Advertisements
Chapters
2: Compound Interest (Without using formula)
3: Compound Interest (Using Formula)
4: Expansions (Including Substitution)
5: Factorisation
6: Simultaneous (Linear) Equations (Including Problems)
7: Indices (Exponents)
8: Logarithms
9: Triangles [Congruency in Triangles]
10: Isosceles Triangles
11: Inequalities
▶ 12: Mid-point and Its Converse [ Including Intercept Theorem]
13: Pythagoras Theorem [Proof and Simple Applications with Converse]
14: Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]
15: Construction of Polygons (Using ruler and compass only)
16: Area Theorems [Proof and Use]
17: Circle
18: Statistics
19: Mean and Median (For Ungrouped Data Only)
20: Area and Perimeter of Plane Figures
21: Solids [Surface Area and Volume of 3-D Solids]
22: Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals]
23: Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios]
24: Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle]
25: Complementary Angles
26: Co-ordinate Geometry
27: Graphical Solution (Solution of Simultaneous Linear Equations, Graphically)
28: Distance Formula
![Selina solutions for Concise Mathematics [English] Class 9 ICSE chapter 12 - Mid-point and Its Converse [ Including Intercept Theorem] Selina solutions for Concise Mathematics [English] Class 9 ICSE chapter 12 - Mid-point and Its Converse [ Including Intercept Theorem] - Shaalaa.com](/images/concise-mathematics-english-class-9-icse_6:b313c06da7fb4b0f885a06c3b5e4e4fa.jpg)
Advertisements
Solutions for Chapter 12: Mid-point and Its Converse [ Including Intercept Theorem]
Below listed, you can find solutions for Chapter 12 of CISCE Selina for Concise Mathematics [English] Class 9 ICSE.
Selina solutions for Concise Mathematics [English] Class 9 ICSE 12 Mid-point and Its Converse [ Including Intercept Theorem] Exercise 12 (A) [Pages 150 - 151]
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
D, E, and F are the mid-points of the sides AB, BC, and CA respectively of ΔABC. AE meets DF at O. P and Q are the mid-points of OB and OC respectively. Prove that DPQF is a parallelogram.
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
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:
- Point P bisects BE,
- PQ is parallel to AB.
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.
D and F are midpoints of sides AB and AC of a triangle ABC. A line through F and parallel to AB meets BC at point E.
- Prove that BDFE is a parallelogram
- Find AB, if EF = 4.8 cm.
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.
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.
Selina solutions for Concise Mathematics [English] Class 9 ICSE 12 Mid-point and Its Converse [ Including Intercept Theorem] Exercise 12 (B) [Pages 153 - 154]
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.
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 = 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.
Solutions for 12: Mid-point and Its Converse [ Including Intercept Theorem]
![Selina solutions for Concise Mathematics [English] Class 9 ICSE chapter 12 - Mid-point and Its Converse [ Including Intercept Theorem] Selina solutions for Concise Mathematics [English] Class 9 ICSE chapter 12 - Mid-point and Its Converse [ Including Intercept Theorem] - Shaalaa.com](/images/concise-mathematics-english-class-9-icse_6:b313c06da7fb4b0f885a06c3b5e4e4fa.jpg)
Selina solutions for Concise Mathematics [English] Class 9 ICSE chapter 12 - Mid-point 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 12 (Mid-point 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.
Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. Selina textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.
Concepts covered in Concise Mathematics [English] Class 9 ICSE chapter 12 Mid-point and Its Converse [ Including Intercept Theorem] are Theorem of Midpoints of Two Sides of a Triangle, Equal Intercept Theorem.
Using Selina Concise Mathematics [English] Class 9 ICSE solutions Mid-point 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.
Get the free view of Chapter 12, Mid-point and Its Converse [ Including Intercept Theorem] Concise Mathematics [English] Class 9 ICSE additional questions for Mathematics Concise Mathematics [English] Class 9 ICSE CISCE, and you can use Shaalaa.com to keep it handy for your exam preparation.