In the above figure, squareXLMT is a rectangle. LM = 21 cm, XL = 10.5 cm. Diameter of the smaller semicircle is half the diameter of the larger semicircle. - Geometry Mathematics 2

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Sum


In the above figure, `square`XLMT is a rectangle. LM = 21 cm, XL = 10.5 cm. Diameter of the smaller semicircle is half the diameter of the larger semicircle. Find the area of non-shaded region.

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Solution


Given: `square`XLMT is a rectangle.

LM = XT and XL = TM   .....(Opposite sides are equal)

∴ XT = 21 cm

TM = 10.5 cm

To find the area of a non-shaded region 

The diameter of a smaller semicircle is half the diameter of a larger semi-circle.

Let the diameter of the larger semicircle is D and the diameter of the smaller semicircle be `1/2 xx D = D/2`.

Now, the diameter of the smaller semicircle + the diameter of the larger semicircle = 21 cm

`D + D/2 = 21`

`(2D + D)/2 = 21`

`(3D)/2 = 21`

`3D = 21 xx 2`

`D = (21 xx 2)/3`

D = 14 

A(smaller semi-circle) = `1/2 pir^2`

= `1/2 xx 22/7 xx 7/2 xx 7/2`

= `77/4 "cm"^2`

A(largerer semi-circle) = `1/2pir^2`

`= 1/2 xx 22/7 xx 14/2 xx 14/2`

= `77  "cm"^2`

A(`square`XLMT) = length × breadth

= LM × XL

= 21 cm × 10.5 cm

= 220.5 cm2

Area of the non-shaded region

= A(`square`XLMT) – A(smaller semi-circle) – A(larger semi-circle)

= 220.5 – `77/4 - 77`

= 220.5 – 19.25 – 77

= 124.25 cm2

  Is there an error in this question or solution?
2018-2019 (March) Set 1

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