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Karolina [17]
2 years ago
8

Work out the area of this trapezium

Mathematics
1 answer:
skelet666 [1.2K]2 years ago
6 0

Step-by-step explanation:

Area = ½ × b × h

= ½ × (7.9 + 10.8) × 6.2

= ½ × 18.7 × 6.2

= ½ × 115.94

= <u>5</u><u>7</u><u>.</u><u>9</u><u>7</u><u> </u><u>cm²</u>

So, the area of that trapezium is 57.97 cm²

<em>Hope </em><em>it </em><em>helpful </em><em>and </em><em>useful </em><em>:</em><em>)</em>

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Solve this thing pls<br> <img src="https://tex.z-dn.net/?f=3%5Cleft%28q-7%5Cright%29%3D27" id="TexFormula1" title="3\left(q-7\ri
evablogger [386]

Answer:

  • Solution of equation ( q ) = <u>1</u><u>6</u>

Step-by-step explanation:

In this question we have given an equation that is <u>3 </u><u>(</u><u> </u><u>q </u><u>-</u><u> </u><u>7</u><u> </u><u>)</u><u> </u><u>=</u><u> </u><u>2</u><u>7</u><u> </u>and we have asked to solve this equation that means to find the value of <u> </u><u>q</u><u> </u><u>.</u>

<u>Solution : -</u>

\quad \: \longmapsto \:  3(q - 7) = 27

<u>Step </u><u>1</u><u> </u><u>:</u> Solving parenthesis :

\quad \: \longmapsto \:3q - 21 = 27

<u>Step </u><u>2</u><u> </u><u>:</u> Adding 21 on both sides :

\quad \: \longmapsto \:3q -  \cancel{ 21} +  \cancel{21} = 27  +  21

On further calculations we get :

\quad \: \longmapsto \:3q = 48

<u>Step </u><u>3 </u><u>:</u> Dividing by 3 from both sides :

\quad \: \longmapsto \: \frac{ \cancel{3}q}{ \cancel{3}}  =  \cancel {\frac{48}{3} }

On further calculations we get :

\quad \: \longmapsto \:   \pink{\boxed{\frak{q = 16}}}

  • <u>Therefore</u><u>,</u><u> </u><u>solution</u><u> </u><u>of </u><u>equation</u><u> </u><u>(</u><u> </u><u>q </u><u>)</u><u> </u><u>is </u><u>1</u><u>6</u><u> </u><u>.</u>

<u>Verifying</u><u> </u><u>:</u><u> </u><u>-</u>

Now we are very our answer by substituting value of q in the given equation . So ,

  • 3 ( q - 7 ) = 27

  • 3 ( 16 - 7 ) = 27

  • 3 ( 9 ) = 27

  • 27 = 27

  • L.H.S = R.H.S

  • Hence, Verified.

<u>Therefore</u><u>,</u><u> </u><u>our </u><u>solution</u><u> </u><u>is </u><u>correct</u><u> </u><u>.</u>

<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>
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