Question

Triangle ABC is right angled at A. AD is perpendicular to BC. IF AB=5 cm, BC = 13 cm and AC = 12 cm, Find the area of Triangle ABC. Also find the length of AD​

Answer 1

Given:-

• Side AB = 5 cm

• Side BC = 13 cm

• Side AC = 12 cm

To Find:-

• Area of ΔABC.

• Length of AD.

Solution:-

Here, All the sides of the triangle is given,

So, We use Heron's formula to find the area of triangle.

Heron's formula = $\sqrt{S (S - a) (S - b) (S - c)}$        where, S = $\dfrac{a + b + c}{2}$

                                                                       a, b and c are sides of triangle.

Now, Put the values in the formula,

At first, S = $\dfrac{5 + 12 + 13}{2}$   = 15

⇒ $\sqrt{15 (15 - 5) (15 - 12) (15 - 13)}$

⇒ $\sqrt{15 (10) (3) (2)}$

⇒ $\sqrt{15 \times 10 \times 3 \times 2}$

⇒ $\sqrt{900}$

⇒ 30 cm²

Hence, Area of ΔABC is 30 cm².

Now, Area of triangle = $\dfrac{1}{2} \times base \times height$

                      30 cm² = $\dfrac{1}{2} \times 13 \ cm \times h$

                        $\dfrac{30 \times 2}{13} = h$

                                $h = 4.6 \ cm$

Hence, Length of AD is 4.6 cm.

Some Important Terms:-

• Area of Equilateral triangle = $\dfrac{\sqrt{3}}{4} \times Side^{2}$

• Area of Square  = $Side^{2}$

• Area of Rectangle = $Length \times Breadth$

Answer 2

$In \: \: \: right \: \: \: angles \: \: triangle \\ BAC, \: \: AB=5cm \: \: and \\ AC=12cm \\ </p><p>Area \: \: \: of \ \\triangle \\ = \frac{1}{2} \times \: \: base \times hight = \frac{1}{2} \times ab \times ac \\ = > \frac{1}{2} \times 5 \times 12 = {30cm}^{2} \\ now \: in \: triangle \: \: abc \\ area \: of \: triangle \: \: of \: abc = \frac{1}{2} \times bc \times ad \\ \\ = > 30 = \frac{1}{2} \times 13 \times ad \\ = > ad = \frac{30 \times 2}{13} = \frac{60}{13}cm</p><p>$