Question

Total area of the given closed figure will be ................. square units.

A .31
B .45
C .25
D .40

Answer 1

Hint: In order to solve this problem, we must remember the formulae of area of the square and area of the right angle triangle along with the proper understanding of the figure that is being provided in the question.

Complete step-by-step answer:
 

In the question it is given two square ABCD and DEFG of side 4cm and 3cm respectively and one triangle CDG. From the figure it is clearly seen that triangle has two sides common one with square of side 4 cm and other one with the square of side 3 cm, hence two side CD and DG of triangle CDG are known, only one side CG is left to find which can be find by Pythagoras theorem.

We know that
Area of a square is defined as the number of square units needed to completely fill a square. Area of a square can be given by the formulae:
Area of a Square = Side × Side
Therefore, the area of square = ${Side^2}$ sq. units

Area of square ABCD with side 4 units = ${4^2}$= 16 sq. units
Similarly

Area of square DEFG with side 3 units = ${3^2}$= 9 sq. units

 A right-angled triangle is the one which has 3 sides, base(CD), hypotenuse(DG) and height(CG) with the angle between base and height being 90°. So in the triangle of given figure base =4 cm and height = 3 cm.

The area of a triangle can be calculated by formulae:
Area of the triangle CDG = $\dfrac{{{\text{Base}} \times {\text{Height}}}}{2}$
On putting the values of base and height
Area of the triangle = $\dfrac{{4 \times 3}}{2}$= 6 sq. units
∴ Area of the given closed figure = Area of the square ABCD + Area of the square DEFG + Area of the triangle CDG
∴ Area of the given closed figure =16+9+6=31 sq. units
Hence Option A is correct.

Note: Whenever we face such types of problems the key concept we have to remember is that always remember the formulas of area of the square and area of the right angle triangle which are stated above, then using these formulas calculate the areas of all polygons and at last add all the areas of different parts, which will be our required answer.