# R.D. Sharma Solutions Class 9th: Ch 14 Quadrilaterals Exercise 14.4

#### Chapter 14 Quadrilaterals R.D. Sharma Solutions for Class 9th Exercise 14.4

**Exercise 14.4**

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

**Solution**

Given that,

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

In △ABC,

F and E are the mid points AB and AC.

∴ EF = 1/2 BC

Similarly,

DF = 1/2 AC and DE = 1/2 AB

Perimeter of △DEF = DE + EF + DF

= 1/2 AB + 1/2 BC + 1/2 AC

= 1/2×7 + 1/2×8 + 1/2×9

= 3.5 + 4 + 4.5 = 12 cm

Perimeter of △DEF = 12 cm.

**Solution**

In △ABC,

D and E are mid points of AB and BC

By mid-point theorem,

DE∥AC, DE = 1/2 AC

F is the midpoint of AC.

Then, DE = 1/2 AC = CF

In quadrilateral DECF,

DE∥AC, DE = CF

Hence, DECF is a parallelogram.

∠C = ∠D = 70° [opposites sides of parallelogram]

Similarly,

BEFD is a parallelogram, ∠B = ∠F = 60°

ADEF is a parallelogram, ∠A = ∠E = 50°

∴ Angles of △DEF are ∠D = 70°, ∠E = 50°, ∠F = 60°

**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 △ABC,

R and P are mid points of AB and BC

RP∥AC, RP = 1/2 AC [By midpoint Theorem]

In a quadrilateral, a pair of side is parallel and equal.

RP∥AQ, RP = AQ

Therefore, RPQA is a parallelogram.

⇒ AR = 1/2 AB = 1/2×30 = 15 cm

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

⇒ RP = 1/2 AC = 1/2×21 = 10.5 cm

RP = AQ = 10.5 cm

Now,

Perimeter of ARPQ = AR + QP + RP + AQ

= 15+15+10.5+10.5 = 51 cm

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

**Solution**

△ABC is given with AD as the median extended to point X such that AD = DX.

Join BX and CX.

We get a quadrilateral ABXC, we need to prove that it’s a parallelogram.

We know that AD is the median.

By definition of median we get:

BD = CD

Also, it is given that

AD = DX

Thus, the diagonals of the quadrilateral ABCX bisect each other.

Therefore, quadrilateral ABXC is a parallelogram.

Hence proved.

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**

ΔABC is given with

*E*and

*F*as the mid points of sides

*AB*and

*AC.*

Also, AP ⊥BC intersecting

*EF*at

*Q*.

We need to prove that AQ = QP

In △ABC,

*E*and

*F*are the mid-points of

*AB*and

*AC*respectively.

Theorem states, 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.

Therefore, we get: EF ⊥BC

Since,

*Q*lies on

*EF*.

Therefore, FQ | | BC

This means,

*Q*is the mid-point of

*AP*.

Thus, AQ = QP (Because, F is the mid point of AC and FQ | | BC )

Hence proved.

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**

7. In Fig. 14.95, triangle ABC is right-angled 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**

8. In Fig. 14.96, M, N and P are the 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 = 3cm, NP = 3.5cm and MP = 2.5cm

To find BC, AB and AC

In △ABC,

M and N are mid-points of AB and AC.

∴ MN = 1/2 BC, MN∥BC [By midpoint theorem]

⇒ 3 = 1/2 BC

⇒ 3×2 = BC

⇒ BC = 6 cm

Similarly,

AC = 2MP = 2×2.5 = 5 cm

AB = 2NP = 2×3.5 = 7 cm

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**

10. In Fig. 14.97, 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

11. In Fig. 14.98, AB = AC and CP || BA and AP is the bisector of exterior ∠CAD of ΔABC. Prove that (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram.

**Solution**

12. 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**

13. 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**

14. ABC is a triangle, D is a point on AB such that AD = 1/4 AB and E is a point on AC such that AE = 1/4 AC. Prove that DE = 1/2 BC.

**Solution**

15. In Fig. 14.99, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ=1/4 AC. If PQ produced meets BC at R, Prove that R is a midpoint of BC.

**Solution**

16. In Fig. 14.100, ABCD and PQRC are rectangle and Q is the midpoint of AC. Prove that

(i) DP = PC

(ii) PR = 1/2 ∠AC

**Solution**

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

**Solution**

18. RM 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**

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

**Solution**

20. 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) Isosceles

(ii) Right triangle

(iii) Parallelogram