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Answer 1 · 2018-02-03T14:18:48

The area of the semi-circle will be = $77 s q . c m .$ $77 sq.cm.$

Given: If a wire is bent into the shape of a square, then the area of the square is 81 sq.cm.

That means each side of the square = $\sqrt{81} = 9 c m$ $sqrt 81 = 9 cm$ ---as, area of square = $s i d e^{2}$ $side^2$

The length $l$ $l$ , of the wire will be equal to the perimeter of this square:

$l = 4 \times 9 = 36 c m$ $l = 4 xx 9 = 36 cm$

Now, when the wire is bent into a semi-circular shape, the perimeter will remain same = $36 c m$ $36cm$

And perimeter of semicircle will be given as:

$(\frac{1}{2} \times 2 \times π \times r) + (2 \times r)$ $(1/2 xx 2xxpi xx r) + (2xxr)$ ------ half the circumference of complete circle + diameter.

SO,
$(\frac{1}{2} \times 2 \times π \times r) + (2 \times r) = 36$ $(1/2 xx 2xxpi xx r) + (2xxr) = 36$

$=> (1/cancel2 xx cancel2xxpi xx r) + (2xxr) = 36$

$=> pi r +2r = 36$ -------assume value of $pi =22/7$

$=> 22/7 xx r +2r =36$

$=> 22/7 xx r +2rxx 7/7 =36$

$=> (22r +14r)/7 =36$

$=> 36 r = 36 xx 7$

$=> r = 7 cm$

Area of this semicircle 'a' will be :

$a =1/2 pi r^2$

$=>a = 1/2 xx 22/7 xx 7^2$

$=> a = 1/cancel2^1 xx cancel22^11/cancel7^1 xx cancel49^7 = 11xx7$

$therefore a = 77 sq.cm.$ will be the area of the semi-circle.