During revision, students quickly go through Class 7 Maths Extra Questions Part 2 Chapter 1 Geometric Twins Class 7 Extra Questions with Answers for clarity.
Class 7 Geometric Twins Extra Questions
Class 7 Maths Chapter 1 Geometric Twins Extra Questions
Geometric Twins Extra Questions Class 7
Question 1.
Without drawing the triangles, state the correspondence between the vertices, sides and the angles of the following pairs of triangles.
(a) ΔABC ≅ ΔPQR
(b) ΔDEF ≅ ΔBCA
(a) Corresponding vertices: A 
P, B Q, C R
Corresponding sides: AB PQ, BC QR, AC PR
Corresponding angles: ∠A ∠P, ∠B ∠Q, ∠C∠R
(b) Corresponding vertices: DB. E C, F A
Corresponding sides: DE BC, EF CA, DFBA
Corresponding angles: ∠D ∠B, ∠E ∠C, ∠F∠A
Question 2.
In the given figure, AD = CD and AB = CB.
(a) State the three pairs of equal parts in ΔABD and ΔCBD.
(b) Is ΔABD ≅ ΔCBD? Why or why not’
(e) Does BD bisect ∠ABC? Give reasons.
Solution:
(a) In ΔABD and ΔCBD.
AB = CB (Qiven)
AD = CD (Given)
and BD = BD (Common side to both triangles)
(b) From part (a) above.
ΔABD ≅ ΔCBD (By SSS congruence condition)
(c) ∠ABD ∠CBD, as they are corresponding parts of congruent triangles.
Hence, BD bisects ΔABC.
Question 3.
In the adjoining figure, DA ⊥ AB,
CB ⊥ AB and AC = BD. State the three pairs of equal parts in ΔABC and ΔBAD.
Also, show that ΔABC ≅ ΔBAD
Solution:
The three pairs of equal parts are:
∠ABC = ∠BAD (Each 90°)
AC = BD (Given)
AB = BA (Common Side)
From the above,
ΔABC = ΔBAD
(By RHS congruence condition)
Question 4.
In the adjoining figure, AB = AC and D is the mid-point of BC.
(a) State the three pairs of equal parts in AADB and A ADC.
(b) Is ΔADB = ΔADC? Give reasons.
(c) Is ∠BAD = ∠CAD? Why?
Solution:
(a) AB = AC (Given)
AD = AD (Common side)
BD = CD (D is the midpoint of BC)
(b) From part (a) above,
ΔADB ≅ ΔADC (By SSS congruence condition)
(c) Yes, ∠BAD ≅ ∠CAD,
as they are corresponding parts of congruent triangles.
Question 5.
Use ASA congruence condition and verify that
ΔAOC ≅ ΔBOD?
Solution:
In the two triangles. AOC and BOD,
∠C = ∠D (Each 70°)
Also, ∠AOC = ∠BOD = 30°
(Vertically opposite angles)
So, in ΔAOC, ∠A= 180° – (70° + 30°) = 80°
(Using angle sum property of a triangle)
Similarly, in ΔBOD, ∠B = 180° – (70° + 30°) = 80°
Also, AC = BD (Each 4 cm)
Thus, we have ∠A = ∠B, AC = BD and ∠C = ∠D
Now, side AC is between ∠A and ∠C and side BD is between ∠B and ∠D.
So, by ASA congruence condition, ΔAOC ≅ ΔBOD.
Question 6.
Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not, by ASA congruence condition. In case of congruence, write it in symbolic form.
| ΔDEF | ΔPQR | |
| (a) | ∠D = 60°, ∠F = 80°, DF = 5 cm |
∠Q = 60°, ∠R = 80°, QR = 5 cm |
| (b) | ∠D = 60°, ∠F = 80°, DF = 6 cm |
∠Q = 60°, ∠R = 80°, QP = 6 cm |
| (c) | ∠E = 80°, ∠F = 30°, EF = 5 cm |
∠P = 80°, PQ = 5 cm, ∠R = 30° |
(a) ∠D = 60°, ∠F = 80°, DF = 5 cm
∠Q = 60°, ∠R = 80°, QR = 5 cm
The two given angles correspond (D Q, F R)
and DF is the side included between ∠D and ∠F while QR is the side included between ∠Q and ∠R.
So, by ASA congruence condition, the triangles are congruent.
Symbolically: ΔDEF = ΔQPR (D→Q, E→P, F→R)
(b) ∠D = 60°, ∠F = 80°, DF = 6 cm
∠Q = 60°, ∠R = 80°, QP = 6 cm
Although the two angles match, the given side QP is not the side included between ∠Q and ∠R (the included side would be QR).
So, ASA congruency condition does not apply → the triangles are not congruent by ASA.
(c) ∠E = 80°, ∠F = 30°, EF = 5 cm
∠P = 80°, ∠R = 30°, PQ = 5 cm
The two angles match (EP, FR), but EF is the side included between those angles while PQ is not the side included between ∠P and ∠R (the included side would be PR).
So, ASA congruency condition does not apply → the triangles are not congruent by ASA.
The post Geometric Twins Class 7 Extra Questions Maths Part 2 Chapter 1 appeared first on Learn CBSE.
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