find the area of a rombus whose side is 25 m and diagonal is 48 m??
Answers
Answer:
Property of Rhombus is that diagonals bisect at right angles.
Hence there is formed a right angle triangle. Side of rhombus becomes Hypotenuse. Hence
Side^2 = (25/2)^2 + (42/2)^2
side^2 = 156.25 + 441
side = 24.4 (approx)
Perimeter = 4 x 24.4 = 97.6 cm
Area of rhombus = 0.5 x (product of diagonals) = 525 cm^2.
Answer:
336 m²
Step-by-step explanation:
USE THE IMAGE BELOW FOR REFERENCE
AD = DC = BC = AB (Sides of rhombus are equal.)
We know that diagonals of a rhombus bisect each other.
Thus, AO = OC and DO = OB.
Let AC = 48 m.
As AO = OC and AO + OC = AC, AO = AC = AC/2
=> 24 m = AO = OC
Now, in △ AOD,
∠ O = 90°
∴ By using Pythagoras' theorem,
AO² + OD² = AD²
=> 24² + OD² = 25²
=> 576 + OD² = 625
=> OD² = 625 - 576 => 49
∴ OD = √49 => 7 m
As OA = OC (shown above) and OA + OC = AC,
OA = 7 m = OC and
OA + OC = AC
=> 7 + 7 = AC
∴ AC = 14 m
Now, we know that area of a rhombus is (D1*D2)/2 where D1 and D2 are its two diagonals.
=> (48*14)/2
=> 48*7 => 336 m²
Thus, the area of a rhombus whose side is 25 m and diagonal is 48 m is 336 m².
