Maddymcg408idn
Answered

Get the answers you've been searching for with IDNLearn.com. Get accurate and timely answers to your queries from our extensive network of experienced professionals.

13. PLEASE HELP ME
The area of a playground is 336 yd2. The width of the playground is 5 yd longer than its length. Find the length and width of the playground.

A. length = 16 yd, width = 21 yd

B. length = 21 yd, width = 26 yd

C. length = 21 yd, width = 16 yd

D. length = 26 yd, width = 21 yd

Sagot :

  • Let length be x
  • width=x+5

We know

[tex]\boxed{\sf Area=Length\times Width}[/tex]

  • Putting values

[tex]\\ \sf\longmapsto 336=x(x+5)[/tex]

[tex]\\ \sf\longmapsto x^2+5x=336[/tex]

[tex]\\ \sf\longmapsto x^2+5x-336=0[/tex]

  • Using mid term split

[tex]\\ \sf\longmapsto x^2-16x+21x-336=0[/tex]

[tex]\\ \sf\longmapsto x(x-16)+21(x-16)=0[/tex]

[tex]\\ \sf\longmapsto (x-16)(x+21)=0[/tex]

[tex]\\ \sf\longmapsto (x-16)=0\:or\:(x+21)=0[/tex]

[tex]\\ \sf\longmapsto x=16\:or\:x=-21[/tex]

  • Ignore negative value

[tex]\\ \sf\longmapsto Length=x=16yd[/tex]

[tex]\\ \sf\longmapsto Width=x+5=16+5=21yd[/tex]

Option a is correct

Briannajohnson6897idn
1. Find out which one has the width 5 yd longer
2. It is A and B
3. Multiply both
16x21= 336. 21x26=546
4. See which one equals 336
5. It is option A
6. Your answer is option A
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.