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# In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)

**Solution:**

As we all know the Basic Proportionality Theorem (B.P.T) or (Thales Theorem)

Two triangles are similar if

- Their corresponding angles are equal
- Their corresponding sides are in the same ratio (or proportion)

In, ΔABC

BC || DE

In ΔABC and ΔADE

∠ABC = ∠ADE [corresponding angles]

∠ACB = ∠AED [ corresponding angles]

∠A = ∠A common

⇒ ΔABC ~ ΔADE

AD/DB = AE/EC

1.5/3 = 1/EC

EC = 3 × 1/1.5

EC = 2 cm

(ii) In ΔABC and ΔADE

∠ABC = ∠ADE [corresponding angles]

∠ACB = ∠AED [ corresponding angles]

∠A = ∠A common

ΔABC ∼ ΔADE

AD/DB = AE/EC

AD/7.2 = 1.8/5.4

AD = (7.2 × 1.8)/5.4

AD = 2.4 cm

**☛ Check: **Class 10 Maths NCERT Solutions Chapter 6

**Video Solution:**

## In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)

Class 10 Maths NCERT Solutions Chapter 6 Exercise 6.2 Question 1

**Summary:**

For the given figure (i) and (ii), if DE || BC the missing sides are EC = 2 cm and AD = 2.4 cm.

**☛ Related Questions:**

- 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 = 0.36 cm
- In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
- In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
- In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

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