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A triangle has an area of 15 cm2. The area of another triangle which has same base length is 20 cm2. By what percentage is the altitude of second triangle is larger than the first triangle?1). 33%2). 44%3). 66%4). 45% |
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Answer» Let the base of the triangle be B, Altitude of first triangle is X and altitude of SECOND triangle is Y Area of first triangle is = (1/2) × B × X = 15 cm2----(1) Area of second triangle = (1/2) × B × Y = 20 cm2----(2) DIVIDING equation (1) by (2), we get ⇒ 15/20 = X/Y ⇒ 3Y = 4X ⇒ Y = (4/3)X = (1 + 1/3)X Difference in two lengths = (1/3)X ∴ % of EXCESS length = 33% |
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