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

Answer:
Yes, they are congruent.

• Both have the same angle measure
• Both arms appear to be the same length
• They have the same shape and size
• One could be superimposed on the other through rotation.

Answer:

(a) Circle
(b) Rectangle
Using this, state how would you check if two-
(a) Circles are congruent?
(b) Rectangles are congruent?
Answer:
To create a congruent figure we will take measurements given below :
(a) Circle
Measure the radius (distance from center to any point on the circle)
Measure the diameter (distance across the circle through the center)
(b) Rectangle
Measure the length (one pair of opposite sides)
Measure the breadth/width (the other pair of opposite sides)

(a) Two circles are congruent if and only if their radii are equal (or their diameters are equal).
Method to check:
Measure the radius of the first circle
Measure the radius of the second circle
If both radii are equal, the circles are congruent

(b) Two rectangles are congruent if and only if their corresponding sides are equal.
Method to check:
Measure the length and breadth of the first rectangle
Measure the length and breadth of the second rectangle
If the length of the first rectangle equals the length of the second rectangle AND the breadth of the first rectangle equals the breadth of the second rectangle, then the rectangles are congruent.

Answer:
The figure consists of three line segments meeting at a point.
To check whether two such figures are congruent, we should:
Measure the lengths of all the arms (the three line segments).
Measure the angles between the arms.
Compare the corresponding arm lengths and corresponding angles of both figures.

Pair 1 (Top-left and Top-right figures)
Observation:
The arm lengths appear equal.
The angles between the arms are also equal.
The figures differ only in orientation.
Conclusion:
These two figures are congruent.

Pair 2 (Bottom-left and Bottom-right figures)
Observation:
One arm length is different.
The angle between the arms is also different.
Conclusion:
These two figures are not congruent.

Answer:
When ∆HEN is congruent to ∆BIG, the order of the vertices matters because it indicates which vertices correspond to each other.
From ∆HEN ≅ ∆BIG, we know:
H corresponds to B
E corresponds to I
N corresponds to G
For a congruence statement to be correct, the corresponding vertices must be in the same positions. We can rearrange the vertices, but the correspondence must remain the same.

Correct ways to express this congruence:

• ∆HEN ≅ ∆BIG
• ∆HNE ≅ ∆BGI
• ∆EHN ≅ ∆IBG
• ∆ENH ≅ ∆IGB
• ∆NEH ≅ ∆GIB
• ∆NHE ≅ ∆GBI

Each of these maintains the correspondence: H↔B, E↔I, N↔G. There are 6 total ways because we can arrange 3 vertices in 3 = 6 different orders.

Answer:
Triangle RED:
RE = 3.5 cm
RD = 6 cm
ED = 5 cm

Triangle JMA:
JM = 6 cm
MA = 5 cm
JA = 3.5 cm

Compare the corresponding sides:
RE = 3.5 cm corresponds to JA = 3.5 cm
RD = 6 cm corresponds to JM = 6 cm
ED = 5 cm corresponds to MA = 5 cm
Since all three pairs of corresponding sides are equal, the triangles are congruent by SSS (Side-Side-Side) congruence.
The congruence can be expressed as ∆RED ≅ ∆JMA

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.

Answer:
Yes, there is a pair of congruent triangles: ∆ABC ≅ ∆ADC
Looking at triangles ABC and ADC, they share side AC and have the following:
AB = AD (given)
CB = CD (given)
AC = AC (common side – reflexive property)
Since all three pairs of corresponding sides are equal, the triangles are congruent by SSS (Side-Side-Side) congruence postulate.
Yes, AC divides both ∠BAD and ∠BCD into two equal parts.

Since ∆ABC ≅ ∆ADC (proven above), all corresponding parts of congruent triangles are congruent (CPCTC).
Therefore:
∠BAC ≅ ∠DAC, which means AC bisects ∠BAD
∠BCA ≅ ∠DCA, which means AC bisects ∠BCD
In other words, AC is the angle bisector of both ∠BAD and ∠BCD because the congruent triangles ensure that the angles on either side of AC are equal.

Answer:
For ∆DFE and ∆GED:
1. DF = DG (given)
2. FE = GE (given)
3. DE = DE (common side – reflexive property)
Now we have:
– DF = DG
– FE = GE
– DE = DE
All three corresponding sides are equal, therefore the triangles are congruent.

Answer:
Comparing the triangles:
AB = XZ = 7 cm
BC = YZ = 5 cm
∠B = ∠Z = 47°
Yes, the triangles are congruent.
Condition used: SAS (Side-Angle-Side) congruence postulate
Congruence statement: ∆ABC ≅ ∆XZY
This correspondence shows:
A ↔ X
B ↔ Z
C ↔ Y

Answer:
Using Parallel Lines:
Since CD ∥ AB, and they are cut by transversal lines, we have alternate interior angles:
∠DCO = ∠BAO (alternate interior angles)
∠CDO = ∠ABO (alternate interior angles)

Now for triangles DOC and AOB:
CD = AB (given)
∠DCO = ∠BAO (alternate interior angles)
∠CDO = ∠ABO (alternate interior angles)
The triangles are congruent by ASA (Angle-Side-Angle) congruence.
Congruence statement: ∆DOC ≅ ∆AOB
Since the triangles are congruent, all corresponding parts are equal:
DO = AO (corresponding sides)
CO = BO (corresponding sides)
∠DCO = ∠BAO (alternate interior angles)
∠CDO = ∠ABO (alternate interior angles)
∠DOC = ∠AOB (vertical angles)
The parallel lines and equal segments CD = AB lead to congruent triangles, which means O is the midpoint of both diagonals (DO = AO and CO = BO).

Answer:
Consider ∆ABC and ∆DBC:
∠ABC = ∠DBC (given)
∠ACB = ∠DCB (given)
BC = BC (common side – reflexive property)
By ASA (Angle-Side-Angle) congruence postulate: ∆ABC ≅ ∆DBC
Since the triangles are congruent, all corresponding parts are congruent (CPCTC).
Therefore, ∠BAC = ∠BDC
Yes the Two Triangles Congruent∆ABC ≅ ∆DBC
We already established this congruence above using:
∠ABC = ∠DBC (given)
BC = BC (common side)
∠ACB = ∠DCB (given)
This satisfies the ASA (Angle-Side-Angle) congruence condition.

Answer:
In △ABC and △DCB:
∠ACB = ∠DBC (given)
∠ABD = ∠DCA (given)
Since ∠ABD is part of ∠ABC, and ∠DBC is the other part, we can write:
∠ABC = ∠ABD + ∠DBC
Similarly, since ∠DCA is part of ∠DCB, and ∠ACB is the other part:
∠DCB = ∠DCA + ∠ACB
Since ∠ABD = ∠DCA and ∠ACB = ∠DBC, we can substitute:
∠ABC = ∠DCA + ∠ACB = ∠DCB
Now we have:
– ∠ACB = ∠DBC
– ∠ABC = ∠DCB
– BC = CB (common side)
By the ASA (Angle-Side-Angle) congruence criterion, △ABC ≅ △DCB.

Therefore, the equal parts are:
Sides:
– AB = DC
– AC = DB
– BC = CB (common/shared side)
Angles:
– ∠ABC = ∠DCB
– ∠ACB = ∠DBC (given)
– ∠BAC = ∠CDB

Answer:
Since ∆AIR ≅ ∆FLY:
Corresponding vertices:
A ↔ F
I ↔ L
R ↔ Y

Corresponding sides:
AI = FL
IR = LY
AR = FY

Corresponding angles:
∠A = ∠F
∠I = ∠L
∠R = ∠Y

(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
Answer:
(a) AB = DE, BC = EF, CA = DF
Congruent: Yes (SSS Side-Side-Side)
∆ABC ≅ ∆DEF

(b) AB = EF, ∠A = ∠E, AC = ED
Congruent: No, The equal sides don’t include the equal angle between them (not SAS).

(c) AB = DF, ∠B = ∠D = 90°, AC = FE
Congruent: Yes, RHS (Right angle-Hypotenuse-Side). AC and FE are hypotenuses.
∆ABC ≅ ∆DFE

(d) ∠A = ∠D, ∠B = ∠E, AC = DF
Congruent: No, AAA doesn’t prove congruence. The equal side AC = DF is not between the equal angles (not ASA).

(e) AB = DF, ∠B = ∠F, AC = DE
Congruent: No, The equal angle is not between the equal sides (not SAS).

[Hint: AD is a transversal for these two lines. Are there any equal alternate angles?]

Answer:
In triangles AOB and DOC:
OA = OD (given)
OB = OC (given)
∠AOB = ∠DOC (vertically opposite angles at point O)
△AOB ≅ △DOC [By the SAS (Side-Angle-Side) congruence criterion]
Since the triangles are congruent
∠OAB = ∠ODC (corresponding angles in congruent triangles)

Now, looking at the transversal AD crossing lines AB and CD:
∠OAB is the angle between AB and AD at point A
∠ODC is the angle between CD and AD at point D
Since ∠OAB = ∠ODC and these are alternate angles with respect to transversal AD.
Hence, AB ∥ CD.

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:
ABCD is a square.
All sides of a square are equal and all angles are 90°.
In triangles ΔABC and ΔADC:
AB = AD (sides of a square)
BC = DC (sides of a square)
AC = AC (common side)
Therefore, by SSS congruence, ΔABC ≅ ΔADC.
Yes, ΔABC is also congruent to ΔCDA.
This is because ∠DAC = ∠DCA.
So, in square ABCD, diagonal AC divides the square into two triangles.
ΔABC ≅ ΔADC and also ΔABC ≅ ΔCDA.

Congruent in six different ways
Yes. Take two equilateral triangles ΔABC and ΔPQR having the same side length.
Since
AB = BC = CA and PQ = QR = RP, every vertex of the first triangle can correspond to any vertex of the second triangle. Therefore there are 3 × 2 × 1 = 6 ways:

1. ΔABC ≅ ΔPQR
2. ΔABC ≅ ΔPRQ
3. ΔABC ≅ ΔQPR
4. ΔABC ≅ ΔQRP
5. ΔABC ≅ ΔRPQ
6. ΔABC ≅ ΔRQP

So, two congruent equilateral triangles provide an example where one triangle is congruent to the other in six different ways.

Answer:
A is the centre of the circle.
So, AB = AC (radii of the same circle).
Therefore, triangle ABC is an isosceles triangle.
Given ∠BAC = 120°.
Since AB = AC, ∠B = ∠C
Let ∠B = ∠C = x
Sum of angles of a triangle = 180°.
∠A + ∠B + ∠C = 180°
⇒ 120° + x + x = 180°
⇒ 2x = 60°
⇒ x = 60° ÷ 2 = 30°
Therefore, ∠B = ∠C = 30°

Answer:
According to the given convention, line segments marked with the same number of marks are equal.
Hence, triangles formed with equal sides are isosceles and angles opposite equal sides are equal. Also, we use the following facts:
Sum of angles of a triangle is 180°.
Angles on a straight line add up to 180°.
Angles around a point add up to 360°.
Vertically opposite angles are equal.
A right angle is 90°.
Using these rules, the missing angles (marked in red) are obtained as follows:

In ΔCUR,
∠1 = 90° [Given]
Also, CR = CU
So, ∠2 = ∠3 = (180° − 90°)/2 = 45°

Similarly, in ΔUEF
∠EUF = 34°
Also, UF = UE [Given]
So, ∠6 = ∠7 = (180° − 34°)/2 = 73°.

In ΔAFU, ∠UAF= 56° and UA = UF, hence
∠UAF = ∠9 = 56°, also
∠8 = 180° − 56° − 56° = 68°.

At point U,
∠8 + 34° + ∠4 + ∠3 = 180° [Forming straight angle]
⇒ 45° + ∠4 + 34° + 68° = 180°,
⇒ ∠4 = 33°.

In ΔRVH
∠RVH = 68°
Also, VR = VH [Given]
So, ∠r = ∠q = (180° − 68°)/2 = 56°.
Hence, ∠r = ∠q = 56°.

At point R,
∠2 + ∠5 + 34° + ∠r = 180° [Forming straight angle]
⇒ 45° + ∠5 + 34° + 56° = 180°,
⇒ ∠5 = 45°.

In ΔRUE
∠REU = 180° − ∠4 − ∠5
⇒ ∠REU = 180° − 33° − 45° = 102°.

In ΔREH
∠w = 180° − 34° − 44° = 102°.

At point E,
∠REU + ∠6 + ∠s + ∠w = 360° [Forming complete angle]
⇒ 102° + 73° + ∠s + 102° = 360°,
⇒ ∠s = 83°.

In ΔHEF
∠s + ∠t + 46° = 180°
⇒ 83° + ∠t + 46° = 180°,
⇒ ∠t = 51°.

In ΔHGF
∠x = 180° − 90° − 56° = 34°.

In ΔAKF
∠10 = 180° − 34° − 44° = 102°,

At point F,
∠KFG + 44° + ∠9 + ∠7 + ∠t + ∠x = 360° [Forming complete angle]
⇒ ∠KFG + 44° + 56° + 73° + 51° + 34° = 360°
⇒ ∠KFG = 360° – 258° = 102°

In ΔKFG
∠a = 180° − ∠KFG − 30°
⇒ ∠a = 180° − 102° − 30° = 48°.
⇒ ∠a = 48°

The ΔBGF is an equilateral triangle, so
∠f = ∠g = ∠h = 60°.
Consequently,
∠e = 90° − ∠f = 90° − 60° = 30°
Also, ∠b = ∠e = 30°

In ΔBGL,
∠d = 180° − ∠BLG − ∠e
⇒ ∠d = 180° − 90° − 30° = 60°.
⇒ ∠d = 60°
Also, ∠c = ∠d = 60°.

At point F,
98° + ∠i +∠g = 180° [Forming straight angle]
⇒ 98° + ∠i + 60° = 180°
⇒ ∠i = 22°.

Around the point G,
∠j + 56° + 30° + ∠c + ∠d + ∠h = 360° [Forming complete angle]
⇒ ∠j + 56° + 30° + 60° + 60° + 60° = 360°
⇒ ∠j = 94°.

In ΔFGH,
∠k = 180° − ∠i − ∠k = 64°
⇒ ∠k = 180° − 22° − 94° = 64°.

At point V,
68° + ∠HVD = 180° [Forming linear pair]
⇒ ∠HVD = 180° − 68° = 112°

In ΔHVD, VH = VD, so ∠p = ∠n
Therefore, ∠p = ∠n = (180° − 112°)/2 = 34°.
Thus, ∠o = 90° − 34° = 56°

In ΔFHD,
∠m = 180° − 98° − ∠m = 26°
⇒ ∠m = 180° − 98° − 56° = 26°.