hence, the sum of any two sides of a triangle is greater than the third side.
In triangle GEF
$GE = 4\ cm$
$EF = 5\ cm$
$GF = 8\ cm$
$GE + EF = 4\ cm + 5\ cm$
$= 9\ cm$
Here, $9 > 8$
Therefore,
$GE + EF > GF$
Hence, the sum of any two sides of a triangle is greater than the third side.
In triangle ABC
$AB = 3\ cm$
$BC = 4\ cm$
$AC = 5\ cm$
$AB + BC = 3\ cm + 4\ cm$
$= 7\ cm$
Here, $7 > 5$
Therefore,
$AB + BC > AC$
Hence, the sum of any two sides of a triangle is greater than the third side.
In triangle XYZ
$XY = 6\ cm$
$XZ = 7\ cm$
$YZ = 9\ cm$
$XY + XZ = 6\ cm + 7\ cm$
$= 13\ cm$
Here, $13 > 9$
Therefore,
$XY + XZ > YZ$
Hence, the sum of any two sides of a triangle is greater than the third side.
In triangle MNS
$MN = 2\ cm$
$NS = 3\ cm$
$MS = 4\ cm$
$MN + NS = 2\ cm + 3\ cm$
$= 5\ cm$
Here, $5 > 4$
Therefore,
$MN + NS > MS$
Hence, the sum of any two sides of a triangle is greater than the third side.
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