# 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.

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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.

Updated on 10-Oct-2022 13:32:08