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olasank [31]
3 years ago
15

Talia took the bus from her home to the bank and then walked back to her home along the same route. The round trip took 0.9 hour

s total. The bus traveled at an average speed of 40 km/h and she walked at an average speed of 5 km/h. Use the table to complete these statements.

Mathematics
2 answers:
sergij07 [2.7K]3 years ago
6 0

Answer:

Step-by-step explanation:

OlgaM077 [116]3 years ago
4 0
The bus traveled<span> at an </span>average speed<span> of </span>40 km/h<span> and </span>she<span> - 2385764. ... </span>bus<span> from </span>her home<span> to the </span>bank <span>and </span>then walked back<span> to </span>her home along<span> the </span>same route<span>. ... To determine the time, x, that it </span>took Talia<span> to</span>walk home<span>, </span>she<span> used the equation </span>40(0.9<span> – x) = 5x. 1 ... Therefore, it </span>took Talia<span> 0.8 </span>hour<span> to </span>walk home<span> or 48 minutes</span>
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\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}

  • Given - <u>A </u><u>trapezium</u><u> </u><u>ABCD </u><u>with </u><u>non </u><u>parallel </u><u>sides </u><u>of </u><u>measure </u><u>1</u><u>5</u><u> </u><u>cm </u><u>each </u><u>!</u><u> </u><u>along </u><u>,</u><u> </u><u>the </u><u>parallel </u><u>sides </u><u>are </u><u>of </u><u>measure </u><u>1</u><u>3</u><u> </u><u>cm </u><u>and </u><u>2</u><u>5</u><u> </u><u>cm</u>

  • To find - <u>Area </u><u>of </u><u>trapezium</u>

Refer the figure attached ~

In the given figure ,

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<u>Construction</u><u> </u><u>-</u>

draw \: CE \: \parallel \: AD \:  \\ and \: CD \: \perp \: AE

Now , we can clearly see that AECD is a parallelogram !

\therefore AE = CD = 13 cm

Now ,

AB = AE + BE \\\implies \: BE =AB -  AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm

Now , In ∆ BCE ,

semi \: perimeter \: (s) =  \frac{15 + 15 + 12}{2}  \\  \\ \implies \: s =  \frac{42}{2}  = 21 \: cm

Now , by Heron's formula

area \: of \: \triangle \: BCE =  \sqrt{s(s - a)(s - b)(s - c)}  \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)}  \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21}  \: cm {}^{2}

Also ,

area \: of \: \triangle \:  =  \frac{1}{2}  \times base \times height \\  \\\implies 18 \sqrt{21} =  \: \frac{1}{\cancel2}  \times \cancel12  \times height \\  \\ \implies \: 18 \sqrt{21}  = 6 \times height \\  \\ \implies \: height =  \frac{\cancel{18} \sqrt{21} }{ \cancel 6}  \\  \\ \implies \: height = 3 \sqrt{21}  \: cm {}^{2}

<u>Since </u><u>we've </u><u>obtained </u><u>the </u><u>height </u><u>now </u><u>,</u><u> </u><u>we </u><u>can </u><u>easily </u><u>find </u><u>out </u><u>the </u><u>area </u><u>of </u><u>trapezium </u><u>!</u>

Area \: of \: trapezium =  \frac{1}{2}  \times(sum \: of \:parallel \: sides) \times height \\  \\ \implies \:  \frac{1}{2}  \times (25 + 13) \times 3 \sqrt{21}  \\  \\ \implies \:  \frac{1}{\cancel2}  \times \cancel{38 }\times 3 \sqrt{21}  \\  \\ \implies \: 19 \times 3 \sqrt{21}  \: cm {}^{2}  \\  \\ \implies \: 57 \sqrt{21}  \: cm {}^{2}

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