A perpendicular height drawn on the base of the triangle is ten - sixth of its base on which it is drawn. If the height is increased by 8 m and the base is decreased by 4 m, the area of the triangle remains unchanged. Find the area of the triangle.

- 480 m
^{2} - 400 m
^{2} - 620 m
^{2} - 560 m
^{2} - 800 m
^{2}

Option 4 : 560 m^{2}

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Let the base of triangle be A cm

⇒ Height of the triangle = 10/6 A

⇒ Area of triangle = 1/2 × A × 10/6 A = 5/6 A^{2}

⇒ By the given condition,

⇒ Area of new triangle = Area of original triangle

⇒ 1/2 × (A – 4) × (10/6 A + 8) = 5/6 A^{2}

⇒ (A – 4) × (10A + 48) = 5A^{2}

⇒ 10A^{2} + 48A – 40A – 192 = 10A^{2}

⇒ 8A = 192

⇒ A = 24

∴ Answer is Area = 1/2 × 24 × 40 = 480 m^{2}

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