SSC CGL 20201)What is the area of a sector of a circle of radius 14 cm and central angle 45 degree? take \(\pi = {22 \over 7} \)

Correct Option: B

77 sqcm

Area of Sector = \(x = \pi r^2 .{\theta \over 360 } = 77 cm^2\)

SSC CGL 20192)In \(\triangle ABC\), D and E are the points on sides AB and BC respectively such that DE||AC . If AD : DB = 5 : 3, then what is the ratio of the area of \(\triangle BDE\) to that of the trapezium ACED?

Correct Option: B

9 : 55

\(\triangle ABC ~ \triangle DEB;\)

(\because DE is parallel to AC);

\(\angle B\) is a common angle.

So, ratio area of the \( \triangle BDE : \triangle ABC \);

\((3)^2 : (3 + 5)^2\) = 9 : 64;

Area of trapezium ACED = area of the \(\triangle ABC - \triangle BDE \)= 64 - 9 = 55;

Ratio of the area of\( \triangle BDE\) to that of the trapezium ACED = 9 : 55

SSC CGL 20193)In a trapezium ABCD, \(DC \parallel AB \), AB = 12 cm and DC = 7.2 cm. What is the length of the line segment joining the midpoints of its diagonals?

Correct Option: C

2.4 cm

By the property, EF = \(\frac{AB - CD}{2} = \frac{12 - 7.2}{2} = \frac{4.8}{2} \)= 2.4 cm

SSC CGL 20194)A field roller, in the shape of a cylinder, has a diameter of 1 m and length of \(1\frac{1}{4}\) m. If the speed at which the roller rolls is 14 revolutions per minute, then the maximum area (in square m ) that it can roll in 1 hour is: (Take \(\pi={22\over7}\) )

Correct Option: C

3300

Area covered in single revolution = \(2 \pi rh\) ;

Area covered in 1 hour = \({2\times {22\over 7} \times {1\over 2} \times {5\over 4} \times 14 \times 60} = 3300 cm^2\)

SSC CGL 20205)A wheel covers a distance of 1,100 cm in one round. The radius of the wheel is :

Correct Option: B

175 cm

A wheel covers a distance of 1,100 cm in one round so,

Perimeter of wheel = 1100 cm;

\(2\pi r\) = 1100;

r = \(550\times\frac{7}{22}\) = 175 cm;

Diameter = 2r = 2 175 = 350 cm

SSC CGL 20206)The perimeter of a square plot is the same as that of a rectangular plot with sides 35 m and 15 m. The side of the square plot is:

Correct Option: C

25 metre

The perimeter of a rectangular plot = 2(length + breadth) = 2(35 + 15) = 100 m;

The perimeter of a square plot = perimeter of a rectangular plot;

The perimeter of a square plot = 100 m;

100 = 4 side;

side = 100/4 = 25 m

SSC CGL 20207)What is the area of a triangle whose sides are 3 cm, 5 cm and 4 cm?

Correct Option: A

\(6\) \(cm^2\)

By triplet 3-4-5, the triangle will be right angle triangle so,

5 will be hypotenuse.

Area of triangle = \({1\over2}\times base\times height\) = \({1\over2}\times3\times4 = 6 \) \(cm^2\)

SSC CGL 20208)Find the area and circumference of a circle if the radius is 14 cm.(Take \(\pi= {22\over7}\))

Correct Option: D

Area = \(616 \)\(cm^2\); circumference = 88 cm

Radius of circle = 14 cm. ;

Area of circle = \(\pi r^2={22\over7}\times{(14)^2}= 616 cm^2\);;

Circumference of circle = \(2\pi r = 2\times{22\over7}\times 14 = 88 cm\)

SSC CGL 20209)If the perimeter of a certain rectangle is 50 units and its area is 150 sq. units, then how many units is the length of its shorter side?

Correct Option: C

10

Let, the length of rectangle= x units; Its width = y units; According to the question, 2(x+y) = 50;

x + y = 25 ---(1) ; and, xy = 150___(2); \((x-y)^2=(x+y)^2-4xy\) \(=(25)^2-4\times150=25\); ⇒ \(x-y=\sqrt{25} =5\)____(3);

By equation (1) - (3) we have; x + y - x + y = 25 - 5 = 20; So y = 10 units

SSC CGL 202010)The perimeter of an isosceles triangle is 50 cm.If the base is 18 cm, then find the length of the equal sides.

Correct Option: A

16 cm

Let Each equal side of isosceles triangle = x cm.

According to the question, x + x + 18 = 50; ⇒ x = 16 cm.

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