Mid-Point Theorem & Intercept Theorem — Multiple Choice Questions

BP and CQ are two medians of a △ABC. If QP = 4 cm, then BC =

1. 2 cm

2. 6 cm

3. 8 cm

4. 9 cm

Answer

Given,

BP is the median.

∴ P is the mid-point of AC

CQ is the median.

∴ Q is the mid-point of AB

By mid-point theorem,

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.

Since, Q and P are the mid-points of AB and AC respectively. Thus,

⇒ QP || BC and QP = $\dfrac{1}{2}$ BC

⇒ BC = 2 × QP

⇒ BC = 2 × 4

⇒ BC = 8 cm.

Hence, option 3 is the correct option.

ABC is a right angled isosceles triangle in which ∠A = 90°. If D and E are the mid-points of AB and AC respectively, then ∠ADE =

1. 30°

2. 45°

3. 60°

4. 90°

Answer

Given,

ABC is a right angled isosceles triangle. Since, hypotenuse is the largest side thus other two sides of triangle will be equal.

AC = AB

⇒ ∠C = ∠B = x (let)

In △ABC,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 90° + x + x = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = $\dfrac{90°}{2}$

⇒ x = 45°

⇒ ∠C = ∠B = 45°

By mid-point theorem,

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.

Since, D and E are the mid-points of AB and AC respectively.

DE || BC

AB is the transversal.

⇒ ∠ABC = ∠ADE (Corresponding angles are equal)

⇒ ∠ADE = 45°.

Hence, option 2 is the correct option.

On the sides AB and AC of a △ABC, D and E are two points such that AD : AB = AE : AC = 1 : 2. If BC = 7 cm, then DE =

1. 3.5 cm

2. 7 cm

3. 14 cm

4. 15 cm

Answer

Given,

AD : AB = 1 : 2

⇒ AB = 2 AD

∴ D is the mid-point of AB

AE : AC = 1 : 2

⇒ AC = 2 AE

∴ E is the mid-point of AC

By mid-point theorem,

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.

Since, D and E are the mid-points of AB and AC respectively.

⇒ DE = $\dfrac{1}{2} BC = \dfrac{1}{2}$ × 7 = 3.5 cm

Hence, option 1 is the correct option.

AD and BE are two medians of a ABC. F is a point on AC such that DF || BE. If AC = 12 cm, then FC =

1. 6 cm

2. 4 cm

3. 3 cm

4. none of these

Answer

In △BCE,

D is the midpoint of BC (As AD is median)

Given,

DF || BE

By converse of mid-point theorem,

A line drawn through the midpoint of one side of a triangle, and parallel to another side, will bisect the third side.

Thus, in triangle BEC,

F is the mid-point of CE

⇒ FC = $\dfrac{1}{2}$ CE ....(1)

Given,

BE is median.

⇒ E is the mid-point of AC

⇒ AE = CE

⇒ CE = $\dfrac{1}{2}$ AC

Substituting value of CE in eq.(1), we get:

⇒ FC = $\dfrac{1}{2} \times \dfrac{1}{2} \text{AC}$

⇒ FC = $\dfrac{1}{4} \times 12$

⇒ FC = 3 cm.

Hence, option 3 is the correct option.

D, E and F are respectively the mid-points of the sides BC, CA and AB of a ABC. If BC = 10 cm, CA = 12 cm and AB = 17 cm, then the perimeter of the DEF is :

1. 13 cm

2. 19.5 cm

3. 39 cm

4. None of these

Answer

By mid-point theorem,

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.

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

⇒ FE || BC and FE = $\dfrac{1}{2} BC = \dfrac{1}{2}$ × 10 = 5 cm

Since, F and D are the mid-points of AB and BC respectively.

⇒ FD || AC and FD = $\dfrac{1}{2} AC = \dfrac{1}{2}$ × 12 = 6 cm

Since, D and E are the mid-points of BC and AC respectively.

⇒ DE || AB and DE = $\dfrac{1}{2}AB = \dfrac{1}{2}$ × 17 = 8.5 cm

Perimeter of DEF = DE + FE + FD = 8.5 + 5 + 6 = 19.5 cm

Hence, option 2 is the correct option.

D, E and F are respectively the mid-points of the sides BC, CA and AB of a ABC. If BC = 12 cm, CA = 15 cm and AB = 18 cm, then the perimeter of the quadrilateral DEAF is :

1. 45 cm

2. 27 cm

3. 30 cm

4. 33 cm

Answer

By mid-point theorem,

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.

Since, D and E are the mid-points of BC and AC respectively.

⇒ DE || AB and DE = $\dfrac{1}{2} AB = \dfrac{1}{2}$ × 18 = 9 cm

Since, F and D are the mid-points of AB and BC respectively.

⇒ FD || AC and FD = $\dfrac{1}{2} AC = \dfrac{1}{2}$ × 15 = 7.5 cm

AE = $\dfrac{1}{2}AC = \dfrac{1}{2} \times 15$ = 7.5 cm

AF = $\dfrac{1}{2}AB = \dfrac{1}{2} \times 18$ = 9 cm

Perimeter of AFED = AF + FD + DE + AE = 9 + 7.5 + 9 + 7.5 = 33 cm.

Hence, option 4 is the correct option.

In the adjoining figure, AD = BD and DE || BC. If AC = 6 cm and DE = 4 cm, then the length of BC is :

1. 4 cm

2. 5 cm

3. 6 cm

4. 8 cm

Answer

Given,

D is the mid-point of AB.

By converse of mid-point theorem,

A line drawn through the midpoint of one side of a triangle, and parallel to another side, will bisect the third side.

In △ABC,

Since, D is the mid-point of AB and DE // BC, thus :

E is mid-point of AC.

By mid-point theorem,

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.

Since, D and E are the mid-points of AB and AC respectively.

DE || BC

⇒ DE = $\dfrac{1}{2}$ BC

⇒ BC = 2 DE

⇒ BC = 2 × 4

⇒ BC = 8 cm.

Hence, option 4 is the correct option.

In the given figure, BP and CQ are two medians of the △ABC. If BC = 12 cm, the length of QP =

1. 4 cm

2. 6 cm

3. 8 cm

4. 10 cm

Answer

We know that,

Median drawn from the vertex bisects the opposite side.

Since, CQ is the median to AB.

⇒ AQ = BQ

Since, BP is the median to AC

⇒ AP = CP

By mid-point theorem,

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.

Since, Q and P are the mid-points of AB and AC respectively.

⇒ QP || BC and QP = $\dfrac{1}{2}$ BC

⇒ QP = $\dfrac{1}{2}$ × 12

⇒ QP = 6 cm.

Hence, option 2 is the correct option.

In △ABC, E is the mid-point of the median AD. BE is joined and produced to meet AC at F. Then, AF = ....AC.

1. 3

2. 2

3. $\dfrac{1}{3}$

4. $\dfrac{1}{2}$

Answer

Draw DY parallel to BF.

Since, BF || DY so, EF || DY

By converse of mid-point theorem,

A line drawn through the midpoint of one side of a triangle, and parallel to another side, will bisect the third side.

In △ADY,

Since, E is the mid-point of AD and EF || DY

⇒ F is the mid-point of AY.

∴ AF = FY

Given,

AD is the median.

In △BCF,

Since, D is the mid-point of BC and BF || DY

⇒ Y is the mid-point of FC.

∴ FY = CY

⇒ AF = FY = CY

From figure,

AC = AF + FY + CY = AF + AF + AF = 3 AF

⇒ AF = $\dfrac{1}{3}$ AC

Hence, option 3 is the correct option.

In the trapezium ABCD, AB || DC and AB > DC. P and Q are the mid-points of the diagonals AC and BD. Then, PQ || AB and PQ = .....(AB - DC).

1. 2

2. 3

3. $\dfrac{1}{2}$

4. $\dfrac{1}{3}$

Answer

In △BQR and △CQD,

⇒ ∠BQR = ∠CQD (Vertically opposite angles are equal)

⇒ ∠BRQ = ∠QCD (Alternate angles are equal)

⇒ BQ = DQ (Q is the mid-point of BD)

∴ △BQR ≅ △CQD

⇒ BR = DC (Corresponding parts of congruent triangles are equal)

⇒ QR = CQ (Corresponding parts of congruent triangles are equal)

Given,

AB || DC and PQ || AB

∴ PQ || AB || DC

In △ARC,

Since, P and Q are the mid-points of AC and CR respectively.

By mid-point theorem,

⇒ PQ = $\dfrac{1}{2}$ AR

⇒ PQ = $\dfrac{1}{2}$ (AB - BR)

⇒ PQ = $\dfrac{1}{2}$ (AB - DC) (∵ BR = DC)

Hence, option 3 is the correct option.