Question 2 - In fig, two circular flower beds have been - Examples (2024)

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Question 2(Method 1)In figure, two circular flower beds have been shown on two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.Area of lawn + Flower bed = Area of lawn(square of side 56 m) + Area of flower bed AB & CDArea of lawn = Area of square of side 56 m = (Side)2 = (56)2 = 3136 m2Now we find area of flower bed ABArea of flower bed AB = Area of segment AB= Area of sector OAB – Area of Δ AOBFirst we find sides OA & OB.We know that diagonals of a square bisects at right angle & are equalSo, ∠ AOB = 90° & OA = OBLet OA = OB = xNow, In right angle triangle AOB Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 (AB)2 = (OA)2 + (OB)2 (56)2 = x2 + x2 (56)2 = 2x2 2x2 = (56)2 x2 = (56 × 56)/2 x2 = 56 × 28Area of sector OAB = θ/(360°)×𝜋𝑟2Here θ = 90° ,r = x Area of sector AOB = (90°)/(360°)×22/7 (x)2Putting value of x2 from (1) = 1/4×22/7×56×28 = 22×56 = 1232 m2 Area of Δ AOBDiagonals of a square divides the square into congruent trianglesHence, Δ AOB ≅ Δ BOC ≅ Δ COD ≅ Δ AOD Congruent triangles have the same areaSo, ar(AOB) = ar(BOC) = ar(COD) = ar(AOD)⇒ ar(AOB) = ar(BOC) = ar(COD) = ar(AOD) = 1/4ar(ABCD)⇒ ar(AOB) = 1/4ar(ABCD)Putting ar(ABCD) = 3136 m2⇒ ar(AOB) = 1/4 × 3136⇒ ar(AOB) = 784 m2Area of flower bed AB = Area of sector OAB – Area of Δ AOB= 1232 – 784= 448 m2 Similarly, by symmetry Area of flower bed CD = 448 m2Area of lawn + Flower bed = Area of lawn(square of side 56 m) + Area of flower bed AB & CD = 3136 + (448 + 448) = 4032 m2Hence, Total area of the lawn and flower beds = 4032 m2Question 2(Method 2)In figure, two circular flower beds have been shown on two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.Area of lawn + Flower bed = Area of sector OAB + Area of sector OCD+ Area of Δ AOD+ Area of Δ BOCArea of square ABCD = (Side)2 = (56)2 = 3136 m2Area of ΔAOD & Area ΔBOCDiagonals of a square divides the square into congruent trianglesHence, Δ AOB ≅ Δ BOC ≅ Δ COD ≅ Δ AOD Congruent triangles have the same areaSo, ar(AOB) = ar(BOC) = ar(COD) = ar(AOD)⇒ ar(AOB) = ar(BOC) = ar(COD) = ar(AOD) = 1/4ar(ABCD)⇒ ar(AOD) = 1/4ar(ABCD)Putting ar(ABCD) = 3136 m2⇒ ar(AOD) = 1/4 × 3136⇒ ar(AOD) = 784 m2So, ar(AOD) = ar(BOC) = 784 m2 Now we find area of sector OABFirst we find sides OA & OB.We know that diagonals of a square bisects at right angle and are equalSo, ∠ AOB = 90° & OA = OBLet OA = OB = xNow, In right angle triangle AOB Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 (AB)2 = (OA)2 + (OB)2 (56)2 = x2 + x2 (56)2 = 2x2 2x2 = (56)2 x2 = (56 × 56)/2 x2 = 56 × 28Area of sector OAB = θ/(360°)×𝜋𝑟2Here θ = 90° ,r = x Area of sector AOB = (90°)/(360°)×22/7 (x)2Putting value of x2 from (1) = 1/4×22/7×56×28 = 22×56 = 1232 m2Similarly, as radius(r) and angle(θ) are sameArea of sector COD = Area of sector AOB = 1232 m2 NowArea of lawn + Flower bed = Area of sector OAB + Area of sector OCD+ Area of Δ AOD + Area of Δ BOC= 1232 + 1232 + 784 + 784 = 4032 m2Hence, Total area of the lawn and flower beds = 4032 m2