Triangles: MCQs with Answers

Theorem or formula used

Similar polygons require equal corresponding angles and proportional corresponding sides.

For rectangles, the length-to-breadth ratio must also match.

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Similarity requires matching corresponding angles as well as proportional corresponding sides.

A square has four right angles. A general rhombus need not have right angles.

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SAS similarity: two pairs of sides must be proportional and their included angles must be equal.

∠C is included between AC and BC; ∠R is included between PR and QR.

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AA similarity: two pairs of equal corresponding angles are sufficient for similarity.

The correspondence is A ↔ D, B ↔ E and C ↔ F; AB and DE are corresponding sides.

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SAS similarity requires equal included angles between the pairs of proportional sides.

AB and BC meet at B; DE and DF meet at D.

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SSS similarity: three proportional pairs of corresponding sides determine the vertex correspondence.

AB ↔ QR, BC ↔ RP and CA ↔ PQ. Matching shared endpoints gives A ↔ Q, B ↔ R and C ↔ P.

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AA similarity: a common angle at A and a pair of right angles establish similarity.

The vertex correspondence for △BAC is B ↔ E, A ↔ A and C ↔ D.

Theorem guide · Team TutorVertex

SAS similarity: the included angles of the pairs of proportional sides must be equal.

AB and AC meet at A; EF and ED meet at E. The correspondence is A ↔ E, B ↔ F and C ↔ D.

Theorem guide · Team TutorVertex

AA similarity follows from two pairs of equal corresponding angles.

A ↔ P, B ↔ Q and C ↔ R. AB and PQ are corresponding sides, with scale factor 2.

Theorem or formula used

For △ABC ∼ △DEF: $\frac{\text{AB}}{\text{DE}}$ = $\frac{\text{BC}}{\text{EF}}$ = $\frac{\text{AC}}{\text{DF}}$.

A ↔ D, B ↔ E and C ↔ F. Therefore AB ↔ DE and BC ↔ EF.

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Corresponding altitudes of similar triangles are in the same ratio as corresponding sides.

AM and PN are perpendicular heights from the corresponding vertices A and P.

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By AA similarity, △AOD ∼ △COB and $\frac{\text{AD}}{\text{CB}}$ = $\frac{\text{AO}}{\text{CO}}$.

O is the intersection of AC and BD; AD and BC are the parallel sides.

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For △ABC ∼ △PQR: $\frac{\text{AB}}{\text{PQ}}$ = $\frac{\text{BC}}{\text{QR}}$ = $\frac{\text{AC}}{\text{PR}}$.

A ↔ P, B ↔ Q and C ↔ R. AB corresponds to PQ and BC to QR.

Theorem or formula used

For similar triangles, the ratio of their perimeters equals the ratio of any pair of corresponding sides.

P denotes the perimeter of △PQR. PQ ↔ AB, QR ↔ BC and PR ↔ AC.

Theorem or formula used

For similar triangles, the ratio of their perimeters equals the ratio of any pair of corresponding sides.

AB and PQ are corresponding sides, taking △ABC ∼ △PQR in the stated order.

Theorem or formula used

For similar triangles, the ratio of their perimeters equals the ratio of any pair of corresponding sides.

DE corresponds to BC because △ADE ∼ △ABC. Take the small-triangle perimeter over the large-triangle perimeter.

Theorem guide · Team TutorVertex

For △ABC ∼ △PQR: $\frac{\text{AB}}{\text{PQ}}$ = $\frac{\text{BC}}{\text{QR}}$ = $\frac{\text{AC}}{\text{PR}}$.

AB ↔ PQ, BC ↔ QR and AC ↔ PR. A perimeter is the sum of the three side lengths.

Theorem guide · Team TutorVertex

Corresponding side ratios and perimeter ratios of similar triangles are equal.

P₁ is the perimeter of △ABC and P₂ is the perimeter of △DEF. Here P₂ = 9 + 6 + 12 = 27 cm. AB ↔ DE, BC ↔ EF and AC ↔ DF.

Theorem guide · Team TutorVertex

For similar triangles, the ratio of their perimeters equals the ratio of any pair of corresponding sides.

Let x be the corresponding side of the second triangle, in centimetres. Keep the first-to-second order in both ratios.

Theorem or formula used

Since PQ ∥ BC, △APQ ∼ △ABC and $\frac{\text{AP}}{\text{AB}}$ = $\frac{\text{AQ}}{\text{AC}}$.

AC = AQ + QC; AQ is the upper part and QC is the remaining part of AC.

Theorem or formula used

Since DE ∥ BC, △ADE ∼ △ABC and $\frac{\text{AD}}{\text{AB}}$ = $\frac{\text{AE}}{\text{AC}}$.

AD ↔ AE are the two upper segments; AB ↔ AC are the corresponding whole sides.

Theorem or formula used

Since PQ ∥ BC, △APQ ∼ △ABC and $\frac{\text{PQ}}{\text{BC}}$ = $\frac{\text{AP}}{\text{AB}}$.

AB = AP + PB is the whole side. PQ corresponds to BC in the similar triangles.

Theorem or formula used

Basic Proportionality Theorem: if D lies on AB, E lies on AC and DE ∥ BC, then $\frac{\text{AD}}{\text{DB}}$ = $\frac{\text{AE}}{\text{EC}}$.

x denotes the length AE, in centimetres. Match AD with AE and DB with EC.

Theorem or formula used

Basic Proportionality Theorem: if D lies on AB, E lies on AC and DE ∥ BC, then $\frac{\text{AD}}{\text{DB}}$ = $\frac{\text{AE}}{\text{EC}}$.

x is the unknown number used in the two length expressions; AD = 2x cm and AE = (x + 2) cm.

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Since DE ∥ BC, △ADE ∼ △ABC and $\frac{\text{AD}}{\text{AB}}$ = $\frac{\text{AE}}{\text{AC}}$.

AE is the upper part of AC; EC = AC − AE.

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Since PQ ∥ BC, △APQ ∼ △ABC and $\frac{\text{AP}}{\text{AB}}$ = $\frac{\text{AQ}}{\text{AC}}$.

AB = AP + PB; AC is the whole side containing AQ.

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Since PQ ∥ AC, △BPQ ∼ △BAC and $\frac{\text{BP}}{\text{BA}}$ = $\frac{\text{BQ}}{\text{BC}}$.

B ↔ B, P ↔ A and Q ↔ C. BA = BP + PA.

Theorem or formula used

Parallel lines cut proportional segments on the two sides: $\frac{\text{AP}}{\text{AQ}}$ = $\frac{\text{PX}}{\text{QY}}$ = $\frac{\text{XB}}{\text{YC}}$.

AP, PX and XB are successive parts of AB; AQ, QY and YC are the matching parts of AC.

Theorem or formula used

If DE ∥ BC, the Basic Proportionality Theorem requires $\frac{\text{AD}}{\text{DB}}$ = $\frac{\text{AE}}{\text{EC}}$.

D and E divide AB and AC respectively. Compare the ratios of the two parts on each side.

Theorem or formula used

Converse of BPT: if $\frac{\text{AX}}{\text{XB}}$ = $\frac{\text{AY}}{\text{YC}}$, then XY ∥ BC.

XB = AB − AX is the remaining part of AB; YC is the remaining part of AC.