Question

The Ratio of Length To Breadth of a Rectangular Field is 4 : 3. If Its Diagonal is 200 m, The Area of The Field in square meters Do it With Proof



a) 18200 b) 19200 c) 18500 d) 18900​

Answer 1

Given :

• The Ratio of Length To Breadth of a Rectangular Field is 4 : 3.
• Diagonal of Rectangle is 200 m .

To Find :

• Area of field .

Solution :

$\longmapsto\tt{Let\:length\:of\:the\:field\:be=4x}$

$\longmapsto\tt{Let\:breadth\:of\:the\:field\:be=3x}$

Using Formula :

$\longmapsto\tt\boxed{Diagonal=\sqrt{{(l)}^{2}+{(b)}^{2}}}$

Putting Values :

$\longmapsto\tt{200=\sqrt{{(4x)}^{2}+{(3x)}^{2}}}$

$\longmapsto\tt{200=\sqrt{{16x}^{2}+{9x}^{2}}}$

$\longmapsto\tt{200=\sqrt{{25x}^{2}}}$

$\longmapsto\tt{200=5x}$

$\longmapsto\tt{\cancel\dfrac{200}{5}=x}$

$\longmapsto\tt\bf{40=x}$

Value of x is 40 .

Therefore :

$\longmapsto\tt{Length\:of\:field=4(40)}$

$\longmapsto\tt\bf{160\:m}$

$\longmapsto\tt{Breadth\:of\:field=4(30)}$

$\longmapsto\tt\bf{120\:m}$

For Area :

Using Formula :

$\longmapsto\tt\boxed{Area\:of\:Rectangle=l\times{b}}$

Putting Values :

$\longmapsto\tt{160\times{120}}$

$\longmapsto\tt\bf{19200\:{m}^{2}}$

Option b) 19200 is Correct .

Answer 2

See the attachment for figure.

________________________________________

Let length and breadth be 4x and 3x respectively.

Given:

PR = 200m(Diagonal)

PQ = 4x(Length)

QR = 3x(Breadth)

To Find:

• The area of the field in m^2

Solution:

In the ∆ PQR, Using pathagoras theorem,

$\sf \pink{\large{(PR)^2 = (PQ)^2 + (QR)^2}}$

$\sf{(200)^{2} = (4x)^{2} + (3x)^{2}}$

$\sf\to{40000 = (4x)^{2} \times + (3x) ^{2} }$

$\sf\to{40000 = 16 {x}^{2} + 9 {x}^{2} }$

$\sf\to{40000 = 25 {x}^{2}}$

$\sf\to{ {x}^{2} = \cancel\frac{40000}{25}}$

$\sf\to{ {x}^{2} = 1600}$

$\sf\to{ {x}^{2} = 1600}$

$\sf \pink{\large\to{x = 40}}$

PQ = 4x = 4 × 40 = 160m^2

QR = 3x = 3 × 40 = 120m^2

A = l × b

A = PQ × QR

A = (160 × 120)m^2

A = 19200m^2

Therefore, area of the field is 19200m^2(Option b)