If a rectangle and a parallelogram have same perimeter, show that area of parallelogram is less than area of rectangle.??
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Step-by-step explanation:
Area of ||gm = base × height, And
Area of rect. = length × width,.
Suppose,W of rect. = B of ||gm
And,L of rect. = H of ||gm
So, Area of rectangle = Area of parallelogram
Here, In area, let height of parallelogram be h,
Breadth be a ,And
length be b.
So, sin(α)=ha
Then h=a sin(α) so Area of the parallelogram = (ab)sin(α)
It will form alpha in parallelogram.
Here, alpha in ||gm is less than 90∘, so sin(α) <1.
Now, we will multiply both sides of that inequality(in text. and ||gm) by ab, and we will get:
(ab)sin(α) < ab
…which means: Area of parallelogram < Area of rectangle.
Note please: sin(a),sin(ab),............ = sina, sinab ........
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