Question

in ∆ABC, AD is perpendicular to BC. If AB =6 cm, BC= 10 cm and AC =8 cm, find the area of ∆ABC. Also find the length of AD. ​

Answer 1

Answer:

In right angles triangle BAC, AB=5cm and AC=12cm

Area of triangle=

2

1

×base×height=

2

1

×AB×AC

=

2

1

×5×12=30cm

2

Now, in ΔABC,

Area of triangle ABC=

2

1

×BC×AD

⇒30=

2

1

×13×AD

⇒ AD=

13

30×2

=

13

60

cm.

Answer 2

Answer:

Area of triangle ABC =

area of triangle ABC=

$\frac{1}{2} \times height \times base$

$\frac{1}{2} \times ac \times ab \\ \frac{1}{2} \times 8 \times 6 = 24cm {}^{2}$length of AD=

$\frac{1}{2} \times base \times height \\ \frac{1}{2} \times 10 \times ad = 24 \\ 24 \times \frac{2}{10} = \frac{24}{5} ( 4 \times \frac{4}{5} )$