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 triangle and theorems 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.

## Download PDF of RD Sharma Solutions for Class 9 Maths Chapter 14 Quadrilaterals Exercise 14.4

### Access Answers to Maths RD Sharma Class 9 Chapter 14 Quadrilaterals Exercise 14.4 Page number 14.55

**Question 1: In a Î”ABC, 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 Î”DEF.**

**Solution:**

Given: AB = 7 cm, BC = 8 cm, AC = 9 cm

In âˆ†ABC,

In a Î”ABC, 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 âˆ†DEF = DE + EF + DF

= 1/2 (AB + BC + AC)

= 1/2 (7 + 8 + 9)

= 12

Perimeter of Î”DEF = 12cm

**Question 2: In a Î”ABC, âˆ A = 50 ^{0}, âˆ B = 60^{0} and âˆ C = 70^{0}. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.**

** Solution:**

In Î”ABC,

D, E and F are mid points of AB,BC and AC respectively.

In a Quadrilateral DECF:

By Mid-point theorem,

DE âˆ¥ AC â‡’ DE = AC/2

And CF = AC/2

â‡’ DE = CF

Therefore, DECF is a parallelogram.

âˆ C = âˆ D = 70^{0}

Similarly,

ADEF is a parallelogram, âˆ A = âˆ E = 50^{0}

BEFD is a parallelogram, âˆ B = âˆ F = 60^{0}

Hence, Angles of Î”DEF are: âˆ D = 70^{0}, âˆ E = 50^{0}, âˆ F = 60^{0}.

**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 Î”ABC,

R and P are mid points of AB and BC

By Mid-point Theorem

RP âˆ¥ AC â‡’ RP = AC/2

In a quadrilateral, ARPQ

RP âˆ¥ AQ â‡’ 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 Î”ABC 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 Î”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:**

In a Î”ABC

E and F are mid points of AC and AB (Given)

EF âˆ¥ FE â‡’ EF = BC/2 and

[By mid-point theorem]In Î”ABP

F is the mid-point of AB, again by mid-point theorem

FQ âˆ¥ BP

Q is the mid-point of AP

AQ = QP

Hence Proved.

**Question 6: In a Î”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:**

Given that,

In Î”BLM and Î”CLN

âˆ BML = âˆ CNL = 90^{0}

BL = CL [L is the mid-point of BC]

âˆ MLB = âˆ NLC [Vertically opposite angle]

By ASA criterion:

Î”BLM â‰… Î”CLN

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 Î”ADE.**

**Solution:**

In Î”ABC,

âˆ B=90^{0 }(Given)

AB = 9 cm, AC = 15 cm (Given)

By using Pythagoras theorem

AC^{2} = AB^{2} + BC^{2}

â‡’15^{2} = 9^{2} + BC^{2}

â‡’BC^{2 }= 225 â€“ 81 = 144

or BC = 12

Again,

AD = DB = AB/2 = 9/2 = 4.5 cm [D is the midâˆ’point of AB

D and E are mid-points of AB and AC

DE âˆ¥ BC â‡’ DE = BC/2 [By midâˆ’point theorem]

Now,

Area of Î”ADE = 1/2 x AD x DE

= 1/2 x 4.5 x 6

=13.5

Area of Î”ADE is 13.5 cm^{2}

**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 midâˆ’point theorem, we have

MNâˆ¥BC â‡’ 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 Î”PQR is double the perimeter of Î”ABC.**

**Solution:**

ABCQ and ARBC are parallelograms.

Therefore, BC = AQ and BC = AR

â‡’AQ = AR

â‡’A is the mid-point of QR

Similarly B and C are the mid points of PR and PQ respectively.

By midâˆ’point theorem, we have

AB = PQ/2, BC = QR/2 and CA = PR/2

or PQ = 2AB, QR = 2BC and PR = 2CA

â‡’PQ + QR + RP = 2 (AB + BC + CA)

â‡’ Perimeter of Î”PQR = 2 (Perimeter of Î”ABC)

Hence proved.

**Question 10: In figure, BE âŠ¥ 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 âˆ PQR = 90 ^{0}.**

**Solution:**

BEâŠ¥AC and P, Q and R are respectively mid-point of AH, AB and BC. (Given)

In Î”ABC, Q and R are mid-points of AB and BC respectively.

By Mid-point theorem:

QR âˆ¥ AC â€¦..(i)

In Î”ABH, Q and P are the mid-points of AB and AH respectively

QP âˆ¥ BH â€¦.(ii)

But, BEâŠ¥AC

From (i) and (ii) we have,

QPâŠ¥QR

â‡’âˆ PQR = 90^{0}

Hence Proved.

## Access other exercise solutions of Class 9 Maths Chapter 14 Quadrilaterals

## RD Sharma Solutions for Class 9 Maths Chapter 14 Exercise 14.4

Class 9 Maths Chapter 14 Quadrilaterals Exercise 14.4 is based on the following topics:

- Useful facts about Triangle
- Theorems results of Triangle

RD Sharma solutions given here include the solutions of all the questions enlisted under exercise 14.4. Students can download pdf and start practising.