Question

A body oscillate with SHM according. to the
equation X= 5 cos( pit+pi/4)m. what is the
maximum amplitude of
Oscillation​

Answer 1

Answer:

Given -

Sides of three triangles.

To find -

Area of triangles.

Formula used -

Heron's formula

Solution -

In the question, we are provided with the sides of 3 triangles, and we need to find it's area. For that, we will apply heron's formula and will find every triangle's area. For that, first we will add up side a , b and c and then, semi - perimeter and then applying heron's formula, we will find their area

Let's do it!

Heron's formula says -

\begin{gathered} \sf \: \sqrt{s \:(s \: - a)(s \: - b)(s \: - c)} \\ \end{gathered}

s(s−a)(s−b)(s−c)

For 1st triangle -

Side a = 9 cm

Side b = 10 cm

Side c = 17 cm

Semi - Perimeter -

\begin{gathered} \sf \: s \: = \dfrac{9 \: + \: 10 \: + \: 17}{2} \\ \\ \sf\: s \: = \cancel\frac{36}{2} \\ \\ \sf\: s \: = 18\end{gathered}

s=

2

9+10+17

s=

2

36

s=18

Area of 1st triangle -

\begin{gathered} \sf \sqrt{18(18 \: - \: 9)(18 \: - \: 10)( 18\: - \: 17)} \\ \\ \sf \: \sqrt{18 \: \times \: 9 \: \times \: 8 \: \times \: 1} \\ \\ \sf \: \sqrt{1296} \\ \\ \sf \: 36 { \: cm}^{2} \\ \end{gathered}

18(18−9)(18−10)(18−17)

18×9×8×1

1296

36cm

2

For 2nd triangle -

Side a = 13 cm

Side b = 5 cm

Side c = 12 cm

Semi - Perimeter -

\begin{gathered} \sf \: s \: = \dfrac{13 \: + \: 5 \: + \: 12}{2} \\ \\ \sf \: s \: = \cancel \dfrac{30}{2} \\ \\ \sf \: s \: = 15 \\ \end{gathered}

s=

2

13+5+12

s=

2

30

s=15

Area of 2nd triangle -

\begin{gathered} \sf \: \sqrt{15(15 \: - \: 13)(15 \: - \: 5)(15 \: - \: 12)} \\ \\ \sf \: \sqrt{15 \: \times 2 \: \times \: 10 \: \times \: 3} \\ \\ \sf \: \sqrt{900} \\ \\ \sf \: 30 { \: cm}^{2} \end{gathered}

15(15−13)(15−5)(15−12)

15×2×10×3

900

30cm

2

For 3rd triangle -

Side a = 35 m

Side b = 45 m

Side c = 50 m

Semi - perimeter -

\begin{gathered} \sf \: s \: = \dfrac{35 \: + \: 45 \: + \: 50}{2} \\ \\ \sf \: s \: = \cancel\frac{130}{2} \\ \\ \sf \: s \: = 65 \\ \end{gathered}

s=

2

35+45+50

s=

2

130

s=65

Area of 3rd triangle -

\begin{gathered} \sf \: \sqrt{65(65 \: - \: 35)(65 \: - \: 45)(65 \: - \: 50)} \\ \\ \sf \: \sqrt{65 \: \times \: 30 \: \times \: 20 \: \times \: 15} \\ \\ \sf \: \sqrt{585000} \\ \\ \sf \: 764.8 \: m^{2} \: (appox) \\ \end{gathered}

65(65−35)(65−45)(65−50)

65×30×20×15

585000

764.8m

2

(appox)