Question

A5. The length of one of the diagonals of a field in the form of a quadrilateral is 40 m. The
perpendicular distances of the other two vertices from this quadrilateral are 12 m and 9 m.
Find the area of the field.​

Answer 1

Given :

• Length of one of the diagonals of the quadrilateral is 40 m
• Perpendicular distances of other two vertices from vertex are 12 m and 9 m

To Find :

• The area of the field

Solution :

Area of a quadrilateral is given by ,

$\\ {\boxed{{\bf{Area = \frac{1}{2} \times d(h + h) }}}}$

Here ,

• d is length of diagonal of the quadrilateral
• h₁ and h₂ are perpendicular distances

We have ,

• d = 40 m
• h₁ = 12 m
• h₂ = 9 m

Substituting the values in the formula ,

$\\ : \implies \sf \: Area = \frac{1}{2} \times 40(9 + 12)$

$\\ : \implies \sf \: Area = \frac{1}{2} \times 40(21)$

$\\ : \implies \sf \: Area = 20\times 21$

$\\ : \implies{\underline {\boxed{\mathfrak{\blue{Area = 420 \: {m}^{2} }}}}} \: \bigstar$

Hence , The area of the given field is 420 m²

Answer 2

Answer:

Given :-

• Length of one of the diagonals of the quadrilateral is 40 m
• Perpendicular distances of other vertices from the vertex are 12 m and 9 m

To Find :-

Area of field

SoluTion :-

Area =½ × Diagonal (h₁ + h₂)

Area = ½ × 40(12 + 9)

Area = 20 (21)

Area = 420 m²

Hence :-

The area of the field is 420 m²

Know More :-

Diagonal = √(length + Breadth)

Area = Length × Height

Perimeter = 2(l + b)