Exercise 7.4: Triangles — NCERT Solutions

Given: $\angle A = 40^\circ$, $\angle B = 60^\circ$.

By the angle sum property: $\angle C = 180^\circ - 40^\circ - 60^\circ = 80^\circ.$

So the angles in increasing order are: $\angle A(40^\circ) < \angle B(60^\circ) < \angle C(80^\circ).$

The side opposite a larger angle is longer. The sides opposite these angles are:

• opposite $\angle A$: $BC$
• opposite $\angle B$: $CA$
• opposite $\angle C$: $AB$

Therefore, in ascending order of length: $BC < CA < AB.$

Given: $PQ = 5$ cm, $QR = 7$ cm, $PR = 6$ cm.

Sides in increasing order: $PQ(5) < PR(6) < QR(7).$

The angle opposite a longer side is larger. The angle opposite each side is:

• opposite $PQ$: $\angle R$
• opposite $PR$: $\angle Q$
• opposite $QR$: $\angle P$

Therefore, in ascending order: $\angle R < \angle Q < \angle P.$

Triangle inequality property: In any triangle, the sum of the lengths of any two sides is greater than the length of the third side. This is because the straight segment between two vertices is the shortest path, so going via a third vertex is always longer.

Check for sides $3$, $4$, $8$:

Test the three inequalities:

1. $3 + 4 = 7$, and $7 < 8$ — fails (must be greater than $8$).
2. $4 + 8 = 12 > 3$ — holds.
3. $3 + 8 = 11 > 4$ — holds.

Since the first inequality fails ($3 + 4 \not> 8$), such a triangle cannot exist.