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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.

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.

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.

It is given that P, Q and R are the mid-points of BC, CA and AB respectively.

Also, we have  ,  and
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.

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.

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.

In  , BM and CN are perpendiculars on any line passing through A.
Also.

We need to prove that
From point L let us draw
It is given that   and
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

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.

We have   as follows:

M, N and P are the mid-points of sides AB ,AC and BC respectively.
Also,  ,   and
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,
And

Hence, the measure for BC, AB and AC is  ,   and   respectively.

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.

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°

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.

krishan