## RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4

Other Exercises

- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

Question 1.

In a ∆ABC, D, E and F are respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC and CA are 7cm, 8cm and 9cm, respectively, find the perimeter of ∆DEF.

Solution:

In ∆ABC, D, E and F are the mid-points of sides,

BC, CA, AB respectively

AB = 7cm, BC = 8cm and CA = 9cm

∵ D and E are the mid points of BC and CA

∴ DE || AB and DE =\(\frac { 1 }{ 2 }\) AB =\(\frac { 1 }{ 2 }\) x 7 = 3.5cm

Similarly,

Question 2.

In a triangle ∠ABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.

Solution:

In ∆ABC,

∠A = 50°, ∠B = 60° and ∠C = 70°

D, E and F are the mid points of sides BC, CA and AB respectively

DE, EF and ED are joined

∵ D, E and F are the mid points of sides BC, CA and AB respectively

∴ EF || BC

DE || AB and FD || AC

∴ BDEF and CDEF are parallelogram

∴ ∠B = ∠E = 60° and ∠C = ∠F = 70°

Then ∠A = ∠D = 50°

Hence ∠D = 50°, ∠E = 60° and ∠F = 70°

Question 3.

In a triangle, P, Q and R are the mid-points of sides BC, CA and AB respectively. If AC = 21 cm, BC = 29cm and AB = 30cm, find the perimeter of the quadrilateral ARPQ.

Solution:

P, Q, R are the mid points of sides BC, CA and AB respectively

AC = 21 cm, BC = 29 cm and AB = 30°

∵ P, Q, R and the mid points of sides BC, CA and AB respectively.

∴ PQ || AB and PQ = \(\frac { 1 }{ 2 }\) AB

Question 4.

In a ∆ABC median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.

Solution:

Given : In ∆ABC, AD is median and AD is produced to X such that DX = AD

To prove : ABXC is a parallelogram

Construction : Join BX and CX

Proof : In ∆ABD and ∆CDX

AD = DX (Given)

BD = DC (D is mid points)

∠ADB = ∠CDX (Vertically opposite angles)

∴ ∆ABD ≅ ∆CDX (SAS criterian)

∴ AB = CX (c.p.c.t.)

and ∠ABD = ∠DCX

But these are alternate angles

∴ AB || CX and AB = CX

∴ ABXC is a parallelogram.

Question 5.

In a ∆ABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.

Solution:

Given : In ∆ABC, E and F are the mid-points of AC and AB respectively.

EF are joined.

AP ⊥ BC is drawn which intersects EF at Q and meets BC at P.

To prove: AQ = QP

proof : In ∆ABC

E and F are the mid points of AC and AB

∴ EF || BC and EF = \(\frac { 1 }{ 2 }\)BC

∴ ∠F = ∠B

In ∆ABP,

F is mid point of AB and Q is the mid point of FE or FQ || BC

∴ Q is mid point of AP,

∴ AQ = QP

Question 6.

In a ∆ABC, BM and CN are perpendiculars from B and C respectively on any line passing through A. If L is the mid-point of BC, prove that ML NL.

Solution:

In ∆ABC,

BM and CN are perpendicular on a line drawn from A. L is the mid point of BC. ML and NL are joined.

Question 7.

In the figure triangle ABC is right-angled at B. Given that AB = 9cm. AC = 15cm and D, E are the mid points of the sides AB and AC respectively, calculate.

(i) The length of BC

(ii) The area of ∆ADC

Solution:

In ∆ABC, ∠B = 90°

AC =15 cm, AB = 9cm

D and E are the mid points of sides AB and AC respectively and D, E are joined.

Question 8.

In the figure, M, N and P are the mid points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5cm and MP = 2.5cm, calculate BC, AB and AC.

Solution:

In ∆ABC,

M, N and P are the mid points of side, AB, AC and BC respectively.

Question 9.

In the figure, 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.

Solution:

Given : In ABC, AB = AC

nd CP || BA, AP is the bisector of exterior ∠CAD of ∆ABC

To prove :

(i) ∠PAC = ∠BCA

(ii) ABCP is a ||gm

Proof : (i) In ∆ABC,

∵ AB =AC

∴ ∠C = ∠B (Angles opposite to equal sides) and ext.

∠CAD = ∠B + ∠C

= ∠C + ∠C = 2∠C ….(i)

∵ AP is the bisector of ∠CAD

∴ 2∠PAC = ∠CAD …(ii)

From (i) and (ii)

∠C = 2∠PAC

∠C = ∠CAD or ∠BCA = ∠PAC

Hence ∠PAC = ∠BCA

(ii) But there are alternate angles,

∴ AD || BC

But BA || CP

∴ ABCP is a ||gm.

Question 10.

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

Solution:

Given : In fne figure, ABCD is a kite in which AB = AD and BC = CD.

P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively.

To prove : PQRS is a rectangle.

Construction : Join AC and BD.

Proof: In ∆ABD,

P and S are mid points of AB and AD

∴ PS || BD and PS = \(\frac { 1 }{ 2 }\) BD …(i)

Similarly in ∆BCD,

Q and R the mid points of BC and CD

∴ QR || BD and

QR = \(\frac { 1 }{ 2 }\) BD …(ii)

∴ Similarly, we can prove that PQ || SR and PQ = SR …(iii)

From (i) and (ii) and (iii)

PQRS is a parallelogram,

∵ AC and BD intersect each other at right angles.

∴ PQRS is a rectangle.

Question 11.

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:

In ∆ABC, AB = AC

D, E and F are the mid points of the sides BC, CA and AB respectively,

AD and EF are joined intersecting at O

To prove : AD and EF bisect each other at right angles.

Construction : Join DE and DF.

Proof : ∵ D, E and F are the mid-points of

the sides BC, CA and AB respectively

∴ AFDE is a ||gm

∴ AF = DE and AE = DF

But AF = AE

(∵ E and F are mid-points of equal sides AB and AC)

∴ AF = DF = DE = AE

∴AFDE is a rhombus

∵ The diagonals of a rhombus bisect each other at right angle.

∴ AO = OD and EO = OF

Hence, AD and EF bisect each other at right angles.

Question 12.

Show that the line segments joining the mid points of the opposite sides of a quadrilateral bisect each other.

Solution:

Given : In quad. ABCD,

P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively.

PR and QS to intersect each other at O

To prove : PO = OR and QO = OS

Construction: Join PQ, QR, RS and SP and also join AC.

Proof: In ∆ABC

P and Q are mid-points of AB and BC

∴ PQ || AC and PQ = \(\frac { 1 }{ 2 }\) AC …(i)

Similarly is ∆ADC,

S and R are the mid-points of AD and CD

∴ SR || AC and SR = \(\frac { 1 }{ 2 }\) AC ..(ii)

from (i) and (ii)

PQ = SQ and PQ || SR

PQRS is a ||gm (∵ opposite sides are equal area parallel)

But the diagonals of a ||gm bisect each other.

∴ PR and QS bisect each other.

Question 13.

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

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

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

Question 14.

ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of ∆PQR is double the perimeter of ∆ABC.

Solution:

Given : In ∆ABC,

Through A, B and C, lines are drawn parallel to BC, CA and AB respectively meeting at P, Q and R.

To prove : Perimeter of ∆PQR = 2 x perimeter of ∆ABC

Proof : ∵ PQ || BC and QR || AB

∴ ABCQ is a ||gm

∴ BC = AQ

Similarly, BCAP is a ||gm

∴ BC = AP …(i)

∴ AQ = AP = BL

⇒ PQ = 2BC

Similarly, we can prove that

QR = 2AB and PR = 2AC

Now perimeter of ∆PQR.

= PQ + QR + PR = 2AB + 2BC + 2AC

= 2(AB + BC + AC)

= 2 perimeter of ∆ABC.

Hence proved

Question 15.

In the figure, BE ⊥ AC. AD is any line from A to BC intersecting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that PQR = 90°.

Solution:

Given: In ∆ABC, BE ⊥ AC

AD is any line from A to BC meeting BC in D and intersecting BE in H. P, Q and R are respectively mid points of AH, AB and BC. PQ and QR are joined B.

To prove : ∠PQR = 90°

Proof: In ∆ABC,

Q and R the mid points of AB and BC 1

∴ QR || AC and QR = \(\frac { 1 }{ 2 }\) AC

Similarly, in ∆ABH,

Q and P are the mid points of AB and AH

∴ QP || BH or QP || BE

But AC ⊥ BE

∴ QP ⊥ QR

∴ ∠PQR = 90°

Question 16.

ABC is a triangle. D is a point on AB such that AD = \(\frac { 1 }{ 4 }\) AB and E is a point on AC such that AE = \(\frac { 1 }{ 4 }\) AC. Prove that DE = \(\frac { 1 }{ 4 }\) BC.

Solution:

Given : In ∆ABC,

D is a point on AB such that

AD = \(\frac { 1 }{ 4 }\) AB and E is a point on AC such 1

that AE = \(\frac { 1 }{ 4 }\) AC

DE is joined.

To prove : DE = \(\frac { 1 }{ 4 }\) BC

Construction : Take P and Q the mid points of AB and AC and join them

Proof: In ∆ABC,

∵ P and Q are the mid-points of AB and AC

Question 17.

In the figure, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = \(\frac { 1 }{ 4 }\) AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.

Solution:

Given : In ||gm ABCD,

P is the mid-point of DC and Q is a point on AC such that CQ = \(\frac { 1 }{ 4 }\) AC. PQ is produced meets BC at R.

To prove : R is mid point of BC

Construction : Join BD

Proof : ∵ In ||gm ABCD,

∵ Diagonal AC and BD bisect each other at O

∴ AO = OC = \(\frac { 1 }{ 2 }\) AC …(i)

In ∆OCD,

P and Q the mid-points of CD and CO

∴ PQ || OD and PQ = \(\frac { 1 }{ 2 }\) OD

In ∆BCD,

P is mid-poiht of DC and PQ || OD (Proved above)

Or PR || BD

∴ R is mid-point BC.

Question 18.

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

Prove that (i) DP = PC (ii) PR = \(\frac { 1 }{ 2 }\) AC.

Solution:

Given : ABCD are PQRC are rectangles and Q is the mid-point of AC.

To prove : (i) DP = PC (ii) PR = \(\frac { 1 }{ 2 }\) AC

Construction : Join diagonal AC which passes through Q and join PR.

Proof : (i) In ∆ACD,

Q is mid-point of AC and QP || AD (Sides of rectangles)

∴ P is mid-point of CD

∴ DP = PC

(ii) ∵PR and QC are the diagonals of rectangle PQRC

∴ PR = QC

But Q is the mid-point of AC

∴ QC = \(\frac { 1 }{ 2 }\) AC

Hence PR = \(\frac { 1 }{ 2 }\) AC

Question 19.

ABCD is a parallelogram, E and F are the mid points AB and CD respectively. GFI is any line intersecting AD, EF and BC at Q P and H respectively. Prove that GP = PH.

Solution:

Given : In ||gm ABCD,

E and F are mid-points of AB and CD

GH is any line intersecting AD, EF and BC at GP and H respectively

To prove : GP = PH

Proof: ∵ E and F are the mid-points of AB and CD

Question 20.

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

Solution:

In ∆ABC,

BM and CN are perpendicular on a line drawn from A.

L is the mid point of BC.

ML and NL are joined.

Hope given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 are helpful to complete your math homework.

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