SSC CGL 2019 objective Ques (3 results)
SSC CGL 20191)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?
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 20192)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?
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 20193)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 \(\pi={22\over7}\) )
3300
Area covered in single revolution = \(2 \pi rh\) ;
Area covered in 1 hour = \({2\times {22\over 7} \times {1\over 2} \times {5\over 4} \times 14 \times 60} = 3300 cm^2\)