Maths WBBSE Class 10 Solutions Chapter 18 Similarity Exercise 18.1
Question 1. But will the two congruent figures be similar? Since the shape and size in the two congruent figures are equal, the two congruent are always 
[similar/not similar].
Answer. Similar.
Question 2. I observe whether the pictures in each group are similar or not.
Answer. I am observing that the pictures of groups 1, 2, and 4 are similar to each other. But the pictures of the grups 3 and 5 are not similar to each other.
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Question 3. Let us write the right answer in
1. All squares are [congruent/similar]
Answer. Similar.
2. All circles are [congruent/similar]
Answer. Similar.
Maths WBBSE Class 10 Solutions
3. All [equilateral/isosceles] triangles are always similar.
Answer. Equilateral.
Question 4. Two quadrilaterals will be similar if their corresponding angles are [equal/proportional] and corresponding sides are [unequal/proportional].
Answer. Equal and Proportional.
Chapter 18 Similarity Exercise 18.1 True or False
1. Any two congruent figures are similar.
Answer: True
2. Any two similar figures are always congruent.
Answer: False
3. The corresponding angles of any two polygonal figures are equal.
Answer: True
4. The corresponding sides of any two polygonal figures are proportional.
Answer: True
5. The square and rhombus are always similar.
Answer: False
Maths WBBSE Class 10 Solutions
Question 5. Let us write examples of one pair of similar.
Solution:
Question 6. Let us draw a pair of figures which are not similar.
Solution:
The two 1 & 2 are similar and the two 3 & 4 are similar. Because their corresponding sides are and their corresponding angles are equal.
Answer. Similar.
Question 7. Now by cutting off ∠AA, and B, and putting them on ABC, I am observing that the two angles are overlapping.
Question 8. In a similar way, I have drawn another triangle and a straight line that has divided any two other sides (or their extended sides) in the same ratio and I am observing that the straight line is parallel to the third side.
Solution: I draw ΔABC, straight line DE, cut AB & AC at D & E proportionately,
AD/BD = AE/EC
∴ DE || BC
Again I draw AABC, straight line DE produced cuts in proportion.
AB/AD = AC/AE
∴ BC || DE
WBBSE Solutions Guide Class 10
Application 1. In adjoining, DE || BC of ΔABC; if AD = 5 cm. DB = 6 cm and AE = 7.5 cm, then let us write by calculating the length of AC.
Solution: In ΔABC, DE BC,
AD/DB = AE/EC [From Thales theorem]
5/6 = 7.5/EC
∴ EC = 7,5 + 6/5 cm
= 9 cm.
∴ AC = AE + EC
= 7.5cm + 9cm
= 16.5cm.
WBBSE Solutions Guide Class 10
Application 2. If BC || DE in AABC, AD/DB = 2/5, and AC = 21 cm, then let us write by calculating the value of AE
Solution: As BC II BE
∴ AD/DB = AE/EC = 2/5
Set AE = x cm
∴ EC = (2x – x) cm
∴ X/21-x = 2/5
Or, 5x = 42 – 2x
or, 7x = 42
∴ x = 6
∴ AE = 6cm
West Bengal Board Class 10 Math Book Solution In English
Application 7. With the help of the converse of Thales theorem, let us prove that the line segment joining two mid-points of a triangle is parallel to its third side.
Solution: In ΔABC,
D & E are the midpoints of AB & AC, respectively
Join D, E
∴ AD/BD = AE/EC = 1/1
∴ BC || DE
∴ The line segment joining two midpoints of a triangle is parallel to its third side.
