The document RD Sharma Solutions Ex-14.4, (Part - 1), Quadrilaterals, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**Q.1.**** In 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 7cm, 8cm and 9cm, respectively, find the perimeter of Î”DEF.**

**Solution:**

Given that,

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

In âˆ†ABC,

F and E are the mid points of AB and AC.

Similarly

Perimeter of âˆ†DEF = DE + EF + DF

=12cm

âˆ´ Perimeterof Î”DEF = 12cm

**Q.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 and E are mid points of AB and BC.

By Mid point theorem,

F is the midpoint of AC

In a Quadrilateral DECF

DEâˆ¥AC,DE = CF

Hence DECF is a parallelogram.

âˆ´âˆ C =âˆ D = 70^{0} [Opposite sides of a parallelogram]

Similarly

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

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

âˆ´ Angles of Î”DEF are

âˆ D = 700,âˆ E = 500,âˆ F = 600

**Q.3. In a triangle, P, Q and R are the mid points of sides BC, CA and AB respectively. If AC = 21cm, BC = 29cm and AB = 30cm, 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]

In a quadrilateral,

[A pair of side is parallel and equal]

RPâˆ¥AQ,RP = AQ

Therefore, RPQA is a parallelogram

AR = QP = 15cm [Opposite sides are equal]

RP = AQ = 10.5cm [Opposite sides are equal ]

Now,

Perimeter of ARPQ = AR + QP + RP +AQ

= 15 +15 +10.5 +10.5

= 51cm

**Q.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, we have

AD = DX [Given]

BD = DC [Given]

So, diagonals AX and BC bisect each other.

Therefore, ABXC is a parallelogram.

**Q.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 AB and AC

[By mid point theorem]

In Î”ABP

F is the mid-point of AB and âˆ´ FQâˆ¥BP[âˆµEFâˆ¥BP]

Therefore, Q is the mid-point of AP [By mid-point theorem]

Hence, AQ = QP.

**Q6. 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”BLMandÎ”CLN

âˆ BML=âˆ CNL= 90^{0}

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

âˆ MLB = âˆ NLC [Vertically opposite angle]

âˆ´Î”BLM = Î”CLN

âˆ´LM = LN [corresponding parts of congruent triangles]

**Q.7. In figure 14.95, Triangle ABC is a right angled triangle 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) TheareaofÎ”ADE.**

**Solution:**

In Î”ABC,âˆ B= 90^{0},

By using Pythagoras theorem

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

â‡’15^{2} = 9^{2} + BC^{2}

D and E are mid-points of AB and AC

[By midâˆ’point theorem]

[âˆµD is the midâˆ’point of AB]]

=13.5cm^{2}

**Q.8. In figure 14.96, M, N and P are mid-points of AB, AC and BC respectively. If MN = 3cm, NP = 3.5cm and MP = 2.5cm, calculate BC, AB and AC.**

**Solution:**

Given MN = 3cm, NP = 3.5cm and MP = 2.5cm.

To find BC, AB and AC

In Î”ABC

M and N are mid-points of AB and AC

[By midâˆ’point theorem]

â‡’ 3âˆ—2 = BC

â‡’ BC = 6cm

Similarly

AC = 2MP = 2 (2.5) = 5cm

AB = 2 NP = 2 (3.5) = 7cm

**Q.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:**

Clearly 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.

â‡’ 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)

**Q.10. In figure 14.97, BEâŠ¥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:**

Given,

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

To prove: âˆ PQR = 90^{0}

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

âˆ´ QRâˆ¥ACâ€¦..(i)

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

âˆ´ QPâˆ¥BHâ€¦.(ii)

But, BEâŠ¥AC

Therefore, from equation (i) and equation (ii) we have,

QPâŠ¥QR

â‡’âˆ PQR = 90^{0}

Hence Proved