RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Exercise 13.4 (Updated for 2024)

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RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4

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Question 1.
In an ∆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, and AB respectively
AB = 7cm, BC = 8cm and CA = 9cm
∵ D and E are the midpoints of BC and CA
∴ DE || AB and DE =12 AB =12 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 midpoints of sides BC, CA, and AB respectively
DE, EF, and ED are joined
∵ D, E, and F are the midpoints 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, and R are the midpoints of sides BC, CA, and AB respectively
AC = 21 cm, BC = 29 cm and AB = 30°
∵ P, Q, R, and the midpoints of sides BC, CA, and AB respectively.
∴ PQ || AB and PQ = 12 AB

Question 4.
In an ∆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 midpoints)
∠ADB = ∠CDX (Vertically opposite angles)
∴ ∆ABD ≅ ∆CDX (SAS criteria)
∴ 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  ∆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 is 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 midpoints of AC and AB
∴ EF || BC and EF = 12BC
∴ ∠F = ∠B
In ∆ABP,
F is midpoint of AB and Q is the midpoint of FE or FQ || BC
∴ Q is midpoint of AP,
∴ AQ = QP

Question 6.
In an ∆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 midpoint 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 midpoints 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 midpoints of sides AB and AC respectively and D, and E are joined.

Question 8.
In the figure, M, N, and P are the midpoints 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 sides, AB, AC, and BC respectively.

Question 9.
In the figure, AB = AC and CP || BA and AP are the bisector of the 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 the 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 midpoints of AB and AD
∴ PS || BD and PS = 12 BD …(i)
Similarly in ∆BCD,
Q and R the midpoints of BC and CD
∴ QR || BD and
QR = 12 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, and 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 midpoints 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 a 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 midpoints of the opposite sides of a quadrilateral bisect each other.
Solution:

Given: In quad. ABCD,
P, Q, R, and S are the midpoints 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 the mid-points of AB and BC
∴ PQ || AC and PQ = 12 AC …(i)
Similarly  ∆ADC,
S and R are the mid-points of AD and CD
∴ SR || AC and SR = 12 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 a 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, and 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 perimeters 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 midpoints of AH, AB, and BC. PQ and QR are joined B.

To prove : ∠PQR = 90°
Proof: In ∆ABC,
Q and R the midpoints of AB and BC 1
∴ QR || AC and QR = 12 AC

Similarly, in ∆ABH,
Q and P are the midpoints 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 = 14 AB and E is a point on AC such that AE = 14 AC. Prove that DE = 14 BC.
Solution:

Given: In ∆ABC,
D is a point on AB such that
AD = 14 AB and E is a point on AC such as 1
that AE = 14 AC
DE is joined.

To prove : DE = 14 BC
Construction: Take P and Q the midpoints 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 = 14 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 = 14 AC. PQ is produced and meets BC at R.

To prove: R is midpoint of BC
Construction: Join BD
Proof : ∵ In ||gm ABCD,

∵ Diagonal AC and BD bisect each other at O
∴ AO = OC = 12 AC …(i)
In ∆OCD,
P and Q the mid-points of CD and CO
∴ PQ || OD and PQ = 12 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 = 12 AC.

Solution:
Given: ABCD is PQRC are rectangles and Q is the mid-point of AC.
To prove : (i) DP = PC (ii) PR = 12 AC
Construction: Join diagonal AC which passes through Q and joins PR.

Proof : (i) In ∆ACD,
Q is the mid-point of AC and QP || AD (Sides of rectangles)
∴ P is the mid-point of the CD
∴ DP = PC
(ii) ∵PR and QC are the diagonals of rectangle PQRC
∴ PR = QC
But Q is the mid-point of AC
∴ QC = 12 AC
Hence PR = 12 AC

Question 19.
ABCD is a parallelogram, and E and F are the midpoints 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 perpendicular 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 midpoint of BC.
ML and NL are joined.

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