CBSE Class 7 Mathematics Congruence Of Triangles Worksheet Set 04

Question 1. In the given figure, AD = CD and AB = CB
(a) State the three pairs of equal sides in \( \Delta \text{ABD} \) and \( \Delta \text{CBD} \).
(b) Is \( \Delta \text{ABD} \cong \Delta \text{CBD} \)? Why?
(c) Does BD bisect \( \angle \text{ABC} \)? Give reasons.
Answer:
(a) In \( \Delta \text{ABD} \) and \( \Delta \text{CBD} \), the three pairs of equal sides are:
- \( \text{AD} = \text{CD} \) (Given)
- \( \text{AB} = \text{CB} \) (Given)
- \( \text{BD} = \text{BD} \) (Common side shared by both triangles)
(b) Yes, \( \Delta \text{ABD} \cong \Delta \text{CBD} \).
Reason: By SSS (Side-Side-Side) congruence rule, all three pairs of corresponding sides are equal.
(c) Yes, BD bisects \( \angle \text{ABC} \).
Reason: Since \( \Delta \text{ABD} \cong \Delta \text{CBD} \), their corresponding angles are equal (CPCT). Therefore, \( \angle \text{ABD} = \angle \text{CBD} \), meaning line BD divides \( \angle \text{ABC} \) into two equal halves.
In simple words: The two triangles have all three sides matching. This proves they are identical, which means line BD splits the bottom angle into two equal parts.

Exam Tip: Always identify the shared side as "common side" and mention CPCT when showing that two angles are equal after proving congruence.

Question 2. In \( \Delta \text{PQR} \) and \( \Delta \text{XYZ} \), PQ = XZ and QR = YZ. What additional information is required to make the two triangles congruent by SSS congruence criterion?
Answer: Under the SSS criterion, all three corresponding sides of two triangles must be equal in length.
We are already given two pairs of matching sides:
- \( \text{PQ} = \text{XZ} \)
- \( \text{QR} = \text{YZ} \)
The remaining third sides of the two triangles are PR and XY.
Therefore, the additional information required is:
\( \text{PR} = \text{XY} \).
In simple words: The SSS rule needs all three sides to match. Since two sides already match, the third sides PR and XY must also be equal.

Exam Tip: For SSS questions, match the side names carefully using the remaining pair of vertices from each triangle.

Question 3. If \( \Delta \text{PQR} \) is an isosceles triangle such that PQ = PR, then prove that the altitude PS from P on QR bisects QR.
Answer: Given that in \( \Delta \text{PQR} \), \( \text{PQ} = \text{PR} \), and PS is an altitude to side QR, meaning \( \text{PS} \bot \text{QR} \) and \( \angle \text{PSQ} = \angle \text{PSR} = 90^\circ \).
To prove: PS bisects QR, so \( \text{QS} = \text{RS} \).
In right triangles \( \Delta \text{PSQ} \) and \( \Delta \text{PSR} \):
- \( \angle \text{PSQ} = \angle \text{PSR} = 90^\circ \) (Altitude PS is perpendicular to QR)
- \( \text{PQ} = \text{PR} \) (Hypotenuse sides given equal)
- \( \text{PS} = \text{PS} \) (Common side)
By the RHS congruence rule:
\( \Delta \text{PSQ} \cong \Delta \text{PSR} \).
Since corresponding parts of congruent triangles are equal (CPCT):
\( \text{QS} = \text{RS} \).
Hence, the altitude PS bisects QR.
In simple words: The height line makes two right triangles with equal hypotenuses and a shared height. By the RHS rule, the base is cut cleanly into two equal parts.

Exam Tip: Be sure to state that the triangles are right-angled before citing the RHS rule, highlighting both the hypotenuse and the common leg.

Question 4. In the given figure, AB \( \bot \) QR, AC \( \bot \) QP and QC = QB.
(a) Is \( \Delta \text{AQB} \cong \Delta \text{AQC} \)?
Give reasons in support of your answer.
(b) Which angle is equal to \( \angle \text{AQB} \)?
Answer:
(a) Yes, \( \Delta \text{AQB} \cong \Delta \text{AQC} \).
Reason: In right-angled triangles \( \Delta \text{AQB} \) and \( \Delta \text{AQC} \):
- \( \angle \text{ABQ} = \angle \text{ACQ} = 90^\circ \) (Given \( \text{AB} \bot \text{QR} \) and \( \text{AC} \bot \text{QP} \))
- \( \text{QA} = \text{QA} \) (Common hypotenuse)
- \( \text{QB} = \text{QC} \) (Given)
Therefore, by RHS congruence rule, \( \Delta \text{AQB} \cong \Delta \text{AQC} \).
(b) Angle equal to \( \angle \text{AQB} \):
Since \( \Delta \text{AQB} \cong \Delta \text{AQC} \), corresponding angles are equal (CPCT).
Therefore, \( \angle \text{AQC} = \angle \text{AQB} \).
In simple words: The two triangles have 90-degree corners, equal base sides, and the same hypotenuse QA. This makes them congruent by RHS, so angle AQB equals angle AQC.

Exam Tip: Notice that segment QA serves as the common hypotenuse because it lies directly opposite the 90-degree angles in both triangles.

Question 5. In the given figure, AB || DC and AB = DC.
(a) Is \( \angle \text{BAC} = \angle \text{DCA} \)? Why?
(b) Is \( \Delta \text{ABC} \cong \Delta \text{CDA} \)?
Answer:
(a) Yes, \( \angle \text{BAC} = \angle \text{DCA} \).
Reason: Since \( \text{AB} \parallel \text{DC} \) and AC acts as a transversal line intersecting them, alternate interior angles are equal.
(b) Yes, \( \Delta \text{ABC} \cong \Delta \text{CDA} \).
In \( \Delta \text{ABC} \) and \( \Delta \text{CDA} \):
- \( \text{AB} = \text{CD} \) (Given)
- \( \angle \text{BAC} = \angle \text{DCA} \) (Alternate interior angles)
- \( \text{AC} = \text{CA} \) (Common side)
Therefore, by SAS (Side-Angle-Side) congruence rule, \( \Delta \text{ABC} \cong \Delta \text{CDA} \).
In simple words: The parallel lines make alternate angles equal across diagonal AC. With matching sides AB and CD and the shared diagonal, the SAS rule proves both triangles are congruent.

Exam Tip: Specify "alternate interior angles" clearly when two parallel lines are cut by a transversal.

Question 6. In the given figure, AB = AC and D is the mid-point of BC.
(a) State the three pairs of equal sides.
(b) Is \( \Delta \text{ADB} \cong \Delta \text{ADC} \)? Give reasons.
Answer:
(a) In \( \Delta \text{ADB} \) and \( \Delta \text{ADC} \), the three pairs of equal sides are:
- \( \text{AB} = \text{AC} \) (Given)
- \( \text{BD} = \text{CD} \) (Given that D is the midpoint of segment BC)
- \( \text{AD} = \text{AD} \) (Common side)
(b) Yes, \( \Delta \text{ADB} \cong \Delta \text{ADC} \).
Reason: By SSS (Side-Side-Side) congruence rule, because all three corresponding sides of the two triangles are equal.
In simple words: The two outer sides are equal, the bottom is divided equally at point D, and line AD is shared. Since all three sides match, the triangles are congruent by SSS.

Exam Tip: Remember that a midpoint divides a line segment into two equal segments, giving you one pair of equal sides immediately.

Exercise 7.2

Question 1. In \( \Delta \text{ABC} \) and \( \Delta \text{DEF} \), AB = DE and BC = EF. What additional information is required to make the two triangles congruent by SAS congruence criterion?
Answer: Under the SAS criterion, the two matching sides must have their included angle equal.
In \( \Delta \text{ABC} \), the angle formed between sides AB and BC is \( \angle \text{B} \) (or \( \angle \text{ABC} \)).
In \( \Delta \text{DEF} \), the angle formed between sides DE and EF is \( \angle \text{E} \) (or \( \angle \text{DEF} \)).
Therefore, the additional information required is:
\( \angle \text{B} = \angle \text{E} \) (or \( \angle \text{ABC} = \angle \text{DEF} \)).
In simple words: The SAS rule needs the angle directly between the two given sides. That means the angle at corner B must match the angle at corner E.

Exam Tip: Never pick any random angle for SAS - it must always be the "included angle" trapped right between the two equal sides.

Question 2. In the given figure, PX and QY are perpendicular to PQ and PX = QY. Show that AX = AY, using AAS congruence criterion.
Answer: In \( \Delta \text{PAX} \) and \( \Delta \text{QAY} \):
- \( \angle \text{XPA} = \angle \text{YQA} = 90^\circ \) (Given that \( \text{PX} \bot \text{PQ} \) and \( \text{QY} \bot \text{PQ} \))
- \( \angle \text{PAX} = \angle \text{QAY} \) (Vertically opposite angles are equal)
- \( \text{PX} = \text{QY} \) (Given)
Therefore, by AAS (Angle-Angle-Side) congruence rule:
\( \Delta \text{PAX} \cong \Delta \text{QAY} \).
Since corresponding parts of congruent triangles are equal (CPCT):
\( \text{AX} = \text{AY} \).
In simple words: The two triangles share right angles, matching crossing angles, and equal outer vertical segments. By the AAS rule, both triangles match completely, so AX equals AY.

Exam Tip: Make sure the side in AAS is a non-included side. Here, side PX is not between the two angles, so AAS is the correct criterion.

Question 3. In the given figure, \( \Delta \text{ABC} \) is a right-angled triangle in which \( \angle \text{B} = 90^\circ \) and D is the mid-point of AC.
Prove that BC = AB
[Hint: Prove that \( \Delta \text{ABD} \cong \Delta \text{CBD} \), using SAS congruence criterion]
Answer: Given that D is the midpoint of AC, so \( \text{AD} = \text{CD} \), and segment BD is perpendicular to AC, meaning \( \angle \text{ADB} = \angle \text{CDB} = 90^\circ \).
In \( \Delta \text{ABD} \) and \( \Delta \text{CBD} \):
- \( \text{AD} = \text{CD} \) (Given that D is the midpoint of AC)
- \( \angle \text{ADB} = \angle \text{CDB} = 90^\circ \) (BD is perpendicular to AC)
- \( \text{BD} = \text{BD} \) (Common side)
Therefore, by SAS (Side-Angle-Side) congruence rule:
\( \Delta \text{ABD} \cong \Delta \text{CBD} \).
Since corresponding parts of congruent triangles are equal (CPCT):
\( \text{BC} = \text{AB} \).
In simple words: Line BD divides the large triangle into two smaller matching ones with a shared side and right angles. Since the smaller triangles are congruent, side AB must equal side BC.

Exam Tip: Follow the given hint step by step: prove the two inner triangles congruent first using SAS, then deduce the required sides using CPCT.

Question 4. In the given figure, AB || to CD. If O is the midpoint of BC, show that O is also the midpoint of AD.
Answer: Given that \( \text{AB} \parallel \text{CD} \) and O is the midpoint of BC, so \( \text{OB} = \text{OC} \).
In \( \Delta \text{AOB} \) and \( \Delta \text{DOC} \):
- \( \angle \text{ABO} = \angle \text{DCO} \) (Alternate interior angles, since \( \text{AB} \parallel \text{CD} \) and BC is transversal)
- \( \text{OB} = \text{OC} \) (Given that O is the midpoint of BC)
- \( \angle \text{AOB} = \angle \text{DOC} \) (Vertically opposite angles)
Therefore, by ASA (Angle-Side-Angle) congruence rule:
\( \Delta \text{AOB} \cong \Delta \text{DOC} \).
Since corresponding parts of congruent triangles are equal (CPCT):
\( \text{OA} = \text{OD} \).
Because OA equals OD, point O is also the midpoint of AD.
In simple words: The parallel lines create equal alternate angles, and the crossing lines give equal vertical angles. By ASA, both triangles are identical, proving that O sits right in the middle of line AD.

Exam Tip: Showing that a point is a midpoint simply means proving the two segments it forms on that line are equal via CPCT.

Question 5. Prove that the three angles of an equilateral triangle are equal.
Answer: Let ABC be an equilateral triangle.
By definition of an equilateral triangle, all three sides are equal:
\( \text{AB} = \text{BC} = \text{CA} \).
1. Since \( \text{AB} = \text{AC} \), the angles opposite to these sides are equal:
\( \angle \text{C} = \angle \text{B} \) ... (Equation 1)
2. Since \( \text{AB} = \text{BC} \), the angles opposite to these sides are equal:
\( \angle \text{C} = \angle \text{A} \) ... (Equation 2)
Comparing Equation 1 and Equation 2:
\( \angle \text{A} = \angle \text{B} = \angle \text{C} \).
Because the sum of all three angles in any triangle is \( 180^\circ \):
\( \angle \text{A} + \angle \text{B} + \angle \text{C} = 180^\circ \)
\( 3 \angle \text{A} = 180^\circ \)
\( \angle \text{A} = 60^\circ \).
Therefore, \( \angle \text{A} = \angle \text{B} = \angle \text{C} = 60^\circ \).
Hence, all three angles of an equilateral triangle are equal.
In simple words: If all sides of a triangle are equal, then all the opposite angles must also be equal. This means every angle in an equilateral triangle is 60 degrees.

Exam Tip: Use the theorem "angles opposite to equal sides of a triangle are equal" twice to show that all three angles are identical.

Question 6. In the given figure, \( \Delta \text{ABC} \) is an isosceles triangle in which AB = AC. If BD \( \bot \) AC and CE \( \bot \) AB, prove that BD = CE.
Answer: In \( \Delta \text{BDC} \) and \( \Delta \text{CEB} \):
- \( \angle \text{BDC} = \angle \text{CEB} = 90^\circ \) (Given that \( \text{BD} \bot \text{AC} \) and \( \text{CE} \bot \text{AB} \))
- \( \angle \text{DCB} = \angle \text{EBC} \) (Angles opposite to equal sides AB and AC in isosceles \( \Delta \text{ABC} \))
- \( \text{BC} = \text{CB} \) (Common side)
Therefore, by AAS (Angle-Angle-Side) congruence rule:
\( \Delta \text{BDC} \cong \Delta \text{CEB} \).
Since corresponding parts of congruent triangles are equal (CPCT):
\( \text{BD} = \text{CE} \).
In simple words: The two triangles share the base BC, have 90-degree corners, and have matching base angles. By the AAS rule, they are congruent, meaning altitudes BD and CE are equal.

Exam Tip: Altitudes drawn to the equal sides of an isosceles triangle are always equal in length. Mention the base angles are equal due to AB = AC.

Question 7. In the given figure, the triangles are congruent. Find the values of x and y.
Answer: We are given that \( \Delta \text{ABC} \cong \Delta \text{PQR} \).
Since corresponding sides and corresponding angles of congruent triangles are equal (CPCT):
1. Comparing corresponding angles at vertices B and Q:
\( \angle \text{B} = \angle \text{Q} \)
\( x - 8 = 65 \)
\( x = 65 + 8 \)
\( x = 73 \)
2. Comparing corresponding angles at vertices C and R:
\( \angle \text{C} = \angle \text{R} \)
\( 25 = y + 3 \)
\( y = 25 - 3 \)
\( y = 22 \)
Thus, the required values are \( x = 73 \) and \( y = 22 \).
In simple words: Since both triangles are congruent, matching angles are equal. Setting the matching corners equal gives x = 73 and y = 22.

Exam Tip: Match the corresponding vertices properly: B corresponds to Q, and C corresponds to R, then set up simple linear equations to solve for x and y.