Mathematics Part I Solutions Solutions for Class 9 Math Chapter 8 Geometric Proportions are provided here with simple step-by-step explanations. These solutions for Geometric Proportions are extremely popular among Class 9 students for Math Geometric Proportions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 9 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Page No 107:

#### Question 1:

Draw a triangle of sides 4, 5 and 6 centimetres. Divide it into two triangles, one with one-third the area, and the other with two-thirds the area, of the original triangle.

#### Answer:

The dimensions of the required triangle are *AB* = 6 cm, *BC* = 5 cm and *CA* = 4 cm.

The steps of construction are as follows:

1) Draw a line segment *AB* of length 6 cm.

2) With *A *as the centre and 4 cm as the radius, draw an arc on the upper side of *AB*.

3) With *B* as the centre and 6 cm as the radius, draw an arc cutting the previously drawn arc at point *C*.

4) Join *AC* and *BC*.

Δ*ABC* is the required triangle.

Let the triangle with area equal to one-third of the area of Δ*ABC* be Δ*ADC*.

Height of both the triangles would be same.

Thus, we need to construct Δ*ADC* such that *AD* = 2 cm.

The steps of construction are as follows:

1) With *A* as the centre and 2 cm as the radius, draw an arc on *AB *and name it as *D*.

2) Join *DC*.

Δ*ADC* so obtained has area equal to one-third the area of Δ*ABC* and Δ*DBC* has area equal to two-third of the area of Δ*ABC*.

#### Page No 107:

#### Question 2:

Draw a triangle and divide it into three triangles of equal area in various ways.

#### Answer:

Let us assume the dimensions of the triangle as *AB* = 6 cm, *BC* = 5 cm and *CA* = 7 cm.

The steps of construction are as follows:

1) Draw a line segment *AB* of length 6 cm.

2) With *A* as the centre and 7 cm as the radius, draw an arc on the upper side of *AB*.

3) With *B* as the centre and 5 cm as the radius, draw an arc cutting the previously drawn arc at point *C*.

4) Join *AC* and *AB*.

Δ*ABC* is the required triangle.

We need to divide the triangle into three triangles of equal area.

Let the triangle with area one-third of the area of Δ*ABC* be Δ*ADC*.

Height of both the triangles would be same.

Thus, we need to construct Δ*ADC* such that *AD* = 2 cm.

Similarly, we need to construct another triangle with the same area on side *AB*.

The steps of construction are as follows:

1) With *A* as the centre and 2 cm as the radius, draw an arc on *AB *and name it as *D*.

2) Join *DC*.

3) With *D* as the centre and 2 cm as the radius, draw an arc on *AB* and name it as *E*.

4) Join *CE*.

Thus, Δ*ADC*, Δ*DEC *and Δ*EBC* are triangles of equal area.

Similarly, we can construct triangles by taking *AC* or *BC* as the base.

#### Page No 107:

#### Question 3:

Draw triangle of sides 6, 7 and 8 centimetres. Divide it into three triangles whose areas are in the ratio 1 : 2 : 3.

#### Answer:

The dimensions of the required triangle are *AB* = 6 cm, *CA* = 7 cm and* BC* = 8 cm.

The steps of construction are as follows:

1) Draw a line segment *AB *of length 6 cm.

2) With *A *as the centre and 7 cm as the radius, draw an arc on the upper side of *AB*.

3) With *B* as the centre and 8 cm as the radius, draw an arc cutting the previously drawn arc at point *C*.

4) Join *AC* and *BC*.

Δ*ABC* is the required triangle.

We need to divide the given triangle into three triangles whose areas are in the ratio 1 : 2 : 3.

Height of all the triangles so formed would be equal to the height of the original triangle.

Let the first triangle be Δ*ADC*.

Area of Δ*ADC* = × Area of Δ*ABC*

× Area of Δ*ABC*

Let the second triangle be Δ*DEC*.

Area of Δ*DEC* = × Area of Δ*ABC*

× Area of Δ*ABC*

Δ*EBC** *so formed, after the construction of the above two triangles, has area equal to half the area of Δ*ABC*.

The steps of construction are as follows:

1) With *A* as the centre and 1 cm as the radius, draw an arc on *AB* and name it as *D*.

2) Join CD.

3) With *D* as the centre and 2 cm as the radius, draw an arc on *AB* and name it as *E*.

4) Join CE.

Thus, Δ*ADC*, Δ*DEC* and Δ*EBC* are the required triangles whose areas are in the ratio 1 : 2 : 3.

#### Page No 113:

#### Question 1:

Draw a line, 8 centimetres long and divide it in the ratio 4 : 5.

#### Answer:

The steps of construction to divide a line segment of length 8 cm in the ratio 4:5 are as follows:

1) Draw a line segment *AB* of length 8 cm.

2) At *B*, draw *BY* perpendicular to *AB*.

3) Taking *A *as the centre and 9 cm (sum of ratios) as the radius, draw an arc on *BY*. The point where the arc cuts *BY* is named as *Q*. Join *AQ*.

4) Taking *A* as the centre and 4 cm as the radius, draw an arc on *AQ* and name it as *P*. The remaining length *XY* = 9 cm − 4 cm = 5 cm.

5) Draw *PC* parallel to *QB*.

We have obtained that *AC*:*CB* = 4:5.

Thus, point *C* divides line segment *AB* of length 8 cm in the ratio 4:5.

#### Page No 113:

#### Question 2:

Draw a line, 10 centimetres long ad divide it in the ratio 3 : 4.

#### Answer:

The steps of construction to divide a line segment of length 10 cm in the ratio 3:4 are as follows:

1) Draw a line segment *AB* of length 10 cm.

2) Draw an acute angle, ∠*YAB*, at point *A* on the upper side of *AB* such that *AY* = 7 cm (sum of ratios).

3) Mark an arc on* AY* at a distance of 3 cm from *A *and name it as *X*. The remaining length *XY* = 7 cm − 3 cm = 4 cm.

4) Join *YB* and draw a line segment *XZ* parallel to* YB*.

We have obtained that *AZ*:*ZB* = 3:4.

Thus, point *Z* divides line segment *AB* of length 10 cm in the ratio 3:4.

#### Page No 113:

#### Question 3:

Draw a line, 6 centimetres long and divide it in the ratio 2 : 3 : 4.

#### Answer:

The steps of construction to divide a line segment of length 6 cm in the ratio 2:3:4 are as follows:

1) Draw a line segment *AB* of length 6 cm.

2) At *B*, draw *BY* perpendicular to *AB*.

3) Taking *A *as the centre and 9 cm (sum of ratios) as the radius, draw an arc on *BY*. The point where the arc cuts *BY* is named as *R*. Join *AR*.

4) Taking *A* as the centre and 2 cm as the radius, draw an arc on *AR* and name it as *P*.

5) Taking *P* as the centre and 3 cm as the radius, draw an arc on *AR* and name it as *Q*. The remaining length *QR *= 9 cm − 2 cm − 3 cm = 4 cm.

6) Draw *PC* and *QD* parallel to *RB*.

We have obtained that *AC*:*CD*:*DB* = 2:3:4.

Thus, points *C* and *D* divide line segment *AB* of length 6 cm in the ratio 2:3:4.

#### Page No 113:

#### Question 4:

Draw a triangle of perimeter 13 centimetres and lengths of sides in the ratio 2 : 3 : 4.

#### Answer:

First, we will divide a line segment of length 13 cm in the ratio 2:3:4.

The steps of construction are as follows:

1) Draw a line segment *AB* of length 13 cm.

2) Draw an acute angle, ∠*YAB*, at point *A* on the upper side of *AB* such that *AY* = 9 cm (sum of ratios).

3) Mark an arc on* AY* at a distance of 2 cm from *A *and name it as *X*.

4) Mark an arc on* AY* at a distance of 3 cm from *X *and name it as *Z*. The remaining length *ZY* = 9 cm − 2 cm − 3 cm = 4 cm.

5) Join *YB* and draw line segments *XC *and *ZD *parallel to* YB*.

We have obtained that *AC*:*CD*:*DB* = 2:3:4.

Thus, points *C* and *D* have divided line segment *AB* of length 13 cm in the ratio 2:3:4.

Now, we need to construct a triangle whose perimeter is 13 cm. Let the base of the required triangle is *CD*.

The steps of construction are as follows:

1) Taking C as the centre and *CA* as the radius, mark an arc on the lower side of *AB*.

2) Taking *D* as the centre and *DB* as the radius, mark an arc cutting the previously drawn arc at point *E*.

3) Join *CE* and *DE*.

Δ*CDE* thus obtained has perimeter 13 cm.

#### Page No 113:

#### Question 5:

Draw a line, 8 centimetres long and divide it into five equal parts.

#### Answer:

The steps of construction to divide a line segment of length 8 cm into five equal parts, i.e., in the ratio 1:1:1:1:1 are as follows:

1) Draw a line segment *AB* of length 8 cm.

2) Draw an acute angle, ∠*YAB*, at point *A* on the upper side of *AB* such that *AY* = 5 cm (sum of ratios).

3) Mark a point *X*_{1} on *AY* at a distance of 1 cm from *A*.

4) Mark a point *X*_{2} on *AY* at a distance of 1 cm from *X*_{1}.

5) Mark a point *X*_{3} on *AY* at a distance of 1 cm from *X*_{2}.

6) Mark a point *X*_{4} on *AY* at a distance of 1 cm from *X*_{3}. The remaining length *X*_{4}*Y* = 5 cm − 1 cm − 1 cm − 1 cm − 1 cm = 1 cm.

7) Join *YB* and draw line segments *X*_{1}*C*, *X*_{2}*D*, *X*_{3}*E* and *X*_{4}*F* parallel to* YB*.

We have obtained that *AC*:*CD*:*DE*:*E *:*FB* = 1:1:1:1:1

Thus, points *C*, *D*, *E* and *F* divide line segment *AB* of length 8 cm into five equal parts.

#### Page No 119:

#### Question 1:

In ∠ABC shown below, the length of AB is 6 centimetres and the length of AC is 5 centimetres. The length of AP is 4 centimetres. The line PQ is parallel to BC.

Find the lengths of AQ and QC.

#### Answer:

Given: *AB *= 6 cm, *AC *= 5 cm, *AP *= 4 cm

Also, *PQ* is parallel to *BC*.

Let the length of* AQ* be* x* cm.

We know that in a triangle, a line parallel to one side divides the other two sides in the same ratio.

#### Page No 119:

#### Question 2:

In ΔABC, a line parallel to BC cuts AB and AC at P and Q. Show that.

#### Answer:

Given: A triangle *ABC* in which *PQ* is parallel to *BC*

We know that in a triangle, a line parallel to one side divides the other two sides in the same ratio.

#### Page No 119:

#### Question 3:

In the figure below, ABC is a right angled triangle. AB = 10 centimetres and AC = 6 centimetres. The midpoint of AB is M.

Compute the lengths of the sides of ΔMBN.

#### Answer:

Given: *AB *= 10 cm, *AC* = 6 cm

Also, *M* is the midpoint of *AB.*

Applying Pythagoras theorem in Δ*ABC**:*

Since ∠*MNB* = ∠*ACB** *= 90°, *MN* || *AC* by converse of corresponding angles axiom.

We know that in a triangle, a line through the midpoint of a side and parallel to another side bisects the third side.

As *M* is the midpoint of *AB* and *MN* || *AC*, *BN* = *NC*.

Also, (As *M* is the midpoint of *AB*)

Applying Pythagoras theorem in Δ*BNM**:*

Thus, the lengths of the sides of Δ*BNM* are *BM* = 5 cm, *BN* = 4 cm, and *MN* = 3 cm.

#### Page No 119:

#### Question 4:

In the figure below, ABC is a right angled triangle and M is the midpoint of AB.

(a) Prove that .

(b) Prove that MC = MA = MB.

#### Answer:

In Δ*ABC*, *M* is the midpoint of *AB*.

⇒ *BM* = *MA* = …(1)

Also, ∠*MNB* = ∠*ACB** *= 90°

By converse of corresponding angle axiom, *MN* || *AC*

We know that in a triangle, a line through the midpoint of a side and parallel to another side bisects the third side.

⇒ *BN *= *NC* = …(2)

Applying Pythagoras theorem in Δ*MNB*:

*MB*^{2} = *MN*^{2}^{ }+ *BN*^{2}

⇒ *MN*^{2} = *MB*^{2} − *BN*^{2}

Applying Pythagoras theorem in Δ*ACB*:

*AB*^{2} = *AC*^{2}^{ }+ *BC*^{2}

⇒ *AC*^{2} = *AB*^{2} − *BC*^{2}^{ }

Putting the value of *AB*^{2} − *BC*^{2}^{ }in equation (3):

Now, in Δ*MNC *and Δ*MNB*:

*BN* = *NC * (From equation (2))

∠*MNB* = ∠*MNC* = 90° (Given)

*MN* = *MN * (Common side)

∴ Δ*MNC *≅ Δ*MNB* (By side angle side criterion of congruence)

⇒ *MC* = *MB* (Corresponding parts of congruent triangles are congruent)

Thus, we have *MC* =* MA* =* MB*.

#### Page No 119:

#### Question 5:

Prove that the centre of the circum circle of a right angled triangle is the midpoint of its hypotenuse.

#### Answer:

Given: A right-angled Δ*ACB* with *O* as the midpoint of hypotenuse* AB*

Construction: Draw *OD* ⊥ *AC*. Join OC.

∠*ACB* = ∠*ADO** *= 90°

So, by converse of corresponding angle axiom, *DO* || *CB*.

We know that in a triangle, a line through the midpoint of a side and parallel to another side bisects the third side.

⇒ *AD* =* DC*

Now, in Δ*AOD* and Δ*COD*:

*DO* =* DO *(Common side)

*AD* = *DC * (Proved above)

∠ADO = ∠CDO = 90°

∴ Δ*AOD* ≅ Δ*COD** *(By side angle side criterion of congruence)

⇒ *AO *= *OC* (Corresponding parts of congruent triangles are congruent)

*O *is the midpoint of *AB*.

∴ *OB* = *AO *

⇒ *OB* = *AO *= *OC*

Thus, *O* is the centre of the circle passing through points *A*, *B* and *C*.

Hence, the centre of the circumcircle of a right-angled triangle is the midpoint of its hypotenuse.

#### Page No 120:

#### Question 1:

In the figure below, ABCD is a parallelogram and X, Y are the midpoints of AB, CD. The lines DX and BY cut AC at P and Q.

Prove that AP = PQ = QC.

#### Answer:

Given: *ABCD* is a parallelogram with *X* and *Y *as the midpoints of *AB* and *CD*.

∴ *AX* = *XB* and *CY* = *YD*

Also, *AB *=* DC *(Opposite sides of a parallelogram are equal)

⇒ *AX* + *XB* = *CY* + *YD*

2*XB* = 2*DY** *

*XB* = *DY *

In quadrilateral *XBYD*, *XB* = *DY*

*XB* || *DY* (As *AB* || *DC*)

We know that if a pair of opposite sides are equal and parallel then the quadrilateral is a parallelogram.

∴ *XBYD* is a parallelogram.

⇒ *DX* || *YB*

In Δ*ABQ*,* X *is the midpoint of *AB* and *XP* || *BQ*. (As *DX* || *YB*)

∴ *AP* = *PQ * …(1)

In Δ*CDP*, *Y* is the midpoint of *CD* and *YQ* || *DP*.

∴ *CQ* = *QP * …(2)

From equation (1) and (2), we have:

*AP* = *PQ *= *QC *

#### Page No 120:

#### Question 2:

In the figure, AB and CD are parallel.

Prove that AP × PC = BP × PD

#### Answer:

Given: *AB* || *CD*

Construction: Draw *XY* parallel to *AB* such that it passes through point *P*.

We have *AB* ||* XY* || *CD*

We know that three parallel lines cut any two lines in the same ratio.

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