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AP Board Class 9 Maths Chapter 8 Quadrilaterals Ex 8.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 9th Maths Chapter 8 Quadrilaterals Ex 8.4 Book Answers |
Andhra Pradesh Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.4 Textbooks Solutions PDF
Andhra Pradesh State Board STD 9th Maths Chapter 8 Quadrilaterals Ex 8.4 Books Solutions with Answers are prepared and published by the Andhra Pradesh Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of AP Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.4 Books Answers Solutions, then you are in the right place. Here is a complete hub of Andhra Pradesh State Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.4 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Andhra Pradesh Board STD 9th Maths Chapter 8 Quadrilaterals Ex 8.4 Textbooks. These Andhra Pradesh State Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.4 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Andhra Pradesh Board.Andhra Pradesh State Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.4 Books Solutions
Board | AP Board |
Materials | Textbook Solutions/Guide |
Format | DOC/PDF |
Class | 9th |
Subject | Maths |
Chapters | Maths Chapter 8 Quadrilaterals Ex 8.4 |
Provider | Hsslive |
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AP Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.4 Textbooks Solutions with Answer PDF Download
Find below the list of all AP Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.4 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:AP State Syllabus 9th Class Maths Solutions 8th Lesson Quadrilaterals Exercise 8.4
Question 1.
ABC is a triangle. D is a point on AB such that AD = 14 AB and E is a point on AC such that AE = 14 AC. If DE = 2 cm find BC.
Solution:
Given that D and E are points on AB and AC.
Such that AD = 14 AB and AE = 14 AC
Let X, Y be midpoints of AB and AC.
Joint D, E and X, Y.
Now in ΔAXY; D, E are the midpoints of sides AX and AY.
∴ DE // XY and DE = 12 XY
⇒ 2 cm = 12 XY
⇒ XY = 2 x 2 = 4cm
Also in ΔABC; X, Y are the midpoints of AB and AC.
∴ XY//BC and XY = 12 BC
4 cm = 12 BC
⇒ BC = 4 x 2 = 8 cm
Question 2.
ABCD is a quadrilateral. E, F, G and H are the midpoints of AB, BC, CD and DA respectively. Prove that EFGH is a parallelogram.
Solution:
Given that E, F, G and H are the midpoints of the sides of quad. ABCD.
In ΔABC; E, F are the midpoints of the sides AB and BC.
∴ EF//AC and EF = 12 AC
Also in ΔACD; HG // AC
and HG = 12 AC
∴ EF // HG and EF = HG
Now in □EFGH; EF = HG and EF // HG
∴ □EFGH is a parallelogram.
Question 3.
Show that the figure formed by joining the midpoints of sides of a rhom¬bus successively is a rectangle.
Solution:
Let □ABCD be a rhombus.
P, Q, R and S be the midpoints of sides of □ABCD
In ΔABC,
P, Q are the midpoints of AB and BC.
∴ PQ//AC and PQ = 12AC …………………..(1)
Also in ΔADC, ,
S, R are the midpoints of AD and CD.
∴ SR//AC and SR = 12AC ………………(2)
From (1) and (2);
PQ // SR and PQ = SR
Similarly QR // PS and QR = PS
∴ □PQRS is a parallelogram.
As the diagonals of a rhombus bisect at right angles.
∠AOB – 90°
∴ ∠P = ∠AOB = 90°
[opp. angles of //gm PYOX] Hence □PQRS is a rectangle as both pairs of opp. sides are equal and parallel, one angle being 90°.
Question 4.
In a parallelogram ABCD, E and F are the midpoints of the sides AB and DC respectively. Show that the line segments AF and EC trisect the diagonal BD.
Solution:
□ABCD is a parallelogram. E and F are the mid points of AB and CD.
∴ AE = 12AB and CF = 12CD
Thus AE = CF [∵ AB – CD]
Now in □AECF, AE = CF and AE ||CF
Thus □AECF is a parallelogram.
Now in ΔEQB and ΔFDP
EB = FD [Half of equal sides of a //gm]
∠EBQ = ∠FDP[alt. int.angles of EB//FD]
∠QEB = ∠PFD
[∵∠QED = ∠QCF = ∠PFD]
∴ ΔEQB ≅ ΔFPD [A.S.A. congruence]
∴ BQ = DP [CPCT] ……………… (1)
Now in ΔDQC; PF // QC and F is the midpoint of DC.
Hence P must be the midpoint of DQ
Thus DP = PQ …………….. (2)
From (1) and (2), DP = PQ = QB
Hence AF and CE trisect the diagonal BD.
Question 5.
Show that the line segments joining the mid points of the opposite sides of a quadrilateral and bisect each other.
Solution:
Let ABCD be a quadrilateral.
P, Q, R, S are the midpoints of sides of □ABCD.
Join (P, Q), (Q, R), (R, S) and (S, P).
In ΔABC; P, Q are the midpoints of AB and BC.
∴ PQ // AC and PQ = 12AC ………….(1)
Also from ΔADC
S, R are the midpoints of AD and CD
SR // AC and SR = 12 AC …………………(2)
∴ From (1) & (2)
PQ = SR and PQ //SR
∴ □PQRS is a parallelogram.
Now PR and QS are the diagonals of □ PQRS.
∴ PR and QS bisect each other.
Question 6.
ABC is a triangle right angled at’C’. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that
i) D is the midpoint of AC
ii) MD ⊥ AC
iii) CM = MA= 12AB
Solution:
Given that in ΔABC; ∠C = 90°
M is the midpoint of AB.
i) If ‘D’ is the midpoints of AC.
The proof is trivial.
Let us suppose D is not the mid point of AC.
Then there exists D’ such that AD’ = D’C
Then D’M is a line parallel to BC through M.
Also DM is a line parallel to BC through M.
There exist two lines parallel to same line through a point M.
This is a contradiction.
There exists only one line parallel to a given line through a point not on the line.
∴ D’ must coincides with D
∴ D is the midpoint of AC
ii) From (i) DM // BC
Thus ∠ADM = ∠ACB = 90°
[corresponding angles]
⇒ MD ⊥ AC
iii) In ∆ADM and ∆CDM
AD = CD [ ∵ D is midpoint from (i)]
∠ADM = ∠MDC (∵ 90° each)
DM = DM (Common side)
∴ ∆ADM = ∆CDM (SAS congruence)
⇒ CM = MA (CPCT)
CM = 12 AB (∵ M is the midpoint of AB)
∴ CM = MA = 12AB
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