Mid-point Theorem — Multiple Choice Questions

In a △ABC, AB = 3 cm, BC = 4 cm and CA = 5 cm. If D and E are mid-points of AB and BC respectively, then the length of DE is

1. 1.5 cm

2. 2 cm

3. 2.5 cm

4. 3.5 cm

Answer

Since, D is midpoint of AB and E is midpoint of BC,

DE || AC and DE = $\dfrac{1}{2}$AC = $\dfrac{1}{2}$ x 5 = 2.5 cm

Hence, Option 3 is the correct option.

In the adjoining figure, ABCD is a rectangle in which AB = 6 cm and AD = 8 cm. If P and Q are mid-points of the sides BC and CD respectively, then the length of PQ is

1. 7 cm

2. 5 cm

3. 4 cm

4. 3 cm

Answer

Length of diagonal (BD) of ABCD = $\sqrt{AB^2 + AD^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100}$ = 10 cm.

In △BCD,

P is midpoint of BC and Q is midpoint of CD.

PQ || BD and PQ = $\dfrac{1}{2}$BD = $\dfrac{1}{2}$ x 10 = 5 cm.

Hence, Option 2 is the correct option.

D and E are mid-points of the sides AB and AC of △ABC and O is any point on the side BC. O is joined to A. If P and Q are mid-points of OB and OC respectively, then DEQP is

1. a square

2. a rectangle

3. a rhombus

4. a parallelogram

Answer

In △ABC,

D and E are respective midpoints of AB and AC.

∴ DE || (BC or PQ) ......(1)

Again in △ABO,

D and P are respective midpoints of AB and BO.

∴ DP || AO ........(2)

In △ACO,

E and Q are respective midpoints of AC and CO.

∴ EQ || AO ........(3)

From 2 and 3 we get,

DP || EQ .......(4)

From 1 and 4 we get,

DE || PQ and DP || EQ

∴ DEQP is a parallelogram.

Hence, Option 4 is the correct option.

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle if

1. PQRS is a parallelogram

2. PQRS is a rectangle

3. the diagonals of PQRS are perpendicular to each other

4. the diagonals of PQRS are equal.

Answer

Let ABCD be a quadrilateral with P, Q, R and S as mid-points of AB, BC, CD and DA respectively.

Let PR ⊥ QS.

In △QRP,

A and B are midpoints of PQ and QR respectively.

∴ AB || PR and AB = $\dfrac{1}{2}PR$ (By midpoint theorem) ........(1)

Similarly in △PRS,

D and C are midpoints of PS and RS respectively.

∴ DC || PR and DC = $\dfrac{1}{2}PR$ (By midpoint theorem) ........(2)

In △PQS,

D and A are midpoints of PS and PQ respectively.

∴ DA || QS and DA = $\dfrac{1}{2}QS$ (By midpoint theorem).........(3)

Similarly in △QRS,

B and C are midpoints of QR and SR respectively.

∴ BC || QS and BC = $\dfrac{1}{2}QS$ (By midpoint theorem).........(4)

From 1 and 2 we get,

AB = DC and AB || DC

From 3 and 4 we get,

DA = BC and DA || BC

Hence, proved that ABCD is a parallelogram.

Since, DA || QS and PR ⊥ QS

∴ DA ⊥ PR.

Since, DC || PR and PR ⊥ QS

∴ DC ⊥ QS.

In OMDN,

∠OMD = ∠SMC (Vertically opposite angle are equal)

In quadrilateral sum of angles = 360°

∠O + ∠M + ∠N + ∠D = 360°

90° + 90° + 90° + ∠D = 360°

∠D = 360° - 270° = 90°.

In parallelogram sum of alternate angles = 180°.

∠D + ∠B = 180°

90° + ∠B = 180°

∠B = 90°

Since, opposite sides are equal and adjacent sides are perpendicular to each other.

Hence, ABCD is a rectangle.

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a rectangle if the diagonals of PQRS are perpendicular to each other.

Hence, Option 3 is the correct option.

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order is a rhombus if

1. ABCD is a parallelogram

2. ABCD is a rhombus

3. the diagonals of ABCD are equal

4. the diagonals of ABCD are perpendicular to each other.

Answer

Let ABCD be a quadrilateral with P, Q, R and S as mid-points of AB, BC, CD and DA respectively.

Let diagonals be of equal length i.e. AC = BD = x

In △BCA,

P and Q are midpoints of AB and BC respectively.

∴ PQ || AC and PQ = $\dfrac{1}{2}$AC = $\dfrac{1}{2}$x (By midpoint theorem) ........(1)

Similarly in △ACD,

S and R are midpoints of AD and CD respectively.

∴ SR || AC and SR = $\dfrac{1}{2}$AC = $\dfrac{1}{2}$x (By midpoint theorem) ........(2)

In △ABD,

S and P are midpoints of AD and AB respectively.

∴ SP || BD and SP = $\dfrac{1}{2}$BD = $\dfrac{1}{2}$x (By midpoint theorem).........(3)

Similarly in △BCD,

Q and R are midpoints of BC and CD respectively.

∴ QR || BD and QR = $\dfrac{1}{2}$BD = $\dfrac{1}{2}$x (By midpoint theorem).........(4)

From 1, 2, 3 and 4 we get,

PQ = SR = SP = QR.

Hence, proved that PQRS is a rhombus.

∴ The quadrilateral formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order is a rhombus if the diagonals of ABCD are equal.

Hence, Option 3 is the correct option.

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if

1. ABCD is a rhombus

2. diagonals of ABCD are equal

3. diagonals of ABCD are perpendicular to each other

4. diagonals of ABCD are equal and perpendicular to each other.

Answer

Let ABCD be a quadrilateral with P, Q, R and S as mid-points of AB, BC, CD and DA respectively.

Let diagonals be of equal length i.e. AC = BD = x and AC ⊥ BD.

In △BCA,

P and Q are midpoints of AB and BC respectively.

∴ PQ || AC and PQ = $\dfrac{1}{2}AC = \dfrac{1}{2}x$ (By midpoint theorem) ........(1)

Similarly in △ACD,

S and R are midpoints of AD and CD respectively.

∴ SR || AC and SR = $\dfrac{1}{2}AC = \dfrac{1}{2}x$ (By midpoint theorem) ........(2)

In △ABD,

S and P are midpoints of AD and AB respectively.

∴ SP || BD and SP = $\dfrac{1}{2}BD = \dfrac{1}{2}x$ (By midpoint theorem).........(3)

Similarly in △BCD,

Q and R are midpoints of BC and CD respectively.

∴ QR || BD and QR = $\dfrac{1}{2}BD = \dfrac{1}{2}x$ (By midpoint theorem).........(4)

From 1, 2, 3 and 4 we get,

PQ = SR = SP = QR.

Hence, proved that PQRS is a rhombus.

Since, SP || BD and AC ⊥ BD

∴ SP ⊥ AC.

Since, SR || AC and AC ⊥ BD

∴ SR ⊥ BD.

In OMSN,

∠OMS = ∠DMR (Vertically opposite angle are equal)

In quadrilateral sum of angles = 360°

∠O + ∠M + ∠N + ∠S = 360°

90° + 90° + 90° + ∠S = 360°

∠S = 360° - 270° = 90°.

Since, in rhombus adjacent angles sum = 180°

∠S + ∠R = 180°

90° + ∠R = 180°

∠R = 90°.

∠Q + ∠R = 180°

90° + ∠Q = 180°

∠Q = 90°.

∠S + ∠P = 180°

90° + ∠P = 180°

∠P = 90°.

Since, PQ = QR = RS = SP and ∠P = ∠Q = ∠R = ∠S = 90°

Hence, proved that PQRS is a square.

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if diagonals of ABCD are equal and perpendicular to each other.

Hence, Option 4 is the correct option.

Consider the following two statements:

Statement 1: The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.

Statement 2: The line through the mid-point of one side of a triangle and parallel to another side bisects the third side.

Which of the following is valid?

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and Statement 2 is false.

4. Statement 1 is false, and Statement 2 is true.

Answer

According to statement 1 :

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.

This statement describes mid-point theorem.

∴ Statement 1 is true.

According to statement 2 :

The line through the mid-point of one side of a triangle and parallel to another side bisects the third side.

This statement describes the converse of mid-point theorem.

∴ Statement 2 is true.

Hence, option 1 is the correct option.