Question

Dm. How much area does stocul 2 Find the area of a quadrilateral ABCD in which AB = 3 cm. BC=4 cm. CD- DA=5 cms and AC-5 cm 3. Radha made​

Answer 1

Heron's formula for the area of a triangle is: Area = √[s(s - a)(s - b)(s - c)]

Where a, b, and c are the sides of the triangle, and s = Semi-perimeter = Half the Perimeter of the triangle

Now, ABCD is quadrilateral shown in the figure.

For ∆ABC, consider

AB^2 + BC^2 = 32 + 42 = 25 = 52

$⇒ 52 = AC^2$

Since ∆ABC obeys the Pythagoras theorem, we can say ∆ABC is right-angled at B.

Therefore, the area of ΔABC = 1/2 × base × height

$\sf \: = 1/2 × 3 cm × 4 cm = 6 cm {}^{2}$

$\sf \: Area \: of \: ΔABC = 6 cm {}^{2}$

Now, In ∆ADC

we have a = 5 cm, b = 4 cm and c = 5 cm

Semi Perimeter: s = Perimeter/2

s = (a + b + c)/2

$\sf \: s = \frac{(5 + 4 + 5)}{2}$

$\sf \: s = \frac{14}{2}$

s = 7 cm

By using Heron’s formula,

$\sf \: Area \: o f \: ΔADC = \bold{ \sqrt{\bigg[s(s - a)(s - b)(s - c) \bigg]}}$

= √[7(7 - 5)(7 - 4)(7 - 5)]

= √[7 × 2 × 3 × 2]

$\sf \: = 2√21 cm {}^{2}$

$\sf \: Area \: of \: ΔADC = 9.2 cm {}^{2} \: (approx.)$

Area of the quadrilateral ABCD = Area of ΔADC + Area of ΔABC

$\sf \: = 9.2 cm {}^{2} + 6 cm {}^{2}$

$\sf \: = 15.2 cm {}^{2}$

Thus, the area of the quadrilateral

$\sf \: ABCD \: is \: 15.2 cm {}^{2}$