NCERT Exemplar Solutions
Class 6 - Mathematics - Unit 2: Geometry - Fill in the Blanks
Question 21

Question. 21

21. In Fig. 2.15, points A, B, C, D and E are collinear such that AB = BC = CD = DE. Then

(a) AD = AB + _____

(b) AD = AC + _____

(c) mid point of AE is _____

(d) mid point of CE is _____

(e) AE = _____ × AB

Answer:

(a) AD = AB + BC + CD

(b) AD = AC + CD

(c) Midpoint of AE is C

(d) Midpoint of CE is D

(e) AE = 4 × AB

Detailed Answer with Explanation:

All five points are on one straight line and the four small segments are equal:

Let

\[ AB = x \]

Then

\[ BC = x,\quad CD = x,\quad DE = x \]


Build longer segments from equal parts

\[ AC = AB + BC = x + x = 2x \]

\[ AD = AB + BC + CD = x + x + x = 3x \]

\[ AE = AB + BC + CD + DE = x + x + x + x = 4x \]


(a) From the diagram, going from A to D passes through B and C:

\[ AD = AB + BC + CD = AB + (\,\underline{BC + CD}\,) \]

(b) First reach C from A, then add one more equal part to reach D:

\[ AC = AB + BC = 2x \]

\[ AD = AC + CD = AC + (\,\underline{CD}\,) \]

(c) Midpoint of \(AE\):

\[ AE = 4x \Rightarrow \text{half of } AE = 2x \]

Starting at A, after two equal parts (\(AB\) then \(BC\)) we reach C.

So, the midpoint of \(AE\) is \(\underline{C}\).

(d) Midpoint of \(CE\):

\[ CE = CD + DE = x + x = 2x \Rightarrow \text{half of } CE = x \]

From C, one equal part (\(CD\)) takes us to D.

So, the midpoint of \(CE\) is \(\underline{D}\).

(e) Compare \(AE\) with \(AB\):

\[ AE = 4x = \underline{4}\times AB \]

Hence the filled answers follow directly from adding equal segments and identifying halfway points.

NCERT Exemplar Solutions Class 6 – Mathematics – Unit 2: Geometry – Fill in the Blanks | Detailed Answers