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QUADRATIC EQUATION - PROBLEM ON MENSURATOION

**Problem On Mensuration –
**

**Example.1) The perimeter of the rectangular field is 104 m and its area is 640 m². Find the dimension of the field.
**

**Ans.) Let the breadth of the field be x meters. **

**Then, 2 (length + breadth) = 104 meter
**

** => (length + breadth) = 52 meter
**

** => (length + x) = 52 meter
**

** => length = (52 – breadth) m
**

**But, as per the condition, area = 640 m²
**

** x (52 – x) = 640
**

** => 52x - x² = 640
**

** => x² - 52x + 640 = 0
**

** => x² - (32 + 20)x + 640 = 0
**

** => x² - 32x – 20x + 640 = 0
**

** => x (x – 32) – 20(x – 32) = 0
**

** => (x – 32) (x – 20) = 0
**

** => (x – 32) = 0 or (x – 20) = 0
**

** => x = 32 or x = 20
**

**So, the length of the field is 32 m and breadth of the field is 20 m (Ans.)**

**Example.2) Within a rectangular garden, 10 m wide and 20 m long, we wish to pave a walk around the borders of uniform width so as to leave an area of 96 m² for flowers. How wide should the walk to ?
**

**Ans.) Length of the garden = 20 m
**

**Breadth of the garden = 10 m
**

**Let the width of the walk be x meters
**

**Length the garden excluding the walk = (20 – 2x) m
**

**Breadth of garden excluding the walk = (10 – 2x) m
**

**Area for flowers = (20 – 2x) (10 – 2x) m² = (4x² - 60x + 200) m²
**

**Now, as per the given condition, we have –
**

** 4x²- 60x + 200 = 96
**

**=> 4x²- 60x + 104 = 0
**

**=> x²- 15x + 26 = 0
**

**=> x²- (13 + 2)x + 26 = 0
**

**=> x²- 13x – 2x + 26 = 0
**

**=> x (x – 13) – 2(x – 13) = 0
**

**=> (x – 13) (x – 2) = 0
**

**=> (x – 13) = 0 or (x – 2) = 0
**

**=> x = 13 or x = 2
**

**=> x = 2 [x ≠ 13, as the width of walk cannot be 10 m]
**

**So, width of the walk is 2 meters. (Ans.)**

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