Question

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Answer 1

Answer:

Step-by-step explanation:

From the figure it is given that, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE.

(i) We have to prove that, EF = FC

From the figure, D is the midpoint of BC and also DF parallel to BE.

So, F is the midpoint of EC

Therefore, EF = FC

= ½ EC

EF = ½ AE

(ii) Now consider the ∆AGE and ∆ADF

Then, (BG or GE) ||DF

Therefore, ∆AGE ~ ∆ADF

So, AG/GD = AE/EF

AG/GD = 1/½

AG/GD = 1 × (2/1)

Therefore, AG: GD = 2: 1

20.

(a) In the figure given below, AB, EF and CD are parallel lines. Given that AB =15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm. Calculate

(i) EF

(ii) AC.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 25

Solution:-

From the figure it is given that, AB, EF and CD are parallel lines.

(i) Consider the ∆EFG and ∆CGD

∠EGF = ∠CGD [Because vertically opposite angles are equal]

∠FEG = ∠GCD [alternate angles are equal]

Therefore, ∆EFG ~ ∆CGD

Then, EG/GC = EF/CD

5/10 = EF/18

EF = (5 × 18)/10

Therefore, EF = 9 cm

(ii) Now, consider the ∆ABC and ∆EFC

EF ||AB

So, ∆ABC ~ ∆EFC

Then, AC/EC = AB/EF

AC/(5 + 10) = 15/9

AC/15 = 15/9

AC = (15 × 15)/9

Therefore, AC = 25 cm

Answer 2

Step-by-step explanation:

Here ur solution.

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