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A Doorway is Decorated as Shown in the Figure. There Are Four Semi-circles. Bc, the Diameter of the Larger Semi-circle is of Length 84 Cm. Centres of the Three Equal Semicircles Lie on Bc. Abc is an Isosceles Triangle with Ab = Ac. If Bo = Oc, Find the Area of the Shaded Region. (Take `Pi = 22/7`) - ICSE Class 10 - Mathematics

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Question

A doorway is decorated as shown in the figure. There are four semi-circles. BC, the diameter of the larger semi-circle is of length 84 cm. Centres of the three equal semicircles lie on BC. ABC is an isosceles triangle with AB = AC. If BO = OC, find the area of the shaded region. (Take `pi = 22/7`)

Solution

Here, the radius of larger semicircle = `84/2 = 42 cm`

And, the radius of smaller semi-circle = `84/(3xx2) = 14 cm`

Area of the shaded region  =` (pi(42)^2)/2 + 3xx(pi(14)^2)/2 - 1/2 xx 84 xx 42`

`= 22/7 [21 xx 42 + 3 xx 14 xx 7] - 42 xx 42`

`= 22[3 xx 42 + 42] - 42 xx 42`

`= 42 xx [88 - 42]`

`= 1932 cm^2`

  Is there an error in this question or solution?

APPEARS IN

 2009-2010 (March) (with solutions)
Question 6.3 | 4.00 marks

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Solution A Doorway is Decorated as Shown in the Figure. There Are Four Semi-circles. Bc, the Diameter of the Larger Semi-circle is of Length 84 Cm. Centres of the Three Equal Semicircles Lie on Bc. Abc is an Isosceles Triangle with Ab = Ac. If Bo = Oc, Find the Area of the Shaded Region. (Take `Pi = 22/7`) Concept: Lines of Symmetry.
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