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40 votes
Find the area and perimeter of a TV set with dimensions 88 cm x 49 cm (to nearest cm). (Hint: find the upper and lower bounds to find the area and perimeter)​

User Hlt
by
3.0k points

2 Answers

17 votes
17 votes

Here

  • Length=L=88cm
  • Breadth=49cm=B

Note the formulas


\boxed{\bf Perimeter=2(L+B)}


\boxed{\bf Area=LB}

Now


\\ \tt\hookrightarrow Perimeter=2(88+49)=2(137)=274cm


\\ \tt\hookrightarrow Area=88(49)=4312cm^2

User Robertp
by
3.1k points
21 votes
21 votes

Answer:

Given below.

Explanation:

For upper bound, the length is 88.5 cm and width is 49.5 cm

So, area of rectangle, TV = length * width

= 88.5 * 49.5

= 4380.75 cm²

So, perimeter = 2( length + width )

= 2( 88.5 + 49.5 )

= 276 cm

For lower bound, the length is 87.5 cm and width is 48.5 cm

So, area of rectangle, TV = length * width

= 87.5 * 48.5

= 4243.75 cm²

So, perimeter = 2( length + width )

= 2( 87.5 + 48.5 )

= 272 cm

User Batool
by
3.1k points