Maddymcg408id
Answered

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

Briannajohnson6897id
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

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