Question

area = 2.4 dm", height = 80 cm
Find the area of the triangle whose sides are 13 cm, 20 cm and 21 cm. Also find the altitude of
the triangle corresponding to the largest side.​

Answer 1

Step-by-step explanation:

As given,

=> a = 13 cm

=> b = 20 cm

=> c = 21 cm

Now, We have to find Semi- Perimeter

=> S = a+b+c/2

=> S = 13+20+21/2

=> S = 54/2

=> S = 27

Now, Area of triangle,

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

s(s−a)(s−b)(s−c)

= > area = \sqrt{27(27 - 13)(27 - 20)(27 - 21)}=>area=

27(27−13)(27−20)(27−21)

= > area = \sqrt{27 \times 14 \times 7 \times 6}=>area=

27×14×7×6

= > area = \sqrt{3 \times 3 \times 3 \times 7 \times 2 \times 7 \times 2 \times 3}=>area=

3×3×3×7×2×7×2×3

= > area = 3 \times 3 \times 7 \times 2=>area=3×3×7×2

= > area = 126 \: {cm}^{2}=>area=126cm

2

Now, We are required to find out Altitude :-

Here, longest side will 21 cm

Area of Triangle = 1/2 x b x h

=> 126 cm^2 = 1/2 x 20 x h

=> 126 cm^2 = 10 h

=> (126/10) cm = h

=> 12.6 cm = h

=> h = 12.6 cm