**1.**What will be the length of the

diagonal of that square plot whose area is equal to the area of a rectangular

port of length 45 meters and breadth 40 meters?

(a) 42.5

meters

meters

(b) 60 meters

(c) Data

inadequate

inadequate

(d) 45 meters

**2.**The length of a rectangle is 20%

more than its breadth. What will be the ratio of the area of a rectangle to

that of a square whose side is equal to the breadth of the rectangle?

(a)

2:1

2:1

(b) 6:5

(c) 3:4

(d) Data inadequate

**3.**If the length of a certain

rectangle is decreased by 4 cm and the width is increased by 3 cm, a square

with the same area as the original rectangle would result. Find the perimeter

of the original rectangle.

(a) 40

cm

cm

(b) 50 cm

(c) 60

cm

cm

(d) 30 cm

**4.**The base of triangular field is

three times its altitude. If the cost of cultivating the field at Rs. 24.68 per

hectare be Rs. 333.18, find its base and height.

(a) 200

m

m

(b) 275 m

(c) 300

m

m

(d) 600 m

**5.**The difference between two

parallel sides of a trapezium is 4 cm. The perpendicular distance between them

is 19 cm. If the area of the trapezium is 475 cm

^{2}, find the lengths

of the parallel sides.

(a) 27 cm & 23

cm

cm

(b) 22 cm & 25cm

(c) 20 cm & 27

cm

cm

(d) 23 cm & 25 cm

**6.**A farmer wishes to start a 100sq

m triangular vegetable garden. Since he has only 30m barbed wire, he fences

three sides of the garden letting his house compound wall act as the fourth

side fencing. The dimension of the garden is.

(a) 15 m * 6.67

m

m

(b) 20 m * 5 m

(c) 30 m *

3.33m

3.33m

(d) 40m * 2.5 m

**7.**The perimeter of 5 squares is 24

cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square

equal in area to the sum of the areas of these squares is:

(a) 31

cm

cm

(b) 62 cm

(c) 124

cm

cm

(d) 961 cm

**8.**If the ratio of areas of two

squares is 225:256, then the ratio of their perimeters is:

(a)

256:225

256:225

(b) 225:256

(c)

16:15

16:15

(d) 15:16

**9.**The sides of a triangle are 3 cm,

4 cm, and 5 cm. The area (in cm

^{2}) of the triangle formed by joining

the mid-points of the sides of this triangle is:

(a)

3/4

3/4

(b) 3/2

(c)

3

3

(d) 6

**10.**One of sides of a right-angled triangle is

twice the other, and the hypotenuse is 10 cm. The area of the triangle is :

(a) 20 cm

^{2 }

(b) 33 1/3 cm

^{2}
(c) 40 cm

^{2 }(d) 50 cm^{2}

**Answer with solution:**

**1.**(b)

Area= (45* 40) m

1800 m

^{2 }=1800 m

^{2,}
(Diagonal)

^{2}/2= 1800,
Diagonal = 60m

**2.**(b)

Let breadth be x meters.

Then length= 120% of x = 120/100 x= 6x/5 m

Then length= 120% of x = 120/100 x= 6x/5 m

Therefore, the required ratio=

{(6x/5)*x*x}/x*x = 6:5

{(6x/5)*x*x}/x*x = 6:5

**3.**(b)

Let x and y be the length and breadth

of the rectangle respectively.

of the rectangle respectively.

Then, x - 4 =y + 3 or x – y = 7

………..(i)

………..(i)

Area of rectangle = x*y,

Area of square = (x-4) * (y+3)

So, (x-4) * (y+3)= x*y,

or 3x – 4y = 12……………………………….(ii)

Solving eq (i) & (ii) x= 16 and y

= 9

= 9

Therefore perimeter of rectangle =

2(x+ y)= 2(16+9)= 50 cm

2(x+ y)= 2(16+9)= 50 cm

**4.**(c)

Area of the field = Total cost/

Rate = 333.18/ 24.68 hectares = 13.5 hectares

Rate = 333.18/ 24.68 hectares = 13.5 hectares

= (13.5 * 1000) m

135000 m

^{2 }=135000 m

^{2}
Let altitude = x meters and base = 3x

meters.

meters.

Then, 1/2 * 3x * x = 135000

x

^{2 = }90000
x= 300 m

**5.**(a)

Let the two parallel sides of the

trapezium be a cm and b cm,

trapezium be a cm and b cm,

Then, a-b = 4……………………….(i)

And, 1/2 *(a+b)*19 = 475

(a+b) = (475*2)/19 =50…………(ii)

Solving eq(i)& (ii) a= 27 cm and b

= 23 cm

= 23 cm

**6.**(b)

According to quest, 2b + L = 30, (as

fencing is only for 3 sides)

fencing is only for 3 sides)

So, L =30 – 2b

Now area of the garden = 100 sq m,

L * b =100, OR b*(30-2b) = 100

b

^{2}-15b+ 50 =0,
now b= 5 & b=10

if b = 5 then L = 20 and if b= 10 then

L= 10

L= 10

since the garden is rectangular so the

dimension would be 20m * 5 m

dimension would be 20m * 5 m

**7.**(c)

The sides of the 5 squares are (24/4),

(32/4), (40/4), (76/4), (80/4) i.e., 6 cm, 8 cm , 10 cm , 19 cm, 20 cm

(32/4), (40/4), (76/4), (80/4) i.e., 6 cm, 8 cm , 10 cm , 19 cm, 20 cm

Area of the new square = [6

8

^{2}+8

^{2}+ (10)^{2}+ (19)^{2}+ (20)^{2}]
= (36 + 64 + 100 + 361 + 400) cm

961 cm

^{2}=961 cm

^{2}
Side of the new square = √961 cm= 31

cm

cm

Perimeter of the new square = (4*31)

cm = 124 cm

cm = 124 cm

**8.**(d)

A

= (15)

_{1 }/ A_{2 }=225/256= (15)

^{2}/(16)^{2}
=15/16

Now 4 A1 / 4A2= 15/16

**9.**(b)

A= 3 cm, b= 4 cm, and c= 5 cm

It is a right-angled triangle with

base = 3 cm and height= 4 cm

base = 3 cm and height= 4 cm

So its area= (1/2 * 3 * 4) cm

6 cm

^{2}=6 cm

^{2}
Area of required triangle= (1/4 * 6)

cm

cm

^{2 }= 3/2 cm^{2}**10.**(a)

Let the sides be a cm and 2a cm

Then, a

(10)

^{2}+ (2a)^{2 }=(10)

^{2}
5 a

^{2}= 100
a

^{2 }= 20 cm^{2}
Area = (1/2 *a *2a) = a

20 cm

^{2}=20 cm

^{2}
Gr8t adda

1-b

2-b

3-b

4-

5-a

6-b

7-c

8-d

9-b

10-a

Prevention of Corrosion

1. The rusting of iron can be prevented by painting, oiling, greasing,galvanising, chrome plating, anodising or

making alloys.

2. Galvanisation is a method of protecting steel

and iron from rusting by coating them with a thin layer of zinc.

3. Alloying is a very good method of improving the

properties of a metal.We can get the desired properties by this

method

4. When iron is mixed with nickel and chromium, we

get stainless steel, which is hard and does not

rust.

5. Gold is alloyed with either silver or copper to make it hard

6. one of the metals in Alloy is mercury, then the

alloy is known as an amalgam.

7. brass, an alloy of copper and zinc (Cu and Zn),

8. bronze, an alloy of copper and tin (Cu and Sn),

9. Solder, an alloy of lead and tin (Pb and Sn),

has a low melting point and is used for welding

electrical wires together.

Dude answer to question 4 is C and to question no 10 is B

Refresh!!!!

G.K.Quiz On Geography

http://www.sscadda.com/2015/04/gkquiz-on-geography.html

Thank u manish

In question no 3 how got the equation 1.plz..explain me.

please do not make the typing mistakes

Thanks

Q no. 6?