# Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

**Solution:**

We divide the quadrilateral into two triangles, and by using Heron’s formula, we can calculate the area of triangles.

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} = 3^{2} + 4^{2 }= 25 = 5^{2}

⇒ 5^{2 }= 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

= 1/2 × 3 cm × 4 cm = 6 cm^{2}

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

s = (5 + 4 + 5)/2

s = 14/2

s = 7 cm

By using Heron’s formula,

Area of ΔADC = √[s(s - a)(s - b)(s - c)]

= √[7(7 - 5)(7 - 4)(7 - 5)]

= √[7 × 2 × 3 × 2]

= 2√21 cm^{2}

Area of ΔADC = 9.2 cm^{2} (approx.)

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

= 9.2 cm^{2} + 6 cm^{2}

= 15.2 cm^{2}

Thus, the area of the quadrilateral ABCD is 15.2 cm^{2}.

**Video Solution:**

## Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

### Class 9 Maths NCERT Solutions - Chapter 12 Exercise 12.2 Question 2:

**Summary:**

We have found that area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm is 15.2 cm^{2}.