Question

14. The diagonal of a quadrilateral shaped field is 24 m and perpendicular dropped on it from the remaining opposite vertices are 6 m and 12 m. Find the area of the field answer is 216 sq m. I need steps​

Answer 1

$\huge \underline\pink{Give : }$

• The diagonal of a quadrilateral shaped field is 24 metre.
• The perpendiculars dropped on it from the remaining opposite vertices are 8 metre and 13

$\huge \underline \pink{To \: find:}$

• The area of the quadrilateral shaped field.

$\huge \underline\green{Solution:}$

Let's assume the □ ABCD to be the quadrilateral shaped field.

AC = Diagonal of the quadrilateral measuring 24 m

BE and FD are two perpendiculars dropped on it from the remaining opposite vertices.

BE = 6 m

FD = 12 m

The field is divided into two triangles, Δ ABC and Δ ACD

About Δ ABC :-

Base = AC = 24 m

Height = BE = 6 m

About Δ ACD :-

Base = AC = 24 m

Height = FD = 12 m

Now using the data for two triangles we will calculate the area of the triangles and then sum their areas to find the area of the field.

Area of Δ ABC,

We know that area of a triangle is calculated using the formula,

Area of triangle=1/2 ×base × height

Plug in the values,

$\bf \: A (Δ ABC) = \large\frac{1}{2}× AC × BE$

$\bf \: A (Δ ABC) = \large\frac{1}{ \cancel{2}} × \cancel{24 } × 6$

$\bf \: A (Δ ABC) = 12 × 6$

$\bf \: A ( Δ ABC) = 72sq. m \\ \\ \bf \: A (Δ ABC) = 72 m²$

Similarly we will calculate the area of Δ ACD using the same formula.

$\bf \: A (Δ ACD) = \large\frac{1}{2} × AC × FD$

Plug in the values,

$\bf \: A (Δ ABC) = \large\frac{1}{{ \cancel2}} × \cancel{24} × 12$

$\bf \: A (Δ ACD) = 12 × 12$

$\bf \: A ( Δ ACD ) = 144 sq. m \\ \\ \bf \: A (Δ ACD) = 144 m²$

Now we will calculate the area of the quadrilateral shaped field.

$\bf \green{Area \: of \: field = Area \: of \: Δ ABC + Area \: of \: Δ ACD}$

$\bf \: Area \: of \: field = 72+ 144 \\ \\ \bf \: Area \: of \: field = 216 \: sq.m$

$\boxed{ \bf \:Area of quadrilateral shaped field is 252 m²}$