Question

Q2The sides of the triangle are 8cm,15cm and 17cm.Find the area of the triangle. Also find the length of the altitude drawn on the side with length 17cm. ​

Answer 1

Step-by-step explanation:

let a, b, c sides of the sides of triangle respectively, that are 8cm, 15cm, 17cm

$s = \frac{a + b + c}{2} \\ = \frac{ 8cm+15cm+17cm }{ 2 } \\ = \frac{40cm}{2} = 20cm$

$area = \sqrt{ s(s-a)(s-b)(s-c) }$

$= \sqrt{ 20cm(20cm-8cm)(20cm-15cm)(20cm-17cm) } \\ = \sqrt{20cm \times 12cm \times 5cm \times 3cm} \\ = \sqrt{ {3600cm}^{4} } = 60 {cm}^{2}$

area = 60cm²

also

$area = \frac{1}{2} bh \\ 60 {cm}^{2} = \frac{1}{2} \times 17cm \times h \\ \frac{60{cm}^{2} \times 2}{17cm} = \frac{120}{17} cm = h$

$altitude \: = h = \frac{120}{17} cm$

Answer 2

Answer:

by using hero's formula

Step-by-step explanation:

perimeter of the triangle=8+15+17

=40

s=40/2

=20

$area = \sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{20(20 - 8)(20 - 15)(20 - 17)}$

$= \sqrt{20 \times 12 \times 5 \times 3}$

$= \sqrt{3600}$

$= 60$

again, area=1/2(base)(altitude)

60=1/2(17)(altitude)

120/17=altitude

altitude=7.06