Answer :
Option B
Step by step explanation :
As we can see in figure, Triangle is right angled at A
In ∆ABC
By Pythagoras theorem,
BC² = AC² + AB²
=> BC² = (3)² + (4)²
=> BC² = 9 + 16
=> BC² = 25
=> BC = 5 cm
We know,
[tex]\:\mathsf{Area \:of\: triangle}[/tex]=[tex]\:\mathsf{\frac{1}{2}\: x\: base\: x\: height}[/tex]
Here,
base = AC = 3cm
height = AB = 4cm
=> [tex]\:\mathsf{Area \:of\: triangle}[/tex]=[tex]\:\mathsf{\frac{1}{2} \:x\: 3cm\: x\:4cm}[/tex]
=> [tex]\:\mathsf{Area \:of\: triangle\: = \:3cm\: x\: 2cm}[/tex]
=> [tex]\:\bf\boxed{Area \:of\: triangle\: = \:6cm²}[/tex]
As BC = DE
And then BD = EC
it shows the property of rectangle
So BD = 6 cm
Now,
[tex]\:\mathsf{Area \:of\: rectangle\: =\: length \:x\: width}[/tex]
=> Area of rectangle = 6cm x 5cm
=> [tex]\:\bf\boxed{Area\: of\: rectangle\: = \:30cm²}[/tex]
Now,
Area of the figure = Area of rectangle + Area of triangle
=> Area of the figure = 30cm² + 6cm²
=> [tex]\:\bf\boxed{Area\: of\: the\: figure\: = \:36cm²}[/tex]
Therefore, option B is the correct answer for this question.
