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QUADRATIC EQUATION - PROBLEM ON GEOMETRIC FIGURE

**PROBLEM ON GEOMETRIC FIGURES –
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**Example.1) In a right angled triangle, the hypotenuse is 2 cm longer than the base and 4 cm longer than the shortest side. Find the lengths of the three sides of the triangle
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**Ans.) Let, the length of the hypotenuse be x cm.
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**Then, base = (x – 2) cm and shortest side as per the given condition would be = (x – 4) cm
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**By, Pythagoras theorem, we have –
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** (x – 2)²+ (x – 4)²= x²
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**=> (x²- 4x + 4) + (x² - 8x + 16) = x²
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**=> 2x²- 12x + 20 - x² = 0
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**=> x²- 12x + 20 = 0
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**=> x²- (10 + 2) x + 20 = 0
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**=> x²- 10x – 2x + 20 = 0
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**=> x (x – 10) – 2(x – 10) = 0
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**=> (x – 10) (x – 2) = 0
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**=> (x – 10) = 0 or (x – 2) = 0
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**=> x = 10 [neglecting x = 2, since side of a triangle cannot be negative (-ve)]
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**Hence hypotenuse = 10 cm, base = (10 -2) cm = 8 cm
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**Shortest side would be = (10 – 6) cm = 4 cm (Ans.)**

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