Chapter 5: Introduction to Euclid’s Geometry

Q1Written answer

Two salesmen make equal sales during the month of August. In September, each salesman doubles his sale of the month of August. Compare their sales in September.

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Worked solution

Let the sale of both the salesmen in August = x.

According to the question, we have,

In September, each salesman doubles his sales of August.

Hence, we have,

In September,

Sales of first salesmen = 2x

And, sales of second salesman = 2x.

According to Euclid’s axioms, things which are double of the same things are equal to one another.

Therefore, in September their sales are again equal.

Q2Written answer

It is known that x + y = 10 and that x = z. Show that z + y = 10?

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Worked solution

According to the question,

We have,

x+y=10 …(i)

And, x=z …(ii)

Applying the Euclid’s axiom,

“if equals are added to equals, the wholes are equal”

We get,

From Eqs. (i) and (ii)

x+y=z+y ….(iii)

From Eqs. (i) and (iii)

z+y=10

Q3Written answer

Look at the Fig. 5.3. Show that length AH > sum of lengths of AB + BC + CD.

NCERT Exemplar Solutions For Class 9 Maths Chapter 5 Exercise 5.3-1

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Worked solution

According to the given figure, we have,

AB+BC+CD =AD

Here, AD is a part of AH.

According to Euclid’s axiom,

“The whole is greater than the part”

i.e., AH > AD

Therefore, length AH > sum of the lengths of AB+BC+CD.

Q4Written answer

In the Fig.5.4, we have AB = BC, BX = BY. Show that AX = CY.

NCERT Exemplar Solutions For Class 9 Maths Chapter 5 Exercise 5.3-2

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Worked solution

According to the question,

We have,

AB =BC …(i)

and BX=BY …(ii)

According to Euclid’s axiom,

“If equals are subtracted from equals, the remainders are equal.”

Subtracting Eq.(ii) from (i),

We get,

AB-BX = BC-BY

⇒ AX = CY [from the given figure]

Q5Written answer

In the Fig.5.5, we have X and Y are the mid-points of AC and BC and AX = CY. Show that AC = BC.

NCERT Exemplar Solutions For Class 9 Maths Chapter 5 Exercise 5.3-3

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Worked solution

Given, X is the mid-point of AC

AX = CX= ½ AC

⇒ 2AX =2CX = AC …(i)

Y is the mid-point of BC.

BY = CY = ½ BC

⇒ 2BY = 2CY= BC …(ii)

According to the question,

We also have,

AX=CY …(iii)

Applying the Euclid’s axiom,

“Things which are double of the same things are equal to one another”.

We get,

From Eq. (iii),

2AX = 2CY

Using Eqs. (i) and (ii), we get,

AC=BC

Hence Proved.

Q6Written answer

In the Fig.5.6, we have BX = ½ AB, BY = ½ BC and AB = BC. Show that BX = BY.

NCERT Exemplar Solutions For Class 9 Maths Chapter 5 Exercise 5.3-4

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Worked solution

According to the question,

We have,

BX = ½ AB and BY = ½ BC

⇒ 2BX = AB ….(i)

⇒ 2BY = BC ….(ii)

It is also given that,

AB = BC …(iii)

Substituting the values from Eqs. (i) and (ii) in eq. (iii), we get,

2BX = 2BY

Applying the Euclid’s axiom, “things which are double of same things are equal to one another”.

BX =BY

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Exercise 5.3