Question

∆ ABC is right angled at A (Fig 9.16). AD is perpendicular to BC. If AB = \(5\) cm, BC = \(13\) cm and AC = \(12\) cm, Find the area of ∆ ABC. Also find the length of AD.

Answer 1

\(Area= \frac{1}{2}\times Base \times Height\)

\(=\frac{1}{2}\times 5\times 12=30\; cm^2\)

Also, area of triangle = \(\frac{1}{2}\times AD\times BC\)

\(30=\frac{1}{2}\times AD\times 13\)

\(\frac{30\times 2}{13}=AD\)

\(AD=4.6\;cm\)

Answer 2

Given:
AB = 5cm and AC = 12cm
Step 1: Calculate the Area of \(\Delta ABC\)
Since \(\Delta ABC\) is a right-angled triangle at A, we can use the lengths of AB and AC to find the area.
The area of a right-angled triangle is given by:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)

Thus, the area is:
\(\text{Area} = \frac{1}{2} \times 5 \, \text{cm} \times 12 \, \text{cm}\)
\(\text{Area} = \frac{1}{2} \times 60 \, \text{cm}^2\)
\(\text{Area} = 30 \, \text{cm}^2\)

Step 2: Calculate the Length of AD
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)
\(30 \, \text{cm}^2 = \frac{1}{2} \times 13 \, \text{cm} \times AD\)

\(30 \, \text{cm}^2 = \frac{13}{2} \times AD\)
\(30 \times 2 = 13 \times AD\)
\(60 = 13 \times AD\)
\(AD = \frac{60}{13}\)
\(AD = 4.62 \, \text{cm}\)

So, the answer is 4.62 cm