Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q.11. In figure 14.98, AB=AC and CP||BA and AP is the bisector of exterior ∠CAD of ∆ABC . Prove that

(i) ∠PAC=∠BCA.

(ii) ABCP is a parallelogram.

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Solution:

Given,

AB = AC and CD∥BA and AP is the bisector of exterior ∠CAD of ΔABC

To prove:

(i) ∠PAC = ∠BCA

(ii) ABCP is a parallelogram.

Proof:

(i) We have,

AB=AC

⇒∠ACB = ∠ABC                              [Opposite angles of equal sides

of triangle are equal]

Now, ∠CAD=∠ABC+∠ACB

⇒∠PAC+∠PAD = 2∠ACB  [∴∠PAC = ∠PAD]

⇒2∠PAC = 2∠ACB

⇒∠PAC = ∠ACB

(ii) Now,

∠PAC = ∠BCA

⇒AP∥BC and CP∥BA                [Given]

Therefore, ABCP is a parallelogram.


Q.12. ABCD is a kite having AB=AD and BC = CD. Prove that the figure found by joining the mid points of the sides, in order, is a rectangle.

Solution:

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Given,

A kite ABCD having AB=AD and BC=CD. P, Q, R, S are the mid-points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.

To prove:

PQRS is a rectangle.

Proof:

In ∆ABC, P and Q are the mid-points of AB and BC respectively.

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

In ∆ADC, R and S are the mid-points of CD and AD respectively.

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

From (i) and (ii) we have

PQ ∥ RS and PQ = RS

Thus, in quadrilateral PQRS, a pair of opposite sides is equal and parallel. So, PQRS is a parallelogram. Now, we shall prove that one angle of parallelogram PQRS is a right angle.

Since AB = AD

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

⇒AP = AS….(iii)    [∵P and S are mid points of AB and AD]

⇒∠1 =∠2….(iv)

Now, in ΔPBQ and ΔSDR,we have

PB = SD    RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

BQ = DR            [Since PB = SD]

And PQ = SR     [Since, PQRS is a parallelogram]

So, by SSS criterion of congruence, we have

ΔPBQ ≅ ΔSDR

⇒∠3 =∠4[CPCT]

Now, ⇒ ∠3+∠SPQ+∠2 = 1800

And ∠1+∠PSR+∠4 = 1800

∴∠3+∠SPQ+∠2 =∠1+∠PSR+∠4

⇒ ∠SPQ  =∠PSR[∠1 = ∠2 and ∠3 =∠4]

Now, transversal PS cuts parallel lines SR and PQ at S and P respectively.

∴∠SPQ+∠PSR = 1800

⇒ 2∠SPQ = 1800

⇒∠SPQ = 900              [∵∠PSR =∠SPQ]

Thus, PQRS is a parallelogram such that ∠SPQ = 900.

Hence, PQRS is a parallelogram.


Q.13. Let ABC be an isosceles triangle in which AB = AC. If D, E, F be the mid points of the, sides BC,CA and AB respectively, show that the segment AD and EF  bisect each other at right angles.

Solution:

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Since D, E and F are mid-points of sides BC, CA and AB respectively.

∴ AB∥DE and AC∥DF

∴ AF∥DE and AE∥DF

ABDE is a parallelogram.

AF = DE and AE = DF

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

DE = DF                                 [Since, AB = AC]

AE = AF = DE = DF

ABDF is a rhombus.

⇒ AD and FE bisect each other at right angle.


Q.14. ABC is a triangle. D is a point on AB such that AD=1/4AB and E is a point on AC such that AE=1/4AC. Prove that DE=1/4BC.

Solution:

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Let P and Q be the mid-points of AB and AC respectively.

Then PQ∥BC

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

In ΔAPQ, D and E are the mid-points of AP and AQ respectively.

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

 

From (i) and (ii):    RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Hence proved.


Q.15. In Figure 14.99, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = 1/4AC. If PQ produced meets BC at R, prove that R is a mid-point of BC. 

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Solution:

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Join B and D.

Suppose AC and BD intersect at O.

Then  RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Now,

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

In ΔDCO, P and Q are mid points of DC and OC respectively.

∴ PQ∥DO

Also in ΔCOB, Q is the mid-point of OC and QR∥OB

Therefore, R is the mid-point of BC.


Q.16. In figure 14.100, ABCD and PQRC are rectangles and Q is the mid-point of AC. Prove that

(i) DP = PC    
RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Solution:

(i) In ΔADC, Q is the mid-point of AC such that PQ∥AD

Therefore, P is the mid-point of DC.

⇒ DP = DC     [Using mid-point theorem]

(ii) Similarly, R is the mid-point of BC

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics    [Diagonal of rectangle are equal, BD = AC]


Q.17. ABCD is a parallelogram; E and f are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC at G, P and H respectively. Prove that GP = PH.

Solution:

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Since E and F are mid-points of AB and CD respectively

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

But, AB = CD

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

⇒ BE = CF

Also, BE∥CF[∵AB∥CD]

Therefore, BEFC is a parallelogram

BC∥EF and BE = PH …..(i)

Now, BC∥EF

⇒AD∥EF  [∵BC∥AD as ABCD is aparallelogram]

Therefore, AEFD is a parallelogram.

⇒ AE = GP

But E is the mid-point of AB.

So, AE = BF

Therefore, GP = PH.


Q.18. BM and CN are perpendiculars to a line passing through the vertex A of triangle ABC. If L is the mid-point of BC, prove that LM = LN.

Solution:

To prove LM = LN

Draw LS as perpendicular to line MN.

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Therefore, the lines BM, LS and CN being the same perpendiculars on line MN are parallel to each other.

According to intercept theorem,

If there are three or more parallel lines and the intercepts made by them on a transversal are equal, then the corresponding intercepts on any other transversal are also equal.

In the figure, MB, LS and NC are three parallel lines and the two transversal lines are MN and BC.

We have, BL = LC                 [As L is the given mid-point of BC]

Using the intercept theorem, we get

MS = SN …. (i)

Now in ΔMLS and ΔLSN

MS = SN using equation (i).

∠LSM =∠LSN=900  [LS⊥MN]

And SL = LS is common.

∴ΔMLS ≅ ΔLSN   [SAS CongruencyTheorem]

∴LM = LN [CPCT]


Q.19. Show that, the line segments joining the mid-points of opposite sides of a quadrilateral bisects each other.

Solution:

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Let ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively.

So, by using mid-point theorem we can say that

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Similarly in ΔBCD

RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

From equations (i) and (ii), we have

SP∥QR and SP = QR

As in quadrilateral SPQR, one pair of opposite sides is equal and parallel to each other.

So, SPQR is a parallelogram since the diagonals of a parallelogram bisect each other.

Hence PR and QS bisect each other.

Q.20. Fill in the blanks to make the following statements correct:

(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is ____________.

(ii) The triangle formed by joining the mid-points of the sides of a right triangle is ____________.

(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is ____________.

Solution:

(i) Isosceles

(ii) Right triangle

(iii) Parallelogram

The document RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
All you need of Class 9 at this link: Class 9
91 docs

Top Courses for Class 9

FAQs on RD Sharma Solutions Ex-14.4, (Part - 2), Quadrilaterals, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What are the properties of a quadrilateral?
Ans. A quadrilateral is a polygon with four sides and four angles. The properties of a quadrilateral include: 1. Sum of interior angles: The sum of the interior angles of a quadrilateral is always 360 degrees. 2. Opposite sides: Opposite sides of a quadrilateral are equal in length. 3. Opposite angles: Opposite angles of a quadrilateral are equal in measure. 4. Diagonals: A quadrilateral has two diagonals, which are line segments connecting opposite vertices. The diagonals of a quadrilateral bisect each other. 5. Types of quadrilaterals: Quadrilaterals can be classified into various types such as parallelogram, rectangle, square, rhombus, trapezium, etc.
2. What is the difference between a parallelogram and a rectangle?
Ans. Both a parallelogram and a rectangle are types of quadrilaterals, but they have some differences: 1. Shape: A parallelogram is a quadrilateral with opposite sides that are parallel, while a rectangle is a quadrilateral with all four angles equal to 90 degrees. 2. Angles: In a parallelogram, opposite angles are equal, but in a rectangle, all angles are equal to 90 degrees. 3. Diagonals: The diagonals of a parallelogram bisect each other, while the diagonals of a rectangle are equal in length and bisect each other. 4. Properties: A rectangle has additional properties like all four sides are equal in length, opposite sides are parallel, and diagonals are equal.
3. How can we prove that a quadrilateral is a parallelogram?
Ans. To prove that a quadrilateral is a parallelogram, we can use any of the following methods: 1. Opposite sides are parallel: If we can prove that the opposite sides of a quadrilateral are parallel, then it is a parallelogram. 2. Opposite angles are equal: If we can prove that the opposite angles of a quadrilateral are equal, then it is a parallelogram. 3. Diagonals bisect each other: If we can prove that the diagonals of a quadrilateral bisect each other, then it is a parallelogram. 4. Consecutive angles are supplementary: If we can prove that the consecutive angles of a quadrilateral are supplementary (add up to 180 degrees), then it is a parallelogram.
4. What is the sum of interior angles of a quadrilateral?
Ans. The sum of the interior angles of a quadrilateral is always 360 degrees. This means that if we measure all the angles of a quadrilateral and add them together, the result will always be 360 degrees.
5. How can we find the area of a quadrilateral?
Ans. The area of a quadrilateral can be found using different formulas based on its type. Here are some common formulas: 1. Parallelogram: Area = base x height 2. Rectangle: Area = length x width 3. Square: Area = side x side 4. Rhombus: Area = (product of diagonals) / 2 5. Trapezium: Area = (sum of parallel sides) x (perpendicular distance between them) By using the appropriate formula, we can calculate the area of a given quadrilateral.
91 docs
Download as PDF
Explore Courses for Class 9 exam

Top Courses for Class 9

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Exam

,

Class 9

,

Quadrilaterals

,

RD Sharma Solutions Ex-14.4

,

past year papers

,

Class 9

,

Sample Paper

,

Previous Year Questions with Solutions

,

mock tests for examination

,

shortcuts and tricks

,

practice quizzes

,

Summary

,

pdf

,

Maths | RD Sharma Solutions for Class 9 Mathematics

,

ppt

,

(Part - 2)

,

(Part - 2)

,

Maths | RD Sharma Solutions for Class 9 Mathematics

,

Extra Questions

,

Maths | RD Sharma Solutions for Class 9 Mathematics

,

Quadrilaterals

,

Free

,

Objective type Questions

,

Important questions

,

Viva Questions

,

Semester Notes

,

RD Sharma Solutions Ex-14.4

,

RD Sharma Solutions Ex-14.4

,

Quadrilaterals

,

video lectures

,

MCQs

,

study material

,

(Part - 2)

,

Class 9

;