Question 1
Question: Say whether the following quadrilateral are similar or not.
(i) 
(ii) 
Answer: (i)ABCD similar to PQRS (ii)ABCD is not similar to PQRS.
Question 2
Question: Give two examples of pairs of similar figures.
Answer: (i) 
(ii) 
Question 3
Question: ABCD and EFGH are similar rectangles. What should be the measure of EF? 
Answer: EF = 2 x 2.5 = 5cm.
Because GF = HE = 2 x BC = 2 x 1.8 = 3.6cm
Question 4
Question: What is the difference between similarity and congruent?
Answer: Similarity: The shapes of two figures are same.
Congruent: The size and shape of two figures are same.
Question 5
Question: If in a D ABC, DE||BC as shown in the figure. [3 Mark]
Show that
i. 
ii. 
Answer: If DE||BC (given)
... (1) (by basic proportionality theorem)
Adding 1 on both sides,
(On simplification)
.... (2)
Thus (i) is proved.
(Reciprocal ratio of (1))
Adding 1 on both sides,
Thus (ii) is proved.
| Note: |
| The following results can be obtained from BPT |
In a D ABC, DE||BC and intersects AB in D and AC in E, then
You should be familiar with these results to use them in the problems.Note:The following results can be obtained from BPT
Question 6
Question: Prove mid-point theorem from converse of Basic Proportionality Theorem. [2 Mark]
Answer: 
Given:
D and E are the mid-points of AB and AC respectively of a triangle ABC.
To prove:
DE||BC
Proof:
.... (1) (Since D is the mid-point of AB, AD = DB)
From (1) and (2), we have
DE||BC (
If a line divides any two sides of a triangle in the same ratio, the line is parallel to the third side)
Question 7
Question: Prove the converse of mid-point theorem using Basic Proportionality Theorem. That is, prove that the line drawn from the mid-point of one side of a triangle, parallel to the second side, bisects the third side. [2 Mark]
Answer: 
Given:
ABC is a triangle where AD = DB and DE||BC.
To prove: AE = EC
Proof:
DE||BC (given)
AD = DB (given)
.... (1)
The Basic Proportionality Theorem states that 'If a line is drawn parallel to one side of a triangle, then other two sides are divided in the same ratio'.
.... (1) (since D is the mid-point of AB, AD = DB)
Since L.H.S of (1) and (2) are equal, their R.H.S are also equal
i.e., DE bisects AC .
Question 8
Question: Given D ABC and DE||BC. [2 Mark]
AD = 4x - 3 DB = 3x - 1
AE = 8x - 7 EC = 5x - 3
Find the value of x.
Answer: Since DE||BC, using Basic Proportionality Theorem, we get
Discard the value 
because the distance becomes negative on substituting this value, which is not possible.
\ x = 1
Question 9
Question: P is the mid-point of side BC of a D ABC. If Q is the mid-point of AP and BQ when produced meets AC in L, then prove that 
[3 Mark]
Answer: 
Given:
In triangle D ABC, BP = PC and AQ = QP.
To prove:
Construction: Through P, draw PM parallel to BL. Proof: In D APM, QL||PM (by construction)
.... (1) (By BPT)
But 
L.H.S of (1) and (2) are equal, therefore R.H.S of (1) and (2) must be equal
(3)
In D CBL, PM||BL
(4) (BPT)
But 
(5) (given)
(From (4) and (5))
(6) From (3) and (6),
AL = LM = CM ...(7)
(from (7))
Question 10
Question: Prove that the diagonals of a trapezium divide each other proportionally. [2 Mark]
Answer: 
Given: ABCD is a trapezium. Diagonals AC and DB intersect at E.
To prove:
Construction:
Through E, draw EF||AB.
Proof:
In 
(AB ll EF (by construction))
(1) (By BPT)
(2) (Taking the reciprocal ratio of (1))
In D DAB,
EF||AB (by construction)
(3) L.H.S of (2) and (3) are equal, therefore R.H.S must be equal
