Selina Concise Class-9th Mid Point and Intercept Theorem

Selina Concise Class-9th Mid Point and Intercept Theorem ICSE Mathematics Solutions Chapter-12. We provide step by step Solutions of Exercise / lesson-12 Mid Point and its Converse ( Including Intercept Theorem ) for ICSE Class-9 Concise Selina Mathematics by R K Bansal.

Our Solutions contain all type Questions with Exe-12 A and Exe-12 B, to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics .

Selina Concise Class-9th Mid Point and Intercept Theorem ICSE Mathematics Solutions Chapter-12


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Exe – 12 A,

Exe -12 B,


Exercise – 12 A, Selina Concise Class-9th Mid Point and its Converse ( Including Intercept Theorem ) ICSE Mathematics Solutions

Question 1

In triangle ABC, M is mid-point of AB and a straight line through M and parallel to BC cuts AC in N. Find the lengths of AN and MN if Bc = 7 cm and Ac = 5 cm.

Answer

The triangle is shown below,

Ans 1 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths

Question 2

Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.

Answer

The figure is shown below,

Ans 2 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths

Similarly 2PQ=2SR=AC and PQ||SR—– (2)

From (1) and (2) we get

PQ=QR=RS=PS

Therefore PQRS is a rhombus.

Hence proved

Question 3

D, E and F are the mid-points of the sides AB, BC and CA of an isosceles ΔABC in which AB = BC. Prove that ΔDEF is also isosceles.

Answer

The figure is shown below

Ans 3 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths

Given that ABC is an isosceles triangle where AB=AC.

Since D,E,F are midpoint of AB,BC,CA therefore

2DE=AC and 2EF=AB this means DE=EF

Therefore DEF is an isosceles triangle an DE=EF.

Hence proved

Question 4

The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:

…………………

Answer

Here from triangle ABD P is the midpoint of AD and PR||AB, therefore Q is the midpoint of BD

Similarly R is the midpoint of BC as PR||CD||AB

From triangle ABD 2PQ=AB …….(1)

From triangle BCD 2QR=CD …..(2)

Now (1)+(2)=>

2(PQ+QR)=AB+CD

Ans 4 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths

Hence proved

Question 5

The figure, given below, shows a trapezium ABCD. M and N are the mid-point of the non-parallel sides AD and BC respectively. Find:

……………

(i) MN, if AB = 11 cm and DC = 8 cm.

(ii) AB, if Dc = 20 cm and MN = 27 cm.

(iii) DC, if MN = 15 cm and AB = 23 cm.

Answer

Let we draw a diagonal AC as shown in the figure below,

Ans 5 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths-compressed (1)

Question 6

The diagonals of a quadrilateral intersect at right angles. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is rectangle.

Answer

The figure is shown below

Ans 6 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths-compressed

Question 7

L and M are the mid-point of sides AB and DC respectively of parallelogram ABCD. Prove that segments DL and BM trisect diagonal AC.

Answer

The required figure is shown below

Ans 7 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths

From figure,

BL=DM and BL||DM and BLMD is a parallelogram, therefore BM||DL

From triangle ABY

L is the midpoint of AB and XL||BY, therefore x is the midpoint of AY.ie AX=XY …..(1)

Similarly for triangle CDX

CY=XY …..(2)

From (1) and (2)

AX=XY=CY and AC=AX+XY+CY

Hence proved

Question 8

ABCD is a quadrilateral in which AD = BC. E, F, G and H are the mid-points of AB, BD, CD and Ac respectively. Prove that EFGH is a rhombus.

……………………..

Answer

Given that AD=BC …..(1)

From the figure,

For triangle ADC and triangle ABD

2GH=AD and 2EF=AD, therefore 2GH=2EF=AD …..(2)

For triangle BCD and triangle ABC

2GF=BC and 2EH=BC, therefore 2GF=2EH=BC …..(3)

From (1),(2),(3) we get,

2GH=2EF=2GF=2EH

GH=EF=GF=EH

Therefore EFGH is a rhombus.

Hence proved

Question 9

A parallelogram ABCD has P the mid-point of Dc and Q a point of Ac such that CQ = 1/4 AC. PQ produced meets BC at R.

Prove that

(i)R is the midpoint of BC

(ii)……….

Answer

For help we draw the diagonal BD as shown below

Ans 9 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths-compressed

Question 10

D, E and F are the mid-points of the sides AB, BC and CA respectively of ABC. AE meets DF at O. P and Q are the mid-points of OB and OC respectively. Prove that DPQF is a parallelogram.

Answer

The required figure is shown below

Ans 10 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths

For triangle ABC and OBC

2DE=BC and 2PQ=BC, therefore DE=PQ …..(1)

For triangle ABO and ACO

2PD=AO and 2FQ=AO, therefore PD=FQ …..(2)

From (1),(2) we get that PQFD is a parallelogram.

Hence proved

Question 11

In triangle ABC, P is the mid-point of side BC. A line through P and parallel to CA meets AB at point Q; and a line through Q and parallel to BC meets median AP at point R. Prove that

(i)AP=2AR

(ii)BC=4QR

Answer

The required figure is shown below

Ans 11 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths-compressed

Question 12

In trapezium ABCD, AB is parallel to DC; P and Q are the mid-points of AD and BC respectively. BP produced meets CD produced at point E. Prove that:

(i) Point P bisects BE,

(ii) PQ is parallel to AB.

Answer

The required figure is shown below

Ans 12 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths

 

(ii)For tiangle ECB PQ||CE

Again CE||AB

Therefore PQ||AB

Hence proved

Question 13

In a triangle ABC, AD is a median and E is mid-point of median AD. A line through B and E meets AC at point F.

Prove that: AC = 3AF

Answer

The required figure is shown below

Ans 13 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths

For help we draw a line DG||BF

Now from triangle ADG, DG||BF and E is the midpoint of AD

Therefore F is the midpoint of AG,ie AF=GF …..(1)

From triangle BCF, DG||BF and D is the midpoint of BC

Therefore G is the midpoint of CF,ie GF=CF …(2)

AC=AF+GF+CF

AC=3AF(From (1) and (2))

Hence proved

Question 14

D and F are mid-points of sides AB and AC of a triangle ABC. A line through F and parallel to AB meets BC at point E.

(i) Prove that BDFE is parallelogram

(ii) Find AB, if EF = 4.8 cm.

Answer

The required figure is shown below

Ans 14 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths

Question 15

In triangle ABC, AD is the median and DE, drawn parallel to side BA, meets AC at point E. Show that BE is also a median.

Answer

Ans 15 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths-compressed

Question 16

In ∆ABC, E is mid-point of the median AD and BE produced meets side AC at point Q. Show that BE : EQ = 3:1.

Answer

Ans 16 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths-compressed

Question 17

In the given figure, M is mid-point of AB and DE, whereas N is mid-point of BC and DF. Show that: EF = AC.

Answer

Ans 17 Exercise - 12 A Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths


Selina Concise Class-9th Mid Point and Intercept Theorem ICSE Mathematics Solutions Exercise – 12 B

Question 1

Use the following figure to find:

(i) BC, if AB = 7.2 cm.

(ii) GE, if FE = 4 cm.

(iii) AE, if BD = 4.1 cm

(iv) DF, if CG = 11 cm.

Answer

According to equal intercept theorem since CD=DE

Therefore AB=BC and EF=GF

(i)BC=AB=7.2cm

(ii)GE=EF+GF=2EF==8cm

Since B,D,F are the midpoint and AE||BF||CG

Therefore AE=2BD and CG=2DF

Ans 1 Exercise - 12 B Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths

Question 2

In the figure, give below, 2AD = AB, P is mid-point of AB, Q is mid-point of DR and PR // BS. Prove that:

(i) AQ // BS

(ii) DS = 3 Rs.

Answer

Given that AD=AP=PB as 2AD=AB and p is the midpoint of AB

(i)From triangle DPR, A and Q are the midpoint of DP and DR.

Therefore AQ||PR

Since PR||BS ,hence AQ||BS

(ii)From triangle ABC, P is the midpoint and PR||BS

Therefore R is the midpoint of BC

Ans 2 Exercise - 12 B Mid Point And Its Converse ( Including Intercept Theorem ) Concise Class-9th Selina ICSE Maths

Question 3

The side AC of a triangle ABC is produced to point E so that CE = 1/2AC. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meet AC at point P and EF at point R respectively. Prove that:

(i) 3DF = EF(ii) 4CR = AB.

Answer

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