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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

#### 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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