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Chapters
1: Rational and Irrational Numbers
Unit 2: Commercial Mathematics
2: Compound Interest (Stage 1) [Basic Concepts]
3: Compound Interest (Stage 2) [Applications]
Unit 3: Algebra
4: Expansions
5: Factorisation
6: Simultaneous (Linear) Equations [Including Problems]
7: Indices [Exponents]
8: Logarithms
Unit 4: Geometry
9: Triangles [Congruency in Triangles]
10: Isosceles Triangles [Including Inequalities]
▶ 11: Mid-point Theorem and Its Converse [Including Intercept Theorem]
12: Pythagoras Theorem [Proof and Simple Applications with Converse]
13: Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]
14: Construction of Polygons (Using ruler and compass only)
15: Area Theorems [Proof and Use]
16: Circle
Unit 5: Statistics and Graph Work
17: Statistics
18: Mean and Median [For Ungrouped Data Only]
Unit 6: Mensuration
19: Area and Perimeter of Plane Figures
20: Solids [Surface Area and Volume of 3-D Solids]
Unit 7: Trigonometry
21: Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals]
22: Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle]
Unit 8: Co-Ordinate
23: Co-ordinate Geometry
24: Graphical Solution [Solution of Simultaneous Linear Equations, Graphically]
25: Distance Formula
![Selina solutions for Concise Mathematics [English] Class 9 ICSE chapter 11 - Mid-point Theorem and Its Converse [Including Intercept Theorem] Selina solutions for Concise Mathematics [English] Class 9 ICSE chapter 11 - Mid-point Theorem and Its Converse [Including Intercept Theorem] - Shaalaa.com](/images/concise-mathematics-english-class-9-icse_6:e09935b48e334a1e8f06ebb2011509f8.jpg)
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Solutions for Chapter 11: Mid-point Theorem and Its Converse [Including Intercept Theorem]
Below listed, you can find solutions for Chapter 11 of CISCE Selina for Concise Mathematics [English] Class 9 ICSE.
Selina solutions for Concise Mathematics [English] Class 9 ICSE 11 Mid-point Theorem and Its Converse [Including Intercept Theorem] Exercise 11 (A) [Pages 168 - 169]
Multiple Choice Type: Choose the correct answer from the options given below.
In the given figure, ABCD is a rectangle. As per the given information, the length of PQ is:

12 cm
14 cm
20 cm
10 cm
The quadrilateral obtained by joining the mid-points (in order) of the sides of quadrilateral ABCD is a ______.
rectangle
rhombus
parallelogram
square
If BC = 12 cm, AB = 14.8 cm. AC = 12.8 cm, the perimeter of quadrilateral BCYX is ______.

31.8 cm
15.9 cm
29.8 cm
32.8 cm
In the given figure, AB = AC, P, Q and R are mid-points of sides BC, CA and AB, respectively, then ΔPQR is ______.

scalene
isosceles
equilateral
obtuse angled
P, Q, R and S are the midpoints of sides AB, BС, CD and DA, respectively, of rectangle ABCD; then quadrilateral PQRS is a ______.
rectangle
rhombus
square
parallelogram
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.
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
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.
The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:
PR = `[1]/[2]` ( AB + CD)

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:

- MN, if AB = 11 cm and DC = 8 cm.
- AB, if DC = 20 cm and MN = 27 cm.
- DC, if MN = 15 cm and AB = 23 cm.
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.
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.
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.

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) PR = `[1]/[2]` DB
Selina solutions for Concise Mathematics [English] Class 9 ICSE 11 Mid-point Theorem and Its Converse [Including Intercept Theorem] Exercise 11 (B) [Pages 171 - 172]
Multiple Choice Tуре: Choose the correct answer from the options given below.
In the given figure, l // m // n and D is mid-point of CE. If AE = 12.6 cm, then BD is ______.

12.6 cm
25.2 cm
6.3 cm
18.9 cm
In a trapezium ABCD, AB//DC, E is midpoint of AD and F is mid-point of BC, then:
2EF = `1/2` (AB + DC)
2EF = AB + DC
EF = AB + DC
EF = `1/2` × AB × DC
The given figure shows a parallelogram ABCD in which E is mid-point of AD and DL//EB. Then, BF is equal to:

AD
BE
AE
AB
In the given figure, AD and BE are medians; then ED is equal to:

2AB
`1/2` AB
`1/4` AB
`1/8` AB
In the quadrilateral ABCD, if AB//CD, E is mid-point of side AD, and F is mid-point of BC. If AB = 20 cm and EF = 16 cm, the length of side DC is:

18 cm
12 cm
24 cm
32 cm
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.

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.

The side AC of a triangle ABC is produced to point E so that CE = AC. 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:
- 3DF = EF
- 4CR = AB
In triangle ABC, the medians BP and CQ are produced up to points M and N respectively such that BP = PM and CQ = QN. Prove that:
- M, A, and N are collinear.
- A is the mid-point of MN.
In triangle ABC, angle B is obtuse. D and E are mid-points of sides AB and BC respectively and F is a point on side AC such that EF is parallel to AB. Show that BEFD is a parallelogram.
In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively.
Prove that:
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.
In triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meets side BC at points M and N respectively. Prove that: BM = MN = NC.
In triangle ABC; M is mid-point of AB, N is mid-point of AC and D is any point in base BC. Use the intercept Theorem to show that MN bisects AD.
If the quadrilateral formed by joining the mid-points of the adjacent sides of quadrilateral ABCD is a rectangle,
show that the diagonals AC and BD intersect at the right angle.
Selina solutions for Concise Mathematics [English] Class 9 ICSE 11 Mid-point Theorem and Its Converse [Including Intercept Theorem] TEST YOURSELF [Pages 172 - 174]
Multiple Choice Type: Choose the correct answer from the options given below.
The mid-points of the sides of a triangle are joined together to get four triangles. These four triangles are ______.
not equal to each other.
congruent to each other.
not congruent to each other.
none of these
In the given figure, AB//CD//EF. If AC = 7 cm, AE = 14 cm and BF = 20 cm. then DF is equal to:

7 cm
14 cm
10 cm
16 cm
In the given figure, AB//DC//EF and E is mid-point of side AD, then:

OE : OF = 11 : 3
OE = OF
OF = 2 × OE
CF = FB
In rhombus PQRS, A, B and C are midpoints of sides PQ. QR and RS, respectively. If ∠P = 60°, the angle PQR is equal to:

60°
90°
120°
none of these
Statement (1): The diagonals of a quadrilateral are perpendicular to each other; P, Q, R and S are mid-points of sides AB, BC, CD and DA, respectively. Then PQRS will be a rectangle.

Statement (2): Quadrilateral PQRS will be a square as each of its angles will be 90°.
Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.
Statement (1): AD is median of triangle ABC and DE is parallel to BA. Then DE will bisect AC.

Statement (2): DE is the median of ΔADC.
Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.
Assertion (A): The figure formed by joining the midpoints of the sides of a quadrilateral ABCD is a square.
Reason (R): Diagonals of the quadrilateral ABCD are not given to be equal and perpendicular to each other.
A is true, R is false.
A is false, R is true.
Both A and R are true, and R is the correct reason for A.
Both A and R are true, and R is the incorrect reason for A.
Assertion (A): R, S, D and E are midpoints of OС, ОВ, АВ, and AC, respectively; then DERS is a parallelogram.

Reason (R): DS//AO//ER and DS = ER = `1/2` AO.
A is true, R is false.
A is false, R is true.
Both A and R are true, and R is the correct reason for A.
Both A and R are true, and R is the incorrect reason for A.
In triangle ABC, D and E are mid-points of the sides AB and AC, respectively. Through E, a straight line is drawn parallel to AB to meet BC at F. Prove that BDEF is a parallelogram. If AB = 8 cm and BC = 9 cm, find the perimeter of the parallelogram BDEF.
P, Q and R are mid-points of sides AВ, ВС and CD, respectively, of a rhombus ABCD. Show that PQ is perpendicular to QR.
The diagonals of a quadrilateral ABCD are perpendicular to each other. Prove that the quadrilateral obtained by joining the midpoints of its adjacent sides is a rectangle.
In ∆ABC, E is the mid-point of the median AD, and BE produced meets side AC at point Q.
Show that BE: EQ = 3: 1.
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.
In triangle ABC, D and E are the mid-points of the sides AB and AC, respectively. Through E, a straight line is drawn parallel to AB to meet BC at F. Prove that BDEF is a parallelogram. If AB = 16 cm, AC = 12 cm and BC = 18 cm, find the perimeter of the parallelogram BDEF.
In the given figure, AD and CE are medians and DF // CE.
Prove that: FB = `1/4` AB.
In parallelogram ABCD, E is the mid-point of AB and AP is parallel to EC which meets DC at point O and BC produced at P.
Prove that:
(i) BP = 2AD
(ii) O is the mid-point of AP.
In trapezium ABCD, sides AB and DC are parallel to each other. E is mid-point of AD and F is mid-point of BC.
Prove that: AB + DC = 2EF.
In Δ ABC, AD is the median and DE is parallel to BA, where E is a point in AC. Prove that BE is also a median.
Adjacent sides of a parallelogram are equal and one of the diagonals is equal to any one of the sides of this parallelogram. Show that its diagonals are in ratio √3:1.
Case-Study Based Question
A school is designing a triangular garden ΔАВС. То construct a walking path inside the garden, the gardener marks the mid-points of two sides: point D is the mid-point of side AB and point E is the mid-point of side AC. The path DE is drawn to connect these mid-points. The length of side BC of the triangular garden is 12 m. AB = 10 m and AC = 10 m.

Based on the above information, answer the following:
- What is the length of path DЕ?
- Assign a special name to Quadrilateral BCED and find its perimeter.
Solutions for 11: Mid-point Theorem and Its Converse [Including Intercept Theorem]
![Selina solutions for Concise Mathematics [English] Class 9 ICSE chapter 11 - Mid-point Theorem and Its Converse [Including Intercept Theorem] Selina solutions for Concise Mathematics [English] Class 9 ICSE chapter 11 - Mid-point Theorem and Its Converse [Including Intercept Theorem] - Shaalaa.com](/images/concise-mathematics-english-class-9-icse_6:e09935b48e334a1e8f06ebb2011509f8.jpg)
Selina solutions for Concise Mathematics [English] Class 9 ICSE chapter 11 - Mid-point Theorem and Its Converse [Including Intercept Theorem]
Shaalaa.com has the CISCE Mathematics Concise Mathematics [English] Class 9 ICSE CISCE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. Selina solutions for Mathematics Concise Mathematics [English] Class 9 ICSE CISCE 11 (Mid-point Theorem and Its Converse [Including Intercept Theorem]) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.
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Concepts covered in Concise Mathematics [English] Class 9 ICSE chapter 11 Mid-point Theorem and Its Converse [Including Intercept Theorem] are .
Using Selina Concise Mathematics [English] Class 9 ICSE solutions Mid-point Theorem and Its Converse [Including Intercept Theorem] exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in Selina Solutions are essential questions that can be asked in the final exam. Maximum CISCE Concise Mathematics [English] Class 9 ICSE students prefer Selina Textbook Solutions to score more in exams.
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