CBSE Class 7 Mathematics Congruence Of Triangles Worksheet Set 05

Question 1. If \( \Delta \text{ABC} \cong \Delta \text{DEF} \), \( \angle \text{A} = 100^\circ \) and \( \angle \text{B} = 45^\circ \) then \( \angle \text{F} = \) ________
Answer: \( 35^\circ \)
In \( \Delta \text{ABC} \), the sum of all three interior angles is \( 180^\circ \).
\( \angle \text{A} + \angle \text{B} + \angle \text{C} = 180^\circ \)
\( 100^\circ + 45^\circ + \angle \text{C} = 180^\circ \)
\( 145^\circ + \angle \text{C} = 180^\circ \)
\( \angle \text{C} = 180^\circ - 145^\circ = 35^\circ \)
Because \( \Delta \text{ABC} \cong \Delta \text{DEF} \), their corresponding angles are equal.
Therefore, \( \angle \text{F} = \angle \text{C} = 35^\circ \).
In simple words: The three angles of any triangle always add up to 180 degrees. That makes angle C equal to 35 degrees. Since both triangles are exact copies, angle F is also 35 degrees.

Exam Tip: Look at the order of letters in the congruence statement to find matching vertices. Here, C matches F, so angle C equals angle F.

Question 2. ABCD is a rhombus. AC is a diagonal
(i) Show three pairs of equal parts giving reasons, in \( \Delta \text{ABC} \) and \( \Delta \text{ADC} \).
(ii) Is \( \Delta \text{ABC} \cong \Delta \text{ADC} \) ? Give reason.
(iii) Is \( \angle \text{BAC} = \angle \text{DAC} \)? Give reason.
Answer:
(i) In \( \Delta \text{ABC} \) and \( \Delta \text{ADC} \), the three pairs of equal parts are:
- \( \text{AB} = \text{AD} \) (All four sides of a rhombus are equal in length)
- \( \text{BC} = \text{DC} \) (All four sides of a rhombus are equal in length)
- \( \text{AC} = \text{AC} \) (Common side shared by both triangles)
(ii) Yes, \( \Delta \text{ABC} \cong \Delta \text{ADC} \).
Reason: By SSS (Side-Side-Side) congruence rule, all three pairs of corresponding sides are equal.
(iii) Yes, \( \angle \text{BAC} = \angle \text{DAC} \).
Reason: Corresponding parts of congruent triangles are equal (CPCT).
In simple words: A rhombus has four equal sides, and both triangles share the middle line AC. Because all three sides match, the triangles are congruent, so the two corner angles must also be equal.

Exam Tip: When writing equal parts, always give a clear geometric reason like "sides of a rhombus" or "common side" to get full marks.

Question 3. ABC is an isosceles triangle with AB = BC and AD \( \bot \) BC.
In \( \Delta \text{ABD} \) and \( \Delta \text{ACD} \)
(i) Show three pairs of equal parts giving reasons.
(ii) Is \( \Delta \text{ADB} \cong \Delta \text{ADC} \) ? Give reason.
(iii) Is \( \angle \text{BAD} = \angle \text{CAD} \) ? Give reason.
Answer:
(i) In \( \Delta \text{ABD} \) and \( \Delta \text{ACD} \), the three pairs of equal parts are:
- \( \angle \text{ADB} = \angle \text{ADC} = 90^\circ \) (Given that \( \text{AD} \bot \text{BC} \))
- \( \text{AB} = \text{AC} \) (Hypotenuse sides marked equal in the given isosceles triangle)
- \( \text{AD} = \text{AD} \) (Common side)
(ii) Yes, \( \Delta \text{ADB} \cong \Delta \text{ADC} \).
Reason: By RHS (Right angle-Hypotenuse-Side) congruence rule.
(iii) Yes, \( \angle \text{BAD} = \angle \text{CAD} \).
Reason: Corresponding parts of congruent triangles are equal (CPCT).
In simple words: Both smaller triangles have a right angle, matching slant sides, and a shared middle side. This makes them identical by the RHS rule, so their top angles match as well.

Exam Tip: Remember that RHS applies only when you use a right angle, the hypotenuse, and one other side of the triangles.

Question 4. In the figure PQ and XY bisect each other at O.
(i) Show three pairs of equal parts in \( \Delta \text{POX} \) and \( \Delta \text{QOY} \)
(ii) Is \( \Delta \text{POX} \cong \Delta \text{QOY} \) Give reasons
(iii) Is \( \angle \text{X} = \angle \text{Y} \) ? Give reasons
Answer:
(i) In \( \Delta \text{POX} \) and \( \Delta \text{QOY} \), the three pairs of equal parts are:
- \( \text{PO} = \text{QO} \) (PQ is bisected at point O)
- \( \angle \text{POX} = \angle \text{QOY} \) (Vertically opposite angles are equal)
- \( \text{XO} = \text{YO} \) (XY is bisected at point O)
(ii) Yes, \( \Delta \text{POX} \cong \Delta \text{QOY} \).
Reason: By SAS (Side-Angle-Side) congruence rule, two sides and their included angle are equal in both triangles.
(iii) Yes, \( \angle \text{X} = \angle \text{Y} \).
Reason: Corresponding parts of congruent triangles are equal (CPCT).
In simple words: Because the two lines cut each other in half, the arms are equal. The angles across the crossing point are also equal, so the SAS rule proves both triangles are identical.

Exam Tip: Be careful to state the vertically opposite angle between the two equal sides to justify the SAS rule correctly.

Question 5. In the figure, O is the midpoint of BC and \( \angle \text{B} = 90^\circ \), \( \angle \text{C} = 90^\circ \)
By using ASA Congruence rule
Show that \( \Delta \text{AOB} \cong \Delta \text{DOC} \)
Answer:
In \( \Delta \text{AOB} \) and \( \Delta \text{DOC} \):
- \( \angle \text{ABO} = \angle \text{DCO} = 90^\circ \) (Given)
- \( \text{BO} = \text{CO} \) (Given that O is the midpoint of segment BC)
- \( \angle \text{AOB} = \angle \text{DOC} \) (Vertically opposite angles are equal)
Thus, by the ASA (Angle-Side-Angle) congruence rule:
\( \Delta \text{AOB} \cong \Delta \text{DOC} \).
In simple words: The two triangles share equal 90-degree angles and equal crossing angles. Since the side between these two angles has the same length in both triangles, they are congruent by ASA.

Exam Tip: For ASA, the equal side must lie between the two chosen angles. Make sure to list the angle-side-angle in that exact order.

Question 6. ABCD is a rectangle. AC is a diagonal (Draw a figure). By using SSS Congruence rule Show that \( \Delta \text{ABC} \cong \Delta \text{CDA} \)
Answer:
In a rectangle, opposite sides are always equal in length.
In \( \Delta \text{ABC} \) and \( \Delta \text{CDA} \):
- \( \text{AB} = \text{CD} \) (Opposite sides of rectangle ABCD are equal)
- \( \text{BC} = \text{DA} \) (Opposite sides of rectangle ABCD are equal)
- \( \text{AC} = \text{CA} \) (Common side)
Therefore, by SSS (Side-Side-Side) congruence rule:
\( \Delta \text{ABC} \cong \Delta \text{CDA} \).
In simple words: The opposite sides of a rectangle are equal, and the diagonal line is shared by both triangles. Since all three sides are equal, the triangles are congruent by SSS.

Exam Tip: Remember that a diagonal divides any rectangle into two congruent triangles. Write the vertices in corresponding order (ABC to CDA).

Question 7. In the figure AD \( \bot \) BC . D is the midpoint of BC
Using SAS Congruence rule show that
\( \Delta \text{ABD} \cong \Delta \text{ACD} \)
Is AB = AC ? Why?
Answer:
In \( \Delta \text{ABD} \) and \( \Delta \text{ACD} \):
- \( \text{BD} = \text{CD} \) (Given that D is the midpoint of BC)
- \( \angle \text{ADB} = \angle \text{ADC} = 90^\circ \) (Given that \( \text{AD} \bot \text{BC} \))
- \( \text{AD} = \text{AD} \) (Common side)
Therefore, by SAS (Side-Angle-Side) congruence rule:
\( \Delta \text{ABD} \cong \Delta \text{ACD} \).
Yes, \( \text{AB} = \text{AC} \).
Reason: Corresponding parts of congruent triangles are equal (CPCT).
In simple words: The two parts BD and CD are equal, the angle between them is 90 degrees on both sides, and AD is shared. This proves congruence by SAS, meaning the outside edges AB and AC must also be equal.

Exam Tip: Whenever a question asks "Why?" for corresponding sides after proving congruence, write "CPCT" (Corresponding Parts of Congruent Triangles).

Question 8. Given the figure,
Prove that \( \Delta \text{ABC} \cong \Delta \text{RQP} \)
If \( \angle \text{A} = 30^\circ \), \( \angle \text{R} = 30^\circ \)
Answer:
In right-angled triangle ABC:
\( \angle \text{B} = 90^\circ \) and \( \angle \text{A} = 30^\circ \).
Using the angle sum property of a triangle:
\( \angle \text{C} = 180^\circ - (90^\circ + 30^\circ) = 180^\circ - 120^\circ = 60^\circ \).
In right-angled triangle RQP:
\( \angle \text{Q} = 90^\circ \) and \( \angle \text{R} = 30^\circ \).
Using the angle sum property of a triangle:
\( \angle \text{P} = 180^\circ - (90^\circ + 30^\circ) = 180^\circ - 120^\circ = 60^\circ \).
Now, compare \( \Delta \text{ABC} \) and \( \Delta \text{RQP} \):
- \( \angle \text{B} = \angle \text{Q} = 90^\circ \) (Both are right angles)
- \( \text{BC} = \text{QP} \) (Given, indicated by equal tick marks)
- \( \angle \text{C} = \angle \text{P} = 60^\circ \) (Proved above)
Therefore, by the ASA (Angle-Side-Angle) congruence rule:
\( \Delta \text{ABC} \cong \Delta \text{RQP} \).
(Note: You can also use the AAS congruence rule directly using \( \angle \text{A} = \angle \text{R} = 30^\circ \), \( \angle \text{B} = \angle \text{Q} = 90^\circ \), and \( \text{BC} = \text{QP} \).)
In simple words: First find the third angle in each triangle, which comes out to 60 degrees. Both triangles have equal angles of 90 degrees and 60 degrees with the same side length in between, so they are congruent by ASA.

Exam Tip: When given two angles and a non-included side, find the third angle first using angle sum property, or apply the AAS congruence criterion directly.