Théodore Gudin
(1802-1880)

French painter (1802-1880) – Théodore Gudin was born in Paris (Capital of France) on August 15th, 1802 and died in Boulogne-Billancourt (French commune in Hauts-de-Seine, Île-de-France) on April 11th, 1880 at the age of 77. Today Théodore Gudin would be 221 years old.

Age

how old was Théodore Gudin when he died?

77

Théodore Gudin died in 1880 at the age of 77.


Biographical data

Birthday August 15th, 1802 (Sunday)
Place of Birth Paris
Capital of France
Death Date April 11th, 1880 (Sunday)
Death place Boulogne-Billancourt
French commune in Hauts-de-Seine, Île-de-France
Birth sign (Zodiac) Leo (The Lion) ♌
Chinese Zodiac Dog 狗

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Also born in 1802

Théodore Gudin

Birthday

When was Théodore Gudin born?

Théodore Gudin was born on August 15th, 1802.

Death

When did Théodore Gudin die?

Théodore Gudin died on April 11th, 1880 at the age of 77 in Boulogne-Billancourt (French commune in Hauts-de-Seine, Île-de-France). Today Théodore Gudin would be 221 years old.

Place of Birth

Where was Théodore Gudin born?

Paris (Capital of France).

Death place

Where did Théodore Gudin die?

Boulogne-Billancourt (French commune in Hauts-de-Seine, Île-de-France).

Zodiac

What is the zodiac sign of Théodore Gudin?

Théodore Gudin was born in the zodiac sign Leo (The Lion).

Chinese Zodiac

Which Chinese zodiac sign does Théodore Gudin have?

Théodore Gudin was born in 1802 in the year of the Chinese zodiac Dog.

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Died on April 11th, 1880:
Anniversaries of death

French painter (1802-1880)

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