Question

To verify the basic proportionality theorem by using parallel lines board, triangle cut outs.​

Answer 1

Answer:

Prerequisite Knowledge

Statement of Basic Proportionality theorem.

Drawing a line parallel to a given line which passes through a given point.

Materials Required

White chart paper, coloured papers, geometry box, sketch pens, fevicol, a pair of scissors, ruled paper sheet (or Parallel line board).

Procedure

Cut an acute-angled triangle say ABC from a coloured paper.

Paste the ΔABC on ruled sheet such that the base of the triangle coincides with ruled line.

ncert-class-10-maths-lab-manual-basic-proportionality-theorem-triangle-1

Mark two points P and Q on AB and AC such that PQ || BC.

ncert-class-10-maths-lab-manual-basic-proportionality-theorem-triangle-2

Using a ruler measure the length of AP, PB, AQ and QC.

Repeat the same for right-angled triangle and obtuse-angled triangle.

Now complete the following observation table.

Observation

ncert-class-10-maths-lab-manual-basic-proportionality-theorem-triangle-3

Result

In each set of triangles, we verified that \frac { AP }{ PB } =\frac { AQ }{ QC }

Learning Outcome

Students will observe that in all the three triangles the Basic Proportionality theorem is verified.

Answer 2

Basic Proportionality: According to the theorem, if a line is resemblant to one of the triangle's sides and crosses the other sides in two distinct locales, the line proportionally splits the triangle's sides.

Assume that ABC is the triangle.

At D and E, line l parallel to BC intersects AB and AC.

To prove:

$\frac{AD}{DB} = \frac{AE}{EC}$

Solution,

The following process can be applied to validate the statement.

Join the line BE and CD.

Draw EF vertical to AB and DG vertical to CA.

The height of the triangles ΔADE and ΔDBE is the length EF.

ΔADE = $\frac{1}{2} * AD * EF$

ΔDBE = $\frac{1}{2} * DB*EF$

Dividing the area of ΔADE by ΔDBE gives $\frac{AD}{DB}$.  (Equation 1)

Analogously, dividing the area of ΔADE by ΔDCE gives $\frac{AE}{EC}$ .  (Equation 2)

ΔDBE and ΔDCE, on the other hand, are the same base DE and are located between the same resemblant straight lines BC and DE.

Area of ΔDBE equal to the area of ΔDCE         (Equation 3)

From all the above equations, we have;

$\frac{AD}{DB} = \frac{AE}{EC}$

(Hence proved.)