1)

Triangle PDC is drawn inside the square ABCD of side 24cm where P lies on AB. What is the area of the triangle ?

SSC CGL 2020

A)

280\(cm^2\)

B)

200\(cm^2\)

C)

298\(cm^2\)

D)

288\(cm^2\)

Correct Option: D

288\(cm^2\)

---

since ABCD is a square, Area of triangle PDC = \( {1 \over 2}\times CD \times PE\) = \( {1 \over 2}\times 24 \times 24 = 288 cm^2\)

2)

In quadrilateral PQRS, \(RM\perp QS\), \(PN\perp QS\) and QS = 6 cm. If RM = 3 cm and PN = 2 cm, then the area of PQRS is

SSC CGL 2020

A)

\(15cm^2\)

B)

\(14cm^2\)

C)

\(13cm^2\)

D)

\(11cm^2\)

Correct Option: A

\(15cm^2\)

---

Area of quadrilateral PQRS = Area of \(\triangle PQS\) + Area of \(\triangle QRS\) =\({1\over2}\times QS\times PN + {1\over2}\times QS\times RM={1\over2}\times QS(PN + RM)={1\over2}\times6(2+3)=15 cm^2\)

3)

In the figure. if \(\angle A = 100^0\) then \(\angle C = ?\)

SSC CGL 2020

A)

\(50^0\)

B)

\(90^0\)

C)

\(80^0\)

D)

\(100^0\)

4)

The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Its area is :

SSC CGL 2020

A)

\(96\space cm^2\)

B)

\(48\space cm^2\)

C)

\(69\space cm^2\)

D)

\(28\space cm^2\)

5)

ABCD is a cyclic quadrilateral which sides AD and BC are produced to meet at P, and sides DC and AB meet at Q when produced. If \(\angle A = 60^\circ\) \( and\space \angle ABC = 72^\circ,\)\( then\space \angle DPC - \angle BQC = ?\)

SSC CGL 2020

A)

\(40^0\)

B)

\(36^0\)

C)

\(24^0\)

D)

\(30^0\)

Correct Option: B

\(36^0\)

---

Since ABCD is a cyclic quadrilateral.  \(\angle A+\angle C = 180 \) & \(\angle B +\angle D = 180^0\) ; 

\(\therefore \angle C = 120^0\)  & \(\angle D = 108^0\) ; 

In \(\triangle PDC\) ⇒ \(\angle PDC = 180-108=72^0\) & \(\angle PCD = 180-120=60^0\) ;  

\(\therefore \angle P = 180-72-60=48^0\) ;

Similarly, In \(\triangle BCQ\) ⇒ \(\angle Q = 12^0\) ;  Now \(\angle P-\angle Q= 48^0-12^0=36^0\)

6)

Sides AB and DC of cyclic quadrilateral ABCD are produced to meet at E, and sides AD and BC are produced to meet at F. If \(\angle ADC = 75^\circ\) and \(\angle BEC =52^\circ\), then the difference between \(\angle BAD\) and \(\angle AFB \) is:

SSC CGL 2020

A)

\(23^0\)

B)

\(22^0\)

C)

\(21^0\)

D)

\(31^0\)

Correct Option: D

\(31^0\)

---

ABCD is a cyclic quadrilateral.  \(\angle ADC = 75^\circ\) ;
 ∴\(\angle ABC + ∠ADC=180^0\) ;
⇒ \(∠ABC = 180^0 − 75^0=105^0\) ;
⇒ \(∠CBE = 180^0 − 105^0=75^0\)  ;
\(\therefore ∠BCE = 180-(52+75)=53^0\); 
\(\angle BCD=180-53=127^0\) ;
\(\because \angle BCD+\angle BAD=180^0\) ; ⇒ \(\angle BAD = 53^0\); 
\(\angle AFB = 180-(105+53)=22^0\) ;
Required difference = \(53^0-22^0=31^0\)