**Dear Readers,**

We are providing you **Important Short Tricks on Mensuration Questions** which are usually asked in **Bank Exams**. Use these below given short cuts to solve questions within minimum time. These shortcuts will be very helpful for your upcoming All **Bank Exam.**

To make the chapter easy for you all, we are providing you with all some **Important Short Tricks to Mensuration Questions** which will surely make the chapter easy for you all.

**About Mensuration:-**

Mensuration questions are an important part of **Quant paper** in banking exam. Mensuration question asked in banking exam related to **Perimeter and Area**.

**About Perimeter & Area:- **

To put simply, Area measures the area of shape i.e. the space that shape takes up. Perimeter is the measurement of the boundary of the figure.

**1.** **Square:-** A square is a four-sided polygon characterized by right angles and sides of **equal length.**

**Area** **= side ^{2},**

**Perimeter** **= 4side**^{}**2.** **Rectangle**:- A four-sided flat shape with straight sides where all interior angles are right angles (90°). Also, the opposite sides are parallel and of equal length.

**Area** **=** **Length × Breadth,**

**Perimeter** = **2(L+B)**

**3. Circle**;- Circle is the locus of points equidistant from a given point, the centre of the circle. The common distance from the centre of the circle to its points is called the radius

**Area** **= π(radius) ^{2},**

**Perimeter** **= 2πradius**

**Let’s look at some questions asked:-****Que 1.** The length of the rectangular plot is 20 m more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per meter is Rs. 5300, what is the length of the plot?**Sol.** Cost = Rate × Perimeter

⇒ Perimeter = 5300/26.50 = 200

⇒ 2 (L + B) = 200

⇒ 2 (L + L – 20) = 200

⇒ L – 10 = 50

⇒ L = 60 m

**Que 2.** If the length is to be increased by 10% and breadth of a rectangle plot to be decreased by 12%, then find % change in area?**Sol:-** **In such questions, use formula: Increment in area** = [L%+B%+{(L%*B%)/100)}]

Increment in area = 10 – 12(because breadth is decreased) {+ 10+(-12) (-120/100)} = -3.2%

**Que 3. **The length of the rectangular plot is increased by 60%. By what percentage should the width be decreased to maintain the same area?**Sol:-** In such questions, use the formula:

**Required % decrease in breadth = [%change in L{100/(100+%change in L)}]**

Required % decrease in breadth = 60 (100/160) = 37.5%

**Que 4. **If the radius of the circle is increased by 5% find the percentage change in its area.

In such questions, use the formula:

**Sol:-****Change in area** = (2x + x^{2}/100) %**Change in area** = 2×5 + 5^{2}/100 = 10 + ¼ = 10.25% increment.

** Note:** In such questions, the negative sign implies decrements while positive sign shows increment.

**Que 5. **The circumference of a circle is 100 cm. Find the side of the square inscribed in the circle.**Sol.** Always use this formula, Side of a square inscribed in a circle of radius **r = Root 2**

Circumference = 100 ⇒ 2πr = 100

r = 50/π

Side of square = **Root 2 (50/π)**

*Note***: Similar formula:**

**1)**Area of largest triangle inscribed in a semi-circle of radius

**r = r**

^{2}

**2)**Area of largest circle that can be drawn in a square of side

**x = π(x/2)**

^{2}**Que 6. **The length and breadth of the floor of the room are 20 by 10 feet respect. Square tiles of 2 feet are to be laid. Black tiles are laid in first row on all sides, white tiles on 1/3^{rd} of remaining sides and blue tiles on the rest. How many blue tiles are required?**Sol:-** Side of a tile = 2 feet ⇒ Area of 1 tile = 2^{2} = 4 sq ft. ----- (1)

Length left after lying black tile on 4 sides = 20 – 4

⇒ Area left after black tiles = (20 – 4)×(10-4) = 96 sq. ft.

Area left after white tiles = 2/3×96 = 64 sq. ft.

⇒ Area for blue tiles = 64 sq. ft.

Number of blue tiles = 64/4 = 16 (using 1)

**Que 7. **A cow is tethered in the middle of the field with a 14 ft long rope. If the cow grazes 100 sq. ft. per day, then approximate time taken to graze the whole field?**Sol:** Here, the rope of cow is like radius

**Area** = π (14)^{2}

No. of days = (Area of field)/Rate of cow = π (14)^{2}/100 = 6 days(approx)

**Que 8.** A circle and a rectangle have same perimeter. Sides of rectangle are 18 by 26 cm. What is the area of circle?**Sol:-** 2πr = 2 (18 + 26) ⇒ r = 14 cm**Area** = π r^{2} = 616 cm^{2}

**Que 9.** What will be the ratio between the area of a rectangle and the area of a triangle with one of the sides of the rectangle as base and a vertex on the opposite side of the rectangle?**Sol:-**

**Area of triangle** = ½ × L × B

**Area of rectangle** = L × B

Area of a rectangle: area of a triangle = L × B: ½ × L × B = 2: 1

**Que 10.** In a rectangular plot, a cow is tied down at a corner with a rope of 14m long. Find the area that cow can graze?**Sol.** The area that cow can graze can be illustrated as shaded area:

Here, the shaded area is a quarter of a circle with radius 14m,

Area of grazed field = ¼ × π (14)^{2 }= 154 m^{2}

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