By using Ganita Prakash Class 7 Solutions and Part 2 Chapter 1 Geometric Twins Class 7 Question Answer, students can improve their problem-solving skills.
Class 7 Maths Ganita Prakash Part 2 Chapter 1 Solutions
Class 7 Maths Geometric Twins Solutions
Class 7 Ganita Prakash Part 2 Chapter 1 Solutions Geometric Twins
Figure it Out (Page : 3-4)
Question 1.
Check if the two figures are congruent.
Solution:
Let’s measure the angles:
Although both figures have the same arm lengths, but their included angles does not coincide. Therefore, the two figures are not congruent.
Question 2.
Circle the pairs that appear congruent.
Solution:
(a) and (d) are congruent. As they can be superimposed exactly.
Question 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?
Solution:
I will place circle and rectangle one over another. If they exactly superimpose, and satisfied following conditions they are congruent.
(a) Two circles are congruent if they have the same radius or diameter.
(b) Two rectangles are congruent if they have the same length and breadth.
Question 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.
Solution:
To check if the two figures are congruent, one would need to measure the lengths of the corresponding line segments and the angle between them.
Yes, both pairs are congruent because their arm lengths 3.3 cm (Horizontal line) and 2.3 cm (Vertical line) and the included angles (82°) between them are equal.
Figure it Out (Page: 8-9)
Question 1.
Suppose △HEN is congruent to △BIG. List all the other correct ways of expressing this congruence.
Solution:
Given
Δ HEN ≅ ΔBIG, then the correspondence is:
H ↔ B
E ↔ I
N ↔ G
NEH ≅ GIB.
All other correct ways to express the same Congruence are:
(i) △HEN ≅ △BIG
(ii) △ENH ≅ △IGB
(iii) △NHE ≅ △GBI
(iv) △HNE ≅ △BGI
(v) △NEH ≅ △GIB
(vi) △EHN ≅ △IBG
(All these six statements gives the same Vertex Correspondence.)
Question 2.
Determine whether the triangles are congruent. If yes, express the congruence.
Solution:
In △RED and △JAM
RE = JA = 3.5 cm (Given)
RD = JM = 6 cm (Given)
ED = AM = 5 cm (Given)
ΔRED ≅ ΔJAM (S.S.S)
∴ Yes, given Δ ‘s are congruent.
Question 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)
BC = DC (Given)
AC = AC (Common side)
△ ABC ≅ △ ADC (S.S.S. condition)
∠BAC ≅ ∠DAC (Corresponding part of Congruent triangles i.e CPCT).
∠BCA =∠DCA (CPCT)
∴ Yes, AC bisects both ∠BAD and ∠BCD into two equal parts.
Question 4.
In the figure below, are △DFE and △GED congruent to each other? It is given that DF = DG and FE = GE.
Solution:
In △DEF and △GED
DF = DG (Given)
FE = EG (Given)
DE = DE (Common)
Hence, △ DEF ≅ △ GED(SSS condition)
Figure it Out (Page: 13-14)
Question 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.
Question 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, Δ OAB ≅ Δ OCD.
Therefore, the corresponding equal parts are OA = OC, OB = OD, ∠AOB =∠COD.
Question 3.
Given that ∠ABC=∠DBC and ∠ACB =∠DCB, show that ∠BAC =∠BDC. Are the two triangles congruent?
Solution:
In △ABC and △DBC
1. ∠ABC = ∠DBC (Given)
2. (BC = BC) (common side)
3. ∠ACB = ∠DCB (Given)
Therefore, △ABC ≅ △ DBC by A.S.A. Congruence criterion
Question 4.
Identify the equal parts in the following figure, given that ∠ABD = ∠DCA and ∠ACB = ∠DBC.
Solution:
∠ABD =∠DCA ……. .(Given)
∠ACB = ∠DCB ……. (Given)
In △ABC and △DCB
∠ABD = ∠DCA …….. (proved above)
BC = CB …….. (common side)
∴ △ ABC ≅ △ DCB (A.S.A. condition)
Therefore, corresponding equal parts are.
∠BAC =∠CDB, AB = DC, AC = DB.
Figure it Out (Page: 20-21)
Question 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
Question 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 = D F
∠B =∠D = 90°
AC = FE
(d) ∠A =∠D
∠B =∠E
AC = DF
(e) AB = DF
∠B =∠F
AC = DE
Solution:
(a) The triangles are Congruent by the Side-by-Side (SSS) Congruence Criterion.
△ ABC ≅ △ DEF
(b) The triangles are Congruent by the Side-Angle-Side (SAS) Congruence Criterion.
△ABC ≅ △ EFD
(c) The triangles are Congruent by the Right-Angle-Hypotenuse-Side (RHS Congruence Criterion.
△ ABC ≅ △ FDE
(d) The triangles are Congruent by the Angle- Angle-Side (AAS) Congruence Criterion.
△ ABC ≅ △ DE
(e) The triangles are not necessarily Congruent. The two equal sides are not corresponding and the angle is not included between them. This is the Angle-Side-Side (ASS) case, which is not a Valid Congruence Criterion.
Question 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:
Consider the △AOB and △DOC
OA = OD …….. (given)
∠AOB =∠DOC (vertically opposite angles are always equal).
OB = OC …….. (given)
By the Side-Angle-Side (SAS), the two triangles are congruent.
∴ △ 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 = CD.
Hence, AB || CD.
Question 4.
ABCD is a square. Show that △ ABC ≅ Δ ADC. Is Δ ABC also congruent to Δ 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, Δ ABC ≅ △ ADC
Yes, △ABC is also congruent to △CDA, because △CDA is the same triangle as Δ ADC written in a different order.
Question 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°.
Question 6.
Find the missing angles. As per the convention that we have been following, all line segments marked with a single ‘ I ‘ are equal to each other and those marked with a double ‘|’ are equal to each other, etc. Angles is the upper Triangle (around point A).
Solution:
In △CUR
∠CUR =∠CRU = x (say)
CU = CR
According to angle sum property of triangle
x + x + 90° = 180°
⇒ 2 x = 180° – 90° = 90°
⇒ x = 45°
∴ ∠CUR = ∠CRU = 45°
In △VRN
∠VRN =∠VNR = a (say)
[∵ Angles opposite to equal sides are equal]
∵ VR = VN
Since sum of angles of a triangle is 180°.
So,
a + a + 68° = 180°
2a = 180° – 68°
2a = 112°
a = 56°
∠VRN = ∠VNR = 56°
In △AUP
∠UAP = ∠UPA (∵ they are equal)
∠UPA = 56°
Sum of angles of a triangle is 180°.
So,
56° + 56° + ∠AUP = 180°
112° + ∠AUP = 180°
∠AUP = 180° – 112° = 68°
△BOF is an equilateral triangle as all sides are equal.
So,OB = OF = BF
∠FOB = ∠FBO =∠OFB = 60°
∠RVN + ∠DVN = 180°
68° + ∠DVN = 180°
∠DVN = 180° – 68°
∠DVN = 112°
∠VND + ∠VDN + ∠NVD = 180°
∵ VN = VD
∴ ∠VND = ∠VDN = c
∴ c + c + 112° = 180°
2c = 180° – 112°
2c = 68°
c = 34°
∠VND =∠VDN = 34°
In △OLB
∠OBL = 90°-60° = 30°
∠LOB = 60°
[∵ LO || BF and BO is transversal]
In △OPN
∠OPN + ∠PON + ∠PNO = 180°
∠OPN + 56°+90° = 180°
∠OPN +146° = 180°
∠OPN = 180°-146°
∠OPN = 34°
Now,
∠APK + ∠KPO + ∠OPN = 180°
[ ∵ Straight angle is 180°]
44°+∠KPO+34° = 180°
∠KPO = 180° – 78°
∠KPO = 102°
In △KPO
∠KPO + ∠POK + ∠PKO = 180°
102°+ 30°+∠PKO = 180°
132°+∠PKO = 180°
∠PKO = 180° – 132°
= 48°
∠KAP + ∠KPA + ∠AKP = 180°
[ ∵ Sum of angles of a triangle is 180°]
34° + 44° + ∠AKP = 180°
78° + ∠AKP = 180°
∠AKP = 180° – 78°
∠AKP = 102° and∠PKO = 48°
So,
∠AKP + ∠PKO + ∠OKL = 180°
102° + 48° + ∠OKL = 180°
150°+ ∠OKL = 180°
∠OKL = 180°-150°
∠OKL = 30°
In △KOL
∠OKL +∠OLK + ∠KOL = 180°
30°+ 90°+∠KOL = 180°
∠KOL = 180° – 120°
= 60°
∠KOL = 60°
Also,
Δ OKL ≅ △ OBL
KL = LB
∠OLK ≅ ∠OLB = 90°
Side angle side condition.