In each of the following cases, given reasons to show that the construction of ∆ABC is not possible:
(i) AB = 6 cm, ∠A = 40° and (BC + AC) = 5.8 cm.
(ii) AB = 7 cm, ∠A = 50° and (BC – AC) = 8 cm.
(iii) BC = 5 cm, ∠B = 80°, ∠C = 50° and ∠A = 60°.
(iv) AB = 4 cm, BC = 3 cm and AC = 7 cm.

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Solution

(i) AB = 6 cm, ∠A = 40° and (BC + AC) = 5.8 cm.
Here BC + AC is not greater than AB so, this triangle is not possible.

(ii) AB = 7 cm, ∠A = 50° and (BC – AC) = 8 cm.
Here BC – AC is not less than AB so this triangle is not possible.

(iii) BC = 5 cm, ∠B = 80°, ∠C = 50° and ∠A = 60°.
Sum of the angles of the given triangle is not equal to 180 $°$ so, given triangle is not possible.

(iv) AB = 4 cm, BC = 3 cm and AC = 7 cm.
Sum of two sides of this triangle not greater than third side so given triangle not possible.

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Similar questions

Δ A B C

A B = 5 c m, A C = 7 c m, B C = 6 c m

Construct a quadrliateral ABCD; if:

(i) AB = 4.3 cm, BC = 5.4, CD = 5 cm, DA = 4.8 cm and $∠$ ABC = $75^{\circ}$ .

(ii) AB = 6 cm, CD = 4.5 cm, BC = AD = 5 cm and $∠ BCD = 60^{\circ}$

(iii) AB = 8 cm, BC = 5.4 cm, AD = 6 cm, $∠ A = 60^{\circ}$ and $∠ B = 75^{\circ}$ .

(iv) AB = 5 cm, BC = 6.5 cm, CD = 4.8 cm, $∠ B = 75^{\circ}$ and $∠ C = 120^{\circ}$

(v) AB = 6 cm = AC, BC = 4 cm, CD = 5 cm and AD = 4.5 cm.

(vi) AB = AD = 5 cm, BD = 7 cm and BC = DC = 5.5 cm

Δ A B C

A B = 4 c m, B C = 3 c m, A C = 5 c m

∠ B

A B C D

A B = 6 c m, B C = 4 c m, C D = 7 c m, A C = 8 c m

B D = 7 c m .

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