ICSE Class 9 Maths Chapter 11 Mid Point and Intercept Theorems

11. Mid-Point and Intercept Theorems

Points To Remember

Theorems.

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

2. The straight line drawn through the mid-point of one side of a triangle and parallel to the other side, bisects the third side.

3. Intercept Theorem: If a transversal makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts.

Exercise 11

Q. 1. In the given figure, D, E, F are the mid-points of the sides BC, CA and AB respectively.

(i) If AB = 6-2 cm, find DE.

(ii) If DF = 3-8 cm, find AC.

(iii) If perimeter of \(\triangle\)ABC is 21 cm, find FE.

Sol. D, E and F are the mid-points of the sides BC, CA and AB respectively.

\(\therefore\) EF \(\parallel\) BC and \(EF = \frac{1}{2}\) BC

Similarly DE \(\parallel\) AB and \(DE = \frac{1}{2}\) AB

and FD \(\parallel\) AC and \(FD = \frac{1}{2}\) AC

(i) Now if AB = 6-2 cm, then

\(DE = \frac{1}{2}\) AB = \(\frac{1}{2}\) \(\times\) 6-2 = 3-1 cm

(ii) DF = 3-8 cm,

But \(DF = \frac{1}{2}\) AC \(\Rightarrow\) AC = 2 DF

\(\therefore\) AC = 2 \(\times\) 3-8 = 7-6 cm

(iii) If perimeter of \(\triangle\)ABC = 21 cm

\(\Rightarrow\) AB + BC + CA = 21 cm

\(\Rightarrow\) 6-2 + BC + 7-6 = 21 cm

\(\Rightarrow\) BC + 13-8 = 21 \(\Rightarrow\) BC = 21 - 13-8

\(\Rightarrow\) BC = 7-2 cm

But \(FE = \frac{1}{2}\) BC

\(= \frac{1}{2} \times 7-2 = 3-6\text{ cm Ans.}\)

Teacher's Note

The mid-point theorem helps us understand how parallel lines create proportional segments, much like how engineers use similar principles to design structures with balanced weight distribution.

Q. 2. In the given figure, LMN is a right triangle in which \(\angle\)M = 90°, P and Q are mid-points of LM and LN respectively. If LM = 9 cm, MN = 12 cm and LN = 15 cm, find:

(i) the perimeter of trapezium MNQP,

(ii) the area of trapezium MNQP.

Sol. In right angled \(\triangle\)LMN, \(\angle\)M = 90°

P and Q are the mid-points of sides LM and LN respectively.

\(\therefore\) PQ \(\parallel\) MN and \(PQ = \frac{1}{2}\) MN

Now LM = 9 cm, MN = 12 cm and LN = 15 cm

\(\therefore\) \(PQ = \frac{1}{2}\) MN = \(\frac{1}{2}\) \(\times\) 12 = 6 cm

\(PM = \frac{1}{2}\) LM = \(\frac{1}{2}\) \(\times\) 9 = 4-5\text{cm}\)

\(QN = \frac{1}{2}\) LN = \(\frac{1}{2}\) \(\times\) 15 = 7-5\text{cm}\)

(i) Perimeter of trapezium MNQP

= PQ + QN + MN + PM

= 6 + 7-5 + 12 + 4-5 = 30 cm.

(ii) Area of trapezium MNQP

\(= \frac{1}{2}\) (PQ + MN) \(\times\) PM (\(\because\) \(\angle\)M = 90°)

\(= \frac{1}{2}\) (6 + 12) \(\times\) 4-5 cm\(^2\)

\(= \frac{1}{2}\) \(\times\) 18 \(\times\) 4-5 cm\(^2\) = 40-5 cm\(^2\). Ans.

Teacher's Note

Understanding how to calculate the area and perimeter of trapezoids is useful in real-world applications like calculating the amount of material needed for trapezoidal-shaped roofs or agricultural fields.

Q. 3. In the given figure, D, E, F are respectively the mid-points of the sides AB, BC and CA of \(\triangle\)ABC. Prove that ADEF is a parallelogram.

Sol. Given: In the \(\triangle\)ABC, D, E and F are the mid-points of sides AB, BC and CA respectively.

To prove: ADEF is a parallelogram.

Const. Join DE, EF and FD.

Proof: \(\therefore\) E and F are the mid-points of sides BC and CA respectively.

\(\therefore\) EF \(\parallel\) AB ... (i)

Similarly D and E are the mid-points of sides AB and BC respectively.

\(\therefore\) DE \(\parallel\) AC ... (ii)

From (i) and (ii)

ADEF is a parallelogram.

Hence proved.

Q. 4. If D, E, F are respectively the mid-points of the sides AB, BC and CA of an equilateral triangle ABC, prove that \(\triangle\)DEF is also an equilateral triangle.

Sol. Given: \(\triangle\)ABC is an equilateral triangle. D, E and F are the mid-points of sides AB, BC and CA respectively. DE, EF and FD are joined.

To prove: \(\triangle\)DEF is also an equilateral triangle.

Proof: \(\therefore\) D and E are the mid-point of AB and BC respectively.

\(\therefore\) DE \(\parallel\) AC and \(DE = \frac{1}{2}\) AC ... (i)

Similarly E and F are the mid-points of BC and CA respectively.

\(\therefore\) EF \(\parallel\) AB and \(EF = \frac{1}{2}\) AB ... (ii)

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

\(\therefore\) FD \(\parallel\) BC and \(FD = \frac{1}{2}\) BC ... (iii)

From (i), (ii) and (iii)

DE = EF = FD

(\(\because\) AB = BC = CA)

\(\therefore\) \(\triangle\)DEF is an equilateral triangle.

Hence proved.