PROBLEMS ON PROPERTIES OF SQUARE

What is square ?

A square is a rhombus with four equal angles of 90.

Properties :

• Opposite sides are parallel.
• All sides are equal in length.
• Diagonals bisect each other at right angles.
• Diagonals bisect the angles at each vertex.
• Diagonals are equal in length.

Problem 1 :

The square ABCD, find each length or angle measure.

Solution :

1)  In square, the diagonal will bisect each other. So, BX = 8.5

2)  Diagonals are equal, AX = 8.5

3)  DB = 2DX, DB = 2(8.5)  ==>  17

4)  Diagonals are equal. So AC = 17

5)  AB = 19

6)  BC = 19

7)  Diagonals will bisect each other at right angle, ∠AXB = 90

8)  Diagonals will bisect the vertex angle, ∠XAB = 45

Problem 2 :

Classify the special quadrilateral. Then find the values of x and y. 

Solution :

Since all sides are equal and angles at vertex is 90 degree. The given shape is square.

(1) ⋅ 5 ==> 25x - 10y = 40

(2) ⋅ 2 ==> 6x - 10y = 2

(1) - (2)

19x = 38

x = 38/19

x = 2

By applying x = 1 in (1), we get

5(2) - 2y = 8

10 - 2y = 8

-2y = 8 - 10

-2y = -2

y = 1

Problem 3 :

For the square ABCD, find the missing values.

Solution :

Diagonals will bisect each other.

18 = 2x + 2y

Dividing by 2 on both sides.

9 = x + y

x + y = 9 ----(1)

All sides will be equal.

AB = AD

30 = 2x + 5y

2x + 5y = 30 ----(2)

(1) ⋅ 2 - (2) ==> (2x + 2y) - (2x + 5y) = 9 - 30

2y - 5y = -21

-3y = -21

y = 7

Problem 4 :

The diagonals of square DEFG intersect at H. Given that EH = 5, find the indicated measure. 

Solution :

1)  ∠GHF = 90 degree

2)  HF = 5

3)  ∠DGH = 45 degree

4)  In triangle DHE,

(DE)² = DH² + HE²

DE² = 5² + 5²

DE² = 25 + 25

DE = √50

DE = 5√2

Problem 5 :

ABCD is a given rectangle with length as 80 cm and breadth as 60 cm. P, Q, R, S are the mid points of sides AB, BC, CD, DA respectively. A circular rangoli of radius 10 cm is drawn at the center as shown in figure given below. Find the area of shaded portion.

Solution :

DC = 80 cm, DR = 40 cm

DA = 60 cm, DS = 30 cm

In triangle DRS,

RS² = DR² + DS²

RS² = 40² + 30²

RS² = 1600 + 900

RS² = 2500

RS = 50

RS = PQ = SP = RQ

Area of square PQRS = 2500

Area of circle = πr²

radius = 5 cm

= 3.14 (5)²

= 3.14(25)

= 78.5

Area of shaded portion = 2500 - 78.5

= 1921.5 cm²

Problem 6 :

A large square is made by arranging a small square surrounded by four congruent rectangles as shown in figure given below. If the perimeter of each of the rectangle is 16 cm, find the area of the large square.

Solution :

Perimeter of the rectangle = 16 inches

2(l + b) = 16

l + b = 8

Side length of the square = l + b

Area of square = (l + b)²

= 8²

= 64

Problem 7 :

ABCD is a square with AB = 15 cm. Find the area of the square BDFE. 

Solution :

ABCD is a square, here BD is the diagonal of the square.

BD² = AB² + AD²

BD² = 15² + 15²

BD² = 450

BD = 15√2

Area of square BDEF = BD²

= (15√2)²

= 225(2)

= 450