CBSE Class 7 Mathematics Congruence of Lines and Angles Assignment Set 02

VII - Mathematics Assignment No-05 - Congruence of Triangles

Multiple Choice Questions (M.C.Q)

Select the correct one from the multiple answers.

Question 1. If two figures are congruent then they have
(i) Same shape
(ii) Same size
(iii) Same angles
(iv) Same shape and same size
Answer: (iv) Same shape and same size
Two geometric figures are congruent if they have exactly the same shape and the same size, meaning one can superimpose on the other so that they coincide entirely.
In simple words: Congruent figures are identical in both shape and size.

Exam Tip: Remember that congruence means exact equality in all dimensions and shape.

Question 2. Two triangles are not congruent by
(i) SSS rule
(ii) SAS rule
(iii) AAA rule
(iv) ASA rule
Answer: (iii) AAA rule
AAA (Angle-Angle-Angle) guarantees similarity of triangles, not congruence, because two triangles can have the same corresponding angles while having different sizes (lengths of sides).
In simple words: AAA only proves that triangles are similar, not congruent.

Exam Tip: Side length information is always required to prove triangle congruence.

Question 3. Two line segments are congruent if they have
(i) Same direction
(ii) Same length
(iii) Opposite direction
(iv) None of the options
Answer: (ii) Same length
Two line segments are congruent if and only if they are equal in length, regardless of their orientation or position in the plane.
In simple words: Line segments with equal lengths are congruent.

Exam Tip: Direction and position do not affect line segment congruence.

Question 4. The method of comparing two figures is known as
(i) Superposition
(ii) Congruency
(iii) Equality
(iv) None of the options
Answer: (i) Superposition
Superposition is the method of placing one figure over another to check if they coincide completely.
In simple words: Superposition means placing one figure on top of another to test for congruence.

Exam Tip: Superposition is the fundamental physical test for congruence.

Question 5. If two triangles are congruent, their area is
(i) Unequal
(ii) May be equal or unequal
(iii) Always negative
(iv) Equal
Answer: (iv) Equal
Since congruent triangles have the exact same shape and size, they occupy the exact same amount of space, and thus their areas are always equal.
In simple words: Congruent triangles always have equal areas.

Exam Tip: Equal area does not necessarily imply congruence, but congruence always implies equal area.

Question 6. Two circles are congruent if they have
(i) Same centre
(ii) Different radius
(iii) Same radius
(iv) Same centre but different radius
Answer: (iii) Same radius
Two circles are congruent if and only if their radii are equal, which means they have the same size.
In simple words: Circles with identical radii are congruent.

Question 7. Two squares are congruent if they have
(i) Same angles
(ii) Same side length
(iii) Opposite sides are equal
(iv) Diagonals are equal
Answer: (ii) Same side length
All squares naturally have right angles, so having the same side length is both necessary and sufficient to make two squares congruent.
In simple words: Squares with equal side lengths are congruent.

Exam Tip: For regular polygons and squares, side length alone determines congruence.

Question 8. If the area of two figures are same then the figures
(i) Need not be congruent
(ii) Always congruent
(iii) Equal
(iv) None of the options
Answer: (i) Need not be congruent
Two figures can have the same area (e.g., a rectangle of area 12 and a triangle of area 12) without having the same shape or being congruent.
In simple words: Equal areas do not guarantee congruent shapes.

Exam Tip: Do not confuse area equality with shape congruence.

Question 9. In an isosceles triangle, how many sides are always equal
(i) None
(ii) Only one side
(iii) All the three sides
(iv) Any two sides
Answer: (iv) Any two sides
By definition, an isosceles triangle has at least two sides of equal length.
In simple words: An isosceles triangle always has two equal sides.

Exam Tip: The angles opposite the equal sides are also equal.

Question 10. In an equilateral triangle, the measure of each angle is
(i) \(180^\circ\)
(ii) \(90^\circ\)
(iii) \(60^\circ\)
(iv) \(30^\circ\)
Answer: (iii) \(60^\circ\)
An equilateral triangle has three equal sides and three equal interior angles. Since the sum of angles in a triangle is \(180^\circ\), each angle equals \(\frac{180^\circ}{3} = 60^\circ\).
In simple words: All three angles of an equilateral triangle are equal to \(60^\circ\).

Exam Tip: Equilateral triangles are also equiangular.

Question 11. RHS Congruency rule states
(i) Two legs are equal
(ii) One-hypotenuse and one leg
(iii) Two legs and one angle
(iv) One leg - one angle - hypotenuse
Answer: (ii) One-hypotenuse and one leg
RHS stands for Right angle-Hypotenuse-Side. Two right-angled triangles are congruent if the hypotenuse and one side (leg) of one triangle are respectively equal to the hypotenuse and one side of the other triangle.
In simple words: RHS rule requires a right angle, equal hypotenuses, and equal legs.

Exam Tip: This rule applies exclusively to right-angled triangles.

Question 12. In two triangles \(ABC\) and \(PQR\): \(AB = 5\text{ cm}\), \(BC = 4\text{ cm}\), \(AC = 3\text{ cm}\), \(PQ = 3\text{ cm}\), \(QR = 4\text{ cm}\) and \(PR = 5\text{ cm}\). Write the congruence relation in symbolic form
(i) \(\Delta ABC \cong \Delta PQR\)
(ii) \(\Delta ABC \cong \Delta PRQ\)
(iii) \(\Delta ABC \cong \Delta QPR\)
(iv) \(\Delta ABC \cong \Delta RPQ\)
Answer: (ii) \(\Delta ABC \cong \Delta PRQ\)
Matching the corresponding sides based on their lengths:
\(AB = 5\text{ cm}\) corresponds to \(PR = 5\text{ cm}\)
\(BC = 4\text{ cm}\) corresponds to \(RQ = 4\text{ cm}\)
\(AC = 3\text{ cm}\) corresponds to \(PQ = 3\text{ cm}\)
Thus, vertex correspondence is \(A \leftrightarrow P\), \(B \leftrightarrow R\), and \(C \leftrightarrow Q\). Therefore, \(\Delta ABC \cong \Delta PRQ\).
In simple words: Match corresponding sides with equal lengths to write the correct vertex order.

Exam Tip: Order of vertices in congruence is critical; corresponding parts must align.

Question 13. To apply RHS congruence rule for two triangles \(\Delta ABC \cong \Delta RPQ\). What additional information is needed, if it is given that \(\angle B = \angle P = 90^\circ\) and \(AB = RP\)?
(i) \(\angle A = \angle R\)
(ii) \(\angle C = \angle R\)
(iii) \(AB = PQ\)
(iv) \(AC = RQ\)
Answer: (iv) \(AC = RQ\)
Given right angles at \(B\) and \(P\), the hypotenuses are \(AC\) (opposite to \(B\)) and \(RQ\) (opposite to \(P\)). For RHS, we need the hypotenuses to be equal along with one leg, so \(AC = RQ\) is required.
In simple words: RHS needs the hypotenuses of both triangles to be equal.

Exam Tip: Always identify the hypotenuse as the side opposite the \(90^\circ\) angle.

Question 14. In a \(\Delta ABC\), it is given that \(D\) is the mid-point of side \(BC\) and \(AD \perp BC\). Name the triangles which are congruent.
(i) \(\Delta ABC \cong \Delta ACB\)
(ii) \(\Delta ABD \cong \Delta ADC\)
(iii) \(\Delta ADB \cong \Delta ADC\)
(iv) \(\Delta BDA \cong \Delta CAD\)
Answer: (iii) \(\Delta ADB \cong \Delta ADC\)
In triangles \(ADB\) and \(ADC\):
\(BD = CD\) (\(D\) is mid-point)
\(\angle ADB = \angle ADC = 90^\circ\) (\(AD \perp BC\))
\(AD = AD\) (Common side)
By SAS criterion, \(\Delta ADB \cong \Delta ADC\).
In simple words: The altitude from the vertex to the midpoint divides the triangle into two congruent right triangles.

Exam Tip: Use SAS or RHS criterion for triangles formed by an altitude on a base.

Question 15. ABC is an isosceles triangle such that \(AB = AC\). \(D\) is any point on \(BC\), \(AD\) is joined. Give the name to the segment \(AD\): \(\Delta ABD \cong \Delta ACD\)
(i) \(AD \perp BC\)
(ii) \(AD\) is the angle bisector of \(\angle A\)
(iii) \(AD\) is the median
(iv) None of the options
Answer: (ii) \(AD\) is the angle bisector of \(\angle A\)
Given \(AB = AC\) and \(\Delta ABD \cong \Delta ACD\), by CPCT (Corresponding Parts of Congruent Triangles), the corresponding angles \(\angle BAD = \angle CAD\), which proves that \(AD\) bisects \(\angle A\).
In simple words: Congruency of the divided triangles shows \(AD\) bisects the vertex angle.

Exam Tip: In an isosceles triangle, the line dividing the base symmetrically also bisects the vertical angle.