Question

The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field

Answer 1

Step-by-step explanation:

Given:

The diagonal of a rectangular field is 16m more than the shorter side.

Longer side is 14m more than the shorter side.

To find :

Lengths of the sides of the field.

Solution :

Let the length of the shorter side be x m.

Length of longer side = x + 14 m.

Length of diagonal = x + 16 m.

Using Pythagoras theorem :

• x² + (x + 14)² = (x + 16)²
• x² + 196 + x² + 28x = 256 + x² + 32
• x² - 4x - 60 = 0
• x² + 6x - 10x - 60 = 0
• x(x + 6) - 10(x + 6) = 0
• (x - 10) (x + 6) = 0
• x = -6, 10 [x ≠ -'ve]
• x = 10

Lengths of the sides of the field :

• x = 10 m
• x + 14 = 24 m

∴ The length of the sides is 10 m and 24 m.

Verification :

• x² + (x + 14)² = (x + 16)²

⇒ (10)² + (24)² = (10 + 16)²

⇒ 100 + 576 = 26²

⇒ 676 = 676

∴ L.H.S = R.H.S

...ッ

Answer 2

Answer:

Provided that:

• The diagonal of a rectangular field is 16 metres more than the shorter side.

• The longer side is 14 metres more than the shorter side.

To calculate:

• The lengths of the sides of the field

Solution:

• The lengths of the sides of the field = 10 m and 24 m respectively.

Assumptions:

• Let the length of shorter side = a metres...Equation 1st

• As the diagonal of a rectangular field is 16 metres more than the shorter side. Therefore,

• Diagonal of rectangle = a + 16 metres.

• As the longer side is 14 metres more than the shorter side. Therefore,

• Longer side = a + 14 metres...Equation 2nd

Using concept:

• We can use phythagoras theorm as here diagonal of rectangle is given.

Using formula:

• (H)² = (P)² + (B)²

Where, H denotes hypotenuse, P denotes perpendicular and B denotes base.

Required solution:

→ (H)² = (P)² + (B)²

→ (AC)² = (AB)² + (BC)²

→ (a + 16)² = (a + 14)² + (a)²

→ a + 16 × a + 16 = a + 14 × a + 14 + a × a

→ a² + 32a + 256 = a² + 28a + 196 + a²

→ Combining like terms.

→ a² + a² - a² + 28a - 32a + 196 - 256 = 0

→ a² - 4a - 60 = 0

→ Splitting middle term.

→ a² - 10a + 6a - 60 = 0

→ a(a-10) + 6(a-10) = 0

→ (a-10) (a+6) = 0

→ a - 10 = 0 , a + 6 = 0

→ a = 0 + 10 , a = 0 - 6

→ a = 10 , a = -6

• Here, we can't take a as -6 as we can't take any length or breadth of rectangle as negative, so we are taking a as 10. Henceforth,

→ a = 10

~ Now let us put value of a as 10 in equation 1st and 2nd respectively.

In equation 1st...

→ a metres

→ 10 metres

In equation 2nd...

→ a + 14

→ 10 + 14

→ 24 metres

• Henceforth, longest side = 24 m and shortest side = 10 m $\:$