Euclid has a triangle in mind. Its longest side has length $20$ units and another of its sides has length $10$ units. Its area is $80$ sq. units, then the exact length of its third side is $\sqrt{260}$ .
If true then enter $1$ and if false then enter $0$

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Solution

$G i v e n t h a t - Δ A B C h a s A B = 20 u n i t s, B C = 10 u n i t s & a r Δ A B C = 80 s q . u n i t s . T o f i n d o u t - t h e l e n g t h o f t h e t h i r d s i d e X = A C . S o l u t i o n - W e c o n s t r u c t E u c l i d^{'} s Δ A B C a s d e s c r i b e d a n d d r a w t h e h e i g h t o f t h e Δ C E = h o n A B . L e t t h e t h i r d s i d e b e X = A C . C E w i l l m e e t A B a t E i n t e r n a l l y s i n c e A B i s t h e b i g g e s t s i d e . S o a r Δ A B C = \frac{1}{2} \times A B \times h ⟹ h = 2 \times \frac{a r Δ A B C}{A B} = 2 \times \frac{80}{20} u n i t s = 8 u n i t s . N o w C E = h ⊥ A B . i . e Δ B E C i s a r i g h t t r i a n g l e . ∴ B E = \sqrt{{B C}^{2} - {C E}^{2}} = \sqrt{10^{2} - 8^{2}} u n i t s = 6 u n i t s . i . e A E = A B - B E = (20 - 6) u n i t s = 14 u n i t s . A g a i n C E ⊥ A E . S o Δ A C E i s a r i g h t t r i a n g l e . ∴ X = A C = \sqrt{{A E}^{2} + {C E}^{2}} = \sqrt{14^{2} + 8^{2}} u n i t s = \sqrt{260} u n i t s . A n s - T h e t h i r d s i d e = \sqrt{260} u n i t s .$

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