Mensuration problems and aptitude related to the Square ~ New math tricks for quantitative aptitude

# New math tricks for quantitative aptitude

## Mensuration problems and aptitude related to the Square

In the mensuration topic quadrilateral properties and formula, we have briefly discussed the various shape, properties and shortcut formula. Square is one of the important figures of the quadrilateral. Today we are going to solve some mensuration problems related to the square. As you have learnt that the square is four equal-armed and right angular figure. Before going to solve square mensuration, please read the opening tutorial of quadrilateral properties and mensuration tricks if you need.

## Square mensuration and math tricks

Mensuration Problem 1: Find the area and perimeter of a 20 metres armed square.
Solution: Arm length of the square is 20m. So the area will be (Arm * Arm) square unit. So area= 20*20 Sq metre =400 Sq metre.
The perimeter is the outer length of the figure. The perimeter of the square is (4*Arm). So, perimeter = 4*20 = 80 metre.
mensuration problem 2: The diagonal of a square is 10√2. Now find the area of the square.
Solution: Just look at below how we find the area of the square.
Mensuration problem 3:  20 meters and 40 meters armed two square land marge into a rectangular land, if the base of the rectangular lad is 50 meters then find the width of the land.
Solution: To solve the mensuration problem, we have to calculate the sum of the area of both square and that is the area of rectangular land.  Look at the below how we solved it.
Problem 4: A house owner wants to cover a courtyard with marvels. The base and the width of the courtyard are 10 and 8 meters respectively. The area of each marvel is 1/4 sq metre. Then how many marvels should have to buy to cover the courtyard?
Solution: We can solve this mensuration problem easily without pen and paper. First of all, we will calculate the area of the courtyard. As the courtyard is rectangular, so the area can be calculated multiplying by base and width. Area= 10*8 sq metre
=80 sq metre. Now we have to divide the area of the courtyard by the area of a marvel tile. 80/(1/4)=80*4
=320 tile needs to cover the courtyard.

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