SSC CGL 2020 objective Ques (21 results)
1)In the given fig. triangle ABC, \(\theta = 80^o\) the measure of each of the other two angle will be:
SSC CGL 2020
\(50^0\)
since the two sides are equal in length, and equal side corresponds equal angle, therefore \(2x + \theta = 180\), x = 50
SSC CGL 20202)Arrange the angles of the triangle from smallest to largest in the triangle, where the sides are AB = 7 cm, AC = 8 cm, and BC = 9 cm.
C,B,A
Measurement of angle is according to the length of opposite line.
So, The order of the angles of the triangle from smallest to largest in the triangle,
C < B < A.
SSC CGL 20203)In the triangle, If AB = AC and \(\angle ABC = 72^0\), then \(\angle BAC\) is:
\(36^0\)
By triangle properties,
When AB = AC then \(\angle ABC = \angle ACB\) so,
In\(\triangle ABC,\)
\(\Rightarrow \angle ABC + \angle ACB + \angle BAC = 180^{0}\);
\(\Rightarrow \angle ABC + \angle ABC + \angle BAC = 180^{0}\);
\(\Rightarrow 72^{0} + 72^{0} + \angle BAC = 180^{0}\);
\(\Rightarrow \angle BAC = 180^{0} - 144^{0} = 36^{0}\)
4)In the given figure, a circle inscribed in triangle PQR touches its sides PQ, QR and RP at points S, T and U,respectively. If PQ = 15 cm, QR= 10 cm, and RP = 12 cm, then find the lengths of PS, QT and RU?
SSC CGL 2020
PS = 8.5 cm, QT = 6.5 cm and RU = 3.5 cm
PQ = 15 cm, QR= 10 cm, and RP = 12 cm. ;
PQ = x + y = 15------------------(i);
QR = y + z = 10----------(ii);
RP = z + x = 12 ------(iii);
Solving equation, we get :
z = 3.5 cm = RU;
y = 6.5 cm = QT;
x = 8.5 cm = PS
SSC CGL 20205)In \(\triangle ABC,\) D is a point on BC such that AD is the bisector of \(\angle A\), AB = 11.7 cm. AC = 7.8 cm. and BC = 13 cm. What is the length (in cm.) of DC?
5.2
AD is bisector of \(\angle A\).
\({AB\over AC}={BD\over DC}\); Let, BD = x cm. ; DC = (13 - x)cm. ;
\({11.7\over7.8}= {x\over13-x}\); x = 7.8 cm. ;
DC = 13 - 7.8 = 5.2 cm
SSC CGL 20206)The length of each equal side of an isosceles triangle is 15 cm and the included angle between those two sides is 90°.Find the area of the triangle.
\({225\over2}cm.^2\)
Area of \(\triangle ABC = {1\over2}\times AB \times AC ={1\over2}\times 15\times 15 ={225\over2} cm^2\)
SSC CGL 20207)In \(\triangle ABC\), D, E and F, are the midpoints of sides AB, BC and CA, respectively. If AB = 12 cm, BC = 20 cm and CA = 15 cm, then the value of \({1\over2}(DE+EF+DF)\) is:
11.75 cm
By mid point theorem,
DF = BC/2 = 20/2 = 10 cm;
DE = AC/2 = 15/2 = 7.5 cm;
EF = AB/2 = 12/2 = 6 cm;
\({1\over2}(DF + DE+EF)={1\over2}(10+7.5+6) =11.75\)
8)In the given figure, the measure of \(\angle BAC\) is:
SSC CGL 2020
\(48^0\)
In a triangle exterior angle = sum of other two interior angles ; \(\angle ACD = \angle BAC+\angle ABC;\) ⇒ \(\angle BAC= \angle ACD-\angle ABC\) = \(110^0-62^0= 48^0\)
9)In the given figure, \(\triangle ABC\) is an isosceles triangle, in which AB = AC, \(AD \perp BC\), BC = 6 cm. and AD = 4 cm. The length of AB is:
SSC CGL 2020
5 cm.
AB = AC; \(\angle ADC=\angle ADB=90^0\); \(\therefore BD = DC\) ; BC = 6 cm. ; BD = \(6\over2\)= 3 cm. ; AD = 4 cm. ; In \(\triangle ABD\), \(AB=\sqrt{AD^2+BD^2}=\sqrt{4^2+3^2}\) = 5 cm.
SSC CGL 202010)In \(\triangle ABC\), if the ratio of angles is in the proportion 3 : 5 : 4, then the difference between the biggest and the smallest angles (in degrees) is:
\(30^0\)
We know that sum of the all angles of the triangle is equal to the \(180^0\) .
Let the angles be 3x, 5x and 4x.
so, 3x + 5x + 4x = 180;
x = 15;
Smallest angle = 3x = 3 x 15 = \(45^0\);
Biggest angle = 5x = 5 x 15 = \( 75^0\);
Difference between the biggest and the smallest angles = 75 - 45 = \(30^0\)