Question

the sides of a quadrangular field taken in order are 26m ,27m , 7m and 4m respectively. the angle contained by the last two sides is a right angle.find it's area.

Answer 1

Hello Dear.

Refers to the attachment for the Answer.

From the attachment, In Δ ABC, it is right angled triangles at B.

∴ Area of the Δ ABC = 1/2 × AB × BC.
   = 1/2 × 4 × 7
   = 2 × 7
   = 14 cm².

Applying Pythagoras Theorem In ΔABC,
AC² = AB² + BC²
⇒ AC² = (4)² + (7)²
⇒ AC² = 16 + 49
⇒ AC² = 65
⇒ AC = √65 cm.

Now, In ΔADC,
Taking, a = AC = √65 cm.= 8.06 cm

b = AD = 26 cm.
c = DC = 27 cm.

Using Heron's formula,
S = (a + b + c) ÷ 2
S = (8.06 + 26 + 27) ÷ 2
S = 30.53 cm.
S ≈ 30.5 cm.

Area of the ΔADC = $\sqrt{S(S - a)(S - b)(S - c)}$
= $\sqrt{30.5(30.5 - 8.06)(30.5 - 26)(30.5 - 27)}$
= $\sqrt{30.5(22.44)(4.5)(3.5)}$
= $\sqrt{10779.615}$
= 103.83 cm².

∵ Area of the Quadrilateral ABCD = Area of Δ ABC + Area of ΔADC
 = 14 + 103.83
 = 117.83 cm².


Hence, the area of the Quadrilateral ABCD is 117.83 cm².



Hope it helps.