Figure it Out (Page 3 – 4)

1. Check if the two figures are congruent.

Solution:

Although both figures have the same arm lengths, their included angles are different. Therefore, the two figures are not congruent

2. Circle the pairs that appear congruent.

Solution:

3. What measurements would you take to create a figure congruent to a given:
(a) Circle
(b) Rectangle
Solution:
(a) To create a circle congruent to a given circle, measure only the radius.
(b) To create a rectangle congruent to a given rectangle, measure both the length and the breadth.

Q. Using this, state how would you check if two —
(a) Circles are congruent?
(b) Rectangles are congruent?
Solution:
(a) Two circles are congruent if they have the same radius.
(b) Two rectangles are congruent if they have the same length and the same breadth.

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

Solution:
If two figures have the same arm lengths and the angle between those arms is equal, then the figures are congruent.

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

Solution:
Both pairs are congruent because their arm lengths and the included angles between them are equal.

Figure it Out (Page 8 – 9)

1. Suppose ∆HEN is congruent to ∆BIG. List all the other correct ways of expressing this congruence.
Solution:
If ∆HEN ≅ ∆BIG, then the correspondence is:
H ↔ B
E ↔ I
N ↔ G
All the six correct ways of expressing the congruence are:
1. ∆HEN ≅ ∆BIG
2. ∆ENH ≅ ∆IGB
3. ∆NHE ≅ ∆GBI
4. ∆EHN ≅ ∆IBG
5. ∆HNE ≅ ∆BGI
6. ∆NEH ≅ ∆GIB

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

Solution:
In △RED and △JAM,
RE = JA = 3.5 cm
ED = AM = 5 cm
RD = JM = 6 cm
Thus, the triangles satisfy the SSS condition.
So, the congruence is written as:
ΔRED ≅ ΔJAM.

3. 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:
In △ABC and △ADC,
AB = AD …….. (Given)
CB = CD …….. (Given)
AC = AC …….. (Common side)
Thus, the traingles satisfy the SSS condition.
Hence, △ABC ≅ △ ADC.
Thererfore, the corresponding equal parts are:
∠BAC = ∠DAC. Hence, AC bisects ∠BAD.
∠BCA = ∠DCA. Hence, AC bisects ∠BCD.

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

Solution:
In ΔDFE and ΔDGE,
DF = DG …….. (Given)
FE = GE …….. (Given)
DE = DE …….. (Common)
Thus, the triangles satisfy the SSS codition.
Hence, ΔDFE ≅ ΔDGE.

Figure it Out (Page 13 – 14)

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

Solution:
In △ABC and △XZY,
AB = XZ = 7 cm ……. (given)
BC = ZY = 5 cm …….. (given)
∠B = ∠Z = 47° ……… (given)
Thus, the triangles satisfy the SAS condition..
So, the congruence is written as:
△ABC ≅ △XZY.

2. 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:
Since AB ∥ CD, the transversal lines AC and BD form alternate interior angles.
So,
∠OAB = ∠OCD ……….. (alternate interior angles)
∠OBA = ∠ODC ………… (alternate interior angles)
Also, it is given that AB = CD.
In triangles △OAB and △OCD,
AB = CD ……. (given)
∠OAB = ∠OCD ……….. (alternate interior angles)
∠OBA = ∠ODC ………… (alternate interior angles)
Thus, the triangles satisfy the ASA condition.
Hence, △OAB ≅ △OCD.

Therefore, the corresponding equal parts are OA = OC, OB = OD, ∠AOB = ∠COD.

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

Solution:
In triangles △ABC and △DBC,
∠ABC = ∠DBC ………. (given)
BC = BC ………… (common side)
∠ACB = ∠DCB ………. (given)
Thus, the triangles satisfy the ASA condition.
Hence, △ABC ≅ △DBC.
Therefore, the corresponding equal parts are ∠BAC = ∠BDC.

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

Solution:
∠ABD = ∠DCA and ∠ACB = ∠DBC …………. (given)
∠ABD + ∠DBC = ∠ABC
∠DCA + ∠ACB = ∠DCB

In △ABC and △DCB,
∠ABC = ∠DCB ……… (proved above)
BC = CB ………… (common side)
∠ACB = ∠DBC ………… (given)
Thus, the triangles satify the ASA condition.
Hence, △ABC ≅ △DCB.
Therefore, the corresponding equal parts are:
AB = DC, AC = DB, ∠BAC = ∠CDB.

Figure it Out (Page 13 – 14)

1. ∆AIR ≅ ∆FLY. Identify the corresponding vertices, sides and angles.
Solution:
In ∆AIR ≅ ∆FLY,
The corresponding vertices are A ↔ F, I ↔ L, R ↔ Y.
The corresponding sides are AI = FL, IR = LY, AR = FY.
The corresponding angles are ∠A = ∠F, ∠I = ∠L, ∠R = ∠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) 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
Solution:
(i) AB = DE, BC = EF, CA = DF
All three corresponding sides are equal.
Thus, the triangles satisfy the SSS condition.
Hence, △ABC ≅ △DEF.

(ii) AB = EF, ∠A =∠E, AC = ED
Two corresponding sides and the included angle are equal.
Thus, the triangles satisfy the SAS condition.
Hence, △ABC ≅ △EFD.

(iii) AB = DF, ∠B = ∠D = 90°, AC = FE
The triangles have equal right angles, equal hypotenuses, and one equal corresponding side.
Thus, the triangles satisfy the RHS condition.
Hence, △ABC ≅ △FDE.

(iv) ∠A = ∠D, ∠B = ∠E, AC = DF
Two corresponding angles and one corresponding side are equal.
Thus, the triangles satisfy the AAS condition.
Hence, △ABC ≅ △DEF.

(v) AB = DF, ∠B = ∠F, AC = DE
Here, two corresponding sides and a non-included angle are equal.
Thus, the triangles satisfy the SSA condition, which is not a valid congruence rule.
Hence, △ABC need not be congruent to △DFE.

3. 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:
In △AOB and △DOC,
OA = OD ……… (given)
OB = OC ………… (given)
∠AOB = ∠DOC ………… (vertically opposite angles)
Thus, the triangles satify the SAS condition.
Hence, △AOB ≅ △DOC.

Therefore, the corresponding equal parts are: ∠OAB = ∠ODC and ∠OBA = ∠OCD.
Since these angles are alternate interior angles formed by the transversal AD and BC with the pair of lines AB and CD.
Hence, AB Ⅱ CD.

4. ABCD is a square. Show that ∆ABC ≅ ∆ADC. Is ∆ABC also congruent to ∆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?
Solution:
In square ABCD, all sides are equal.
So,
In △ABC and △ADC,
AB = CD
BC = AD
Diagonal AC is common.
Thus, the triangles follow the SSS condition.
Hence, △ABC ≅ △ ADC
Yes, △ABC is also congruent to △CDA, because △CDA is the same triangle as △ADC written in a different order.

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

Solution:
In △ABC,
AB = AC (radius of circle)
∠C = ∠B (angles opposite to equal sides are equal)
Also,
∠C + ∠B + ∠BAC = 180° (sum of angles of triangle)
∠B + ∠B + 120° = 180°
2∠B = 180° – 120°
2∠B = 60°
∠B = 60°/2 = 30°
Therefore, ∠C = ∠B = 30°.

6. 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:
Do it yourself.