RD Sharma Class 9 Mathematics Chapter 14 Exercise 14.4 solutions for Quadrilaterals are provided here. In this exercise, students will learn about useful facts about triangles and theorem results of triangle.
Some important results:
1. The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.
2. The line drawn through the mid-point of one side of a triangle, parallel to another side, intersects the 3rd side at its mid-point.
RD Sharma Solutions for Class 9 Maths Chapter 14 Quadrilaterals Exercise 14.4
Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 14 Quadrilaterals Exercise 14.4 Page number 14.55
Question 1: In a ΔABC, D, E and F are, respectively, the mid points of BC, CA and AB. If the lengths of sides AB, BC and CA are 7 cm, 8 cm and 9 cm, respectively, find the perimeter of ΔDEF.
Solution:
Given: AB = 7 cm, BC = 8 cm, AC = 9 cm
In ∆ABC,
In a ΔABC, D, E and F are, respectively, the mid points of BC, CA and AB.
According to Midpoint Theorem:
EF = 1/2BC, DF = 1/2 AC and DE = 1/2 AB
Now, Perimeter of ∆DEF = DE + EF + DF
= 1/2 (AB + BC + AC)
= 1/2 (7 + 8 + 9)
= 12
Perimeter of ΔDEF = 12cm
Question 2: In a ΔABC, ∠A = 500, ∠B = 600 and ∠C = 700. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.
Solution:
In ΔABC,
D, E and F are mid points of AB,BC and AC respectively.
In a Quadrilateral DECF:
By Mid-point theorem,
DE ∥ AC ⇒ DE = AC/2
And CF = AC/2
⇒ DE = CF
Therefore, DECF is a parallelogram.
∠C = ∠D = 700
[Opposite sides of a parallelogram]Similarly,
ADEF is a parallelogram, ∠A = ∠E = 500
BEFD is a parallelogram, ∠B = ∠F = 600
Hence, Angles of ΔDEF are: ∠D = 700, ∠E = 500, ∠F = 600.
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 = 29 cm and AB = 30 cm, find the perimeter of the quadrilateral ARPQ.
Solution:
In ΔABC,
R and P are mid points of AB and BC
By Mid-point Theorem
RP ∥ AC ⇒ RP = AC/2
In a quadrilateral, ARPQ
RP ∥ AQ ⇒ RP = AQ
[A pair of side is parallel and equal]Therefore, ARPQ is a parallelogram.
Now, AR = AB/2 = 30/2 = 15 cm
[AB = 30 cm (Given)]AR = QP = 15 cm
[ Opposite sides are equal ]And RP = AC/2 = 21/2 = 10.5 cm
[AC = 21 cm (Given)]RP = AQ = 10.5cm
[ Opposite sides are equal ]Now,
Perimeter of ARPQ = AR + QP + RP +AQ
= 15 +15 +10.5 +10.5
= 51
Perimeter of quadrilateral ARPQ is 51 cm.
Question 4: In a ΔABC median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.
Solution:
In a quadrilateral ABXC,
AD = DX [Given]
BD = DC [Given]
From figure, Diagonals AX and BC bisect each other.
ABXC is a parallelogram.
Hence Proved.
Question 5: In a ΔABC, 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:
In a ΔABC
E and F are mid points of AC and AB (Given)
EF ∥ BC ⇒ EF = BC/2 and
[By mid-point theorem]In ΔABP
F is the mid-point of AB, again by mid-point theorem
FQ ∥ BP
Q is the mid-point of AP
AQ = QP
Hence Proved.
Question 6: In a ΔABC, 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:
Given that,
In ΔBLM and ΔCLN
∠BML = ∠CNL = 900
BL = CL [L is the mid-point of BC]
∠MLB = ∠NLC [Vertically opposite angle]
By ASA criterion:
ΔBLM ≅ ΔCLN
So, LM = LN [By CPCT]
Question 7: In figure, triangle ABC is a right-angled triangle at B. Given that AB = 9 cm, AC = 15 cm and D, E are the mid-points of the sides AB and AC respectively, calculate
(i) The length of BC (ii) The area of ΔADE.
Solution:
In ΔABC,
∠B=900 (Given)
AB = 9 cm, AC = 15 cm (Given)
By using Pythagoras theorem
AC2 = AB2 + BC2
⇒152 = 92 + BC2
⇒BC2 = 225 – 81 = 144
or BC = 12
Again,
AD = DB = AB/2 = 9/2 = 4.5 cm [D is the mid−point of AB
D and E are mid-points of AB and AC
DE ∥ BC ⇒ DE = BC/2 [By mid−point theorem]
Now,
Area of ΔADE = 1/2 x AD x DE
= 1/2 x 4.5 x 6
=13.5
Area of ΔADE is 13.5 cm2
Question 8: In figure, M, N and P are mid-points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate BC, AB and AC.
Solution:
Given: MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm.
M and N are mid-points of AB and AC
By mid−point theorem, we have
MN∥BC ⇒ MN = BC/2
or BC = 2MN
BC = 6 cm
[MN = 3 cm given)Similarly,
AC = 2MP = 2 (2.5) = 5 cm
AB = 2 NP = 2 (3.5) = 7 cm
Question 9: 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 ΔPQR is double the perimeter of ΔABC.
Solution:
ABCQ and ARBC are parallelograms.
Therefore, BC = AQ and BC = AR
⇒AQ = AR
⇒A is the mid-point of QR
Similarly B and C are the mid points of PR and PQ respectively.
By mid−point theorem, we have
AB = PQ/2, BC = QR/2 and CA = PR/2
or PQ = 2AB, QR = 2BC and PR = 2CA
⇒PQ + QR + RP = 2 (AB + BC + CA)
⇒ Perimeter of ΔPQR = 2 (Perimeter of ΔABC)
Hence proved.
Question 10: In 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 = 900.
Solution:
BE⊥AC and P, Q and R are respectively mid-point of AH, AB and BC. (Given)
In ΔABC, Q and R are mid-points of AB and BC respectively.
By Mid-point theorem:
QR ∥ AC …..(i)
In ΔABH, Q and P are the mid-points of AB and AH respectively
QP ∥ BH ….(ii)
But, BE⊥AC
From (i) and (ii) we have,
QP⊥QR
⇒∠PQR = 900
Hence Proved.
Access other exercise solutions of Class 9 Maths Chapter 14 Quadrilaterals
RD Sharma Solutions for Class 9 Maths Chapter 14 Quadrilaterals Exercise 14.4
Class 9 Maths Chapter 14 Quadrilaterals Exercise 14.4 is based on the following topics:
- Useful facts about Triangles
- Theorems results of Triangles
RD Sharma Solutions given here include the solutions of all the questions enlisted under Exercise 14.4. Students can download the PDF and start practising.
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