---
title: "NCERT Solutions for Class 7 Maths Chapter 9 Geometric Twins"
url: https://www.swavid.com/maths/class/7/chapter/geometric-twins/ncert-solutions
dateModified: 2026-10-07T15:09:01+00:00
---

# NCERT Solutions for Class 7 Maths Chapter 9 Geometric Twins

This chapter explores the concept of geometric congruence through various shapes and triangles. It guides students to identify and verify congruent figures using side lengths, angles, and specific geometric conditions.

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## 1.1 Geometric Twins

### Question 1

*3 marks · Short answer*

How do we do it?

**Solution**

1. One way is to trace the outline of the symbol on tracing paper to reconstruct the figure.
2. Tracing is difficult for big symbols.
3. We need to take proper measurements to recreate the figure.

**Answer:** We can either trace the outline of the symbol or take suitable measurements to recreate it.

> Common mistake: Thinking that tracing paper is the only method available.

### Question 2

*3 marks · Short answer*

Can we take some measurements that would allow us to exactly recreate this figure? If yes, what measurements should we take?

**Solution**

1. Yes, we can take measurements to exactly recreate the figure.
2. We need the lengths of the arms and the measure of the angle between them to fix the shape and size.

**Answer:** Yes, we can take measurements of the arm lengths and the included angle to recreate the figure.

> Common mistake: Stating that only side lengths are sufficient.

### Question 3

*3 marks · Short answer*

Are the arm lengths AB and BC sufficient to exactly recreate this figure?

**Solution**

1. No, the arm lengths AB and BC alone are not sufficient to exactly recreate this figure.
2. With the same arm lengths, we can construct several non-congruent symbols by changing the angle between the arms.

**Answer:** No, arm lengths AB and BC are not sufficient because different angles can be formed with the same arm lengths.

> Common mistake: Assuming that knowing side lengths alone determines a multi-segment shape.

### Question 4

*3 marks · Short answer*

To get the exact replica, would it help to take any other measurement?

**Solution**

1. Yes, it would help to take the measure of the angle between the two arms.
2. Fixing the angle along with the arm lengths determines the exact shape and size of the figure.

**Answer:** Yes, measuring the angle between the arms helps to get the exact replica.

> Common mistake: Forgetting that angle is required along with side lengths.

### Question 5

*3 marks · Short answer*

Can you draw the symbol if it is known that AB = 4 cm, BC = 8 cm, and ∠ABC = 80°?

**Solution**

1. Yes, we can draw the symbol using the given measurements.
2. We draw line segment AB = 4 cm, make an angle of $80^\circ$ at B, and cut an arc of length BC = 8 cm.

**Answer:** Yes, we can draw the symbol using AB = 4 cm, BC = 8 cm, and $\angle ABC = 80^\circ$.

> Common mistake: Drawing the given angle at the wrong vertex.

### Question 6

*3 marks · Short answer*

If it is known that both symbols have the same arm lengths, can it be concluded that the two symbols are congruent?

**Solution**

1. No, it cannot be concluded that the two symbols are congruent.
2. Several non-congruent figures can exist with the same arm lengths if the angles between them are different.

**Answer:** No, same arm lengths do not guarantee congruence unless the angle between them is also equal.

> Common mistake: Confusing same side lengths with congruence of general open figures.

## Figure it Out

### Question 1

*3 marks · Short answer*

Check if the two figures are congruent.

**Solution**

1. Figures that are exact copies of each other or have the same shape and size are said to be congruent.
2. A figure can be rotated or flipped to check for congruence by superimposition.
3. Upon careful observation of the two given figures on page 3, they do not have the same size and shape, so they cannot be superimposed exactly. Hence, the two figures are not congruent.

**Answer:** No, the two figures are not congruent.

> Common mistake: Assuming figures that look somewhat similar are congruent without checking if they can be superimposed after rotation or flipping.

### Question 2

*3 marks · Short answer*

Circle the pairs that appear congruent.

**Solution**

1. Examine all the given pairs of figures in the question figure.
2. Compare their shapes, sizes, and proportions visually or by tracing.
3. Identify and circle the pairs that have the exact same shape and size.

**Answer:** The pairs of figures that have identical shape and size are circled.

> Common mistake: Confusing similar shapes of different sizes with congruent figures.

### Question 3

*3 marks · Short answer*

What measurements would you take to create a figure congruent to a given: (a) Circle (b) Rectangle

**Part (a)**

1. A circle is completely determined by its radius.
2. Measure the radius of the given circle to create an exact copy.

Answer (a): Radius or diameter

**Part (b)**

1. A rectangle is completely determined by the lengths of its adjacent sides.
2. Measure the length and breadth of the given rectangle to create an exact copy.

Answer (b): Length and breadth

**Answer:** (a) Radius or diameter (b) Length and breadth

> Common mistake: Forgetting that a rectangle requires two independent measurements (length and breadth), whereas a circle requires only one.

## 1.1 Geometric Twins (continued)

### Question 1

*3 marks · Short answer*

Using this, state how would you check if two — (a) Circles are congruent? (b) Rectangles are congruent?

**Part (a)**

1. Measure the radii or diameters of the two circles.
2. If the radii of both circles are equal, they can be superimposed to fit each other exactly.
3. Hence, two circles are congruent if and only if their radii are equal.

Answer (a): Two circles are congruent if they have equal radii.

**Part (b)**

1. Measure the length and breadth of both rectangles.
2. If the length of one rectangle equals the length of the other, and their breadths are also equal, they coincide completely.
3. Hence, two rectangles are congruent if their lengths and breadths are equal respectively.

Answer (b): Two rectangles are congruent if their lengths and breadths are equal respectively.

**Answer:** Two circles are congruent if they have equal radii, and two rectangles are congruent if their lengths and breadths are equal.

> Common mistake: Stating that circles or rectangles are congruent without specifying that all corresponding dimensions (radius or length and breadth) must be equal.

### Question 4

*3 marks · Short answer*

How would we check if two figures like the one below are congruent?

**Solution**

1. To check if two figures like the given symbol are congruent, we must ensure that all their corresponding arm lengths are equal.
2. We also need to measure the angle between the arms, as fixing the arm lengths and the included angle uniquely determines the shape and size of the figure.
3. If all corresponding arm lengths and the angle between them are equal, the two figures are congruent and can be superimposed exactly.

**Answer:** We check congruence by measuring the arm lengths and the angle between them; if both arm lengths and the included angle are equal, the figures are congruent.

> Common mistake: Checking only the arm lengths while forgetting to verify the angle between the arms.

### Question 3

*3 marks · Short answer*

What do you think they can do?

**Solution**

1. The frame is too big to be traced directly onto a paper using tracing paper.
2. Instead of tracing, the girls can measure certain dimensions of the triangular frame, such as the lengths of its three sides using a measuring tape.
3. These measured sidelengths can then be used to construct an exact replica or a scale model of the triangle on paper.

**Answer:** They can measure the sidelengths of the triangular frame using a measuring tape to construct an identical triangle.

> Common mistake: Thinking that the physical object must always be traced directly even when it is too large.

### Question 4

*3 marks · Short answer*

Do you agree with Meera?

**Solution**

1. Yes, we agree with Meera.
2. Meera states that the three sidelengths of a triangle are sufficient to create a triangle congruent to the given one.
3. This corresponds to the SSS (Side Side Side) condition for congruence, which guarantees that triangles with the same sidelengths are congruent.

**Answer:** Yes, we agree with Meera because the three sidelengths are sufficient to construct a congruent triangle by the SSS condition.

> Common mistake: Believing that angle measurements are also strictly necessary to construct a congruent triangle.

### Question 5

*3 marks · Short answer*

Instead of the lengths being 40 cm, 60 cm, and 80 cm, suppose the sidelengths had been 4 cm, 6 cm, 8 cm (this triangle can fit on our page). Is this information sufficient to replicate the triangle with the same size and shape? If yes, can you do so?

**Solution**

1. Yes, this information is sufficient to replicate the triangle with the same size and shape.
2. According to the SSS condition, knowing all three sidelengths completely determines a unique triangle up to congruence.
3. We can construct it by drawing a line segment of one given length and using compasses with radii equal to the other two lengths to find the third vertex.

**Answer:** Yes, the information is sufficient, and the triangle can be constructed using a ruler and compasses by applying the SSS condition.

> Common mistake: Assuming that side lengths alone cannot uniquely fix the shape of a triangle.

### Question 6

*3 marks · Short answer*

Examine whether ΔABE and ΔABF are congruent.

**Solution**

1. Given the construction, the side $AB$ is common to both $\Delta ABE$ and $\Delta ABF$.
2. The other two sides are equal by construction ($AE = AF$ and $BE = BF$, being radii of equal circles or arcs).
3. Since all three pairs of corresponding sides are equal, by the SSS condition, $\Delta ABE$ and $\Delta ABF$ are congruent.

**Answer:** $\Delta ABE$ and $\Delta ABF$ are congruent by the SSS condition.

> Common mistake: Overlooking the common side $AB$ when identifying the three pairs of equal sides.

### Question 7

*3 marks · Short answer*

This has to be done so that the equal sides overlap. Figure out how.

**Solution**

1. Observe the two triangles $\triangle ABC$ and $\triangle XYZ$ given in the figure in the textbook.
2. Align the equal sides by matching the side $AB$ with $XY$, $BC$ with $YZ$, and $AC$ with $XZ$.
3. Overlap vertex $A$ over vertex $X$, vertex $B$ over vertex $Y$, and vertex $C$ over vertex $Z$ so that the triangles fit exactly over each other.

**Answer:** Overlap vertex A over vertex X, vertex B over vertex Y, and vertex C over vertex Z.

> Common mistake: Overlapping non-corresponding vertices which causes the equal sides to not match up.

### Question 8

*3 marks · Short answer*

Are there other ways of overlapping the vertices so that the triangles fit exactly over each other?

**Solution**

1. Check if the triangle is a general scalene triangle or a special triangle like equilateral or isosceles.
2. For a general triangle with all sides of different lengths, there is only one way to overlap the vertices so that equal sides coincide.
3. Therefore, no other ways of overlapping the vertices exist for general triangles without equal sides.

**Answer:** No, for a general triangle with unequal sides, there is only one way to overlap the vertices.

> Common mistake: Assuming all triangles can be overlapped in multiple ways regardless of side lengths.

### Question 9

*3 marks · Short answer*

Can you identify a pair of congruent triangles below? Why are they congruent?

**Solution**

1. Consider the rectangle $ABCD$ with diagonal $BD$ as shown in Fig. 1.1 in the textbook.
2. In $\triangle ABD$ and $\triangle CDB$, the opposite sides of the rectangle are equal, so $AB = CD$ and $AD = CB$.
3. The diagonal $BD$ is a common side for both triangles, satisfying the SSS condition, hence $\triangle ABD \cong \triangle CDB$.

**Answer:** $\triangle ABD$ and $\triangle CDB$ are congruent by the SSS condition because their corresponding sides are equal and $BD$ is a common side.

> Common mistake: Forgetting to state that the diagonal is a common side shared by both triangles.

### Question 10

*Activity*

Verify this by superimposing paper cutouts of the triangles obtained from the rectangle ABCD (Fig. 1.1).

**Solution**

1. Draw a rectangle $ABCD$ on paper and cut out the two triangles formed by the diagonal $BD$.
2. Superimpose one triangular cutout over the other by matching the sides.
3. Observe that the cutouts fit exactly over each other, verifying their congruence.

**Answer:** Activity verified by superimposing paper cutouts.

### Question 11

*3 marks · Short answer*

Identify the correct correspondence of vertices and express the congruence between the two triangles.

**Solution**

1. Examine the rectangle $ABCD$ and the triangles $\triangle ABD$ and $\triangle CDB$.
2. Match the equal sides: $AB = CD$, $AD = CB$, and the common side $BD = DB$.
3. The correct correspondence of vertices is $A$ to $C$, $B$ to $D$, and $D$ to $B$, written as $\triangle ABD \cong \triangle CDB$.

**Answer:** Vertex A corresponds to C, B to D, and D to B, expressed as $\triangle ABD \cong \triangle CDB$.

> Common mistake: Writing the vertex correspondence in the wrong order, such as matching A to B instead of C.

## Figure it Out

### Question 1

*3 marks · Short answer*

Suppose ΔHEN is congruent to ΔBIG. List all the other correct ways of expressing this congruence.

**Solution**

1. Given that $\Delta \text{HEN} \cong \Delta \text{BIG}$.
2. The corresponding vertices are H and B, E and I, N and G.
3. Any permutation of the corresponding vertices in both triangles gives another correct way: $\Delta \text{HNE} \cong \Delta \text{BGI}$, $\Delta \text{EHN} \cong \Delta \text{IBG}$, $\Delta \text{ENH} \cong \text{IGB}$, $\Delta \text{NHE} \cong \text{GBI}$, and $\Delta \text{NEH} \cong \text{GIB}$.

**Answer:** $\Delta \text{HNE} \cong \Delta \text{BGI}$, $\Delta \text{EHN} \cong \Delta \text{IBG}$, $\Delta \text{ENH} \cong \Delta \text{IGB}$, $\Delta \text{NHE} \cong \Delta \text{GBI}$, $\Delta \text{NEH} \cong \Delta \text{GIB}$

> Common mistake: Listing permutations without maintaining the correct correspondence of vertices.

### Question 2

*3 marks · Short answer*

Determine whether the triangles are congruent. If yes, express the congruence.

**Solution**

1. Observe the side lengths of the two given triangles: one triangle has sides $3.5~\text{cm}$, $5~\text{cm}$, and $6~\text{cm}$, and the other triangle also has sides $3.5~\text{cm}$, $5~\text{cm}$, and $6~\text{cm}$.
2. Since all three corresponding sides of the two triangles are equal, the SSS (Side Side Side) condition is satisfied.
3. Therefore, the triangles are congruent, and we express it as $\Delta \text{RED} \cong \Delta \text{JMA}$.

**Answer:** Yes, the triangles are congruent: $\Delta \text{RED} \cong \Delta \text{JMA}$.

> Common mistake: Writing wrong vertex correspondence when expressing congruence.

### Question 3

*3 marks · Short answer*

In the figure below, AB = AD, CB = CD. Can you identify any pair of congruent triangles? If yes, explain why they are congruent. Does AC divide ∠BAD and ∠BCD into two equal parts? Give reasons.

**Solution**

1. In $\Delta \text{ABC}$ and $\Delta \text{ADC}$, we have $\text{AB} = \text{AD}$ and $\text{CB} = \text{CD}$ as given.
2. Side $\text{AC}$ is common to both triangles, so by the SSS condition, $\Delta \text{ABC} \cong \Delta \text{ADC}$.
3. Since the triangles are congruent, their corresponding angles are equal, so $\angle \text{BAC} = \angle \text{DAC}$ and $\angle \text{BCA} = \angle \text{DCA}$, which shows that $\text{AC}$ divides $\angle \text{BAD}$ and $\angle \text{BCD}$ into two equal parts.

**Answer:** Yes, $\Delta \text{ABC} \cong \Delta \text{ADC}$ by SSS, and AC divides $\angle \text{BAD}$ and $\angle \text{BCD}$ into two equal parts.

> Common mistake: Forgetting to mention the common side AC.

### Question 4

*3 marks · Short answer*

In the figure below, are ΔDFE and ΔGED congruent to each other? It is given that DF = DG and FE = GE.

**Solution**

1. In $\Delta \text{DFE}$ and $\Delta \text{GED}$, we are given that $\text{DF} = \text{DG}$ and $\text{FE} = \text{GE}$.
2. The side $\text{DE}$ is common to both triangles, so $\text{DE} = \text{ED}$.
3. By the SSS condition, all three pairs of corresponding sides are equal, therefore $\Delta \text{DFE} \cong \Delta \text{GED}$.

**Answer:** Yes, $\Delta \text{DFE} \cong \Delta \text{GED}$ by the SSS condition.

> Common mistake: Not identifying the common side DE properly with correct vertex correspondence.

## 1.1 Geometric Twins (continued)

### Question 1

*3 marks · Short answer*

Suppose the angles are 30°, 70°, and 80°. Can we create an exact copy of the frame with this?

**Solution**

1. We know that two triangles can have the exact same set of angles and yet have different sizes.
2. With angles of $30^\circ$, $70^\circ$, and $80^\circ$, we can draw many triangles of different sizes that are similar but not congruent.
3. Therefore, knowing only the three angles is not sufficient to create an exact copy of the frame.

**Answer:** No, because triangles with the same set of angles can have different sizes and need not be congruent.

> Common mistake: Thinking that equal angles guarantee congruence, forgetting that triangles can be scaled up or down.

### Question 2

*3 marks · Short answer*

ΔABC and ΔXYZ are two triangles such that AB = XY = 6 cm, AC = XZ = 5 cm, and ∠A = ∠X = 30°. Are they congruent?

**Solution**

1. We are given two triangles $\triangle ABC$ and $\triangle XYZ$ with $AB = XY = 6\text{ cm}$, $AC = XZ = 5\text{ cm}$, and $\angle A = \angle X = 30^\circ$.
2. The two sides and the included angle of $\triangle ABC$ are equal to the corresponding two sides and the included angle of $\triangle XYZ$.
3. By the SAS (Side Angle Side) condition for congruence, the two triangles are congruent.

**Answer:** Yes, $\triangle ABC$ and $\triangle XYZ$ are congruent by the SAS condition.

> Common mistake: Confusing the included angle with a non-included angle.

### Question 3

*3 marks · Short answer*

Construct a triangle having the above measurements. Compare it with the triangles constructed by your classmates. Are the triangles all congruent? Explain why all such triangles with these measurements are congruent.

**Solution**

1. This activity demonstrates the construction of a triangle using the Side-Angle-Side (SAS) measurements.
2. Observation: When two sides and their included angle are fixed, all constructed triangles are identical in shape and size and can be superimposed exactly, showing that they are congruent.

**Answer:** The triangles constructed by all students are congruent because the SAS condition uniquely determines a triangle.

### Question 4

*3 marks · Short answer*

ΔABC and ΔXYZ are two triangles such that AB = XY = 6 cm, AC = XZ = 4 cm, and ∠B = ∠Y = 30°. Are they congruent?

**Solution**

1. We are given $\triangle ABC$ and $\triangle XYZ$ with $AB = XY = 6\text{ cm}$, $AC = XZ = 4\text{ cm}$, and $\angle B = \angle Y = 30^\circ$.
2. Here, the angle is not included between the two given sides (it is an SSA condition).
3. As demonstrated by triangle construction using arcs and lines, such measurements can lead to two non-congruent triangles.

**Answer:** No, they are not necessarily congruent because the SSA condition does not guarantee congruence.

> Common mistake: Assuming that any combination of two sides and an angle guarantees congruence.

### Question 5

*3 marks · Short answer*

Can there exist non-congruent triangles having these measurements? Construct and find out.

**Solution**

1. This activity involves constructing a triangle given two sides and a non-included angle (SSA).
2. Observation: The arc drawn from one vertex intersects the ray of the non-included angle at two different points, producing two distinct non-congruent triangles that satisfy the given measurements.

**Answer:** Yes, two non-congruent triangles can be constructed with these measurements.

### Question 6

*3 marks · Short answer*

How do we find the required triangle from this figure?

**Solution**

1. We draw the base segment $PQ$ of length $6\text{ cm}$ and construct a line $l$ making an angle of $30^\circ$ at $P$.
2. We draw an arc of radius $4\text{ cm}$ from point $Q$ to intersect the line $l$.
3. Since the arc intersects the line $l$ at two different points ($R$ and $S$), both points give the third vertex, resulting in two possible triangles $\triangle PQR$ and $\triangle PQS$.

**Answer:** The points of intersection of the arc with the line $l$ give the third vertex of the required triangles.

> Common mistake: Stopping after finding only one intersection point when an arc cuts a ray.

### Question 7

*3 marks · Short answer*

ΔABC and ΔXYZ are two triangles with, BC = YZ = 5 cm, ∠B = ∠Y = 50° and ∠C = ∠Z = 30°. Are they congruent?

**Solution**

1. Given two triangles $\triangle ABC$ and $\triangle XYZ$ such that $BC = YZ = 5~\text{cm}$, $\angle B = \angle Y = 50^\circ$, and $\angle C = \angle Z = 30^\circ$.
2. We observe that the two angles and the included side of one triangle are respectively equal to the two angles and the included side of the other triangle.
3. Therefore, by the ASA (Angle Side Angle) condition for congruence, the two triangles are congruent, that is, $\triangle ABC \cong \triangle XYZ$.

**Answer:** Yes, the triangles are congruent by the ASA condition.

> Common mistake: Confusing the included side with a non-included side.

### Question 8

*Activity*

Can there exist non-congruent triangles having these measurements? Construct and find out.

**Solution**

1. Construct a line segment $BC = 5~\text{cm}$.
2. At point $B$, draw an angle of $50^\circ$ and at point $C$, draw an angle of $30^\circ$ using a protractor.
3. The rays meet at a unique point $A$, showing that all triangles with these measurements are unique and congruent.

**Answer:** All triangles constructed with these measurements are congruent, so non-congruent triangles cannot exist.

### Question 9

*3 marks · Short answer*

In the figure, Point O is the midpoint of AD and BC. What can one say about the lengths AB and CD?

**Solution**

1. We are given that point $O$ is the midpoint of $AD$ and $BC$, so $AO = OD$ and $BO = OC$.
2. In $\triangle AOB$ and $\triangle DOC$, we also have $\angle AOB = \angle DOC$ as they are vertically opposite angles.
3. By the SAS condition, $\triangle AOB \cong \triangle DOC$, which implies that their corresponding sides are equal, so $AB = CD$.

**Answer:** The lengths AB and CD are equal.

> Common mistake: Not stating the reason for vertically opposite angles clearly.

### Question 10

*3 marks · Short answer*

Are there any other equal sides or angles?

**Solution**

1. From the congruence of $\triangle AOB$ and $\triangle DOC$, the corresponding vertices are $A$ and $D$, $O$ and $O$, and $B$ and $C$.
2. Therefore, the other equal sides are $AB = CD$.
3. The other equal angles are $\angle OAB = \angle ODC$ and $\angle OBA = \angle OCD$.

**Answer:** The other equal angles are $\angle OAB = \angle ODC$ and $\angle OBA = \angle OCD$.

> Common mistake: Writing incorrect corresponding vertices.

## Figure it Out

### Question 1

*3 marks · Short answer*

Identify whether the triangles below are congruent. What conditions did you use to establish their congruence? Express the congruence.

**Solution**

1. 1. Consider the two triangles $\triangle ABC$ and $\triangle XYZ$ given in the figure in the textbook (page 13).
2. 2. We are given that $AB = XY = 7\text{ cm}$, $BC = YZ = 5\text{ cm}$, and the included angle $\angle B = \angle Y = 47^\circ$.
3. 3. Since two sides and the included angle of one triangle are equal to the corresponding two sides and included angle of the other triangle, the triangles are congruent by the SAS condition, and we express the congruence as $\triangle ABC \cong \triangle XYZ$.

**Answer:** The triangles are congruent by the SAS condition, and $\triangle ABC \cong \triangle XYZ$.

> Common mistake: Writing the corresponding vertices incorrectly in the congruence statement.

### Question 2

*3 marks · Short answer*

Given that CD and AB are parallel, and AB = CD, what are the other equal parts in this figure? (Hint: When the lines are parallel, the alternate angles are equal. Are the two resulting triangles congruent? If so, express the congruence.)

**Solution**

1. 1. Consider the figure in the textbook (page 13) where $CD$ and $AB$ are parallel lines and $AB = CD$.
2. 2. Since $CD \parallel AB$ and $AD$ is a transversal, the alternate interior angles are equal, so $\angle ODC = \angle OBA$ and $\angle OCD = \angle OAB$.
3. 3. By the ASA condition, $\triangle OAB \cong \triangle ODC$, which implies that the other equal parts are $OA = OD$, $OB = OC$, and $\angle AOB = \angle DOC$.

**Answer:** The other equal parts are $OA = OD$, $OB = OC$, and $\angle AOB = \angle DOC$, and the triangles are congruent ($\triangle OAB \cong \triangle ODC$).

> Common mistake: Forgetting to use the alternate interior angles formed by the transversal line.

### Question 3

*3 marks · Short answer*

Given that ∠ABC = ∠DBC and ∠ACB = ∠DCB, show that ∠BAC = ∠BDC. Are the two triangles congruent?

**Solution**

1. 1. Consider $\triangle ABC$ and $\triangle DBC$ given in the figure in the textbook (page 14).
2. 2. We are given $\angle ABC = \angle DBC$ and $\angle ACB = \angle DCB$, with the common side $BC = BC$.
3. 3. By the ASA condition, $\triangle ABC \cong \triangle DBC$, which means their corresponding parts are equal, so $\angle BAC = \angle BDC$.

**Answer:** Yes, the two triangles are congruent by the ASA condition, hence $\angle BAC = \angle BDC$.

> Common mistake: Not identifying the common side properly in the two triangles.

### Question 4

*3 marks · Short answer*

Identify the equal parts in the following figure, given that ∠ABD = ∠DCA and ∠ACB = ∠DBC.

**Solution**

1. 1. Consider the quadrilateral $ABCD$ divided into triangles by diagonals or sides as shown in the textbook (page 14).
2. 2. We are given that $\angle ABD = \angle DCA$ and $\angle ACB = \angle DBC$.
3. 3. Using the given angle equalites and common sides in the figure, the equal parts are the corresponding sides and angles of the congruent triangles formed.

**Answer:** The equal parts are $\angle ABD = \angle DCA$, $\angle ACB = \angle DBC$, and the corresponding sides opposite to these angles.

> Common mistake: Mixing up the alternate or corresponding angles from the given figure.

## 1.1 Geometric Twins (continued)

### Question 1

*3 marks · Short answer*

The following triangles ΔABC and ΔXYZ are such that ∠A = ∠X = 35°, ∠C = ∠Z = 75°, and BC = YZ = 4 cm. Are the triangles congruent? Give a reason.

**Solution**

1. Given that $\angle A = \angle X = 35^\circ$, $\angle C = \angle Z = 75^\circ$, and $BC = YZ = 4\text{ cm}$.
2. The sum of angles in a triangle is $180^\circ$, so finding the third angle gives $\angle B = \angle Y = 70^\circ$.
3. Since two angles and the included side are equal, the triangles are congruent by the ASA (or AAS) condition, so $\Delta ABC \cong \Delta XYZ$.

**Answer:** Yes, the triangles are congruent by the AAS condition.

> Common mistake: Assuming two angles and a non-included side are not sufficient without calculating the third angle.

### Question 2

*3 marks · Short answer*

What are the measures of ∠B and ∠Y?

**Solution**

1. We know that the sum of the angles of a triangle is $180^\circ$.
2. For triangle ABC, $\angle B + 35^\circ + 75^\circ = 180^\circ$, which gives $\angle B + 110^\circ = 180^\circ$, so $\angle B = 70^\circ$.
3. Similarly, for triangle XYZ, we get $\angle Y = 70^\circ$.

**Answer:** $\angle B = 70^\circ$ and $\angle Y = 70^\circ$

> Common mistake: Subtracting incorrectly from $180^\circ$.

### Question 3

*3 marks · Short answer*

Does this help in showing that ΔABC and ΔXYZ are congruent?

**Solution**

1. After finding $\angle B = \angle Y = 70^\circ$, we now have equal angles at $B$ and $Y$ and at $C$ and $Z$, along with the side $BC = YZ = 4\text{ cm}$.
2. These measurements satisfy the ASA condition where the side is included between the two known angles.
3. Therefore, this helps in showing that $\Delta ABC \cong \Delta XYZ$.

**Answer:** Yes, it helps by satisfying the ASA congruence condition.

> Common mistake: Forgetting that the side must be between the two angles for ASA.

### Question 4

*3 marks · Short answer*

ΔABC and ΔXYZ are right-angled triangles such that BC = YZ = 4 cm, ∠B = ∠Y = 90° and AC = XZ=5cm. Are they congruent?

**Solution**

1. Given that $\Delta ABC$ and $\Delta XYZ$ are right-angled triangles with $\angle B = \angle Y = 90^\circ$.
2. The sides given are $BC = YZ = 4\text{ cm}$ and the hypotenuses are $AC = XZ = 5\text{ cm}$.
3. Since the hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and corresponding side of another right-angled triangle, the RHS condition is satisfied.

**Answer:** Yes, the triangles are congruent by the RHS condition.

> Common mistake: Confusing the hypotenuse with the other sides of a right triangle.

### Question 5

*3 marks · Short answer*

Can there exist non-congruent triangles having these measurements? Construct and find out.

**Solution**

1. Draw the base $QR$ of length $4\text{ cm}$.
2. Construct a perpendicular line $l$ at $Q$.
3. Cut an arc of radius $5\text{ cm}$ from $R$ to intersect line $l$ at $P$, and join $P$ to $R$ to form $\Delta PQR$.

**Answer:** Activity-based construction of a right-angled triangle.

> Common mistake: Not drawing the perpendicular correctly at the right-angled vertex.

### Question 6

*3 marks · Short answer*

Consider the downward extension of line l below QR. Would the arc from R meet this line downwards as well (as in the case of triangle construction when the sidelengths are given)? If so, would this lead to a triangle whose size and shape are different from ΔPQR, and yet has the given measurements?

**Solution**

1. The arc from $R$ would indeed meet the downward extension of line $l$ at a point, say $P'$.
2. This would form another triangle $\Delta QP'R$ below $QR$.
3. Due to symmetry, $\Delta QP'R$ is congruent to $\Delta PQR$, meaning no new non-congruent triangle is formed with different size and shape.

**Answer:** Yes, the arc meets the downward line, but it forms a congruent triangle, not a different one.

> Common mistake: Assuming that a second intersection point always yields a non-congruent triangle.

### Question 7

*3 marks · Short answer*

ΔABC is isosceles with AB = AC, and ∠A = 80. What can we say about ∠B and ∠C?

**Solution**

1. In $\Delta ABC$, it is given that $\text{AB} = \text{AC}$.
2. Construct the altitude from $\text{A}$ to $\text{BC}$ meeting $\text{BC}$ at $\text{D}$, which forms two right-angled triangles $\Delta ADB$ and $\Delta ADC$.
3. By applying the $\text{RHS}$ congruence condition to $\Delta ADB$ and $\Delta ADC$, we get $\Delta ADB \cong \Delta ADC$, which shows that $\angle B = \angle C$ as corresponding parts of congruent triangles.

**Answer:** $\angle B = \angle C$

> Common mistake: Confusing the base angles and assuming $\angle A = \angle B$.

### Question 8

*3 marks · Short answer*

Can you use this fact to find ∠B and ∠C?

**Solution**

1. We know that in $\Delta ABC$, $\angle A = 80^\circ$ and $\angle B = \angle C$ since $\text{AB} = \text{AC}$.
2. Using the angle sum property of a triangle, the sum of all three angles is $180^\circ$, so $\angle A + \angle B + \angle C = 180^\circ$.
3. Substituting the known values gives $80^\circ + 2\angle B = 180^\circ$, which means $2\angle B = 100^\circ$, resulting in $\angle B = 50^\circ$ and $\angle C = 50^\circ$.

**Answer:** $\angle B = 50^\circ$ and $\angle C = 50^\circ$

> Common mistake: Dividing the remaining $100^\circ$ incorrectly between $\angle B$ and $\angle C$.

### Question 9

*3 marks · Short answer*

What can we say about their angles?

**Solution**

1. In an equilateral triangle, all three sides are equal in length, so $\text{AB} = \text{BC} = \text{AC}$.
2. Since sides $\text{AB} = \text{AC}$, the angles opposite to them are equal, giving $\angle B = \angle C$.
3. Similarly, since sides $\text{AB} = \text{BC}$, we have $\angle A = \angle C$, which shows that all three angles of an equilateral triangle are equal.

**Answer:** All three angles of an equilateral triangle are equal.

> Common mistake: Assuming that equal sides do not imply equal angles.

### Question 10

*3 marks · Short answer*

What could be their measures?

**Solution**

1. We know that all three angles of an equilateral triangle are equal, let each angle be $x$.
2. Using the angle sum property of triangles, the sum of the angles is $180^\circ$, so $x + x + x = 180^\circ$ or $3x = 180^\circ$.
3. Solving for $x$ gives $x = \frac{180^\circ}{3} = 60^\circ$, so each angle measures $60^\circ$.

**Answer:** Each angle measures $60^\circ$.

> Common mistake: Dividing $180^\circ$ by 2 instead of 3.

### Question 11

*Activity*

Verify this by construction.

**Solution**

1. Draw a line segment of any desired length using a ruler.
2. Use a compass to draw arcs of the same radius from both endpoints of the line segment so that they intersect.
3. Join the intersection point to both endpoints to form an equilateral triangle, and measure each interior angle with a protractor to verify that it is $60^\circ$.

**Answer:** Verification of equilateral triangle angles as $60^\circ$ through construction.

### Question 12

*3 marks · Short answer*

Describe the congruent triangles you see in each picture.

**Solution**

1. Bridges, pyramids, and dome designs use triangles because triangles are rigid and stable geometric figures.
2. In structures like the Howrah Bridge and the Egyptian Pyramids, pairs of triangles sharing common sides or angles are constructed to be exact replicas of one another.
3. These pairs satisfy congruence conditions such as SSS or SAS, ensuring balanced distribution of weight and structural symmetry.

**Answer:** Real-life structures use congruent triangles for rigidity, stability, and symmetry.

> Common mistake: Stating that all triangles in structures are arbitrary without checking congruence conditions.

## Figure it Out

### Question 1

*3 marks · Short answer*

ΔAIR ≅ ΔFLY. Identify the corresponding vertices, sides and angles.

**Solution**

1. Corresponding vertices: A and F, I and L, R and Y.
2. Corresponding sides: AB (or AI) and FL, IR and LY, AR and FY.
3. Corresponding angles: $\angle A$ and $\angle F$, $\angle I$ and $\angle L$, $\angle R$ and $\angle Y$.

**Answer:** Vertices: A$\leftrightarrow$F, I$\leftrightarrow$L, R$\leftrightarrow$Y; Sides: AI$\leftrightarrow$FL, IR$\leftrightarrow$LY, AR$\leftrightarrow$FY; Angles: $\angle A\leftrightarrow\angle F$, $\angle I\leftrightarrow\angle L$, $\angle R\leftrightarrow\angle Y$.

> Common mistake: Writing corresponding vertices out of order.

### Question 2

*3 marks · Short answer*

Each of the following cases contains certain measurements taken from two triangles. Identify the pairs in which the triangles are congruent to each other, with reason. Express the congruence whenever they are congruent. (a) AB = DE, BC = EF, CA = DF (b) AB = EF, ∠A = ∠E, AC = ED (c) AB = DF, ∠B = ∠D = 90°, AC = FE (d) ∠A = ∠D, ∠B = ∠E, AC = DF (e) AB = DF, ∠B = ∠F, AC = DE

**Part (a)**

1. We are given $AB = DE$, $BC = EF$, and $CA = DF$.
2. All three corresponding sides are equal, so the SSS condition is satisfied.
3. Therefore, the triangles are congruent.

Answer (a): Congruent by SSS condition

**Part (b)**

1. We are given $AB = EF$, $\angle A = \angle E$, and $AC = ED$.
2. The given angle is not between the two given sides in both triangles.
3. Therefore, congruence cannot be established.

Answer (b): Not congruent

**Part (c)**

1. We are given $AB = DF$, $\angle B = \angle D = 90^\circ$, and $AC = FE$.
2. Here, the hypotenuse and one side of right-angled triangles are equal.
3. Therefore, the triangles are congruent by the RHS condition.

Answer (c): Congruent by RHS condition

**Part (d)**

1. We are given $\angle A = \angle D$, $\angle B = \angle E$, and $AC = DF$.
2. Two angles and a non-included side are equal, which satisfies the AAS condition.
3. Therefore, the triangles are congruent.

Answer (d): Congruent by AAS condition

**Part (e)**

1. We are given $AB = DF$, $\angle B = \angle F$, and $AC = DE$.
2. The given sides and angle do not satisfy any standard congruence condition because the side $AC$ is not between $\angle B$ and $AB$.
3. Therefore, congruence cannot be established.

Answer (e): Not congruent

**Answer:** Triangles are congruent in cases (a), (c), and (d).

> Common mistake: Confusing SAS with SSA or AAS conditions.

### Question 3

*3 marks · Short answer*

It is given that OB = OC, and OA = OD. Show that AB is parallel to CD. [Hint: AD is a transversal for these two lines. Are there any equal alternate angles?]

**Solution**

1. Consider $\triangle AOB$ and $\triangle DOC$. We are given $OB = OC$ and $OA = OD$.
2. Also, $\angle AOB = \angle DOC$ since they are vertically opposite angles.
3. By the SAS condition, $\triangle AOB \cong \triangle DOC$.
4. Therefore, corresponding angles $\angle OAB = \angle ODC$, which are alternate interior angles for lines $AB$ and $CD$ with transversal $AD$.
5. Since alternate interior angles are equal, $AB$ is parallel to $CD$.

**Answer:** $AB$ is parallel to $CD$ because alternate interior angles $\angle OAB$ and $\angle ODC$ are equal due to congruence of $\triangle AOB$ and $\triangle DOC$.

> Common mistake: Failing to state the correct alternate interior angles.

### Question 4

*3 marks · Short answer*

ABCD is a square. Show that ΔABC ≅ ΔADC. Is ΔABC also congruent to ΔCDA?

**Solution**

1. In square ABCD, opposite sides and all angles are equal, so $AB = CD$, $AD = BC$, and $\angle B = \angle D = 90^\circ$.
2. Consider $\triangle ABC$ and $\triangle ADC$: $AB = AD$ (sides of square), $BC = DC$ (sides of square), and diagonal $AC = CA$ (common side).
3. By SSS condition (or SAS condition), $\triangle ABC \cong \triangle ADC$.
4. Yes, $\triangle ABC$ is also congruent to $\triangle CDA$ when vertices are matched appropriately: $\triangle ABC \cong \triangle CDA$.

**Answer:** $\triangle ABC \cong \triangle ADC$ by SSS condition. Yes, $\triangle ABC$ is also congruent to $\triangle CDA$.

> Common mistake: Incorrectly matching the vertices when writing $\triangle CDA$.

### Question 5

*3 marks · Short answer*

Give more examples of two triangles where one triangle is congruent to the other in two different ways, as in the case above. Can you give an example of two triangles where one is congruent to the other in six different ways?

**Solution**

1. An example of a triangle congruent to itself in two different ways is an isosceles triangle where two sides are equal.
2. For an equilateral triangle, all three sides and all three angles are equal.
3. Because of this symmetry, an equilateral triangle can be mapped onto itself in 6 different ways (3 rotations and 3 reflections).

**Answer:** An equilateral triangle is congruent to itself in six different ways.

> Common mistake: Confusing triangle congruence with self-congruence symmetries.

### Question 5

*3 marks · Short answer*

Find ∠B and ∠C, if A is the centre of the circle.

**Solution**

1. In the circle with center A, $AB$ and $AC$ are radii, so $AB = AC$.
2. Since $AB = AC$, $\triangle ABC$ is an isosceles triangle.
3. In an isosceles triangle, angles opposite to equal sides are equal, so $\angle B = \angle C$.
4. Given $\angle A = 120^\circ$, using the angle sum property of triangles, $\angle B + \angle C + 120^\circ = 180^\circ$.
5. $2\angle B = 60^\circ$, which gives $\angle B = 30^\circ$ and $\angle C = 30^\circ$.

**Answer:** $\angle B = 30^\circ$ and $\angle C = 30^\circ$

> Common mistake: Assuming all angles in the triangle are equal.

### Question 6

*3 marks · Short answer*

Find the missing angles. As per the convention that we have been following, all line segments marked with a single ‘|’ are equal to each other and those marked with a double ‘|’ are equal to each other, etc.

**Solution**

1. Identify the triangles and use the given markings to find equal sides, which imply that angles opposite to equal sides are equal.
2. Use the angle sum property of a triangle, which states that the sum of the three angles of a triangle is $180^\circ$.
3. Proceed step-by-step through the interconnected triangles in the given figure to determine all the missing angles.

**Answer:** The missing angles in the figure are found using properties of isosceles triangles and the angle sum property.

> Common mistake: Mixing up the single and double markings for side lengths or misapplying the angles opposite to equal sides property.

## Expression Engineer!

### Question 1

*Activity*

Draw lines and split the region consisting of white squares into 6 smaller congruent regions.

**Solution**

1. Examine the grid of white squares surrounding the central green square on page 23.
2. Divide the 24 white squares into 6 groups such that each group contains 4 squares of the same shape and size.
3. Draw lines along the grid boundaries to separate the grid into 6 identical L-shaped or T-shaped congruent regions.

**Answer:** The region consisting of white squares can be split into 6 smaller congruent regions of 4 squares each.

## Frequently asked questions

### How many questions are there in Class 7 Maths Chapter 9 Geometric Twins?

This chapter in the new NCERT book for the 2026-27 session contains a total of 57 questions spread across various sections like Geometric Twins, Figure it Out, and Expression Engineer. You can find step-by-step solutions for all these questions in the free PDF available on this page.

### What topics are covered in Class 7 Maths Chapter 9 Geometric Twins?

The chapter covers important concepts such as arm lengths sufficiency, congruence of figures, superimposition, SSS, ASA, SAS, and AAS congruence conditions, along with properties of isosceles and equilateral triangles. All these topics are thoroughly explained in the SwaVid free PDF on this page.

### Which are the hardest question types in this chapter and how should I approach them?

Questions involving ASA and AAS conditions, SSA ambiguity activities, and complex triangle construction are generally found to be the trickiest by students. To approach them, you should carefully verify corresponding parts and use the detailed step-by-step explanations provided in the free PDF on this page.

### How can I write answers for full marks in Class 7 Maths Chapter 9?

To secure full marks, you need to clearly state the congruence criteria used, mention corresponding vertices in correct order, and show all construction steps logically. Referring to the structured answers in SwaVid's free PDF on this page will help you learn the right presentation style.

### Is the free PDF for Class 7 Maths Chapter 9 Geometric Twins available?

Yes, the complete and free PDF containing accurate answers for all 57 questions in this chapter is available right on this page. SwaVid provides these reliable solutions aligned with the new NCERT book for the 2026-27 session to help you revise effectively.

## Related pages

- [Class 7 Maths chapters](https://www.swavid.com/maths/class/7)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student