Question

0.03 You are Samay sharma head boy of New vision public school Ratpur. Write a notice for the school
Notice board in about 80-100 words announcing the inter house karate and music competition to be held
on 10 December 2020​

Answer 1

Format of a Notice :

[ Notice should be kept in a box ]

• Name of the Insititution/Organisation
• NOTICE
• Date
• Heading
• Body
• Regards
• Name
• Designation

Required Notice :

⠀⠀⠀⠀⠀⠀⠀New Vision Public school

⠀⠀⠀⠀⠀⠀⠀⠀ NOTICE

06/12/2020

Inter house karate and music competetion

This is to inform the students that there will be an ' Inter house karate and music competetion ' on 10th december 2020. Intersted students can give their names.Participations will be appreciated

Regards ,

Samay sharma ,

Head boy.

Answer 2

Answer:

Given :

Coordinates of vertices of the triangle are (1,2) , (5,7) and (5,-3)

To Find :

The Area of triangle formed

Knowledge Required :

The area of the triangle formed when the coordinates of vertices of the triangle are (x₁ , y₁) , (x₂ , y₂) and (x₃ , y₃) is given by ,

\begin{gathered} \\ \star \: \boxed{\purple{\sf{Area = \frac{1}{2} | x_1(y_2 - y_3) + x_2( y_3 - y_1) + x_3(y_1 - y_2)| .}}}\end{gathered}

⋆

Area=

2

1

∣x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)∣.

Solution :

By comparing the coordinates of the given vertices with the formula we get ,

x₁ = 1, x₂ = 5 , x₃ = 5

y₁ = 2 , y₂ = 7 , y₃ = -3

Substituying the values ,

\begin{gathered} \\ : \implies \sf \: a = \frac{1}{2} | 1(7 - ( - 3)) +5( - 3 - 2) +5(2 - 7) | \\ \\ \end{gathered}

:⟹a=

2

1

∣1(7−(−3))+5(−3−2)+5(2−7)∣

\begin{gathered} \\ : \implies \sf \: a = \frac{1}{2} | 1(7 + 3) +5( - 5) +5( - 5) | \\ \\ \end{gathered}

:⟹a=

2

1

∣1(7+3)+5(−5)+5(−5)∣

\begin{gathered} \\ : \implies \sf \: a = \frac{1}{2} | 1(10) + ( - 25) + ( - 25) | \\ \\ \end{gathered}

:⟹a=

2

1

∣1(10)+(−25)+(−25)∣

\begin{gathered} \\ : \implies \sf \: a = \frac{1}{2} | 10 - 25 - 25 | \\ \\ \end{gathered}

:⟹a=

2

1

∣10−25−25∣

\begin{gathered} \\ : \implies \sf \: a = \frac{1}{2} | 10 -50 | \\ \\ \end{gathered}

:⟹a=

2

1

∣10−50∣

\begin{gathered} \\ : \implies \sf \: a = \frac{1}{2} | - 40| \\ \\ \end{gathered}

:⟹a=

2

1

∣−40∣

\begin{gathered} \\ : \implies{\boxed{\pink{\sf {\: a = 20 \: sq. \: units }}}} \: \bigstar\\ \\ \end{gathered}

:⟹

a=20sq.units

★

\begin{gathered} \\ \therefore {\underline{\sf{Hence , The \: area \: of \: the \: Triangle \: formed \: is \: \bold{ 20 \: sq.units}}}}\end{gathered}

∴

Hence,TheareaoftheTriangleformedis20sq.units