Class 10 Basic Proportionality Theorem — practice questions with answers

Geometry — Similarity · 5 sample questions from the school syllabus. Try each one, then open the answer and the worked solution.

1. The Basic Proportionality Theorem (Thales' Theorem) states that if a line is drawn parallel to one side of a triangle, then it divides the other two sides in the same _____.

Aratio
Bangle
Carea
Dlength

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Answer: A) ratio

BPT states the line parallel to one side divides the other two sides in the same ratio.

2. In triangle ABC, points D on AB and E on AC satisfy AD = 3 cm, AB = 9 cm, AE = 2 cm, AC = 6 cm. Is DE parallel to BC? Choose the best justification.

AYes, since $\frac{A D}{A B} = \frac{A E}{A C} = \frac{1}{3}$
BNo, since $\frac{A D}{A B}$ is not equal to $\frac{A E}{A C}$
CYes, since AD = AE
DCannot be determined without angles

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Answer: A) Yes, since

\frac{A D}{A B} = \frac{A E}{A C} = \frac{1}{3}

\frac{A D}{A B} = \frac{3}{9} = \frac{1}{3}

and

\frac{A E}{A C} = \frac{2}{6} = \frac{1}{3}

. Equal, so by converse of BPT DE || BC.

3. In triangle ABC, DE || BC. Surveyor Anita reads AD = (3x + 4) m, DB = (x + 1) m, AE = (4x + 2) m and EC = (2x - 1) m. Find x (positive value) using BPT.

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Answer: A) 2

\frac{3 x + 4}{x + 1} = \frac{4 x + 2}{2 x - 1}

. Cross: (3x+4)(2x-1) = (x+1)(4x+2). LHS = 6x² -3x +8x -4 = 6x² +5x -4. RHS = 4x² +2x +4x +2 = 4x² +6x +2. So 6x²+5x-4 = 4x²+6x+2, 2x² - x - 6 = 0, (2x+3)(x-2)=0, x = 2.

4. Inside triangle PQR a point O is marked, and D is a point on the segment PO. Through D a line is drawn parallel to OQ meeting PQ at E, and another line is drawn parallel to OR meeting PR at F. What can be said about EF?

AEF is parallel to PQ, since D lies on PO
BNothing follows unless O is the centroid of triangle PQR
C $EF$ is parallel to QR, since $\frac{PE}{EQ} = \frac{P D}{D O} = \frac{PF}{FR}$ and the converse of BPT then applies
DEF is parallel to OD, since DE and DF both start at D

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Answer: C)

EF

is parallel to QR, since

\frac{PE}{EQ} = \frac{P D}{D O} = \frac{PF}{FR}

and the converse of BPT then applies

Apply BPT twice. In triangle POQ, DE is parallel to OQ, so

\frac{PE}{EQ}

\frac{P D}{D O}

. In triangle POR, DF is parallel to OR, so

\frac{PF}{FR}

\frac{P D}{D O}

. The common middle ratio gives

\frac{PE}{EQ}

\frac{PF}{FR}

, and now the converse of BPT in triangle PQR makes EF parallel to QR. The position of O is irrelevant, so requiring O to be the centroid is unnecessary.

5. In triangle ABC, D lies on AB and E lies on AC, and it is given that

\frac{A D}{A B} = \frac{A E}{A C} = \frac{3}{7}

. What follows about DE and BC?

A $D E = \frac{3}{4} BC$
B $D E = \frac{9}{49} BC$
C $D E = \frac{3}{7} BC$
DDE cannot be compared with BC from this data

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Answer: C)

D E = \frac{3}{7} BC

Equal ratios on the two sides mean, by the converse of BPT, that DE is parallel to BC. Triangle ADE is then similar to triangle ABC, so every pair of corresponding sides is in the ratio

\frac{3}{7}

, giving DE =

\frac{3}{7}

BC. The option

\frac{3}{4}

uses AD : DB (= 3 : 4) instead of AD : AB;

\frac{9}{49}

squares the ratio, which belongs to areas, not lengths; and the data are quite enough to settle the comparison.

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