NCERT Class 10 Maths Lab Manual – Basic Proportionality Theorem for a Triangle
Objective
To verify the basic proportionality theorem by using parallel lines board, triangle cut outs.
Basic Proportionality Theorem
If a line is drawn parallel to one side of a triangle, to intersect the other two sides at distinct points, the other two sides are divided in the same ratio.
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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.
- Mark two points P and Q on AB and AC such that PQ || BC.
- 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
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.
Activity Time
- Find x if DE | | BC.
- Is PQ | | AB ?
- Find x if PQ | | BC
Viva Voce
Question 1.
Is there any other name for B.P.T. (Basic Proportionality Theorem) ?
Answer:
Yes, Thales Theorem
Question 2.
Name the mathematician who gave B.P.T.
Answer:
Greek mathematician Thales
Question 3.
What is the statement of B.P.T. ?
Answer:
If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio.
Question 4.
Can we prove Mid-point theorem by using B.P.T. ?
Answer:
Yes
Question 5.
Is the B.P.T. applicable for a scalene triangle ?
Answer:
Yes
Question 6.
What is the converse of B.P.T. ?
Answer:
If a line divides any two sides of a triangle in the same ratio, the line is parallel to the third side of the triangle.
Question 7.
Give two different examples of pairs of similar figures.
Answer:
Pair of squares, pair of circles
Question 8.
What are the conditions for two polygons of same number of sides to be similar ?
Answer:
- Their corresponding angles are equal.
- Their corresponding sides are proportional.
Multiple Choice Questions
Question 1.
In ∆ABC, if DE || BC,AD = 3.2, DB = 1.6, AE = x and EC = 2.1, then x is
(a) 4.2
(b) 3.2
(c) 1.6
(d) 4.8
Question 2.
In the given fig., LM || QR. Find LQ
(a) 3.1 cm
(b) 2.5 cm
(c) 3 cm
(d) None of these
Question 3.
In the given fig., DE || BC. If \(\frac { AE }{ AC } =\frac { 2 }{ 5 } \) and AB = 15 cm ,find AD.
(a) 6 cm
(b) 5 cm
(c) 4 cm
(d) 7 cm
Question 4.
What value of p will make ST || QR in the given fig.?
(a) 2
(b) 3
(c) 5
(d) None of these
Question 5.
Find x, if DC || AB.
(a) 7
(b) 3
(c) 5
(d) None of these
Question 6.
In the given fig., AB || DE and BD || EF. Find the correct relation.
(a) DC2 = CF x AC
(b) CF2 = DC x AC
(c) AC2 = DC x CF
(d) None of these
Question 7.
If LM || CB and LN || CD, then choose the correct answer
(a) \(\frac { AM }{ AB } =\frac { AN }{ AD } \)
(b) \(\frac { AM }{ AB } =\frac { AD }{ AN } \)
(c) \(\frac { AB }{ AM } =\frac { AN }{ AD } \)
(d) \(\frac { AM }{ AB } \neq \frac { AN }{ AD } \)
Question 8.
In the given figure, DE || OQ and DF || OR, then which is the correct relation ?
(a) EF = \(\frac { 1 }{ 2 } \) QR
(b) EF ≠ QR
(c) EF = QR
(d) EF||QR
Question 9.
In the given figure DE || BC, then EC is
(a) 2 cm
(b) 1.5 cm
(c) 1 cm
(d) 3 cm
Question 10.
In the given figure ABC and AMP are two right triangles, right angled at B and M respectively. Then tick the correct answer.
(a) \(\frac { CA }{ PA } =\frac { BC }{ MP }\)
(b) \(\frac { CA }{ PA } \neq \frac { BC }{ MP }\)
(c) \(\frac { CA }{ PA } =\frac { MP }{ BC }\)
(d) \(\frac { CA }{ PA } \neq \frac { MP }{ BC }\)
Answers
- (a)
- (c)
- (a)
- (a)
- (d)
- (a)
- (a)
- (d)
- (a)
- (a)
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