Chapter 14 Quadrilaterals RD Sharma Solutions Exercise 14.4 Class 9 Maths
Chapter Name  RD Sharma Chapter 14 Quadrilaterals Exercise 14.4 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 14.4 Solutions
1. In a Î”ABC, D, E and F are, respectively, the mid  point of BC, CA and AB. If the length of side AB, BC and CA are 7cm, 8cm and 9cm, respectively, find the perimeter of Î”DEF.
Solution
2. In a triangle ∠ABC, ∠A = 50°, ∠B = 60° and ∠C = 70° . Find the measure of the angles of the triangle formed by joining the mid  points of the sides of this triangle .
Solution
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 = 30 cm, find the perimeter of the quadrilateral ARPQ.
Solution
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 BD = DC [Given]
∴ ABXC is a parallelogram
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 Î”ABC
E and F are midpoints of AB and AC
∴ EF  FE, 1/2 BC = FE [∵ By mid  point theorem]
In Î”ABP,
F is the midpoint of AB and FQ  BP [∵ EF  BC]
∴ Q is the midpoint of AP [By converse of midpoint theorem]
∴ Q is the midpoint of AP [By converse of midpoint theorem]
Hence, AQ = QP
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
7. In Fig. below, triangle ABC is right  angled at B. Given that AB = 9cm, 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
8. In fig. below, 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
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
10. In Fig. below, 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. below, AB = AC and CP  BA and AP is the bisector of exterior ∠CAD of Î”ABC.
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 midpoints of the sides, in order, is a rectangle.
Solution
Given,
A kite ABCD having AB = AD and BC = CD. P,Q,R,S are the midpoint of sides AB, BC, CD and DA respectively PQ, QR, RS and spare joined
To prove :
PQRS is a rectangle
Proof:
In Î”ABC, P and Q are the midpoints of AB and BC respectively.
∴ PQ  AC and PQ = 1/2 AC ...(i)
In Î”ADC, R and S are the midpoint of CD and AD respectively .
In Î”ADC, R and S are the midpoint of CD and AD respectively .
∴ RS  AC and RS = 1/2 AC ...(ii)
From (i) and (ii), we have
From (i) and (ii), we have
PQ  RS and PQ = RS
Thus, in quadrilateral PQRS, a pair of opposite sides are equal and parallel. So PQRS is a parallelogram. Now, we shall prove that one angle of parallelogram PQRS it is a right angle
Since AB = AD
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.
∴ The lines BM, LS and CN being the same perpendiculars, on line MN are parallel to each other.
Let ABCD is a quadrilateral in which P, Q, R and S are midpoints of sides
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 = 1/4 BC.
Let P and Q be the midpoints of AB and AC respectively.
Solution
Let P and Q be the midpoints of AB and AC respectively.
Then PQ  BC such that
PQ = 1/2 BC ...(i)
In Î”APQ, D and E are the midpoint of AP and AQ are respectively
In Î”APQ, D and E are the midpoint of AP and AQ are respectively
15. In below Fig, 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 mid  point of BC.
Solution
16. In the below Fig, ABCD and PQRC are rectangles and Q is the mid  point of AC. Prove that
(i) DP = PC
(ii) PR = 1/2 AC
(i) DP = PC
(ii) PR = 1/2 AC
Solution
(i) In Î”ADC, Q is the midpoint of AC such that PQ  AD
∴ P is the midpoint of DC
⇒ DP = DC [Using converse of midpoint theorem]
(ii) Similarly, R is the midpoint of BC
(ii) Similarly, R is the midpoint of BC
∴ PR = 1/2 BD
PR = 1/2 AC [Diagonal of rectangle are equal ∴ BD = AC]
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.
PR = 1/2 AC [Diagonal of rectangle are equal ∴ BD = AC]
Solution
18. BM 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
To prove LM = LN
Draw LS perpendicular to line MN
∴ The lines BM, LS and CN being the same perpendiculars, on line MN are parallel to each other.
According to intercept theorem,
If there are three or more parallel lines and the intercepts made by them on a transversal or equal. Then the corresponding intercepts on any other transversal are also equal.
In the drawn figure, MB and LS and NC are three parallel lines and the two transversal line are MN and BC
In the drawn figure, MB and LS and NC are three parallel lines and the two transversal line are MN and BC
We have, BL = LC (As L is the given midpoint of BC)
∴ using intercept theorem, we get
∴ using intercept theorem, we get
MS = SN ...(i)
Now in Î”MLS and LSN
Now in Î”MLS and LSN
MS = SN [using ...(i)]
∠LSM = ∠LSN = 90° (LS ⊥ MN and SL = LS common)
∴ Î”MLS ≅ Î”LSN (SAS congruency theorem)
∴ LM = LN (CPCT)
∴ LM = LN (CPCT)
19. Show that the line segment joining the mid  points of the opposite sides of a quadrilateral bisect each other.
Solution
Let ABCD is a quadrilateral in which P, Q, R and S are midpoints of sides
AB, BC, CD and DA respectively join PQ, QR, RS, SP and BD
In Î”ABD, S and P are the midpoints of AD and AB respectively.
So, by using midpoint theorem we can say that
So, by using midpoint theorem we can say that
SP  BD and SP = 1/2 BD ...(1)
Similarly in Î”BCD
QR  BD and QR = 1/2BD ...(2)
From equation (1) and (2) we have
Similarly in Î”BCD
QR  BD and QR = 1/2BD ...(2)
From equation (1) and (2) we have
SP  QR and SP = QR
As in quadrilateral SPQR one pair of opposite side are equal and parallel to each other.
As in quadrilateral SPQR one pair of opposite side are equal and parallel to each other.
So, SPQR is parallelogram
Since, diagonals of a parallelogram bisect each other.
Hence PR and QS bisect each other.
20. Fill in the blanks to make the following statements correct:
(i) The triangle formed by joining the midpoints of the sides of an isosceles triangle is ____
(ii) The triangle formed by joining the midpoints of the sides of a right triangle is ______
(iii) The figure formed by joining the midpoints of consecutive sides of a quadrilateral is _____
Solution
(i) Isosceles
(ii) Right triangle
(iii) Parallelogram