Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 3 Parallel Lines Extra Questions and Answers Notes to clear their doubts.
Kerala SCERT Class 9 Maths Chapter 3 Solutions Parallel Lines
Parallel Lines Class 9 Kerala Syllabus Questions and Answers
Kerala State Syllabus 9th Standard Maths Chapter 3 Parallel Lines Solutions Questions and Answers
Class 9 Maths Chapter 3 Kerala Syllabus – Equal Division
Intext Questions And Answers
Question 1.
Divide a 7 cm long line into three equal pieces.
Answer:
Draw a 6 cm long line and draw on it, perpendiculars 2 centimeters apart.
From any point on the perpendicular, draw an arc of radius 7 centimeters and mark the point where it cuts the last perpendicular.
Join these points, we get a 7 cm long line cut into three equal pieces.
Question 2.
Draw a line 10 cm long and divide it into three equal parts.
Answer:
First, draw two lines as shown below:
Join endpoints. Divide the top line into three equal parts and draw parallels through these points:
Thus the 10 cm long line is cut into three equal parts.
We can draw an equilateral triangle of perimeter 10 centimeters by joining these pieces.
We can also draw a square of perimeter 10 centimeters by dividing the line into four equal parts and joining these pieces.
Textual Questions And Answers (Page 39)
Question 1.
Draw an equilateral triangle of perimeter 11 centimeters.
Answer:
Draw line of length 11 cm and 9 cm line slanted on it. Join endpoints. Divide the top line into three equal parts and draw parallels through these points. Thus, 11 cm long line is cut into three equal parts. Draw equilateral triangle by joining these parts.
Question 2.
Draw a square of perimeter 15 centimeters.
Answer:
Draw a 15 cm long line and a 12 cm long line slanted on it. Join endpoints. Divide the top line into three equal parts and draw parallels through these points. Thus, the 15 cm long line is cut into four equal parts. Now, join these parts to get the square.
Question 3.
Draw a regular hexagon of perimeter 20 centimeter. .
Answer:
Draw 10 cm long line and a 9 cm long line slanted on it. Join endpoints. Divide top line into 3 equal parts. Draw parallel lines through these points. Thus, 10 cm line cut into 3 equal parts. Draw a circle of radius as the length of these piece. Complete the regular hexagon.
Or
Draw a line of 20 cm. Divide it into 6 equal parts and draw the regular hexagon as follows:
Class 9 Maths Kerala Syllabus Chapter 3 Solutions – Unequal Division
Intext Questions And Answers
Question 1.
Draw a rectangle of perimeter 30 centimeters and sides in the ratio 5: 3.
Answer:
Perimeter = 30 cm
2(length + breadth) = 30
length+breadth = 15
So, draw a 15 cm long line and divide it in the ratio 5: 3 as shown:
Draw the rectangle with bottom pieces as sides.
Textual Questions And Answers
Question 1.
Draw a rectangle of perimeter 15 centimeters and sides in the ratio 3:4.
Answer:
Perimeter = 15 cm
2(length + breadth) =15
length + breadth = 7.5
So, draw a 7.5 cm long line and divide it in the ratio 3:4.
Draw the rectangle with bottom pieces as sides.
Question 2.
Draw a triangle of each of the types below, of perimeter 13 centimetres:
(i) Equilateral
(ii) Sides in the ratio 3: 4: 5
(iii) Isosceles with lateral sides one and a half times the base.
Answer:
(i) Draw a 13 cm long line. Divide into three equal parts. Draw triangle with sides as these parts.
(ii) Draw 13 cm long line. Divide it in the ratio 3: 4: 5. Draw triangle with sides as these parts.
(iii)Draw a 13 cm long line. Lateral sides one and a half times the base. So, sides are in the ratio \(\frac{3}{2}\) : 1 : \(\frac{3}{2}\) = 3 : 2 : 3
Divide the line in the ratio 3: 2: 3.
Question 3.
Prove that in any trapezium, the diagonals cut each other in the same ratio.
Answer:
ABCD is a trapezium. AC and BD are diagonals. Through O, draw EO || AB, meets AD at E.
In ΔADC, EO || DC
Therefore,
\(\frac{A E}{E D}=\frac{A O}{O C}\) ……..(1)
In ΔDAB, EO || AB
Therefore,
\(\frac{A E}{E D}=\frac{B O}{O D}\) ………(2)
From 1 and 2, we get
\(\frac{A O}{O C}=\frac{B O}{O D}\)
Hence, Proved.
SCERT Class 9 Maths Chapter 3 Solutions – Triangle Division
Textual Questions And Answers
Question 1.
In the picture below, the green line is parallel to the right side of the blue triangle.
Calculate the lengths of the pieces into which this line cuts the left side.
Answer:
The green line is parallel to the right side of the blue triangle. So, the ratio of bottom side and the left side is the same.
Ratio of bottom side = 3: 5
∴ Ratio of left side = 3: 5
∴ Length of shorter piece = 6 × \(\frac{3}{8}=\frac{9}{4}\) = 2.25 cm
Length of longer piece = 6 × \(\frac{5}{8}=\frac{15}{4}\) = 3.75 cm
Question 2.
In the parallelogram ABCD, the line through the point P on AB, parallel to BC meets AC at Q. The line through Q parallel to AB meets AD at R:
(i) Prove that \(\frac{A P}{P B}=\frac{A R}{R D}\)
(ii) Prove that \(\frac{A P}{A B}=\frac{A R}{A D}\)
Answer:
(i) In ∆ABC,PQ || BC
⇒ \(\frac{A Q}{Q C}=\frac{A P}{P B}\) ……… (1)
In ∆ADC, RQ || AP ⇒ RQ\\DC
⇒ \(\frac{A Q}{Q C}=\frac{A R}{R D}\) ……… (2)
From (1) and (2),
Hence proved.
(ii) In ∆ABC, PQ || BC
⇒ \(\frac{A Q}{A C}=\frac{A P}{A B}\) …… (1)
In ∆ADC, RQ || AP ⇒ RQ || DC
⇒ \(\frac{A Q}{A C}=\frac{A R}{A D}\) …… (2)
From (1) and (2),
\(\frac{A P}{A B}=\frac{A R}{A D}\)
Hence proved.
Question 3.
In the picture below, two corners of a parallelogram are joined to the midpoints of two sides.
Prove that these lines cut the diagonal in the picture into three equal parts.
Answer:
ABCD is a parallelogram. P, Q are midpoints of AB and CD respectively.
∴ PB = DQ.
PBQD is a parallelogram.
∴ PR || BS and DR || QS
In ∆ABS, PR || BS ⇒ \(\frac{A P}{P B}=\frac{A R}{R S}=\frac{1}{1}\) [∵ P is the midpoint of AB]
In ∆CDR, DR || QS ⇒ \(\frac{C Q}{Q D}=\frac{S C}{R S}=\frac{1}{1}\) [∵ Q is the midpoint of CD]
From above to results,
AR = RS = SC
Hence, lines PD and BQ cut the diagonal into three equal parts.
Question 4.
In triangle ABC, the line parallel to AC through the point P on BC, meets AB at Q. The line parallel to AP through Q meets BC at R:
Prove that \(\frac{B P}{P C}=\frac{B R}{R P}\)
Answer:
In ∆ABP, AP || RQ ⇒ \(\frac{B Q}{Q A}=\frac{B R}{R P}\)
In ∆ABC, AC || PQ ⇒ \(\frac{B Q}{Q A}=\frac{B P}{P C}\)
From the above two results,
\(\frac{B P}{P C}=\frac{B R}{R P}\)
Hence proved.
Parallel Lines Class 9 Kerala Syllabus – Midsection
Textual Questions And Answers
Question 1.
In the picture, the midpoints of a triangle are joined to form a smaller triangle inside:
Answer:
Since, the midpoints of large triangle are joined to form a small triangle inside, all the four triangles are equal.
Perimeter of ΔABC = AB + BC + AC
Perimeter of ΔEDF = EF + DF + ED
E, D, F are midpoints of BC, AB, AC respectively.
So, DF || BC mdDF = \(\frac{B C}{2}\)
ED || AC and ED = \(\frac{A C}{2}\)
EF || AB and EF = \(\frac{A B}{2}\)
∴ Perimeter of A EDF = \(\frac{A B}{2}+\frac{B C}{2}+\frac{A C}{2}=\frac{1}{2}\) (AB + BC + AC)
Hence, perimeter of small triangle is \(\frac{1}{2}\) times the perimeter of large triangle.
Since the four triangles are equal, their area are also equal.
Hence,
Area of small triangle is \(\frac{1}{4}\) times the area of large triangle.
Question 2.
See these pictures:
The first picture is that of a triangle cut out from a sheet of paper. The second one shows it with the small triangle in the middle, joining the midpoints of the sides, cut off from the large triangle.
The third picture shows such middle pieces cutoff from each of the small triangles in the second picture.
(i) What fraction of the area of the paper in the first picture is the area of the paper in the second picture?
(ii) What about in the third picture?
Answer:
(i) The small triangle in the middle in the second picture is formed by joining the midpoints of the sides of large triangle in first picture.
So, the four triangles formed in the second picture are equal and hence have equal area. So, area of each triangle is \(\frac{1}{4}\) of the large triangle.
In the second picture middle triangle cut off from the large triangle. The remaining part is \(\frac{3}{4}\). Hence fraction of the area of the paper in the second picture is \(\frac{3}{4}\)of the area of paper in the first picture.
(ii) Similarly, in third picture such middle pieces cutoff from each small triangles in second picture. \(\frac{1}{4}\) part in the second picture gets divided into 4 equal parts. So, each part is \(\frac{1}{16}\).
From this one part cutoff. Remaining \(\frac{3}{16}\)
Totally 3 such parts cut off. So, remaining is 3 x \(\frac{3}{16}\) = \(\frac{9}{16}\)
Question 3.
A quadrilateral is drawn with the midpoints of the sides of a right angle triangle and its square corner as vertices.
(i) Prove that this quadrilateral is a rectangle.
(ii) What fraction of the area of the triangle is the area of the rectangle?
Answer:
(i)
The quadrilateral is formed by joining midpoints of sides of right triangle.
∠B = 90°
P, Q are midpoints of AC and BC respectively.
So, PQ || AB and PQ = \(\frac{1}{2}\) × AB
PQ || AB ⇒ PQ || RB
Since R is the midpoint of AB, RB = \(\frac{1}{2}\) × AB
PQ = RB
Similarly, BQ = RP
PQ || RB , ∠Q = 90°
Similarly, ∠R = ∠P = 90°
So, opposite sides are equal and parallel, all angles are 90°.
Hence, PQBR is a rectangle.
(ii) Area of ∆ABC = \(\frac{1}{2}\) × BC × AB
Area of rectangle RPQB = BQ × RB
= \(\frac{1}{4}\) × BC × AB
∴ Area of rectangle is \(\frac{1}{4}\) of the area of triangle.
Question 4.
In the two pictures below, the first one shows two triangles formed by joining the ends of a line to two points above it. The second one shows the quadrilateral formed by joining the midpoints of the left and right sides of these triangles:
(i) Prove that this quadrilateral is a parallelogram.
(ii) Describe the positions of the points on top for this quadrilateral to be:
(a) Rhombus
(b) Rectangle
(c) Square
(iii) Would we get all these, if one point is taken above the line and one below?
Answer:
(i)
In the figure, AB is a line segment and C and D are any two points on one side of the segment. E, F, G, and H are the midpoints of the sides AC, BC, BD and AD respectively
EFGH is a quadrilateral obtained by joining these midpoints.
Clearly, HG = \(\frac{1}{2}\)AB and HG| |AB
Also EF = \(\frac{1}{2}\) AB and EF || AB
Hence HG = EF and HG || EF
Hence EFGH is a parallelogram.
(ii)
(a) In figure, AB is a line segment and C and D are any two points on one side of the segment such that AB = CD.
From (i), EFGH is a parallelogram.
Clearly, HG = EF = \(\frac{1}{2}\)AB
And, in triangles, ∆ADC and ∆BDC,
EH = FG = \(\frac{1}{2}\)CD
Since, AB = CD, HG = GF = EF = EH.
Hence EFGH is a rhombus.
Hence to get a rhombus, the distance between the points should be same as the length of the line segment.
(b)
In figure, AB is a line segment and C and D are any two points on the perpendicular bisector of AB on one side of the segment.
From (i), EFGH is a parallelogram.
Also, in triangles, ∆ADC and ∆BDC, EH || CD and FG || CD.
Since CJ is perpendicular to AB, both EH and FG will be perpendicular to HG. Hence EFGH is a rectangle.
Hence to get a rectangle, the points should be on the perpendicular bisector of the line segment AB.
In figure, AB is a line segment and C and D are any two points on the perpendicular bisector of AB on one side of the segment such that AB = CD.
From (b), EFGH is a rectangle.
Clearly, HG = EF = \(\frac{1}{2}\) AB
And EH = FG = \(\frac{1}{2}\) CD
Since, AB = CD, HG = GF = EF = EH.
Hence EFGH is a square.
Hence to get a square, the points should be on the perpendicular bisector and also the distance between the points should be same as the length of the line segment.
(iii)
Yes.
- For any two points C and D, EFGH will be a parallelogram.
- If the distance between C and D is equal to the length of AB, then EFGH will become a rhombus.
- If C and D are any two points on the perpendicular bisector of AB, then EFGH will become a rectangle.
- If C and D are on the perpendicular bisector of AB and also if CD = AB then EFGH will become a square.
Question 5.
(i) Prove that the quadrilateral formed by joining the mid points of any quadrilateral is a parallelogram.
(ii) Explain what should be the original quadrilateral to get the inner quadrilateral as:
(a) Rhombus
(b) Rectangle
(c) Square
Answer:
(i)
Consider the quadrilateral ABCD.
Quadrilateral PQRS is formed by joining the midpoints of sides of quadrilateral ABCD.
In ∆ABD, P, Q are midpoints of AD, AB respectively. So, PQ || BD and PQ = \(\frac{B D}{2}\)
Similarly in A ABD, S, R are midpoints of CD, CB respectively.
So, SR || BD and SR = \(\frac{B D}{2}\)
⇒ PQ || SR and PQ = SR
So, opposite sides are equal and parallel.
Hence, quadrilateral PQRS is a parallelogram.
(ii) a) Rectangle
For it to be a rhombus (non- square), the diagonal is must be equal.
Rectangle satisfies this property.
b) Rhombus
For it to be a rectangle, the diagonals must be perpendicular bisectors. In this case rhombus satisfies this property.
c) Square
For it to be a square, the diagonals must be equal and perpendicular. So here square satisfies this property.
Class 9 Maths Parallel Lines Kerala Syllabus – Triangle Centre
Textual Questions And Answers (Page No. 58, 59, 60)
Question 1.
Draw a right triangle and draw the perpendicular from the midpoint of the hypotenuse t& the base:
(i) Prove that this perpendicular is half the vertical side of the large triangle.
(ii) Prove that the distances from the midpoint of the hypotenuse to the three vertices of the large triangle are equal.
(iii) Prove that the circumcentre of a right triangle is the midpoint of its hypotenuse,
Answer:
(i) Given that AB and MN are perpendicular to BC. So, these lines are parallel.
M is the midpoint of AC. Therefore, N is the midpoint of BC.
MA: MC = NB: NC = 1:1
The line joining the midpoints of two sides of a triangle is parallel to the third side and half the length of third side.
MN = \(\frac{1}{2}\) × AB
(ii) Given that MA = MC,
We have seen that NB = NC and MN is perpendicular to BC. ‘ .
We know that ends of a line are equidistant from a point on its perpendicular bisector, ie, MC = MB Hence, MA = MB =MC.
(iii) Since M is equidistant from the vertices A, B, C of the triangle, the circle with centre M and radius, the distance to the vertices passes through all the vertices. It is the circumcircle of triangle ABC.
Question 2.
Prove that in any equilateral triangle, the circumcentre, orthocentre and the centroid coincide.
Answer:
Consider the equilateral triangle ABC inscribed in a circle.AD is the median to the side BC. ••• BD = CD Consider AABC,
AB = AC (Sides of equilateral triangle)
AD = AD (Common)
BD = CD
∴ ∆ABD ≅ ∆ACD
∴ ∠ADB = ∠ADC
Also, ∠ADB + ∠ADC = 180° (Linear pair)
∴ ∆ADB = ∆ADC = 90°
⇒ AD ⊥ BC .
∵ AD ⊥ BC and BD = CD, AD is the altitude of triangle ABC from A to BC.
Similarly medians CF, BE also altitudes of the triangle and the point where all the three meet is the orthocentre, circumcentre and centroid.
i.e, in any equilateral triangle, the circumcentre, orthocentre and the centroid coincide.
Question 3.
In the picture below, the medians of the triangle divide it into six small triangles:
Prove that all six triangles have the same area.
Answer:
AQ, BR, CP are medians. Let the area of triangle APG = x
Then the area of triangle BPG = x (Same base and same height)
Similarly, if the area of triangle BQG = y
Then area of triangle CQG = y
Also, if the area of triangle CGR = z
Then the area of triangle AGR = z
Since triangle ABQ and ACQ have the same base (BQ = QC) and same height, their areas are equal.
i.e, x + x + y = y + z + z
2x = 2z
x = z
Like this, if we take triangle BCR and BAR,
y + y + z = x + x + z
2y = 2x
y = x
i.e, x = y = z, all six triangles are equal and hence their area also equal.
Question 4.
In the picture below, the triangle is divided into three small triangles by joining the centroid to the three vertices:
Prove that all three triangles have the same area. ;
Answer:
Let the area of triangle BGP = x
We know AG = 2 GP and triangles ABG and BGP have the same height.
So, area of triangle ABG = 2x
Also, we know BP = CP (AP is the median)
So, area of triangle CPG is also x.
Then area of triangle AGC = 2x
This means, .
Area of triangle ABG = 2x
Area of triangle BGC = x + x = 2x
Area of triangle AGC = 2x
∴ all three triangles have the same area.
Question 5.
In the picture below, the midpoints of the sides of the blue triangle are joined to make the smaller green triangle. The red line is a median of the large triangle.
(i) Prove that this median bisects the top side of the small triangle.
(ii) Prove that the centroid of the large and small triangles are the same.
Answer:
(i) Since P,Q,R are the mid points of the sides APRQ is a parallelogram. So, its diagonals bisect each other.
i.e, PS = SQ
i.e, the median AR bisected PQ
(ii) Draw BQ and CR. As we did earlier, we can show that PM = MR and RN = NQ
ie PN, QM, SR are the medians of triangle PQR, then G is the centroid of triangle PQR.
Already P, Q, R are the mid points of sides AB, AC and BC. Then AR, BQ, CP are medians. These are along SR, QM and PN. So, AR, BQ and CP also pass through the point G. This means G is the centroid of both triangles.
Question 6.
In the picture below, H is the orthocentre of triangle ABC.
Prove that A is the orthocentre of triangle HBC.
Answer:
Given H is the orthocentre of triangle ABC. That means,
We have to show that A is the orthocentre of triangle HBC.
For that, we have to draw perpendicular from each vertices of triangle HBC to its opposite sides. If we do this, we can see BP is the perpendicular from B to side HC of triangle HBC.
This is because ∠BHC is obtuse.
On extending BP, we get BA. Like this CA is the perpendicular from C. In figure BA and CA pass through A. The third perpendicular HQ also pass through A as AQ. This means A is the orthocentre of triangle HBC.
Parallel Lines Class 9 Extra Questions and Answers Kerala Syllabus
Question 1.
On a base AB, 7cm in length, draw a triangle CAB. Mark X on AB such that AX = 4.2 cm. Through X draw a line XY parallel to BC to meet AC at Y.
(a) Draw a rough diagram
(b) Calculate
(i) \(\frac{A X}{B X}\) and \(\frac{A Y}{C Y}\)
(ii) \(\frac{A B}{A X}\) and \(\frac{A c}{A Y}\)
(iii) \(\frac{A B}{B X}\) and \(\frac{A C}{C Y}\)
Answer:
(a)
Question 2.
Construct a square of perimeter 10.5 cm.
Answer:
Draw a line of length 10.5cm. Divide this line into four equal parts. Draw the square as given below:
Question 3.
Construct a right triangle having perimeter 15 cm and sides are in the ratio 3:4:5
Answer:
Construct a right triangle having perimeter 15cm and sides are in the ratio 3:4:5
Since 3, 4, 5 is a Pythagorean triple triangles with ratio of sides 3:4:5 will be a right triangle.
Draw a line of length 15cm , divide it in the ratio 3:4:5.
Construct the triangle.
Question 4.
In triangle ABC, the line parallel to BC meet AB and AC at D and E respectively. AE = 4.5 cm, \(\frac{A D}{D B}=\frac{2}{5}\). Then find EC.
Answer:
\(\frac{A D}{D B}=\frac{A E}{E C}\)
\(\frac{2}{5}=\frac{4.5}{E C}\)
EC = 4.5 × \(\frac{5}{2}\) = 11.25 cm.
Question 5.
In the figure D, E, F are midpoints of sides of the triangle ABC.
(a) If BC = 8 cm, find DF.
(b) If the perimeter of triangle ABC is 20 cm, find the perimeter of triangle DEF.
(c) If the area of triangle ABC is 16 cm , find the area of triangle DEF.
Answer:
(a) DF = \(\frac{B C}{2}=\frac{8}{2}\) = 4 cm
(b) Perimeter of triangle ABC = 20 cm
Perimeter of triangle DEF = \(\frac{20}{2}\) = 10 cm.
(c) Area of triangle ABC = 16 cm²
∴ Area of triangle DEF \(\frac{16}{4}\) = 4cm²
Question 6.
ABC is a triangle right angled at B. P, Q, R the mid points of the sides of ∆ ABC. PR = 3 cm, PQ = 4 cm.
(a) What is the length QR?
(b) What are the sides of triangle ABC?
(c) Suggest a suitable name to PQBR?
Answer:
(a) ∆PQR is a right triangle.
QR = \(\sqrt{P Q^2+P R^2}\)
= \(=\sqrt{4^2+3^2}=\sqrt{25}\) = 5 cm
(b) Sides of triangle ABC,
AB = 2 × PQ = 2 × 4 = 8 cm
BC = 2 × PR = 2 × 3 = 6 cm
AC = 2 × QR = 2 × 5 = 10 cm
(c) Rectangle
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