Question

# Two Rectangles ABCD and DBEF are as shown in the figure. The area of Rectangle DBEF is

- 10 cm
^{2}
- 12 cm
^{2}
- 14 cm
^{2}
- 15 cm
^{2}

^{2}^{2}^{2}^{2}Hint:

### The internal angles of a rectangle, which has four sides, are all exactly 90 degrees. A triangle in geometry is a three-sided polygon with three edges and three vertices. Here we have given two rectangles and we have to find ABCD and DBEF and we have to find the area of Rectangle DBEF.

## The correct answer is: 12 cm^{2}

### A triangle is a form of polygon with three sides; the intersection of the two longest sides is known as the triangle's vertex. There is an angle created between two sides. One of the crucial elements of geometry is this.

A quadrilateral with parallel sides that are equal to one another and four equal vertices is known as a rectangle. It is also known as an equiangular quadrilateral for this reason.

We have given two Rectangles ABCD and DBEF are as shown in the figure.

Now lets first find DB, we can use Pythagoras theorem to find DB, we get:

AB^{2}+AD^{2}=DB^{2}

4^{2}+3^{2}=DB^{2}^{}

DB^{2}=16+9

DB^{2}=25

Taking square root on both sides, we get:

DB=5 cm.

Area of ABCD = 2 x area of BCD (ad diagonal divdes area equally)

area of BCD=3x4/2

area of BCD=6 cm^{2}

Now we have triangle BCD, we know that if a triangle and rectangle lies on same base and is in between same parallel lines, then the area of rectangle is twice of area of triangle, so:

area of triangle BCD x 2 = area of rectangle DBEF

6 x 2 = area of rectangle DBEF

area of rectangle DBEF= 12 cm^{2}

So here we used the concept of rectangle and triangle, and we understood the relation between them to solve this question. The total of the triangles' individual areas makes up the rectangle's surface area.So the area of rectangle DBEF is 12 cm^{2}.

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