Listen NCERT Audio Books to boost your productivity and retention power by 2X.
The radius of a circle with centre O is 7 cm. Two radii OA and OB are drawn at right angles to each other. Find the areas of minor and major segments.
Consider the above figure,
From here we can conclude that the portion or the segment below the chord AB is the minor segment and the segment above AB is major segment.
Also we know,
Area of minor segment = Area of sector – Area of ∆AOB → eqn1
Now, Area of sector
Where R = radius of the circle and θ = central angle of the sector
Given, R = 7 cm and θ = 90°
Putting these values in the equation 2, we get
⇒ Area of sector = 38.5 cm2→ eqn3
Area of △AOB = 1/2 × base × height
As triangle is isosceles therefore height and base both are 7 cm.
⟹ Area of △AOB = 1/2×7×7
= 24.5 cm2→ eqn4
Putting values of equation 2 and 4 in equation 1 we get
Area of minor segment = 38.5 – 24.5
⇒ Area of minor segment = 14 cm2
Area of major segment = πR2 – Area of minor segment → eqn5
Put the value of R, and Area of minor segment in equation 5
= π(72) – 14
= 49π - 14
= (22×7) - 14
= 154 - 14
= 140 cm2
Area of minor segment is 14 cm2 and of major segment is 140 cm2.
ABCD is a field in the shape of a trapezium, AD || BC, ∠ABC = 90° and ∠ADC = 60°. Four sectors are formed with centres A, B, C and D, as shown in the figure. The radius of each sector is 14 m.
Find the following:
(i) total area of the four sectors,
(ii) area of the remaining portion, given that AD = 55 m, BC = 45 m and AB = 30 m.
A child draws the figure of an aeroplane as shown. Here, the wings ABCD and FGHI are parallelograms, the tail DEF is an isosceles triangle, the cockpit CKI is a semicircle and CDFI is a square. In the given figure, BP ⊥ CD, HQ ⊥ FI and EL ⊥ DF. If CD = 8 cm, BP = HQ = 4 cm and DE = EF = 5 cm, find the area of the whole figure. [Take π = 3.14.]