RD Chapter 14 Quadrilaterals Ex 14.4 Solutions
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.
Answer - 1 : -
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 triangle ∠ABC, ∠A = 50°, ∠B = 60° and C =∠70°. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.
Answer - 2 : - It is given that D, E and F be the mid-points of BC , CA and AB respectively.
Then,
,
and 
.
Now, and transversal CB and CA intersect themat D and E respectively.
Therefore,
[
(Given)]
and 
[
(Given)]
Similarly,
Therefore,
[
(Given)]
and 
[(Given)]
Similarly,
Therefore,
[(Given)] and 
[(Given)]
Now BC is a straightline.
Similarly, 
and 
Hencethe measure of angles are 
,
and
.
Answer - 3 : -
It is given that P, Q and R are the mid-points of BC, CA and AB respectively.
We need to find the perimeter of quadrilateral ARPQ
In
, P and R are the mid-points of CB and AB respectively.Theorem states, the line segment joining the mid-points of any two sides of a traingle is parallel to the third side and equal to half of it.
Therefore, we get:
Similarly, we get
We have Q and R as the mid points of AC and AB respectively.
Therefore,
And
Perimeter of
Hence, the perimeter of quadrilateral ARPQ is
.
Question - 4 : - In a ΔABC median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.
Answer - 4 : -
is given with AD as the median extended to point X such that
.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:
Also, it is given that
Thus, the diagonals of the quadrilateral ABCX bisect each other.
Therefore, quadrilateral ABXC is a parallelogram.
Hence proved.
Question - 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.
Answer - 5 : -
is given with E and F as the mid points of sides AB and AC. Also,
intersecting EF at Q.We need to prove that
In
, 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:
Since, Q lies on EF.
Therefore,
This means,
Q is the mid-point of AP.
Thus,
(Because, F is the mid point of AC and )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.
Answer - 6 : -
In
, BM and CN are perpendiculars on any line passing through A.Also.
We need to prove that
From point L let us draw
Therefore,
Since, L is the mid points of BC,
Therefore intercepts made by these parallel lines on MN will also be equal
Thus,
Now in
,
And
. Thus, perpendicular bisects the opposite sides.Therefore,
is isosceles.Hence
Hence proved.
Question - 7 : - In the given figure, 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
Answer - 7 : -
We have
right angled at B.It is given that
and 
D and E are the mid-points of sides AB and AC respectively.
(i) We need to calculate length of BC.
In
right angled at B:By Pythagoras theorem,
Hence the length of BC is
(ii) We need to calculate area of
.In
right angled at B, D and E 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,
.Thus,
(Corresponding angles of parallel lines are equal)And
.
area of
D is the mid-point of side AB .
Therefore, area of
Hence the area of
is 
.
Question - 8 : - In the given figure, 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.
Answer - 8 : -
We have
as follows:M, N and P are the mid-points of sides AB ,AC and BC respectively.
We need to calculate BC, AB and AC.
In
, M and N 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,
Similarly,
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.
Answer - 9 : -
We have
as follows:Through A,B and C lines are drawn parallel to BC,CA and AB respectively intersecting at P,Q and R respectively.
We need to prove that perimeter of
is double the perimeter of .
and
Therefore,
is a parallelogram.Thus,
Similarly,
is a parallelogram.
Thus,
Therefore,
Then, we can say that A is the mid-point of QR.
Similarly, we can say that B and C are the mid-point of PR and PQ respectively.
In
, 
Theorem states, the line drawn through the mid-point of any one side of a triangle is parallel to the another side, intersects the third side at its mid-point.
Therefore,
Similarly,
Perimeter of
is double the perimeter of Hence proved.
Question - 10 : - In the given 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 = 90°
Answer - 10 : -
is given with
AD is any line from A to BC intersecting BE in H.
P,Q and R respectively are the mid-points of AH,AB and BC.
We need to prove that
Let us extend QP to meet AC at M.
In
, R and Q are the mid-points of BC and AB 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:
…… (i) Similarly, in
,
…… (ii)
From (i) and (ii),we get:
and We get,
is a parallelogram.Also,
Therefore,
is a rectangle.Thus,
Or,
Hence proved.