N/2-2n/16=3/8 what is the value of n?
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Answer:
The area of the trapezium is 216 cm²
Step-by-step explanation:
The parameters of the trapezium are;
The point of intersection of the diagonals <em>ac</em> and <em>bd</em> <em> </em>= point <em>o</em>
The area of ΔCOD = 24 cm²
The area of ΔAOB = 96 cm²
Let <em>h</em> represent the height of the trapezium and let <em>x</em> represent the height of triangle ΔCOD, we have;
The height of ΔAOB = h - x
Therefore;
The area of ΔCOD = (1/2) × The length of CD × x = 24
∴ The length of CD = 24/((1/2) × x) = 48/x
The area of ΔAOB = (1/2) × The length of AB × (h - x) = 96
The length of AB = 96/((1/2) × (h - x)) = 192/(h - x)
The area of the trapezium, A = ((48/x + 192/(h - x))/2) × h
We note that ΔCOD ~ ΔAOB, therefore;
(AB)²/(CD)² = (h - x)²/x² = (arΔAOB)/(arΔCOD) = 96/24 = 4
∴ √((h - x)²/x²) = (h - x)/x = √4 = 2
∴ h - x = 2·x
h = 2·x + x = 3·x
A = ((48/x + 192/(h - x))/2) × h
∴ A = ((48/x + 192/(2·x))/2) × 3·x = ((48 + 192/(2))/2) × 3 = 216
The area of the trapezium, A = 216 cm².
