Question

prove that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude​

Answer 1

Answer:

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Step-by-step explanation:

tommorow

Answer 2

Answer:

Step-by-step explanation:

➡ Given :-  → A ∆ABC in which AB = BC = CA and AD ⟂  BC .

➡ To prove :- → 3AB² = 4AD².

➡ Proof :-  In ∆ADB and ∆ADC, we have

→ AB = AC. [ Given ]

→ B = C = 60° .

→ ADB = ADC = 90° .

•°• ∆ADB  ∆ADC . [ AAS - Congruence ]

•°• BD = DC = ½BC

▶ From right ∆ADB, we have

AB² = AD² + BD² . [ By Pythagoras' theorem ]

= AD² + ( ½ BC )² .

= AD² + ¼ BC² .

=> 4AB² = 4AD² + BC² .

=> 3AB² = 4AD² . [ °•° BC = AB ]

3AB² = 4AD²

HENCE WE HAVE PROVED THAT THREE TIMES THE SQUARE OF ANY SIDE OF AN EQUILATERAL TRIANGLE IS EQUAL TO FOUR TIMES THE SQUARE OF THE ALTITUDE