Question

The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 12 m. Find the area of the field.

Answer 1

Given,

• BD = 24m
• AE = 12m
• CF = 8m

To Find,

• The Area Of Field.

Solution,

→ Area of Quadrilateral Field

Area of Quadrilateral Field = Area Of ∆ABD + Area of ∆BCD

→ Area of Quadrilateral Field

= ½ × base × height + ½ × base × height

→ Area of Quadrilateral Field

= ½ × 24m × 12m + ½ × 24m × 8m

→ Area of Quadrilateral Field

= 24m × 6m + 24m × 4m

→ Area of Quadrilateral Field

= 24m × ( 6m + 4m )

→ Area of Quadrilateral Field

= 24m × 10m

→ Area of Quadrilateral Field = 240m²

Required Answer,

Area of Quadrilateral Field = 240m²

Answer 2

Required Answer :-

• The area of the quadrilateral-shaped field is 240m².

Step-by-step explanation:

To Find :-

• The area of quadrilateral-shaped field

Solution:

Given that,

• The diagonal of the quadrilateral-shaped field is 24m

• The perpendiculars dropped on it from the remaining opposite vertices are 8m and 12m.

Therefore,

As we know that,

$\bigstar \: \: \boxed{\sf{ \bigg \lgroup \: Area \: of \: quadrilateral = \dfrac{1}{2} \times d \times ({h}_{1} + {h}_{2})} \bigg\rgroup \: sq. \: units}$

Where,

• d = diagonal of quadrilateral.
• h = sum of the perpendiculars dropped on diagonal.

According the question,

$\longrightarrow \: \sf{Area = \dfrac{1}{2} \times d \times ({h}_{1} + {h}_{2})}$

By putting the values,

$\longrightarrow \bigg \lgroup \sf{Area = \dfrac{1}{2} \times 24 \times (8 + 12)\bigg\rgroup \: {m}^{2}}$

By adding 8 and 12,

$\longrightarrow \: \bigg \lgroup \sf{Area = \dfrac{1}{2} \times 24 \times 20 \: \bigg\rgroup \: {m}^{2}}$

By dividing 2 and 20,

$\longrightarrow \: \bigg \lgroup \sf{Area = \dfrac{1}{ \cancel{2}} \times 24 \times \cancel{20}\bigg\rgroup \: {m}^{2}}$

$\longrightarrow \: \bigg \lgroup \sf{Area = 1 \times 24 \times 10 \: \bigg\rgroup \: {m}^{2}}$

By multiplying,

$\longrightarrow \: \bigg \lgroup \sf{Area = 24 \times 10\bigg\rgroup \: {m}^{2}}$

$\longrightarrow \: \sf{ \purple{Area = 240 {m}^{2}}}$

Hence,

The area of quadrilateral-shaped field is 240m².