A closed figure formed by three intersecting lines is called

1

(a)

circle

(b)

square

(c)

triangle

(d)

rhombus

A triangle has

1

(a)

2 verticles

(b)

3 verticles

(c)

4 verticles

(d)

5 verticles

The symbol for congruence is

1

(a)

=

(b)

~

(c)

⁰

(d)

≅

The symbol for correspondence is

1

(a)

⟶

(b)

⇔

(c)

↔

(d)

≡

Two circles are congruent.If the radius of one circle is 3cm, what is the radius of the other circle?

1

(a)

3 cm

(b)

6 cm

(c)

1.5 cm

(d)

1 cm

ΔABC ≌ ΔPQR, then which of the following is true:

1

(a)

A↔R

(b)

AB=QR

(c)

AC=PQ

(d)

AB=PQ

ΔABC ≅ ΔPQR.If AB = 5cm, ㄥB = 40⁰and ㄥA = 80⁰, then which of the following is true?

1

(a)

QP = 5cm, ㄥP = 60⁰

(b)

QP = 5cm, ㄥR = 60⁰

(c)

QR = 5cm, ㄥR = 60⁰

(d)

QR = 5cm, ㄥQ = 40⁰

Two triangles are congruent, if two sides and the included angle of one triangle are equal to two sides and the included angle of the other triangle. This rule is known as

1

(a)

SAS congruence rule

(b)

ASA congruence rule

(c)

SSS congruence rule

(d)

RHS congruence rule.

Two triangles are congruent, if two angles and the included side of one triangle are equal to two angles and the included side of other triangle. This rule is known as

1

(a)

SAS congruence rule

(b)

ASA congruence rule

(c)

SSS congruence rule

(d)

AAS congruence rule.

Two triangles are congruent, if any two pairs of angles and one pair of corresponding sides are equal. This rule is known as

1

(a)

SAS congruence rule

(b)

ASA congruence rule

(c)

AAS congruence rule

(d)

SSS congruence rule

In \(\triangle\)ABC and \(\triangle\)DEF, AB=DE, \(\angle\)A=\(\angle\)D. What will be the condition in which the two triangles will be congruent by SAS axiom?

1

(a)

Since AB = DE, \(\angle\)A =\(\angle\)D and \(\triangle\)ABC\(\cong \)\(\triangle\)DEF by SAS.
Therefore AC = DF.

What do we call a triangle if the angles are in the ratio 5 : 3 : 7?

1

(a)

Let the angles of triangle are 5x, 3x and 7x, then
5x + 3x + 7x = 180°
\(\Rightarrow\) 15x = 180°
Thus, x = 12°
\(\therefore\) Angles are 60°, 36°, 84°
\(\therefore\) Each angle is less than 90°
\(\therefore\) The triangle is an acute-angled triangle.

In given fig., AD = BC and \(\angle\)BAD =\(\angle\)ABC, then prove that \(\angle\)ACB = \(\angle\)BDA.

1

(a)

AD = BC  (Given)
\(\angle\)BAD =\(\angle\)ABC  (Given)
AB = AB   (Common)
\(​​\triangle DAB\cong \triangle CBA\) (By SAS)
\(\angle\)BDA =\(\angle\)ACB    (By c.p.ct)

\(\triangle PQR\cong \triangle ABC\), if PQ = 5 cm, \(\angle\)Q = 40° and \(\angle\)P = 80°, calculate the value of \(\angle\)C.

1

(a)

\(\angle\)R= 180° - 80° - 40° = 60°
\(\triangle PQR\cong \triangle ABC\)
\(\therefore\) \(\angle\)R = \(\angle\)C = 60°

Is it possible to construct a triangle, when its sides are 5.4 cm, 2.3 cm, 3.1 cm?

1

(a)

No, Because, 2.3 + 3.1 = 5.4 cm (third side)
\(\therefore\) Not possible to construct a triangle.

In ΔABC, if ㄥA = 50° and ㄥB = 60°, determine the shortest and the longest side of the triangle.

2

(a)

In ΔABC, ㄥA = 60°, ㄥB = 40°, which side of this triangle is the smallest? Give reasons for your answer.

2

(a)

In the figure below, the diagonal AC of quadrilateral ABCD bisects \(\angle\)BAD and \(\angle\)BCD.Prove that BC = CD.

2

(a)

PS is an altitude of an isosceles triangle PQR in which PQ = PR. Show that PS bisects \(\angle\)P.

2

(a)

In the figure, given AC > AB and AD is the bisector of \(\angle\)A. Show that \(\angle\)ADC > \(\angle\)ADB.

2

(a)

ABCD is a quadrilateral in which AD = BC and \(\angle DAB=\angle CBA\) (see fihure).Prove that:

(i) \(\triangle ABD\cong \triangle BAC\)
(ii) BD = AC
(iii) \(\angle ABD=\angle BAC\)

3

(a)

l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that \(\triangle ABC\cong \triangle CDA\)

3

(a)

Show that the angles of an equilateral triangle are 60° each.

3

(a)

In a rectangle ABCD, E is a point which bisects BC Prove that AE = ED

4

(a)

Prove that the medians of an equilateral triangle are equal.

4

(a)

In the given figure, AD is the bisector of \(\angle\)BAC and \(\angle\)CPD = \(\angle\)BPD. Prove that\(\triangle CAP\cong \triangle BAP\) and CP = BP.

5

(a)