Test (elaborations) Maths

Access Answers for SAT Maths Chapter 6 – Triangles
Exercise 6.1
1. Fill in the blanks using correct word given in the brackets: -
(i) All circles are __________. (Congruent, similar)
Answer: Similar
(ii) All squares are __________. (Similar, congruent)
Answer: Similar
(iii) All __________ triangles are similar. (Isosceles, equilateral)

Answer: Equilateral
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are
__________ and (b) their corresponding sides are __________. (Equal, proportional)
Answer: (a) Equal
(b) Proportional
2. Give two different examples of pair of
(i) Similar figures
(ii) non-similar figures
Solution:




3. State whether the following quadrilaterals are similar or not:

,Solution:
From the given two figures, we can see their corresponding angles are different or unequal.
Therefore, they are not similar.




Exercise 6.2
1. In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).




Solution:
(i) Given, in △ ABC, DE∥BC
∴ AD/DB = AE/EC [Using Basic proportionality theorem]
⇒1.5/3 = 1/EC
⇒EC = 3/1.5
EC = 3×10/15 = 2 cm
Hence, EC = 2 cm.
(ii) Given, in △ ABC, DE∥BC

,∴ AD/DB = AE/EC [Using Basic proportionality theorem]
⇒ AD/7.2 = 1..4
⇒ AD = 1.8 ×7.2/5.4 = (18/10)×(72/10)×(10/54) = 24/10
⇒ AD = 2.4
Hence, AD = 2.4 cm.
2. E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following
cases, state whether EF || QR.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm
Solution:
Given, in ΔPQR, E and F are two points on side PQ and PR respectively. See the figure below;




(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm
Therefore, by using Basic proportionality theorem, we get,
PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3
And PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5
So, we get, PE/EQ ≠ PF/FR
Hence, EF is not parallel to QR.
(ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm
Therefore, by using Basic proportionality theorem, we get,
PE/QE = 4/4.5 = 40/45 = 8/9
And, PF/RF = 8/9
So, we get here,
PE/QE = PF/RF
Hence, EF is parallel to QR.

, (iii) Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
From the figure,
EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm
So, PE/EQ = 0.18/1.10 = 18/110 = 9/55…………. (i)
And, PE/FR = 0.36/2.20 = 36/220 = 9/55………… (ii)
So, we get here,
PE/EQ = PF/FR
Hence, EF is parallel to QR.
3. In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD




Solution:
In the given figure, we can see, LM || CB,
By using basic proportionality theorem, we get,
AM/AB = AL/AC……………………..(i)
Similarly, given, LN || CD and using basic proportionality theorem,
∴AN/AD = AL/AC……………………………(ii)
From equation (i) and (ii), we get,
AM/AB = AN/AD
Hence, proved.
4. In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC