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 Meritnation’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.

ΔABC is the required triangle.

Let the triangle with area equal to one-third of the area of ΔABC be ΔADC.

Height of both the triangles would be same.

Thus, we need to construct ΔADC 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.

ΔADC so obtained has area equal to one-third the area of ΔABC and ΔDBC has area equal to two-third of the area of ΔABC.

 

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.

ΔABC 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 ΔABC be ΔADC.

Height of both the triangles would be same.

Thus, we need to construct ΔADC 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, ΔADC, ΔDEC and ΔEBC 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

ΔABC 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 ΔADC.

Area of ΔADC = × Area of ΔABC 

× Area of ΔABC

Let the second triangle be ΔDEC.

Area of ΔDEC = × Area of ΔABC

× Area of ΔABC

ΔEBC so formed, after the construction of the above two triangles, has area equal to half the area of ΔABC.

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, ΔADC, ΔDEC and ΔEBC 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.

ΔCDE 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 X1 on AY at a distance of 1 cm from A

4) Mark a point X2 on AY at a distance of 1 cm from X1

5) Mark a point X3 on AY at a distance of 1 cm from X2.

6) Mark a point X4 on AY at a distance of 1 cm from X3. The remaining length X4Y = 5 cm 1 cm 1 cm 1 cm 1 cm = 1 cm.

7) Join YB and draw line segments X1C, X2D, X3E and X4F 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 ΔABC:

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 ΔBNM:

Thus, the lengths of the sides of ΔBNM 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 ΔABC, 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 ΔMNB:

MB2 = MN2 + BN2

MN2 = MB2 BN2

Applying Pythagoras theorem in ΔACB:

AB2 = AC2 + BC2

AC2 = AB2 BC2  

Putting the value of AB2 BC2 in equation (3):

 

Now, in ΔMNC and ΔMNB:

BN = NC (From equation (2))

MNB = MNC = 90° (Given)

MN = MN (Common side) 

∴ ΔMNC ΔMNB (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 ΔACB 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 ΔAOD and ΔCOD:

DO = DO (Common side)

AD = DC (Proved above)

ADO = CDO = 90°

∴ ΔAOD ΔCOD (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

2XB = 2DY  

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 ΔABQ, X is the midpoint of AB and XP || BQ. (As DX || YB)

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

AP = PQ …(1)

In ΔCDP, 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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