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TRIANGLES - Q&A

EXERCISE 6.1

1. Fill in the blanks using the correct word given in brackets :

(i) All circles are __________. (congruent, similar)

Answer: similar

(ii) All squares are __________. (similar, congruent)

Answer: similar

(iii) All __________ triangles are similar. (isosceles, equilateral)

Answer: equilateral

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

Answer: (a) equal, (b) proportional


2. Give two different examples of pair of :

(i) similar figures.

Answer:
1. Two equilateral triangles with sides 1 cm and 2 cm.
2. Two circles with radii 2 cm and 3 cm.

(ii) non-similar figures.

Answer:
1. A square and a rhombus.
2. A triangle and a parallelogram.


3. State whether the following quadrilaterals are similar or not:

Answer:
The sides of quadrilateral PQRS and ABCD are proportional (Ratio 1.5/3 = 1/2).
However, the corresponding angles are not equal (ABCD has 90° angles, while PQRS does not).
Therefore, the quadrilaterals are not similar.

EXERCISE 6.2

1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution:
(i) Given DE || BC.
According to the Basic Proportionality Theorem (Thales Theorem):
AD / DB = AE / EC
Substituting the given values:
1.5 / 3 = 1 / EC
1/2 = 1 / EC
EC = 2 cm.
Answer: EC = 2 cm

(ii) Given DE || BC.
According to the Basic Proportionality Theorem:
AD / DB = AE / EC
Substituting the given values:
AD / 7.2 = 1.8 / 5.4
AD / 7.2 = 1 / 3
AD = 7.2 / 3 = 2.4 cm.
Answer: AD = 2.4 cm


2. E and F are points on the sides PQ and PR respectively of a Δ PQR. For each of the following cases, state whether EF || QR :

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Solution:
Calculate the ratios of the sides:
PE / EQ = 3.9 / 3 = 1.3
PF / FR = 3.6 / 2.4 = 36 / 24 = 1.5
Since PE / EQ ≠ PF / FR, EF is not parallel to QR.

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

Solution:
PE / QE = 4 / 4.5 = 40 / 45 = 8 / 9
PF / RF = 8 / 9
Since PE / QE = PF / RF, by the converse of Basic Proportionality Theorem, EF || QR.

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Solution:
PE / PQ = 0.18 / 1.28 = 18 / 128 = 9 / 64
PF / PR = 0.36 / 2.56 = 36 / 256 = 9 / 64
Since PE / PQ = PF / PR, by the converse of Basic Proportionality Theorem, EF || QR.


3. In Fig. 6.18, if LM || CB and LN || CD, prove that AM / AB = AN / AD.

Solution:
In Δ ABC, LM || CB.
By Basic Proportionality Theorem:
AM / AB = AL / AC ... (i)

In Δ ADC, LN || CD.
By Basic Proportionality Theorem:
AN / AD = AL / AC ... (ii)

From (i) and (ii), we get:
AM / AB = AN / AD (Hence Proved).


4. In Fig. 6.19, DE || AC and DF || AE. Prove that BF / FE = BE / EC.

Solution:
In Δ ABC, DE || AC.
By Basic Proportionality Theorem:
BD / DA = BE / EC ... (i)

In Δ ABE, DF || AE.
By Basic Proportionality Theorem:
BD / DA = BF / FE ... (ii)

From (i) and (ii), we get:
BF / FE = BE / EC (Hence Proved).


5. In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

Solution:
In Δ POQ, DE || OQ.
By Basic Proportionality Theorem:
PE / EQ = PD / DO ... (i)

In Δ POR, DF || OR.
By Basic Proportionality Theorem:
PF / FR = PD / DO ... (ii)

From (i) and (ii):
PE / EQ = PF / FR.

In Δ PQR, since the line segment EF divides the sides PQ and PR in the same ratio,
By the Converse of Basic Proportionality Theorem:
EF || QR (Hence Proved).


6. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:
In Δ OPQ, AB || PQ.
Therefore, OA / AP = OB / BQ ... (i)

In Δ OPR, AC || PR.
Therefore, OA / AP = OC / CR ... (ii)

From (i) and (ii):
OB / BQ = OC / CR.

In Δ OQR, the ratio of sides is equal.
By Converse of Basic Proportionality Theorem:
BC || QR (Hence Proved).


7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Solution:
Let Δ ABC be the triangle. Let D be the mid-point of AB, so AD = DB.
A line DE is drawn parallel to BC, intersecting AC at E.
We need to prove that E is the mid-point of AC.
By Basic Proportionality Theorem (Theorem 6.1):
AD / DB = AE / EC
Since D is the midpoint, AD = DB, so AD / DB = 1.
Therefore, 1 = AE / EC
⇒ AE = EC.
This implies E is the midpoint of AC.
Hence proved.


8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Solution:
Let Δ ABC be the triangle. D is the mid-point of AB and E is the mid-point of AC.
So, AD = DB and AE = EC.
Therefore, AD / DB = 1 and AE / EC = 1.
This gives AD / DB = AE / EC.
By the Converse of Basic Proportionality Theorem (Theorem 6.2):
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Therefore, DE || BC.
Hence proved.


9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO / BO = CO / DO.

Solution:
Draw a line EF through O parallel to DC (and AB).
In Δ ADC, EO || DC.
By BPT: AE / ED = AO / CO ... (i)

In Δ ABD, EO || AB.
By BPT: AE / ED = BO / DO ... (ii)

From (i) and (ii):
AO / CO = BO / DO
Rearranging the terms:
AO / BO = CO / DO (Hence Proved).


10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO / BO = CO / DO. Show that ABCD is a trapezium.

Solution:
Given: AO / BO = CO / DO.
Rearranging: AO / CO = BO / DO.
Draw a line OE || AB in Δ ABD, meeting AD at E.
By BPT: AE / ED = BO / DO.
Using the given relation, AE / ED = AO / CO.
In Δ ADC, since AE / ED = AO / CO, by Converse of BPT, OE || DC.
Since OE || AB and OE || DC, it implies AB || DC.
A quadrilateral with one pair of opposite sides parallel is a trapezium.
Therefore, ABCD is a trapezium.

EXERCISE 6.3

1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

Solution:
(i) In ΔABC and ΔPQR:
∠A = ∠P = 60°, ∠B = ∠Q = 80°, ∠C = ∠R = 40°.
Criterion: AAA Similarity.
Symbolic form: ΔABC ~ ΔPQR.

(ii) In ΔABC and ΔQRP:
AB/QR = 2/4 = 1/2; BC/RP = 2.5/5 = 1/2; CA/PQ = 3/6 = 1/2.
Criterion: SSS Similarity.
Symbolic form: ΔABC ~ ΔQRP.

(iii) In ΔLMP and ΔDEF:
MP/DE = 2/4 = 1/2; LP/DF = 3/6 = 1/2; but LM/EF = 2.7/5 ≠ 1/2.
Answer: Triangles are not similar.

(iv) In ΔMNL and ΔQPR:
∠M = ∠Q = 70°.
Sides containing the angle: MN/QP = 2.5/5 = 1/2; ML/QR = 5/10 = 1/2.
Criterion: SAS Similarity.
Symbolic form: ΔNML ~ ΔPQR (Note vertex correspondence: M↔Q).

(v) In ΔABC and ΔFDE:
∠A = 80° is not included between the known sides. ∠F = 80° is included.
Answer: Triangles are not similar (Side ratios match, but the angle is not the included angle in ΔABC).

(vi) In ΔDEF and ΔPQR:
In ΔDEF, ∠F = 180° - (70°+80°) = 30°.
In ΔPQR, ∠P = 180° - (80°+30°) = 70°.
So, ∠D = ∠P = 70°, ∠E = ∠Q = 80°, ∠F = ∠R = 30°.
Criterion: AAA Similarity.
Symbolic form: ΔDEF ~ ΔPQR.


2. In Fig. 6.35, Δ ODC ~ Δ OBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

Solution:
1. ∠ DOC forms a linear pair with ∠ BOC.
∠ DOC + 125° = 180°
∠ DOC = 55°.

2. In Δ ODC, sum of angles is 180°.
∠ DCO + ∠ CDO + ∠ DOC = 180°
∠ DCO + 70° + 55° = 180°
∠ DCO + 125° = 180°
∠ DCO = 55°.

3. Since Δ ODC ~ Δ OBA, corresponding angles are equal.
∠ OAB = ∠ OCD (or ∠ DCO)
∠ OAB = 55°.


3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA / OC = OB / OD.

Solution:
In Δ OAB and Δ OCD:
1. ∠ AOB = ∠ COD (Vertically opposite angles)
2. ∠ OAB = ∠ OCD (Alternate interior angles, as AB || DC)
Therefore, Δ OAB ~ Δ OCD (by AA Similarity).
Since triangles are similar, ratios of corresponding sides are equal:
OA / OC = OB / OD.
Hence Proved.


4. In Fig. 6.36, QR / QS = QT / PR and ∠ 1 = ∠ 2. Show that Δ PQS ~ Δ TQR.

Solution:
In Δ PQR, ∠ 1 = ∠ 2.
Therefore, side opposite to equal angles are equal: PQ = PR.
Given: QR / QS = QT / PR.
Substitute PR = PQ in the given ratio:
QR / QS = QT / PQ
Taking reciprocal and rearranging: PQ / QT = QS / QR.
Or simply: QS / QR = PQ / QT.
In Δ PQS and Δ TQR:
1. QS / QR = PQ / QT (Proved above)
2. ∠ Q = ∠ Q (Common angle)
By SAS Similarity Criterion:
Δ PQS ~ Δ TQR (Hence Proved).


5. S and T are points on sides PR and QR of Δ PQR such that ∠ P = ∠ RTS. Show that Δ RPQ ~ Δ RTS.

Solution:
In Δ RPQ and Δ RTS:
1. ∠ RPQ = ∠ RTS (Given)
2. ∠ R = ∠ R (Common angle)
By AA Similarity Criterion:
Δ RPQ ~ Δ RTS (Hence Proved).


6. In Fig. 6.37, if Δ ABE ≅ Δ ACD, show that Δ ADE ~ Δ ABC.

Solution:
Given Δ ABE ≅ Δ ACD.
By CPCT (Corresponding Parts of Congruent Triangles):
AB = AC ... (i)
AE = AD ... (ii)
Dividing (ii) by (i):
AD / AB = AE / AC.
In Δ ADE and Δ ABC:
1. AD / AB = AE / AC (Proved above)
2. ∠ A = ∠ A (Common angle)
By SAS Similarity Criterion:
Δ ADE ~ Δ ABC (Hence Proved).


7. In Fig. 6.38, altitudes AD and CE of Δ ABC intersect each other at the point P. Show that:

(i) Δ AEP ~ Δ CDP

Solution:
∠ AEP = ∠ CDP = 90°.
∠ APE = ∠ CPD (Vertically opposite angles).
By AA Similarity: Δ AEP ~ Δ CDP.

(ii) Δ ABD ~ Δ CBE

Solution:
∠ ADB = ∠ CEB = 90°.
∠ B = ∠ B (Common).
By AA Similarity: Δ ABD ~ Δ CBE.

(iii) Δ AEP ~ Δ ADB

Solution:
∠ AEP = ∠ ADB = 90°.
∠ A = ∠ A (Common).
By AA Similarity: Δ AEP ~ Δ ADB.

(iv) Δ PDC ~ Δ BEC

Solution:
∠ PDC = ∠ BEC = 90°.
∠ C = ∠ C (Common).
By AA Similarity: Δ PDC ~ Δ BEC.


8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ Δ CFB.

Solution:
In parallelogram ABCD, ∠ A = ∠ C (Opposite angles).
In Δ ABE and Δ CFB:
1. ∠ A = ∠ C (Opposite angles of parallelogram)
2. ∠ AEB = ∠ CBF (Alternate interior angles, as AE || BC)
By AA Similarity Criterion:
Δ ABE ~ Δ CFB (Hence Proved).


9. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) Δ ABC ~ Δ AMP

Solution:
In Δ ABC and Δ AMP:
1. ∠ ABC = ∠ AMP = 90°
2. ∠ A = ∠ A (Common)
By AA Similarity Criterion:
Δ ABC ~ Δ AMP.

(ii) CA / PA = BC / MP

Solution:
Since Δ ABC ~ Δ AMP (Proved in (i)), corresponding sides are proportional.
CA / PA = BC / MP.
Hence Proved.


10. CD and GH are respectively the bisectors of ∠ ACB and ∠ EGF such that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If Δ ABC ~ Δ FEG, show that:

(i) CD / GH = AC / FG

Solution:
Given Δ ABC ~ Δ FEG.
So, ∠ C = ∠ G.
Since CD and GH are bisectors, ∠ ACD = ∠ FGH (= half of equal angles).
In Δ ACD and Δ FGH:
1. ∠ A = ∠ F (Given from similar triangles)
2. ∠ ACD = ∠ FGH
So, Δ ACD ~ Δ FGH.
Therefore, CD / GH = AC / FG.

(ii) Δ DCB ~ Δ HGE

Solution:
In Δ DCB and Δ HGE:
1. ∠ B = ∠ E (Given from similar triangles)
2. ∠ BCD = ∠ EGH (Half of equal angles C and G)
By AA Similarity:
Δ DCB ~ Δ HGE.

(iii) Δ DCA ~ Δ HGF

Solution:
Already proved in part (i) as Δ ACD ~ Δ FGH, which is the same as Δ DCA ~ Δ HGF.
Hence Proved.


11. In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that Δ ABD ~ Δ ECF.

Solution:
Given AB = AC, so ∠ ABD = ∠ ECF (Angles opposite to equal sides).
In Δ ABD and Δ ECF:
1. ∠ ADB = ∠ EFC = 90° (Given altitudes)
2. ∠ ABD = ∠ ECF (Proved above)
By AA Similarity Criterion:
Δ ABD ~ Δ ECF.


12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Δ PQR (see Fig. 6.41). Show that Δ ABC ~ Δ PQR.

Solution:
Given: AB/PQ = BC/QR = AD/PM.
Since AD and PM are medians, BC = 2BD and QR = 2QM.
So, AB/PQ = 2BD/2QM = AD/PM.
⇒ AB/PQ = BD/QM = AD/PM.
By SSS Similarity, Δ ABD ~ Δ PQM.
Therefore, ∠ B = ∠ Q (Corresponding angles).
Now in Δ ABC and Δ PQR:
1. AB/PQ = BC/QR (Given)
2. ∠ B = ∠ Q (Proved)
By SAS Similarity Criterion:
Δ ABC ~ Δ PQR.


13. D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ BAC. Show that CA2 = CB.CD.

Solution:
In Δ ADC and Δ BAC:
1. ∠ ADC = ∠ BAC (Given)
2. ∠ C = ∠ C (Common)
By AA Similarity, Δ ADC ~ Δ BAC.
So, sides are proportional:
CA / CB = CD / CA
Cross multiplying:
CA2 = CB.CD.


14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that Δ ABC ~ Δ PQR.

Solution:
Given: AB/PQ = AC/PR = AD/PM.
Extend AD to E such that AD=DE, join CE. Extend PM to L such that PM=ML, join RL.
By construction and congruence, we can show AB=CE and PQ=RL.
This transforms the ratio to CE/RL = AC/PR = AE/PL (since AE=2AD, PL=2PM).
By SSS, Δ ACE ~ Δ PRL.
So ∠ CAE = ∠ RPL. Similarly ∠ BAE = ∠ QPL.
Adding them, ∠ BAC = ∠ QPR (Angle A = Angle P).
Now in Δ ABC and Δ PQR:
1. AB/PQ = AC/PR (Given)
2. ∠ A = ∠ P (Proved)
By SAS Similarity:
Δ ABC ~ Δ PQR.


15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution:
Let height of tower be h.
The triangle formed by pole and shadow is similar to triangle formed by tower and shadow (Sun's elevation is same).
Height of Pole / Shadow of Pole = Height of Tower / Shadow of Tower
6 / 4 = h / 28
h = (6 / 4) * 28
h = 6 * 7 = 42 m.
Answer: Height of tower is 42 m.


16. If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ~ Δ PQR, prove that AB / PQ = AD / PM.

Solution:
Given Δ ABC ~ Δ PQR.
So, AB / PQ = BC / QR and ∠ B = ∠ Q.
Since AD and PM are medians, BC = 2BD and QR = 2QM.
So, AB / PQ = 2BD / 2QM = BD / QM.
In Δ ABD and Δ PQM:
1. AB / PQ = BD / QM
2. ∠ B = ∠ Q
By SAS Similarity, Δ ABD ~ Δ PQM.
Therefore, corresponding sides are proportional:
AB / PQ = AD / PM (Hence Proved).

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Quick Review Flashcards - Click to flip and test your knowledge!
Question
What characteristic defines two figures as being congruent?
Answer
They possess the same shape and the same size.
Question
Two figures that have the same shape but not necessarily the same size are described as _____.
Answer
Similar figures
Question
What is the relationship between congruent figures and similar figures?
Answer
All congruent figures are similar, but similar figures are not necessarily congruent.
Question
Why are all circles considered similar to one another?
Answer
They all share the same shape, regardless of their varying radii.
Question
Name two types of polygons that are always similar regardless of their side lengths.
Answer
Squares and equilateral triangles.
Question
State the first necessary condition for two polygons with the same number of sides to be similar.
Answer
All of their corresponding angles must be equal.
Question
State the second necessary condition for two polygons with the same number of sides to be similar.
Answer
All of their corresponding sides must be in the same ratio (or proportion).
Question
In the context of similar polygons, what is meant by the term 'scale factor'?
Answer
The constant ratio between the corresponding sides of the polygons.
Question
Is a square always similar to a rectangle? Give the primary reason.
Answer
No, because while their angles are equal, their corresponding sides are not necessarily in the same ratio.
Question
Why is a square not similar to a rhombus despite both having four sides of equal length?
Answer
Their corresponding angles are not equal.
Question
What specific term describes triangles that have all their corresponding angles equal?
Answer
Equiangular triangles
Question
According to the mathematician Thales, what is always true about the ratio of corresponding sides in equiangular triangles?
Answer
The ratio of any two corresponding sides is always the same.
Question
State Theorem 6.1, also known as the Basic Proportionality Theorem.
Answer
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Question
In $\triangle ABC$, if a line $DE$ is parallel to $BC$ and intersects $AB$ at $D$ and $AC$ at $E$, what is the ratio relation?
Answer
$\frac{AD}{DB} = \frac{AE}{EC}$
Question
Which mathematical concept is primarily used to prove the Basic Proportionality Theorem?
Answer
The area of a triangle ($Area = \frac{1}{2} \times base \times height$).
Question
In the proof of the Basic Proportionality Theorem, why is $ar(BDE)$ equal to $ar(DEC)$?
Answer
They are on the same base $DE$ and between the same parallel lines $BC$ and $DE$.
Question
State Theorem 6.2, which is the converse of the Basic Proportionality Theorem.
Answer
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Question
In $\triangle ABC$, if $D$ and $E$ are points on $AB$ and $AC$ such that $\frac{AD}{DB} = \frac{AE}{EC}$, what can be concluded about line $DE$?
Answer
Line $DE$ is parallel to side $BC$ ($DE \parallel BC$).
Question
If a line $DE$ is parallel to side $BC$ of $\triangle ABC$, prove the ratio $\frac{AD}{AB} = \frac{AE}{AC}$.
Answer
By Theorem 6.1, $\frac{DB}{AD} = \frac{EC}{AE}$; adding 1 to both sides yields $\frac{DB+AD}{AD} = \frac{EC+AE}{AE}$, which simplifies to $\frac{AB}{AD} = \frac{AC}{AE}$.
Question
In a trapezium $ABCD$ where $AB \parallel DC$, if line $EF$ is parallel to $AB$, how does it divide the non-parallel sides $AD$ and $BC$?
Answer
It divides them in the same ratio: $\frac{AE}{ED} = \frac{BF}{FC}$.
Question
According to the AAA (Angle-Angle-Angle) similarity criterion, when are two triangles similar?
Answer
Two triangles are similar if their corresponding angles are equal, which implies their corresponding sides are in the same ratio.
Question
Explain the AA (Angle-Angle) similarity criterion for triangles.
Answer
Two triangles are similar if two angles of one triangle are respectively equal to two angles of another triangle.
Question
State the SS (Side-Side-Side) similarity criterion for triangles.
Answer
If the sides of one triangle are proportional to the sides of another triangle, then their corresponding angles are equal and the triangles are similar.
Question
What are the requirements for the SAS (Side-Angle-Side) similarity criterion?
Answer
One angle of one triangle must equal one angle of the other, and the sides including these angles must be proportional.
Question
In $\triangle ABC$ and $\triangle DEF$, if $\angle A = \angle D$ and $\frac{AB}{DE} = \frac{AC}{DF}$, are the triangles similar?
Answer
Yes, by the SAS similarity criterion.
Question
If $\triangle ABC \sim \triangle DEF$, what is the correct symbolic way to denote corresponding vertices?
Answer
The order of vertices must match the correspondence, e.g., $A$ to $D$, $B$ to $E$, and $C$ to $F$.
Question
In Figure 6.29, if $PQ \parallel RS$, which criterion proves $\triangle POQ \sim \triangle SOR$?
Answer
The AAA similarity criterion (or AA similarity criterion).
Question
If $OA \cdot OB = OC \cdot OD$, what can be said about the relationship between $\triangle AOD$ and $\triangle COB$?
Answer
They are similar ($\triangle AOD \sim \triangle COB$) by the SAS similarity criterion.
Question
In Figure 6.30, if sides are $AB=3.8$, $BC=6$, $CA=3\sqrt{3}$ and $RQ=7.6$, $QP=12$, $PR=6\sqrt{3}$, which triangles are similar?
Answer
$\triangle ABC \sim \triangle RQP$ by the SSS similarity criterion.
Question
How can the height of a lamp-post be used to find a person's shadow length using similarity?
Answer
By creating similar triangles using the vertical post, the vertical person, and the common angle of light elevation.
Question
What is the relationship between the mid-point theorem and Theorem 6.1?
Answer
The mid-point theorem is a special case of Theorem 6.1 where the ratio of side segments is $1:1$.
Question
If $\triangle ABC \sim \triangle DEF$, and the scale factor is 1, what can be said about the triangles?
Answer
The triangles are congruent.
Question
In a quadrilateral $ABCD$, if the diagonals intersect at $O$ such that $\frac{AO}{BO} = \frac{CO}{DO}$, what type of figure is $ABCD$?
Answer
A trapezium.
Question
What is the belief regarding how Thales measured the height of Mount Everest or the distance to the moon?
Answer
Using the principle of indirect measurements based on the similarity of figures.
Question
In $\triangle PQR$, if $E$ and $F$ are points on $PQ$ and $PR$ such that $PE=3.9$, $EQ=3$, $PF=3.6$, $FR=2.4$, is $EF \parallel QR$?
Answer
No, because $\frac{PE}{EQ} = 1.3$ and $\frac{PF}{FR} = 1.5$ (ratios are not equal).
Question
If $LM \parallel CB$ and $LN \parallel CD$ in a quadrilateral $ABCD$ with diagonal $AC$, prove $\frac{AM}{AB} = \frac{AN}{AD}$.
Answer
By applying BPT to $\triangle ABC$ and $\triangle ADC$ separately and equating the common ratio $\frac{AL}{AC}$.
Question
In Figure 6.19, if $DE \parallel AC$ and $DF \parallel AE$, what ratio identity is proven?
Answer
$\frac{BF}{FE} = \frac{BE}{EC}$.
Question
In $\triangle PQR$, if $A, B, C$ are points on $OP, OQ, OR$ such that $AB \parallel PQ$ and $AC \parallel PR$, what is the relation between $BC$ and $QR$?
Answer
$BC$ must be parallel to $QR$ ($BC \parallel QR$).
Question
True or False: If two triangles are equiangular, their corresponding sides are proportional.
Answer
True.
Question
What does the symbol '$\sim$' represent in geometry?
Answer
'is similar to'
Question
When applying Theorem 6.3 (AAA), if only two corresponding angles are known to be equal, why are the triangles still similar?
Answer
The third angles must be equal due to the angle sum property of triangles (AA criterion).
Question
In the shadow problem (Example 7), if a girl walks at $1.2 \text{ m/s}$ for $4 \text{ seconds}$, what distance $BD$ has she covered?
Answer
$4.8 \text{ metres}$ ($1.2 \times 4$).
Question
In the SAS similarity proof, what construction is made on $\triangle DEF$ to compare it to $\triangle ABC$?
Answer
Points $P$ and $Q$ are marked such that $DP = AB$ and $DQ = AC$, then $PQ$ is joined.
Question
Is the condition of equal corresponding angles alone sufficient to prove similarity in triangles?
Answer
Yes, for triangles, equal angles imply proportional sides (unlike other polygons).
Question
Is the condition of proportional sides alone sufficient to prove similarity in triangles?
Answer
Yes, for triangles, proportional sides imply equal corresponding angles.
Question
In Figure 6.16, if $\frac{PS}{SQ} = \frac{PT}{TR}$ and $\angle PST = \angle PRQ$, what specific type of triangle is $PQR$?
Answer
An isosceles triangle.
Question
Why are all equilateral triangles similar?
Answer
Their angles are always $60^{\circ}$ (equiangular) and their sides are always proportional.
Question
Define 'indirect measurement' as used in Chapter 6.
Answer
A method of finding heights and distances using the principle of similarity rather than direct tape measurement.
Question
If a line intersects sides $AB$ and $AC$ of $\triangle ABC$ at $D$ and $E$ and is parallel to $BC$, find $AD$ if $DB=7.2$, $AE=1.8$, $EC=5.4$.
Answer
$2.4$ (using $\frac{AD}{7.2} = \frac{1.8}{5.4}$).
Question
In similar triangles, corresponding angles are equal, and corresponding sides are _____.
Answer
Proportional (or in the same ratio).
Question
Theorem 6.1: $\frac{AD}{DB} = \frac{AE}{EC}$. If $AD=1.5, DB=3, AE=1$, calculate $EC$.
Answer
$2$.
Question
What is the Representative Fraction (RF) in map making synonymous with in geometry?
Answer
The scale factor.
Question
Can a triangle and a square be similar? Why?
Answer
No, because similar figures must have the same number of sides and the same shape.
Question
If two polygons are similar, and the second is similar to a third, what is the relation between the first and the third?
Answer
The first polygon is similar to the third polygon.
Question
What does Theorem 6.4 (SSS) conclude about the angles of two triangles with proportional sides?
Answer
Their corresponding angles are equal.
Question
In Figure 6.21, if $AB \parallel PQ$, which triangle ratio is established at point $O$?
Answer
$\frac{OA}{AP} = \frac{OB}{BQ}$ (or $\frac{OA}{OP} = \frac{OB}{OQ}$).
Question
When proves $\triangle POQ \sim \triangle SOR$ with $PQ \parallel RS$, which angle pairs are alternate angles?
Answer
$\angle P = \angle S$ and $\angle Q = \angle R$.
Question
If the ratio of sides of two triangles is $1:1$, they are similar and _____.
Answer
Congruent
Question
In Example 5, why is $\angle P$ found to be $40^{\circ}$?
Answer
Because $\triangle ABC \sim \triangle RQP$, and $\angle C = 180^{\circ} - 80^{\circ} - 60^{\circ} = 40^{\circ}$, thus $\angle P = \angle C$.