# The sides of a quadrilateral field, taken in order are 26 m,

Question:

The sides of a quadrilateral field, taken in order are 26 m, 27 m, 7 m, 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.

Solution:

Here the length of the sides of the quadrilateral is given as

AB = 26 m, BC = 27 m, CD = 7 m, DA = 24 m

Diagonal AC is joined.

By applying Pythagoras theorem

$A C^{2}=A D^{2}+C D^{2}$

$A C^{2}=14^{2}+7^{2}$

AC = 25 m

Now area of triangle ABC

Perimeter = 2s = AB + BC + CA

2s = 26 m + 27 m + 25 m

s = 39 m

By using Heron's Formula

The area of a triangle $=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$

$=\sqrt{39 \times(39-26) \times(39-27) \times(39-25)}$

$=291.84 \mathrm{~m}^{2}$

Thuc the area of a trianglo ic $201 \mathrm{~g} 4 \mathrm{~m}^{2}$

Now for area of triangle ADC

Perimeter = 2S = AD + CD + AC

= 25 m + 24 m + 7 m

S = 28 m

By using Heron's Formula

The area of a triangle $=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$

$=\sqrt{28 \times(28-24) \times(28-7) \times(28-25)}$

$=84 \mathrm{~m}^{2}$

Thus, the area of a triangle is $84 \mathrm{~m}^{2}$

Therefore, Area of rectangular field ABCD

= Area of triangle ABC + Area of triangle ADC

= 291.84 + 84

$=375.8 \mathrm{~m}^{2}$