Download SSC CGL Geometry questions with answers PDF based on previous papers very useful for SSC CGL exams. Very important Geometry questions for SSC exams.
Question 1: In a cyclic quadrilateral ∠A+∠C=∠B+∠D=?
Question 2: The height of an equilateral triangle is 15 cm. The area of the triangle is
Question 3: If the interior angles of a five-sided polygon are in the ratio of 2 : 3 : 3 : 5 : 5, then the measure of the smallest angle is
Question 4: If the lengths of the sides of a triangle are in the ratio 4 : 5 : 6 and the inradius of the triangle is 3 cm, then the altitude of the triangle corresponding to the largest side as base is :
Question 5: The ratio of inradius and circumradius of a square is :
Question 6: AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and distance between them is 17 cm, then the radius of the circle is :
Question 7: A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6 . What is the fraction ?
Question 8: If the circumradius of an equilateral triangle ABC be 8 cm, then the height of the triangle is
Question 9: Let ABC be an equilateral triangle and AX, BY, CZ be the altitudes. Then the right statement out of the four given responses is
Question 10: Two supplementary angles are in the ratio 2 : 3. The angles are
Answers & Solutions:
1) Answer (D)
Sum of opposite angle of cyclic quadrilateral is 180° (D)
2) Answer (C)
AD = 15 cm and ABC is equilateral triangle
=> $tan \angle ACD = \frac$
=> BC = 2*DC = $10\sqrt$
Area of $\triangle$ ABC = $\frac> * side^2$
3) Answer (C)
Let the angles of the pentagon be 2x,3x,3x,5x and 5x
Sum of angles of a pentagon = $(n-2) * 180$°
=> Smallest angle = 2*30 = 60°
4) Answer (A)
Let the sides of triangle be $4x , 5x$ and $6x$
Inradius(r) = 3 cm and semi perimeter(s) = $\frac = 7.5x$
=> Area of triangle = $r * s$ = $22.5x$
Let altitude be $h$
Also area = $\frac * 6x *h = 22.5x$
5) Answer (A)
Let the side of square be $a$
6) Answer (C)
Given : AB = 10 , CD = 24 and EF = 17 cm
To find : OB = OD = $r$ = ?
Solution : A line perpendicular to the chord from the centre of the circle bisects the chord.
Similarly, $CE = ED = 12$
Let OF = $x$ => OE = $(17-x)$
In right $\triangle$OFB
Now, in right $\triangle$OED
=> $x^2 + 25 = x^2 – 34x + 289 + 144$
7) Answer (A)
Let the fraction be $\frac$
=> $11x + 22 = 9y + 18$
Solving above equations, we get $x = 7$ and $y = 9$
=> Required fraction = $\frac$
8) Answer (D)
Let ABC be the equilateral triangle and O be the circumcentre. AO extended meet BC at D.
In an equilateral triangle, the centroid, orthocentre, incentre and circumcentre, all lie on the same point, => the median and height are the same lines.
=> O is also the centroid of the triangle.
Since, the centroid divides the median in the ratio 2 : 1
It is given that OA = 8 cm
=> Height AD = $8 * \frac$ = 12 cm
9) Answer (A)
In an equilateral $\triangle$ABC
$\angle$A = $\angle$B = $\angle$ C = 60 °
=> AB = BC = CA and hence AX = BY = CZ
10) Answer (C)
Let the angles be $2x$ and $3x$
Since, the angles are supplementary
=> Angles are 72° and 108°
we hope this Geometry Questions for SSC CGL is helpful to you.
