Question

18.A field is in the form of a quadrilateral and one of its diagonal is 140 m.
The perpendicular distances from the opposite vertices to this diagonal are
45 m and 55 m, then area of the field is ?​

Answer 1

$\bf{\underline{Question:-}}$

A field is in the form of a quadrilateral and one of its diagonal is 140 m.

The perpendicular distances from the opposite vertices to this diagonal are

45 m and 55 m, then area of the field is ?

$\bf{\underline{Given:-}}$

• Diagonal of quadrilateral = 140m
• The perpendicular distances from the opposite vertices to this diagonal are 45 m and 55 m,

$\bf{\underline{To\: Find:-}}$

• Area of Quadrilateral = ?

$\bf{\underline{Solution:-}}$

Area of quad. (ABCD) = Area of ∆ ABC + Area of ∆ ADC

$\sf → \frac{1}{2}×base×height + \frac{1}{2}×base×height$

$\sf →\frac{1}{2}×AC×BQ + \frac{1}{2}×AC×DP$

• Taking 1/2 × AC ( Common )

$\sf → \frac{1}{2}×AC ( BQ + DP )$

$\sf → \frac{1}{2}×140 ( 45 + 55)$

$\sf → 1× 70 × 100$

$\sf → 7000m^2$

$\bf{\underline{Hence:-}}$

• Area of quadrilateral = 7000m²