Question

A rectangular park is to be designed whose breadth is 3m less than its length . its area is to be 4 sq. m. more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12m . find its length and breadth.
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)

Answer 1

Let breadth of a rectangular Park= x m

Length of a rectangular Park=(x+3) m

Area of rectangle= Length × Breadth

Area of rectangular Park= Length × Breadth

= x(x+3)= (x²+3x)

Area of rectangular Park= (x²+3x) m²

Base of a rectangular Park=  Breadth of the rectangular Park   (Given)

Base of a rectangular Park= x m

Altitude of Triangular Park= 12 m (GIVEN)

Area of triangle= ½× base×  altitude

Area of Triangular Park= 1/2× x×12= 6x m²

Area of Triangular Park= 6x m²

ATQ
Area of rectangular Park=  4 + area Triangular Park

x²+3x= 4+ 6x

x²+3x-6x-4=0
x²-3x-4=0
x²-4x+x -4=0

[By middle term splitting]

x(x-4)+1(x-4)=0

(x-4)(x+1)=0

x=4, x= -1

Breadth cannot be negative ,so neglect x= - 1

Breadth (x)=4
Hence,  breadth of a rectangular Park=  4 m and length of the rectangular Park= x+3= 4+3= 7 m.

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Answer 2

Let length of the rectangular park = x m,

breadth of the rectangular park = (x -3)m

∴ Area of the rectangular park = x(x – 3)m2… (i)

Base of an isosceles triangle = (x – 3)m

Altitude of an isosceles triangle = 12 m

∴ Area of isosceles triangle

= 1/2 × base × altitude

= 1/2 × (x – 3) × 12

= 6(x – 3) …(ii)

According to the question,

Ar.(rectangle) – Ar.(isosceles ∆) = 4 m2

⇒ x(x – 3) – 6(x – 3) = 4 … [From (i) & (ii)

⇒ x2 – 3x – 6x + 18 – 4 = 0

⇒ x2 – 9x + 14 = 0

⇒ x2 – 7x – 2x + 14 = 0

⇒ x(x – 7) – 2(x – 7) = 0

⇒ (x – 2) (x – 7) = 0

⇒ x – 2 = 0 or x – 7 = 0

⇒ x = 2 or x = 7

When x = 2, breadth of rectangle becomes -ve, so this is not possible.

∴ Length of the rectangular park, x = 7 m

and Breadth = (x – 3) = 4 m.