Radius of circle = 32 Centroid O divides median AD into 2:1, therefore A0:0D = 2:1 = AO/OD = 2/1 ⇒ 32/OD = 2/1 ⇒ OD = 16 Therefore, AD = 32 + 16 = 48 cm In AABD, AB² = AD² + BD² = (48)² + (AB/2)² ⇒ 3/4 AB² = (48)² ⇒ AB = (48 × 2)/√3 = (96√3)/3 = 32√3 Area of equilateral triangle ABC = √3/4 (32√3)² =Read more
Radius of circle = 32
Centroid O divides median AD into 2:1, therefore A0:0D = 2:1
= AO/OD = 2/1 ⇒ 32/OD = 2/1 ⇒ OD = 16
Therefore, AD = 32 + 16 = 48 cm
In AABD,
AB² = AD² + BD² = (48)² + (AB/2)²
⇒ 3/4 AB² = (48)² ⇒ AB = (48 × 2)/√3 = (96√3)/3 = 32√3
Area of equilateral triangle ABC
= √3/4 (32√3)² = 768√3 cm²
Area of circle
= πr² = π(32)² = 22/7 × 32 × 32 = 22528/7 cm²
Area of design = Area of circle – Area of equilateral triangle ABC
= (22528/7 – 768√3) cm²
The circles drawn taking A, B, C and D form quadrants of radius 7 cm in square. Radius of each quadrant = 7 cm Area of each quadrant = 90°/360° × πr² = 1/4 × π(7)² = 1/4 × 22/7 × 7 × 7 = 77/2 cm² Area of square = (Side)² = (14)² = 196 cm² Area of shaded region = Area of square - Area of 4 quadrantsRead more
The circles drawn taking A, B, C and D form quadrants of radius 7 cm in square.
Radius of each quadrant = 7 cm
Area of each quadrant
= 90°/360° × πr² = 1/4 × π(7)²
= 1/4 × 22/7 × 7 × 7 = 77/2 cm²
Area of square = (Side)² = (14)² = 196 cm²
Area of shaded region = Area of square – Area of 4 quadrants
= 196 – 4 × 77/2 = 196- 154 = 42 cm²
In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Figure. Find the area of the design.
Radius of circle = 32 Centroid O divides median AD into 2:1, therefore A0:0D = 2:1 = AO/OD = 2/1 ⇒ 32/OD = 2/1 ⇒ OD = 16 Therefore, AD = 32 + 16 = 48 cm In AABD, AB² = AD² + BD² = (48)² + (AB/2)² ⇒ 3/4 AB² = (48)² ⇒ AB = (48 × 2)/√3 = (96√3)/3 = 32√3 Area of equilateral triangle ABC = √3/4 (32√3)² =Read more
Radius of circle = 32
See lessCentroid O divides median AD into 2:1, therefore A0:0D = 2:1
= AO/OD = 2/1 ⇒ 32/OD = 2/1 ⇒ OD = 16
Therefore, AD = 32 + 16 = 48 cm
In AABD,
AB² = AD² + BD² = (48)² + (AB/2)²
⇒ 3/4 AB² = (48)² ⇒ AB = (48 × 2)/√3 = (96√3)/3 = 32√3
Area of equilateral triangle ABC
= √3/4 (32√3)² = 768√3 cm²
Area of circle
= πr² = π(32)² = 22/7 × 32 × 32 = 22528/7 cm²
Area of design = Area of circle – Area of equilateral triangle ABC
= (22528/7 – 768√3) cm²
In this Figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
The circles drawn taking A, B, C and D form quadrants of radius 7 cm in square. Radius of each quadrant = 7 cm Area of each quadrant = 90°/360° × πr² = 1/4 × π(7)² = 1/4 × 22/7 × 7 × 7 = 77/2 cm² Area of square = (Side)² = (14)² = 196 cm² Area of shaded region = Area of square - Area of 4 quadrantsRead more
The circles drawn taking A, B, C and D form quadrants of radius 7 cm in square.
Radius of each quadrant = 7 cm
Area of each quadrant
= 90°/360° × πr² = 1/4 × π(7)²
= 1/4 × 22/7 × 7 × 7 = 77/2 cm²
Area of square = (Side)² = (14)² = 196 cm²
Area of shaded region = Area of square – Area of 4 quadrants
= 196 – 4 × 77/2 = 196- 154 = 42 cm²
Here is the Video explanation 😀👇
See lessFigure depicts a racing track whose left and right ends are semicircular.
(i)The distance around the track along its inner edge = AB + arcBEC + CD + arcDFA = 106 + 1/2 × 2πr + 106 + 1/2 × 2πr = 106 + 1/2 × 2 × 22/7 × 30 + 106 + 1/2 × 2 × 22/7 × 30 = 212 + 2 × 22/7 × 30 = 212 + 1320/7 = 2804/7 (ii) Area of Track = (Area of GHIJ - Area of ABCD) + (Area of semicircle HKI - ARead more
(i)The distance around the track along its inner edge
See less= AB + arcBEC + CD + arcDFA
= 106 + 1/2 × 2πr + 106 + 1/2 × 2πr
= 106 + 1/2 × 2 × 22/7 × 30 + 106 + 1/2 × 2 × 22/7 × 30
= 212 + 2 × 22/7 × 30 = 212 + 1320/7 = 2804/7
(ii) Area of Track = (Area of GHIJ – Area of ABCD) + (Area of semicircle HKI – Area of semicircle BRC) + (Area of semicircle GLJ – Area of semicircle AFD)
= (106 × 80 – 106 × 60) + 1/2 × 22/7 × [(40)² – (30)²] + 1/2 × 22/7 × [(40)² – (30)²]
= 106(80 – 60) + 1/2 × 22/7 × (700) + 1/2 × 22/7 × (700) = 2120 + 22/7 × (700) = 2120 + 2200 = 4320