4. In the given fig. , ABC is a Δrt. ∠d at B, with AB = 14 cm & BC = 24 cm. With the vertices A, B and C as centres, arcs are drawn each of radius 7 cm.
Find the area of the shaded region. Use   π = 22 7



5. In fig., ABC is a right angled triangle, right angled at A. Semicircles are drawn on AB, AC & BC as diameters. Find the area of the shaded region.
(Ans. 6 sq. units)

Dear student
4The area of shaded region=Area of ABC-area of 3 sectors.Now area of ABC  right angled at B=12×base×height=12×BC×AB=12×24×14   Given: AB=14 cm and BC=24 cm=168 cm2Consider ABC,by angle sum property we haveA+B+C=180° ...1Now we know that area of sector=θ360°×πr2 , where r is the radius of the circleArea of 3 sectors=A360°×πr2 +B360°×πr2 +C360°×πr2 =πr2360°A+B+C=1360°×72×227  A+B+C       Given r=7 cm=180°360°×7×22   using 1=77 cm3So, area of shaded region=Area of ABC-area of 3 sectors=168 cm2-77 cm2=91 cm2
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wait..
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14) ar triangle -ar of 3 sector 
1/2*14*24 =168
now in tri abc 
ang A+ang B+angC=180
angA+angC=180-90
angA+angc=90
ar of 3 sectors =1/4*22/7*72*2=77
ar of shaded region =168-77=91
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