myhelper: RD Sharma 2020 solution class 9 chapter 13 Quadrilaterals Exercise 13.4

Exercise 13.4

Page-13.62

Question 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.

Answer 1:

is given with D,E and F as the mid-points of BC , CA and AB respectively as shown below:

Also, , and .

We need to find the perimeter of

In , E and F are the mid-points of CA 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:

Similarly, we get

And

Perimeter of

Hence, the perimeter of is .

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 them at D and E respectively.

Therefore,

[(Given)]

and

[(Given)]

Similarly,

Therefore,

[(Given)]

and

[(Given)]

Similarly,

Therefore,

[(Given)] and

[(Given)]

Now BC is a straight line.

Similarly,

and

Hence the measure of angles are , and.

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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.

Answer 3:

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.

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

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

(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.

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:

In the given figure, 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.

Answer 9:

We have the following given figure:

We have and and AP is the bisector of exterior angle of .

(i) We need to prove that

In ,

We have (Given)

Thus, (Angles opposite to equal sides are equal)

By angle sum property of a triangle, we get:

…… (i)

Now,

(AP is the bisector of exterior angle )

(Linear Pair)

…… (ii)

From equation (i) and (ii),we get:

(ii) We need to prove that is a parallelogram.

We have proved that

This means,

Also it is given that

We know that a quadrilateral with opposite sides parallel is a parallelogram.

Therefore, is a parallelogram.

Question 10:

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.

Answer 10:

ABCD is a kite such that and

Quadrilateral PQRS is formed by joining the mid-points P,Q,R and S of sides AB,BC,CD and AD respectively.

We need to prove that Quadrilateral PQRS is a rectangle.

In , P and Q are the mid-points of AB and BC respectively.

Therefore,

and

Similarly, we have

and

Thus,

and

Therefore, PQRS is a parallelogram.

Now,

But, P and S are the mid-points of AB and AD

…… (I)

In : P and S are the mid-point of side AB and AD

By mid-point Theorem, we get:

Or,

In , P is the mid-point of side AB and

By Using the converse of mid-point theorem, we get:

M is the mid-point of AO

Thus,

…… (II)

In and , we have:

(Common)

[From (I)]

[From (II)]

By SSS Congruence theorem, we get:

By corresponding parts of congruent triangles property, we get:

But,

and

Therefore,

(, Corresponding angles should be equal)

Or,

We have proved that

Similarly, .

Then we can say that and

Therefore, is a parallelogram with

Or, we can say that is a rectangle.

We get:

Also, PQRS is a parallelogram.

Therefore, PQRS is a rectangle.

Hence proved.

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Question 11:

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.

Answer 11:

, an isosceles triangle is given with D,E and F as the mid-points of BC, CA and AB respectively as shown below:

We need to prove that the segment AD and EF bisect each other at right angle.

Let’s join DF and DE.

In , D and E are the mid-points of BC 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: Or

Similarly, we can get

Therefore, AEDF is a parallelogram

We know that opposite sides of a parallelogram are equal.

and

Also, from the theorem above we get

Thus,

Similarly,

It is given that , an isosceles triangle

Thus,

Therefore,

Also,

Then, AEDF is a rhombus.

We know that the diagonals of a rhombus bisect each other at right angle.

Therefore, M is the mid-point of EF and

Hence proved.

Question 12:

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

Answer 12:

Figure can be drawn as:

Let ABCD be a quadrilateral such that P,Q ,R and S are the mid-points of side AB,BC,CD and DA respectively.

In , P and Q are the mid-points of AB and BC respectively.

Therefore,

and

Similarly, we have

and

Thus,

and

Therefore, PQRS is a parallelogram.

Since, diagonals of a parallelogram bisect each other.

Therefore, PR and QS bisect each other.

Hence proved.

Question 13:

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 ..........

Answer 13:

(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is isosceles.

Explanation:

Figure can be drawn as: A

, an isosceles triangle is given.

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

Therefore,

…… (I)

Similarly,

…… (II)

And

…… (III)

Now, is an isosceles triangle.

From equation (II) and (III), we get:

Therefore, in two sides are equal.

Therefore, it is an isosceles triangle.

(ii) The triangle formed by joining the mid-points of the sides of a right triangle is right triangle.

Explanation:

Figure can be drawn as: A

right angle at B is given.

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

Therefore,

…… (I)

Similarly,

…… (II)

And

…… (III)

Now, DE || AB and transversal CB and CA intersect them at D and E respectively.

Therefore,

and

Similarly,

Therefore,

and

Similarly,

Therefore,

Now AC is a straight line.

Now, by angle sum property of ,we get:

Therefore,

But,

Then we have:

(iii) The figure formed by joining the mid-points of the consecutive sides of a quadrilateral is parallelogram.

Explanation:

Figure can be drawn as:

Let ABCD be a quadrilateral such that P, Q, R and S are the mid-points of side AB, BC, CD and DA respectively.

In , P and Q are the mid-points of AB and BC respectively.

Therefore,

and

Similarly, we have

and

Thus,

and

Therefore, PQRS is a parallelogram.

Question 14:

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 14:

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 15:

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 15:

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.

Question 16:

ABC is a triangle, D is a point on AB such that AD = $\frac{1}{4}$ AB and E is a point on AC such that AE = $\frac{1}{4}$ AC. Prove that DE = $\frac{1}{4}$ BC.

Answer 16:

is given with D a point on AB such that .

Also, E is point on AC such that.

We need to prove that

Let P and Q be the mid points of AB and AC respectively.

It is given that

and

But, we have taken P and Q as the mid points of AB and AC respectively.

Therefore, D and E are the mid-points of AP and AQ respectively.

In , P and Q 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 and …… (i)

In , D and E are the mid-points of AP and AQ respectively.

Therefore, we get and …… (ii)

From (i) and (ii),we get:

Hence proved.

Question 17:

In the given figure, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = $\frac{1}{4}$ AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.

Answer 17:

Figure is given as follows:

ABCD is a parallelogram, where P is the mid-point of DC and Q is a point on AC such that

PQ produced meets BC at R.

We need to prove that R is a mid-point of BC.

Let us join BD to meet AC at O.

It is given that ABCD is a parallelogram.

Therefore, (Because diagonals of a parallelogram bisect each other)

Also,

Therefore,

In , P and Q are the mid-points of CD and OC 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:

Also, in , Q is the mid-point of OC and

Therefore, R is a mid-point of BC.

Hence proved.

Question 18:

In the given figure, ABCD and PQRC are rectangles and Q is the mid-point of AC. Prove that

(i) DP = PC

(ii) PR = $\frac{1}{2}$ AC

Answer 18:

Rectangles ABCD and PQRC are given as follows:

Q is the mid-point of AC.

In , Q is the mid-point of AC such that

Using the converse of mid-point theorem, we get:

P is the mid-point of DC

That is;

Similarly, R is the mid-point of BC.

Now, in , P and R are the mid-points of DC and BC respectively.

Then, by mid-point theorem, we get:

Now, diagonals of a rectangle are equal.

Therefore putting ,we get:

Hence Proved.

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Question 19:

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.