Question

Hi! Answer this correctly using Herons formula .​

Answer 1

The area of the farm is 2170m².

[Refer to the attachment for naming]

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Considering ΔABC,

∠ABC = 90°, therefore using the Pythagoras Theorem we get —

⇒ Hypotenuse² = Altitude² + Base²

⇒ (AC)² = (BC)² + (AB)²

⇒ AC² = (7)² + (24)²

⇒ AC² = 49 + 576

⇒ AC² = 625

On taking square root on both sides we get,

⇒ √(AC²) = √(625)

⇒ AC = √(625)

⇒ AC = 25 meters

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Therefore, the length of the side AC is 25 meters.

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Now that we have the value of all three sides of ABC, we can use the 'Heron's formula' to find out its area.

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$\boxed{\sf Heron's \ Formula = \sqrt{s(s - a)(s - b)(s - c)}}$

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Here,

• a = AB = 24 meters

• b = BC = 7 meters

• c = AC = 25 meters

• s = semi-perimeter = (a + b + c)/2 = (24 + 7 + 25)/2 = 56/2 = 28 meters.

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Therefore,

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$\sf \Longrightarrow \Delta ABC = \sqrt{s(s - a)(s - b)(s - c)}$

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$\sf \Longrightarrow \Delta ABC = \sqrt{28(28 - 24)(28 - 7)(28 - 25)}$

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$\sf \Longrightarrow \Delta ABC = \sqrt{28(4)(21)(3)}$

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$\sf \Longrightarrow \Delta ABC = \sqrt{(2 \times 2 \times 7) \times (2 \times 2) \times (7 \times 3 \times 3)}$

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$\sf \Longrightarrow \Delta ABC = \sqrt{2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 3 \times 3}$

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$\sf \Longrightarrow \Delta ABC = \sqrt{2^2 \times 2^2 \times 7^2 \times 3^2}$

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$\sf \Longrightarrow \Delta ABC = 2 \times 2 \times 7 \times 3$

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$\sf \Longrightarrow \Delta \textsf{\textbf{ABC = 84m}}^2$

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Therefore, the area of triangle ABC is 84m².

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[Join E to C] Considering ΔEAC,

∠EAC = 90°, therefore using the Pythagoras Theorem we get —

⇒ Hypotenuse² = Altitude² + Base²

⇒ (EC)² = (EA)² + (AC)²

⇒ EC² = (60)² + (25)²

⇒ EC² = 3600 + 625

⇒ EC² = 4225

On taking square root on both sides we get,

⇒ √(EC²) = √(4225)

⇒ EC = √(4225)

⇒ EC = 65 meters

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Since we have the value of EC, we'll be able to find the area of EAC using the 'Herons formula'.

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$\sf \Longrightarrow \Delta EAC = \sqrt{s(s - a)(s - b)(s - c)}$

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Here,

• a = EA = 60 meters

• b = AC = 25 meters

• c = EC = 65 meters

• s = semi-perimeter = (a + b + c)/2 = (60 + 25 + 65)/2 = 150/2 = 75 meters.

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$\sf \Longrightarrow \Delta EAC = \sqrt{75(75 - 60)(75 - 25)(75 - 65)}$

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$\sf \Longrightarrow \Delta EAC = \sqrt{75(15)(50)(10)}$

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$\sf \Longrightarrow \Delta EAC = \sqrt{(3 \times 5 \times 5)(5 \times 3)(5 \times 5 \times 2)(5 \times 2)}$

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$\sf \Longrightarrow \Delta EAC = \sqrt{3 \times 3 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 2 \times 2}$

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$\sf \Longrightarrow \Delta EAC = \sqrt{3^2 \times 5^2 \times 5^2 \times 5^2 \times 2^2}$

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$\sf \Longrightarrow \Delta EAC = 3 \times 5 \times 5 \times 5 \times 2$

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$\sf \Longrightarrow \Delta \textsf{\textbf{EAC = 750m}}^2$

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Therefore, the area of triangle EAC is 750m².

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Now, we'll try to find the area of ΔEDC —

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$\sf \Longrightarrow \Delta EDC = \sqrt{s(s - a)(s - b)(s - c)}$

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Here,

• a = ED = 55 meters

• b = DC = 50 meters

• c = EC = 65 meters

• s = semi-perimeter = (a + b + c)/2 = (55 + 50 + 65)/2 = 170/2 = 85 meters.

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$\sf \Longrightarrow \Delta EDC = \sqrt{85(85 - 55)(85 - 50)(85 - 65)}$

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$\sf \Longrightarrow \Delta EDC = \sqrt{85(30)(35)(20)}$

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$\sf \Longrightarrow \Delta EDC = \sqrt{(5 \times 17)(2 \times 3 \times 5)(5 \times 7)(2 \times 2 \times 5)}$

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$\sf \Longrightarrow \Delta EDC = \sqrt{(5 \times 5 \times 5 \times 5)(2 \times 2 \times 2)(3 \times 7 \times 17)}$

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$\sf \Longrightarrow \Delta EDC = (5 \times 5 \times 2)\sqrt{2 \times 3 \times 7 \times 17}$

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$\sf \Longrightarrow \Delta EDC = 50\sqrt{714}$

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$\sf \Longrightarrow \Delta EDC = 50 \times 26.72$ (approximate)

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$\sf \Longrightarrow \Delta \textsf{\textbf{EDC = 1336m}}^2$ (approximate)

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Therefore, the area of triangle EDC is 1336m².

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Since we have the area of all three segments of the farm, we can find out its area.

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⇒ Area of the farm = ΔABC + ΔEAC + ΔEDC

⇒ Area of the farm = 84 + 750 + 1336

⇒ Area of the farm = 2170m²

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Therefore, the area of the farm is 2170m².

Answer 2

Step-by-step explanation:

The area of the farm is 2170m².

[Refer to the attachment for naming]

‎

Considering ΔABC,

∠ABC = 90°, therefore using the Pythagoras Theorem we get —

⇒ Hypotenuse² = Altitude² + Base²

⇒ (AC)² = (BC)² + (AB)²

⇒ AC² = (7)² + (24)²

⇒ AC² = 49 + 576

⇒ AC² = 625

On taking square root on both sides we get,

⇒ √(AC²) = √(625)

⇒ AC = √(625)

⇒ AC = 25 meters

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Therefore, the length of the side AC is 25 meters.

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Now that we have the value of all three sides of ABC, we can use the 'Heron's formula' to find out its area.

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Heron 's Formula= ✓s(s−a)(s−b)(s−c)

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Here,

• a = AB = 24 meters
• b = BC = 7 meters
• c = AC = 25 meters
• s = semi-perimeter = (a + b + c)/2 = (24 + 7 + 25)/2 = 56/2 = 28 meters.

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Therefore,

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⟹ΔABC= ✓s(s−a)(s−b)(s−c)

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⟹ΔABC= ✓28(28−24)(28−7)(28−25)

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⟹ΔABC= ✓28(4)(21)(3)

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⟹ΔABC= ✓(2×2×7)×(2×2)×(7×3×3)

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⟹ΔABC= 2×2×2×2×7×7×3×3

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⟹ΔABC= ✓2 ²×2 ² ×7 ² ×3 ²

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⟹ΔABC=2×2×7×3

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⟹ΔABC = 84m ²

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Therefore, the area of triangle ABC is 84m².

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[Join E to C] Considering ΔEAC,

∠EAC = 90°, therefore using the Pythagoras Theorem we get —

⇒ Hypotenuse² = Altitude² + Base²

⇒ (EC)² = (EA)² + (AC)²

⇒ EC² = (60)² + (25)²

⇒ EC² = 3600 + 625

⇒ EC² = 4225

On taking square root on both sides we get,

⇒ √(EC²) = √(4225)

⇒ EC = √(4225)

⇒ EC = 65 meters

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Since we have the value of EC, we'll be able to find the area of EAC using the 'Herons formula'.

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⟹ΔEAC= ✓s(s−a)(s−b)(s−c)

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Here,

• a = EA = 60 meters
• b = AC = 25 meters
• c = EC = 65 meters
• s = semi-perimeter = (a + b + c)/2 = (60 + 25 + 65)/2 = 150/2 = 75 meters.

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⟹ΔEAC= ✓75(75−60)(75−25)(75−65)

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⟹ΔEAC= ✓75(15)(50)(10)

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⟹ΔEAC= ✓(3×5×5)(5×3)(5×5×2)(5×2)

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⟹ΔEAC= ✓3×3×5×5×5×5×5×5×2×2

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⟹ΔEAC= ✓3 ² ×5 ² ×5 ² ×5 ² ×2²

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⟹ΔEAC=3×5×5×5×2

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⟹ΔEAC = 750m²

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Therefore, the area of triangle EAC is 750m².

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Now, we'll try to find the area of ΔEDC —

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⟹ΔEDC= ✓s(s−a)(s−b)(s−c)

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Here,

a = ED = 55 meters

b = DC = 50 meters

c = EC = 65 meters

s = semi-perimeter = (a + b + c)/2 = (55 + 50 + 65)/2 = 170/2 = 85 meters.

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⟹ΔEDC= ✓85(85−55)(85−50)(85−65)

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⟹ΔEDC= ✓85(30)(35)(20)

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⟹ΔEDC= √(5×17)(2×3×5)(5×7)(2×2×5)

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⟹ΔEDC= ✓(5×5×5×5)(2×2×2)(3×7×17)

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⟹ΔEDC=(5×5×2) ✓2×3×7×17

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⟹ΔEDC=50 √714

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⟹ΔEDC=50×26.72 (approximate)

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⟹ΔEDC = 1336m²

(approximate)

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Therefore, the area of triangle EDC is 1336m².

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Since we have the area of all three segments of the farm, we can find out its area.

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⇒ Area of the farm = ΔABC + ΔEAC + ΔEDC

⇒ Area of the farm = 84 + 750 + 1336

⇒ Area of the farm = 2170m²

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Therefore, the area of the farm is 2170m².