At each corner of a triangular field of sides 26 m, 28 m and 30 m, a cow is tethered by a rope of length 7 m. What is the area (in m 2 ) unglazed by the cows ?
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In order to find the area of ungrazed portion, we need to find the areas of the ∆ular plot and the circular portion of land grazed by the cow (i.e. Area of ∆ABC-Area of Circular Plot). [AREA OF ∆ABC]--- Let sides 'a', 'b', 'c' be of measure 26m, 28m, 30m respectively. Now, Semiperimeter (s) = (a+b+c)÷2 = (26+28+30)÷2 = 84÷2 = 42m. Now, using Heron's Formula, we shall determine the area of ∆ABC, i.e. √s(s-a)(s-b)(s-c) = √42(42-26)(42-28)(42-30) = √42×16×14×12 = 7×3×2⁴ = 336 m². [AREA OF CIRCULAR PLOT]----- r=measure of the cow's rope = 7m. So, area of circular plot = πr² = 22/7×7² = 154m². [AREA OF UNGRAZED PLOT] = (area of ∆ular plot)-(area of circular plot) = (336-154)m² =182m².
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Answer:
259 m^2
Step-by-step explanation:
Semi perimeter
s= (26+28+30) / 2 = 42
Then
Area of field
√42(42-26)(42-28)(42-30) = 336 sqm
total sum of angles of ∆ = 180°
area of circle (360°) grazed by 1 cow = πr^2 = 22/7 (7x7) = 154 m^2
grazed area of traingle i.e. 180° = 154/2 = 77 m^2
ungrazed area = 336-77 = 259 m^2
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