Triangles Assertion–Reason PYQs Class 10 CBSE (With Answers)

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
Equiangular triangles have equal corresponding angles.
Therefore, they are similar by the AAA similarity criterion.

Reason: False.
Equilateral triangles with sides 2 cm and 4 cm are similar.
Their corresponding sides are unequal, so they are not congruent.

Link: A is true by AAA similarity. R incorrectly claims that every pair of similar triangles must also be congruent. Answer: C.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
Congruent triangles have equal corresponding sides.
Hence the three ratios of corresponding sides are all 1.
The triangles are therefore similar by SSS similarity.

Reason: True.
Each pair of corresponding sides has equal length.
Their ratio is 1 : 1.

Link: R establishes proportional corresponding sides with common ratio 1, which explains A. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
Congruent triangles have equal corresponding angles and sides.
Thus they satisfy the conditions for similarity.

Reason: False.
Consider right triangles with perpendicular sides 3 cm, 4 cm and 2 cm, 6 cm respectively.
Their areas are $\frac{\text{1}}{\text{2}}$ × 3 × 4 = 6 cm² and $\frac{\text{1}}{\text{2}}$ × 2 × 6 = 6 cm².
Their side lengths are different, so they are not congruent.

Link: Equal areas alone do not ensure congruence. A is true and R is false. Answer: C.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
Equal corresponding angles give similarity by the AAA criterion.

Reason: True.
For similar triangles, the ratio of their areas equals the square of the ratio of corresponding sides.
Equal areas give a side-ratio square of 1.
Lengths are positive, so the side ratio is 1.
All corresponding sides are equal; hence the triangles are congruent by SSS congruence.

Link: R describes when triangles already known to be similar are congruent. It does not explain why equal corresponding angles establish similarity in A. Answer: B.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
This is the Basic Proportionality Theorem.
If D and E lie on AB and AC with DE ∥ BC, then $\frac{\text{AD}}{\text{DB}}$ = $\frac{\text{AE}}{\text{EC}}$.

Reason: False.
Through a point on one side of a triangle, a line parallel to another side can be drawn.
For example, a line through an interior point D of AB can be drawn parallel to BC.

Link: A states the Basic Proportionality Theorem correctly. R denies a construction that is possible, so R is false. Answer: C.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
EF ∥ AB and AB ∥ DC, so EF ∥ DC.
The trapezium result proved in R applies to this figure and gives $\frac{\text{AE}}{\text{ED}}$ = $\frac{\text{BF}}{\text{FC}}$.

Reason: True. This follows from the Basic Proportionality Theorem.
Let diagonal AC meet EF at G.
In △ADC, EG ∥ DC gives $\frac{\text{AE}}{\text{ED}}$ = $\frac{\text{AG}}{\text{GC}}$.
In △CAB, GF ∥ AB gives $\frac{\text{CG}}{\text{GA}}$ = $\frac{\text{CF}}{\text{FB}}$.
Taking reciprocals, $\frac{\text{AG}}{\text{GC}}$ = $\frac{\text{BF}}{\text{FC}}$.
Hence $\frac{\text{AE}}{\text{ED}}$ = $\frac{\text{BF}}{\text{FC}}$.

Link: R is the general proportionality result applied to the trapezium in A, so it explains A. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
By the Basic Proportionality Theorem, $\frac{\text{AD}}{\text{DB}}$ = $\frac{\text{AE}}{\text{EC}}$.
Taking reciprocals and adding 1 gives $\frac{\text{DB}}{\text{AD}}$ + 1 = $\frac{\text{EC}}{\text{AE}}$ + 1.
Since AB = AD + DB and AC = AE + EC, $\frac{\text{AB}}{\text{AD}}$ = $\frac{\text{AC}}{\text{AE}}$.
Taking reciprocals, $\frac{\text{AD}}{\text{AB}}$ = $\frac{\text{AE}}{\text{AC}}$.

Reason: True.
R states the Basic Proportionality Theorem.

Link: Applying R and rewriting each whole side as the sum of its two parts gives the proportion in A. R explains A. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
By the distance formula, DE = √[(−3 − 3)² + (−3 − 5)²]
= √(36 + 64)
= 10 units.
By the midpoint theorem, DE = $\frac{\text{1}}{\text{2}}$BC.
Thus BC = 2 × 10 = 20 units.

Reason: True.
R states the midpoint theorem.

Link: R gives BC = 2DE, which converts the calculated distance DE into the length stated in A. R explains A. Answer: A.

Related: Coordinate Geometry: distance-formula practice

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
Let D and E be the midpoints of AB and AC in △ABC.
Then AD = DB and AE = EC.
Therefore, $\frac{\text{AD}}{\text{DB}}$ = $\frac{\text{AE}}{\text{EC}}$ = 1.
By the converse of the Basic Proportionality Theorem, DE ∥ BC.

Reason: True.
R states the converse of the Basic Proportionality Theorem.

Link: The midpoint condition makes the two side-division ratios equal. Applying R proves A. Answer: A.

Official CBSE answer · Explanation by Team TutorVertex

Assertion: True.
A states the converse of the Basic Proportionality Theorem.

Reason: True.
Let D be the midpoint of AB and let DE ∥ BC meet AC at E.
By the Basic Proportionality Theorem, $\frac{\text{AE}}{\text{EC}}$ = $\frac{\text{AD}}{\text{DB}}$ = 1.
Thus AE = EC, so E is the midpoint of AC.

Link: R concerns a line already known to be parallel and the special ratio 1 : 1. It does not establish A, which deduces parallelism from equal division ratios. Answer: B.