The Total solution for NCERT class 6-12
On a squarehandkerchief, nine circular designs each of radius 7 cm are made (see Fig.12.29). Find the area of the remaining portion of the handkerchief.
Answer - 11 : -
In Fig. 12.30, OACBis a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm,
Answer - 12 : -
find the area of the
(i) quadrant OACB, (ii) shaded region.
In Fig. 12.31, asquare OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area ofthe shaded region. (Use π = 3.14)
Answer - 13 : -
Solution:
Side of square = OA = AB = 20 cm
Radius of the quadrant = OB
OAB is right angled triangle
By Pythagoras theorem in ΔOAB,
OB2 = AB2+OA2
⇒ OB2 = 202 +202
⇒ OB2 = 400+400
⇒ OB2 = 800
⇒ OB= 20√2 cm
Area of the quadrant = (πR2)/4 cm2 =(3.14/4)×(20√2)2 cm2 = 628cm2
Area of the square = 20×20 = 400 cm2
Area of the shaded region = Area of thequadrant – Area of the square
= 628-400 cm2 = 228cm2
Answer - 14 : -
Radius of the larger circle, R = 21 cm
Radius of the smaller circle, r = 7 cm
Angle made by sectors of both concentriccircles = 30°
Area of the larger sector = (30°/360°)×πR2 cm2
= (1/12)×(22/7)×212 cm2
= 231/2cm2
Area of the smaller circle = (30°/360°)×πr2 cm2
= 1/12×22/7×72 cm2
=77/6 cm2
Area of the shaded region = (231/2) – (77/6)cm2
= 616/6 cm2 = 308/3cm2
In Fig. 12.33, ABCis a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC asdiameter. Find the area of the shaded region.
Answer - 15 : -
Calculate the areaof the designed region in Fig. 12.34 common between the two quadrants ofcircles of radius 8 cm each.
Answer - 16 : -