Find the area of the shaded and unshaded part of this figure.
![Find the area of the shaded and unshaded part of this figure class=](https://us-static.z-dn.net/files/d6e/33b6073567f17e214fe5679318e27a56.jpg)
Answer:
area of trapezium=½{a+b}h
½(10+12)6
22*3
66cm²
area of rectangle= l*b
6*5
30cm²
area of unshaded portion= area of trapezium - area of rectangle
(66-30)cm²
36cm²
Area of shaded part is 30cm² and area of unshaded part is 36cm² for the given figure
The given figure is a rectangle inside a trapezium.
The rectangle is the shaded part while the trapezium is the sum of the shaded and the unshaded part of the figure.
We know that Area of rectangle is given by
A = L x B
where L=Length of rectangle
B=Breadth of rectangle
Given rectangle has L=6cm and B=5cm
Area of rectangle = L x B
= 6 x 5
= 30cm²
which also gives the area of shaded part of the given figure.
Now, area of trapezium
=Height(h)*{(Length of smaller side(a) + length of longer side(b)) ÷ 2}
=h*{(a+b) ÷ 2}
=6*{(10+12) ÷ 2)
= 66cm²
which gives the area of trapezium.
Area of unshaded part = Area of trapezium - Area of rectangle
= 66 - 30
= 36cm²
Hence, area of shaded part is 30cm² and area of unshaded part is 36cm² for the given figure.
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