1)In the given fig. triangle ABC, \(\theta = 80^o\) the measure of each of the other two angle will be:

SSC CGL 2020

Correct Option: D

\(50^0\)

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.

Correct Option: C

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:

Correct Option: A

\(36^0\)

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

Correct Option: D

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?

Correct Option: D

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.

Correct Option: A

\({225\over2}cm.^2\)

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:

Correct Option: C

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

Correct Option: D

\(48^0\)

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

Correct Option: C

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:

Correct Option: A

\(30^0\)

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 = \(30^0\)

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