Mensuration 2D

 Mensuration in its literal meaning is to measure. It is generally used where geometrical figures are concerned, where one has to determine various physical quantities such as perimeter, area, volume or length. Measuring these quantities is called Mensuration

In this study notes we discuss some important questions on Mensuration 2D

Question 1

 The length of a rectangular floor is more than its breadth by 200%. If Rs. 324 is required to paint the floor at the rate of Rs. 3 per sq m, then what would be the length of the floor ?

(a)      27 m

(b)     24 m

(c)      18 m

(d)     21 m

Solution:

Let the length and the breadth of the floor be L and B respectively.

L = B + 200% of B    =>   L = B + 2B = 3B

Total cost for painting =  (amount per sq m) x ( total area of thr floor )
Area of the floor = 324/3 = 108 sq m

L*B = 108   i.e., L * L/3 = 108

L² = 324  =>  L = 18

Question 2

 If the circumference of a circle is equal to the perimeter of a square, what is the ratio of the area of the circle to the area of the square?

(a)      22:7

(b)     14:11

(c)      11:7

(d)     4:1

Solution:

Circumference of a Circle = 2*π*R

Perimeter of a Square = 4a

A.T.Q :-

2*π*R = 4a     =>     a = π*R/2 

Required Ratio = Area of Circle/Area of a Square = π*R² / a² 

=>  π*R² / [π*R/2]²

=> 14 :11  [ ByTaking Pi = 22/7]

Question 3

If the sides of a equilateral triangle is increased by 10% , 30% and 60% then a new triangle is formed. By what % perimeter of the triangle is increased ?

(a)      40.50%

(b)     32.45%

(c)      33.33%

(d)     35.67%

Solution:

Let a = side of the triangle

Perimeter = 3a

New perimeter = a*110/100 + a*130/100 + a*140/100

= a(11+13+16)/10 = 4a

% increased in perimeter = (4a – 3a/3a)*100

= 100/3 = 33.33%

Question 4

The area of right angle triangle is 100sq cm. The ratio of base to its height is 4:5.Find the length of the hypotenuse ?

(a)      20cm

(b)     22cm

(c)      28cm

(d)     30cm

Solution:

Area = 1/2(4x*5x)

100 = 20x^2/2

X² = 100/10 =10

x = 3.16

4x:5x = 12.64:15.8

Hypotenuse = √12.642+15.82

= √409.40

= 20.23 = 20

Question 5

One of sides of a right-angled triangle is thrice the other, and the hypotenuse is 12 cm. The area of the triangle is ?

(a)      25.4 cm²

(b)     24.6 cm²

(c)      22.5 cm²

(d)     21.6 cm²

Solution:

x² + (3x)² = (12)²
10 x²= 144
x² = 14.4 cm²
Area = 1/2 (x *3x)
=3(14.4)/2
= 21.6 cm²

Question 6

The perimeter of a rectangular plot is 210 m. Find the cost of gardening 2 m broad boundary around rectangular plot whose perimeter The cost of gardening is Rs 14 per m² ?

(a)      Rs 6450

(b)     Rs 6400

(c)      Rs 6480

(d)     Rs 6104

Solution:

Given 2(l+b) = 210

2 m broad boundary means increase in l and b by 4 m

So area of the boundary will be [(l+4)(b+4) – lb] = 4(l+b) + 16 = 2*[2(l+b)] + 16 = 2*210 + 16 = 436 m²

So cost of gardening = 436*14 = 6104

Question 7

The area of the garden formed by two concentric circles with circumferences 44m and 176 m respectively is?

(a)      2390m²

(b)     2590m²

(c)      2310m²

(d)     2490m²

Solution:

2πR₁ = 176

R₁ = 28m

2πR₂ = 44

R₂ = 7m

Area  of the garden = π(R1² – R2²) = 22/7(784 – 49) =2310 m²

Question 8

The ratio of the length and the breadth of a rectangle is 4 : 3 and the area of the rectangle is 6912 sq cm. Find the ratio of the breadth and the area of the rectangle ?

(a)      1 : 96

(b)     1 : 48

(c)      1 : 84

(d)     1 : 68

Solution:

Let the length and the breadth of the rectangle be 4x cm and 3x respectively.

(4x)(3x) = 6912

12x² = 6912

x² = 576 = 4 * 144 = 2² * 12² (x > 0)

=> x = 2 * 12 = 24

Ratio of the breadth and the areas = 3x : 12x² = 1 : 4x = 1: 96.

Question 9

The ratio of the length and breadth of a plot is 4 : 3. If the breadth is 40 m less than the length, What is the perimeter of the plot?

(a)      560 m

(b)     540 m

(c)      500 m

(d)     none of these

Solution:

Let the length be 4x metres. Then, breadth = 3x metres.

Then, 4x - 3x = 40 => x = 40

length l = (4 × 40) = 160 m

breadth b = (3 × 40) = 120 m

Perimeter = 2(160 + 120) = 2(280) = 560 m

Question 10

The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. Find the area of the triangle.

(a)      48

(b)     45

(c)      46

(d)     42

Solution:

Let ABC be the isosceles triangle and AD be the altitude

Let AB = AC = x. Then, BC = (32 - 2x).

Since, in an isosceles triangle, the altitude bisects the base,

so BD = DC = (16 - x).

In triangle ADC,(AC)²=(AD)²+(DC)²

⇒x²=(8)²+(16−x)²⇒x=10

BC = (32- 2x) = (32 - 20) cm = 12 cm.

Hence, required area =  1/2*BC*AD=1/2*12*8=48cm ²

Question 11

What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ?

(a)      424

(b)     814

(c)      758

(d)     756

Solution:

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.

Area of each tile = (41 x 41) sq.cm

Required number of tiles =1517 x 902/(41 x 41) = 814.

Question 12

If the area of a square is equal to the area of that rectangle whose width is double of the one side of the square then the ratio of the length to the breadth of the rectangle will be?

(a)      1 : 2

(b)     1 : 4

(c)      1 : 6

(d)     1 : 8

Solution:

b = 2a

a = b/2

Area of square = b²/4 = Area of rectangle

l * b = b²/4 => l = b/4

l / b =(b/4)/b => 1:4

Question 13

A circular wire of diameter 42cm is folded in the shape of a rectangle whose sides are in the ratio 6:5. Find the ratio of the enclosed rectangle ?

(a)  1080cm²

(b)  1181cm²

(c)  1250cm²

(d)  1800cm²

Solution:

Radius = 42/2 = 21cm

Circumference = 2*22*21/7 = 132cm

L:b = 6x:5x

Perimeter of the rectangle = 2(6x+5x) = 22x

22x = 132

X = 6

Area = 6x * 5x = 30x*x

=30*6*6 = 1080cm²

Question 14

A park is in the form of a square one of whose sides is 50 m. The area of the park excluding the circular lawn in the centre of the park is 1884 m². The radius of the circular lawn is ?

(a) 21 m

(b)  31 m

(c)  41 m

(d)  14 m

Solution:

Area of park = 50 x 50 = 2500 m²

Area of circular lawn = Area of park – area of park excluding circular lawn

= 2500 – 1884

= 616

Area of circular lawn = (22/7) x r² = 616 m²

⇒ r² = (616 x 7) / 22

= 28 x 7

= 2 x 2 x 7 x 7

∴ r = 14 m

Question 15

The area of a rectangle is equal to the area of a square whose diagonal is 12√2 metre. The difference between the length and the breadth of the rectangle is 7 metre. What is the perimeter of rectangle ? (in metre).

(a)  68 metre

(b)  50 metre

(c)  62 metre

(d)  64 metre

Solution:

d = a√2
12√2 = a√2
a = 12
l * b = a² = (12²) = 144
l – b = 7 ; l = b + 7
(b + 7)*(b) = 144
b² + 7b – 144 = 0
b² + 16b –9b - 144 = 0
b(b+16) -9(b+16) =0 
(b+16)(b-9) =0 
=> b= 9 
b = 9; l = 16
2(l + b) = 2(16 + 9) = 50m