Areas of Combination of Plane figures
Trending Questions
- True
- False
- 1
- 2
- 3
- 4
- (Area of I + Area of III) + (Area of II + Area of IV)
- (Area of I + Area of IV) + (Area of II + Area of III)
- (Area of I + Area of II) × 2
- (Area of I + Area of II) + (Area of III + Area of IV)
The area of an equilateral ΔABC is 17320.5 cm2. A circle is drawn taking the vertex of the triangle as centre. The radius of the circle is half the length of the side of triangle. Find the area of the shaded region (in cm2) . (π = 3.14 , √3 =1.73205)
How many tiles whose length and breadth are respectively will be needed to fit in a rectangular region whose length and breadth are respectively?
Given below is a semicircle of diameter 12cm which encloses two other semicircles. Find the area of shaded region.
- 492
- 352
- 1194
- False
- True
Consider the following steps. Which would be the correct order to solve the above-mentioned question?
1. Find the area of each unshaded region.
2. Identify the unshaded regions.
3. Subtract the sum total of areas of unshaded regions from the area of the square.
1, 2, 3
2, 1, 3
3, 2, 1
2, 3, 1
- 30∘
- 90∘
- 45∘
- 60∘
- zero
- πr21
- πr22
- πr22-πr12
- Area of the square - Area of quarter circle
- Area of square + Area of quarter circle
- Area of square - (4 × Area of one quarter circle)
- Area of square + (4 × Area of quarter circle)
State true or False.
Areas of the shaded regions in figures 1 and 2 are equal. The corresponding lengths and breadths of the rectangles are equal, and the radius of the quarter circles in figure 1 is half the breadth of the rectangle.
False
True
Compare area of region I and II.
- Area of region I = Area of region II
- Area of region I > Area of region II
- Area of region I < Area of region II
- Area of region I = 2×Area of region II