NCERT Solutions 10th Maths Chapter 6 Triangles Exercise 6.2

NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 we will restart our exploration of the world of Triangles. Thus, we are providing you Chapter 6 Triangles NCERT Class 10 Maths Solutions that will help in achieving more marks. You don't have to wander and waste your precious time in finding best CBSE NCERT Solutions.

Exercise 6.2
1. In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Answer
(i) (i) In △ ABC, DE∥BC (Given)
∴ AD/DB = AE/EC [By using Basic proportionality theorem]
⇒ 1.5/3 = 1/EC
⇒ Σ EC = 3/1.5
EC = 3×10/15 = 2 cm
Hence, EC = 2 cm.
(ii) In △ ABC, DE∥BC (Given)
∴ AD/DB = AE/EC [By using Basic proportionality theorem]
⇒ AD/7.2 = 1.8/5.4
⇒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 ?PQR. 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
Answer

In △PQR, E and F are two points on side PQ and PR respectively.
(i) PE = 3.9 cm, EQ = 3 cm (Given)
PF = 3.6 cm, FR = 2,4 cm (Given)
∴ PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3 [By using Basic proportionality theorem]
And, PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5
So, PE/EQ ≠ PF
Hence, EF is not parallel to QR.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8cm, RF = 9cm
∴ PE/QE = 4/4.5 = 40/45 = 8/9 [By using Basic proportionality theorem]
And, PF/RF = 8/9
So, PE/QE = PF/RF
Hence, EF is parallel to QR.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm (Given)
Here, 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)
∴ PE/EQ = PF/FR.
Hence, EF is parallel to QR.
3. In the fig 6.20, DE||OQ and DF||OR, show that EF||QR.
Answer
In △PQO, DE || OQ (Given)
∴ PD/DO = PE/EQ ...(i) [By using Basic Proportionality Theorem]
In △PQO, DE || OQ (Given)
∴ PD/DO = PF/FR ...(ii) [By using Basic Proportionality Theorem]
From equation (i) and (ii), we get
PE/EQ = PF/FR
In ΔPQR, EF || QR. [By converse of Basic Proportionality Theorem]