If the area of a square is $90$ cm $^{2}$ , what is its side-length, rounded to the nearest integer?

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Solution

Since $A$ $= l^{2}$ , we have $l^{2} = 90$ . But $81 < 90 < 100$ and $81$ is nearer to $90$ than $100$ . Hence the nearest integer to $\sqrt{90}$ is $\sqrt{81}$ = 9.

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If the area of a square is 90 cm2, what is its side-length, rounded to the nearest integer?

Since A =l2, we have l2=90. But 81<90<100 and 81 is nearer to 90 than 100. Hence the nearest integer to √90 is √81 = 9.

If the area of a square is $90$ cm $^{2}$ , what is its side-length, rounded to the nearest integer?

Since $A$ $= l^{2}$ , we have $l^{2} = 90$ . But $81 < 90 < 100$ and $81$ is nearer to $90$ than $100$ . Hence the nearest integer to $\sqrt{90}$ is $\sqrt{81}$ = 9.