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HERE IS YOUR ANSWER. .
Let ABCD be the trapezium, such that AD = 11 m, BC = 25 m, AB = 13 m and CD = 15 m.
Draw AM perpendicular from A on to BC. Let BM = x.
Draw DN perpendicular from D on to BC. Let NC = 14-x.
Let AM = DN = h.
In RAT, ABM, 13^2 = h^2+x^2 …(1)
In RAT, DCN, 15^2 = h^2+(14-x)^2 …(2)
Subtract (1) from (2)
15^2–13^2 = 56 = 196–28x, or
140 = 28x, or x = 5
From (1), 13^2 = 169 = h^2+25, or
h^2 = 144, or h = 12 m
Area of ABCD = (11+15)*12/2 = 26*6 = 156 sq m.
HOPE IT HELPS YOU! !!
Let ABCD be the trapezium, such that AD = 11 m, BC = 25 m, AB = 13 m and CD = 15 m.
Draw AM perpendicular from A on to BC. Let BM = x.
Draw DN perpendicular from D on to BC. Let NC = 14-x.
Let AM = DN = h.
In RAT, ABM, 13^2 = h^2+x^2 …(1)
In RAT, DCN, 15^2 = h^2+(14-x)^2 …(2)
Subtract (1) from (2)
15^2–13^2 = 56 = 196–28x, or
140 = 28x, or x = 5
From (1), 13^2 = 169 = h^2+25, or
h^2 = 144, or h = 12 m
Area of ABCD = (11+15)*12/2 = 26*6 = 156 sq m.
HOPE IT HELPS YOU! !!
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