NCERT Solutions for Class 7 Maths Part 2 Chapter 1 - Geometric Twins

Class 7 Maths Part 2 Chapter 1 Solutions PDF Free Download - Ganita Prakash Book 2

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NCERT Solutions for Class 7 Maths Ganita Prakash Part 2 Chapter 1 Exercise Questions and Answers

Here are the NCERT Solutions for Class 7 Maths Part 2 Chapter 1 Geometric Twins question answers with clear and detailed solutions.

1. Check if the two figures are congruent.

Answer:

Length of the two sides seems similar or equal. But angles seem different.

So, two figures are not congruent.

2. Circle the pairs that appear congruent.

Answer:

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

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

Answer:

3(a):

To create a circle congruent to a given circle, you need to measure its radius or diameter.

3(b):

To create a rectangle congruent to a given rectangle, you need to measure its length and width.

3 (a):

Two circles are congruent if they have the same radius or the same diameter.

3 (b):

Two rectangles are congruent if they have the same length and the same width.

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

Use this to identify whether each of the following pairs are congruent.

Answer:

If two figures have the same side lengths and the angle between those sides is equal, then the figures are congruent.

All the pairs are congruent from the pictures, as all have the same length and the same angles

1. Suppose $\triangle \mathrm{HEN}$ is congruent to $\triangle \mathrm{BIG}$. List all the other correct ways of expressing this congruence.

Answer:

If $\triangle \mathrm{HEN} \cong \triangle \mathrm{BIG}$, then the corresponding vertices are:

• Vertex $H$ corresponds to vertex $B$.

• Vertex $E$ corresponds to vertex $I$.

• Vertex $N$ corresponds to vertex $G$.

Keeping this correspondence the same, the other correct ways to express the congruence are:

• $\triangle \mathrm{ENH} \cong \triangle \mathrm{IGB}$

• $\triangle \mathrm{NHE} \cong \triangle \mathrm{GBI}$

• $\triangle \mathrm{HNE} \cong \triangle \mathrm{BGI}$

• $\triangle \mathrm{NEH} \cong \triangle \mathrm{GIB}$

• $\triangle \mathrm{EHN} \cong \triangle \mathrm{IBG}$

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

Answer:

In triangles RED and MJA,
RE = JA = 3.5 cm
ED = AM = 5 cm
RD = MJ = 6 cm

So, by Side-Side-Side (SSS) congruence condiiton, $\triangle$RED $\cong \triangle$MJA

3. In the figure below, $\mathrm{AB}=\mathrm{AD}, \mathrm{CB}=\mathrm{CD}$.

Can you identify any pair of congruent triangles? If yes, explain why they are congruent.
Does AC divide $\angle \mathrm{BAD}$ and $\angle \mathrm{BCD}$ into two equal parts? Give reasons.

Answer:

In the triangles $\triangle A B C$ and $\triangle A D C$,

We can verify their congruence by comparing their corresponding sides:
$A B=A D$ (Given)
$C B=C D$ (Given)
$A C=A C$ (Common side shared by both triangles)
Using the Side-Side-Side (SSS) congruence criterion, we get,
$ \triangle A B C \cong \triangle A D C$

4. In the figure below, are $\triangle \mathrm{DFE}$ and $\triangle \mathrm{GED}$ congruent to each other? It is given that $\mathrm{DF}=\mathrm{DG}$ and $\mathrm{FE}=\mathrm{GE}$.

Answer:

In $\triangle \mathrm{DFE}$ and $\triangle \mathrm{GED}$,

$\mathrm{DF}=\mathrm{DG}$ and $\mathrm{FE}=\mathrm{GE}$ [Given]
$\mathrm{DE}$ is a common side.

So, according to the Side-Side-Side (SSS) criterion, $\triangle \mathrm{DFE} \cong\triangle \mathrm{GED}$

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

Answer:

In triangles ABC and XYZ,
AB = XZ = 7 cm
BC= YZ = 5 cm
$\angle$ B = $\angle$ Z= 47°

According to the Side-Angle-Side (SAS) congruence condition,
$\triangle$ABC $\cong \triangle$XYZ

2. Given that CD and AB are parallel, and $\mathrm{AB}=\mathrm{CD}$, 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.)

Answer:

In $\triangle A O B$ and $\triangle D O C$,
$A B=C D$ (Given)
$\angle O A B=\angle O D C$ (Alternate interior angles, since horizontal lines $A B \| C D$ are intersected by the transversal line $A D$ )
$\angle O B A=\angle O C D$ (Alternate interior angles, since horizontal lines $A B \| C D$ are intersected by the transversal line $B C$ )

According to the Angle-Side-Angle (ASA) congruence criterion,:
$\triangle A O B \cong \triangle D O C$

Because the triangles are congruent, all their other corresponding components must be equal by the property of Corresponding Parts of Congruent Triangles (CPCT):

Corresponding Equal Sides

• $OA = OD$ (Point $O$ bisects the line segment $AD$)

• $OB = OC$ (Point $O$ bisects the line segment $BC$)

Corresponding Equal Angles

• $\angle AOB = \angle DOC$ (These are vertically opposite angles at the intersection point $O$)

3. Given that $\angle \mathrm{ABC}=\angle \mathrm{DBC}$ and $\angle \mathrm{ACB}=\angle \mathrm{DCB}$, show that $\angle \mathrm{BAC}=\angle \mathrm{BDC}$. Are the two triangles congruent?

Answer:

in $\triangle A B C$ and $\triangle D B C$,
$\angle A B C=\angle D B C$ (Given)
$B C=B C$ (Common side shared by both triangles)
$\angle A C B=\angle D C B$ (Given)
Using the Angle-Side-Angle (ASA) congruence criterion, where the side is included between the two equal angles:
$\triangle A B C \cong \triangle D B C$
Hence, the two triangles are congruent.

Since we have proven that $\triangle ABC \cong \triangle DBC$, all corresponding parts of these two triangles must be equal by the property of Corresponding Parts of Congruent Triangles (CPCT).

So, $\angle BAC = \angle BDC$

4. Identify the equal parts in the following figure, given that $\angle A B D= \angle \mathrm{DCA}$ and $\angle \mathrm{ACB}=\angle \mathrm{DBC}$.

Answer:

$\angle ABC = \angle ABD + \angle DBC$
$\angle DCB = \angle DCA + \angle ACB$

We are given two sets of equal angles:
$\angle ABD = \angle DCA$ (Indicated by matching single arcs)
$\angle DBC = \angle ACB$ (Indicated by matching double arcs)
By adding these equal equations together:
$\angle ABD + \angle DBC = \angle DCA + \angle ACB$
$\Rightarrow \angle ABC = \angle DCB$

In triangles $\triangle ABC$ and $\triangle DCB$:
$\angle ACB = \angle DBC$ (Given in the problem)
$BC = CB$ (Common base side shared by both triangles)
$\angle ABC = \angle DCB$
Using the Angle-Side-Angle (ASA) congruence criterion:
$\triangle ABC \cong \triangle DCB$

Because $\triangle ABC \cong \triangle DCB$, all other corresponding components must be equal by the property of Corresponding Parts of Congruent Triangles (CPCT):

Corresponding Equal Sides

• $AB = DC$

• $AC = DB$

Corresponding Equal Angles

• $\angle BAC = \angle CDB$

1. $\triangle \mathrm{AIR} \cong \triangle \mathrm{FLY}$. Identify the corresponding vertices, sides and angles.

Answer:

Corresponding Vertices

• Vertex $A =$ Vertex $F$

• Vertex $I =$ Vertex $L$

• Vertex $R =$ Vertex $Y$

Corresponding Sides

• Side $A I =$ Side $F L$

• Side $I R =$ Side $L Y$

• Side $R A =$ Side $Y F$

Corresponding Angles

• $\angle A = \angle F$

• $\angle I = \angle L$

• $\angle R = \angle Y$

2. 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)

$
\begin{aligned}
& \mathrm{AB}=\mathrm{DE} \\
& \mathrm{BC}=\mathrm{EF} \\
& \mathrm{CA}=\mathrm{DF}
\end{aligned}
$

(b)

$
\begin{aligned}
& \mathrm{AB}=\mathrm{EF} \\
& \angle \mathrm{~A}=\angle \mathrm{E} \\
& \mathrm{AC}=\mathrm{ED}
\end{aligned}
$

(c)

$
\begin{aligned}
& \mathrm{AB}=\mathrm{DF} \\
& \angle \mathrm{~B}=\angle \mathrm{D}=90^{\circ} \\
& \mathrm{AC}=\mathrm{FE}
\end{aligned}
$

(d)

$
\begin{aligned}
& \angle \mathrm{A}=\angle \mathrm{D} \\
& \angle \mathrm{~B}=\angle \mathrm{E} \\
& \mathrm{AC}=\mathrm{DF}
\end{aligned}
$

(e)

$
\begin{aligned}
& \mathrm{AB}=\mathrm{DF} \\
& \angle \mathrm{~B}=\angle \mathrm{F} \\
& \mathrm{AC}=\mathrm{DE}
\end{aligned}
$

Answer:

2(a): $\Delta ABC \cong \Delta DEF$

Explanation:

All three pairs of corresponding sides are equal in length.
$\therefore$ By the Side-Side-Side (SSS) congruence criterion, the two triangles are congruent.

2(b): $\Delta ABC \cong \Delta EFD$

Explanation:

Two pairs of sides and their included angle (the angle formed between those two sides) are equal.
$\therefore$ By the Side-Angle-Side (SAS) congruence criterion, the two triangles are congruent.

2(c): $\Delta ABC \cong \Delta DFE$

Explanation:

Both triangles are right-angled ($\angle B = \angle D = 90^\circ$). The hypotenuse of the first triangle ($AC$) equals the hypotenuse of the second ($FE$), and one pair of corresponding legs ($AB = DF$) is equal.
$\therefore$ By the Right Angle-Hypotenuse-Side (RHS) congruence criterion, the two triangles are congruent.

2(d): $\Delta ABC \cong \Delta DEF$

Explanation:

Two pairs of angles and a non-included side are equal. Since two pairs of angles are equal, the third pair must also be equal ($\angle C = \angle F$), making the given side included between them.
$\therefore$ By the Angle-Angle-Side (AAS) criterion (or Angle-Side-Angle (ASA)), the two triangles are congruent.

2(e): Not congruent

Explanation:

Here, two pairs of sides are equal ($AB=DF$ and $AC=DE$), but the given equal angle ($\angle B = \angle F$) is not the included angle between those sides.
$\therefore$ Side-Side-Angle (SSA) is not a valid condition for establishing congruence generally.

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

Answer:

In triangles $\triangle A O B$ and $\triangle D O C$,
$O A=O D$ (Given)
$\angle A O B=\angle D O C$ (Vertically opposite angles formed by the intersecting lines $A D$ and $B C$ at point $O$ )
$O B=O C$ (Given)
Using the Side-Angle-Side (SAS) congruence criterion:
$\triangle A O B \cong \triangle D O C$

Since $\triangle AOB \cong \triangle DOC$, their corresponding angles must be equal by the property of Corresponding Parts of Congruent Triangles (CPCT).

$\Rightarrow \angle OAB = \angle ODC$

$\angle OAB$ (or $\angle DAB$) and $\angle ODC$ (or $\angle ADC$) form a pair of alternate interior angles.
Since these alternate interior angles are equal ($\angle OAB = \angle ODC$), the lines must be parallel.

$\Rightarrow AB \parallel CD$

4. ABCD is a square. Show that $\triangle \mathrm{ABC} \cong \triangle \mathrm{ADC}$. Is $\triangle \mathrm{ABC}$ also congruent to $\triangle \mathrm{CDA}$ ?

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?

Answer:

Let $ABCD$ be a square.
Therefore,
$AB = BC = CD = DA$
$\angle A = \angle B = \angle C = \angle D = 90^\circ$

In triangle ABC and ADC,
$AB = AD$ (Sides of a square are equal)
$BC = DC$ (Sides of a square are equal)
$AC = AC$ (Common side shared by both triangles)

Using the Side-Side-Side (SSS) congruence criterion:
$\triangle ABC \cong \triangle ADC$

In triangle ABC and CDA,

AB = CD [Since $AB$ and $CD$ are opposite sides of the same square]
BC = DA [Since $BC$ and $DA$ are opposite sides of the same square]
AC = CA [Shared diagonal]

Using the Side-Side-Side (SSS) congruence criterion:
$\triangle ABC \cong \triangle CDA$

Consider two equilateral triangles $\triangle P Q R$ and $\triangle X Y Z$, where all sides are equal to 4 cm :

$ P Q=Q R=R P=4 \mathrm{~cm}$

$ X Y=Y Z=Z X=4 \mathrm{~cm}$

Since any side of $\triangle P Q R$ can match any side of $\triangle X Y Z$, every single mathematical permutation of the vertices creates a valid congruence statement. There are $3!=3 \times 2 \times 1=6$ total ways to arrange the correspondences:

1. $\triangle P Q R \cong \triangle X Y Z$ (Mapping: $P \rightarrow X, Q \rightarrow Y, R \rightarrow Z$ )

2. $\triangle P Q R \cong \triangle X Z Y$ (Mapping: $P \rightarrow X, Q \rightarrow Z, R \rightarrow Y$ )

3. $\triangle P Q R \cong \triangle Y X Z$ (Mapping: $P \rightarrow Y, Q \rightarrow X, R \rightarrow Z$ )

4. $\triangle P Q R \cong \triangle Y Z X$ (Mapping: $P \rightarrow Y, Q \rightarrow Z, R \rightarrow X$ )

5. $\triangle P Q R \cong \triangle Z X Y$ (Mapping: $P \rightarrow Z, Q \rightarrow X, R \rightarrow Y$ )

6. $\triangle P Q R \cong \triangle Z Y X$ (Mapping: $P \rightarrow Z, Q \rightarrow Y, R \rightarrow X$ )

5. Find $\angle \mathrm{B}$ and $\angle \mathrm{C}$, if A is the centre of the circle.

Answer: 30° and 30°

Explanation:

Here, AB = AC = Radius

So, $\angle$ B = $\angle$ C = $x$ [Let]

In triangle ABC,

$\angle$ A + $\angle$ B + $\angle$ C = 180°

$⇒120°+x+x=180°$

$⇒2x=60°$

$\therefore x=30°$

Hence, $\angle$ B = $\angle$ C = 30°

6. Find the missing angles. As per the convention that we have been following, all line segments marked with a single ' $\mid$ ' are equal to each other and those marked with a double ' $\mid$ ' are equal to each other, etc.

Answer:

From the large outer rectangle ABCD:
$AC \perp CD$ and $\angle C = 90^\circ$, $\angle A = 90^\circ$, $\angle B = 90^\circ$, $\angle D = 90^\circ$.

In $\triangle$CRU,
CU = CR
So, $\angle CUR = \angle CRU=x$ [Let]
So, $x+x+90°=180°$
$⇒2x=90°$
$\therefore x= 45°$
So, $\angle CUR = \angle CRU=45°$

Key Concepts Explained in Geometric Twins - Ganita Prakash Book 2

Topics you will learn in NCERT Solutions for Class 7 Maths Part 2 Chapter 1 Geometric Twins include:

• 1.1 Geometric Twins

• 1.2 Congruence of Triangles

• 1.3 Angles of Isosceles and Equilateral Triangles

NCERT Solutions for Class 7 Maths Chapter Wise

We at Careers360 compiled all the NCERT Class 7 Maths solutions in one place for easy student reference. The following links will allow you to access them.

NCERT Books and NCERT Syllabus

Before planning a study schedule, always analyse the latest syllabus. Here are the links to the latest NCERT syllabus and some important books that will help students in this endeavour.

• NCERT Syllabus Class 7 Maths

• NCERT Books for Class 7 Maths

• NCERT Books Class 7

• NCERT Syllabus Class 7

• NCERT Books for Class 7 English

• NCERT Books for Class 7 Social Science

• NCERT Books for Class 7 Science

• NCERT Books for Class 7 Hindi

• NCERT Solutions for Class 7 Science

Chapter Summary: NCERT Solutions for Class 7 Maths Ganita Prakash Part 2 Chapter 1

Chapter 1 of Class 7 Part 2, Geometric Twins, introduces the concept of congruence and explains when two geometric figures are exactly alike in shape and size. Students learn to identify congruent figures by comparing their corresponding sides and angles. This chapter focuses on the congruence of triangles and the conditions required to prove that two triangles are congruent. The chapter explores the properties of isosceles triangles, including the equality of their base angles. Students also study the characteristics of equilateral triangles and learn that all their sides and angles are equal.

Regular practice strengthens spatial visualisation, reasoning, and proof-writing skills. Mastering this chapter provides a strong foundation for higher geometry topics and theorem-based problem-solving.

Expert Review of NCERT Solutions for Class 7 Maths Part 2 Chapter 1

These NCERT solutions provide clear, step-by-step explanations for every exercise and theorem in the chapter. Here are some more points about this chapter and solutions.

1. Students learn to apply congruence criteria confidently while solving geometric problems.

2. The concepts covered in this chapter form the basis for advanced topics such as triangle properties, similarity, trigonometry, and coordinate geometry.

3. Overall, Chapter 1 is an essential geometry chapter that builds the foundation for higher mathematical reasoning and geometric proofs.

Important concepts and topics related to the latest NCERT Solutions for Class 7 Maths Ganita Prakash Part 2 Chapter 1 Geometric Twins in higher classes are provided below.

Class 8-10

• Congruence and similarity of triangles

• Properties of triangles

• Basic Proportionality Theorem (Thales' Theorem)

• Triangle constructions

• Circle theorems

• Coordinate geometry

• Trigonometry fundamentals

• Geometric proofs and reasoning

Class 11-12

• Coordinate geometry

• Trigonometry and triangle properties

• Vector algebra

• Three-dimensional geometry

• Euclidean geometry

• Geometric transformations

• Analytical geometry

• Proof-based mathematical reasoning

JEE (Main & Advanced)

• Triangle geometry

• Congruence and similarity concepts

• Coordinate geometry

• Trigonometry

• Circle and conic section geometry

• Vector geometry

• Geometry-based logical reasoning

• Multi-concept problem-solving involving triangles and geometric proofs