Class X · Mathematics Ch.6 · CBSE Pattern · 30 Questions
10
MCQ (1 mark)
3
Assertion-Reason
12
Short Answer (2-3 m)
5
Long Answer (5 m)
47
Total Marks
Section A — Multiple Choice Questions (1 Mark Each)
MCQ 1
Which of the following pairs of figures is always similar?
✓ C is correct. All equilateral triangles have equal angles (60° each) and proportional sides, so they are always similar. Rectangles/rhombuses/isosceles triangles need not be.
MCQ 2
In ΔABC, DE ∥ BC. If AD = 2 cm, DB = 3 cm and AE = 4 cm, then EC =
✓ B is correct. By BPT: AD/DB = AE/EC → 2/3 = 4/EC → EC = 6 cm.
MCQ 3
If ΔABC ~ ΔPQR and AB/PQ = 3/4, then ar(ABC)/ar(PQR) =
✓ B is correct. Ratio of areas = (ratio of sides)² = (3/4)² = 9/16.
MCQ 4
Which set of numbers forms a Pythagorean triplet?
✓ B is correct. 8² + 15² = 64 + 225 = 289 = 17². The others fail a²+b²=c².
MCQ 5
Two triangles are similar if their corresponding:
✓ B is correct. Equal corresponding angles (AA/AAA) guarantee similarity. Equal sides would make them congruent.
MCQ 6
The areas of two similar triangles are 25 cm² and 36 cm². The ratio of their corresponding sides is:
✓ A is correct. Side ratio = √(25/36) = 5/6.
MCQ 7
In a right triangle, the hypotenuse is 25 cm and one side is 7 cm. The other side is:
✓ B is correct. Other side = √(25² − 7²) = √(625 − 49) = √576 = 24 cm (a 7-24-25 triplet).
MCQ 8
If ΔABC ~ ΔDEF and their areas are equal, then the triangles are:
✓ B is correct. Equal areas → area ratio = 1 → side ratio = 1 → congruent.
MCQ 9
A line joining the midpoints of two sides of a triangle is:
✓ B is correct. By the Midpoint Theorem (a special case of BPT), the segment is parallel to and half the length of the third side.
MCQ 10
Which is NOT a valid criterion for similarity of triangles?
✓ D is correct. SSA is not a valid similarity criterion. Only AA, SSS and SAS prove similarity.
Section B — Assertion & Reason (1 Mark Each)
A-R 1
Choose the correct option about the statements below.
ASSERTION (A)If two triangles are similar with side ratio 2:3, their area ratio is 4:9.
REASON (R)The ratio of areas of two similar triangles equals the square of the ratio of corresponding sides.
✓ A is correct. (2:3)² = 4:9, so A is true, and R is exactly the theorem that explains it.
A-R 2
Choose the correct option about the statements below.
ASSERTION (A)All congruent triangles are similar.
REASON (R)All similar triangles are congruent.
✓ C is correct. A is true (congruence is similarity with k=1). R is false — similar triangles are congruent only when k = 1.
A-R 3
Choose the correct option about the statements below.
ASSERTION (A)A triangle with sides 9 cm, 12 cm and 15 cm is right-angled.
REASON (R)By the converse of Pythagoras, if the square of the longest side equals the sum of squares of the other two, the triangle is right-angled.
✓ A is correct. 9²+12² = 81+144 = 225 = 15², so A is true, and R (converse of Pythagoras) explains exactly why.
Section C — Short Answer Questions (2–3 Marks Each)
SA 1
In ΔABC, DE ∥ BC. If AD = 4 cm, DB = 6 cm and AE = 5 cm, find AC.
[2 marks]
By BPT: AD/DB = AE/EC.
4/6 = 5/EC → EC = (5×6)/4 = 7.5 cm.
AC = AE + EC = 5 + 7.5 = 12.5 cm.
SA 2
The areas of two similar triangles are 81 cm² and 144 cm². If the smaller side of the first triangle is 9 cm, find the corresponding side of the other.
[3 marks]
Ratio of areas = 81/144 → ratio of sides = √(81/144) = 9/12 = 3/4.
So 9/x = 3/4 → x = (9×4)/3 = 12 cm.
SA 3
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.
[2 marks]
By Pythagoras: base² = 10² − 8² = 100 − 64 = 36.
base = √36 = 6 m (a 6-8-10 triplet).
SA 4
In ΔABC and ΔPQR, ∠A = ∠P and AB/PQ = AC/PR. Which similarity criterion proves ΔABC ~ ΔPQR?
[2 marks]
One pair of equal angles (∠A = ∠P) with the two including sides proportional (AB/PQ = AC/PR).
This is the SAS similarity criterion.
SA 5
In ΔABC, D and E are points on AB and AC. AD = 2, DB = 3, AE = 3, EC = 4.5. Is DE ∥ BC?
[2 marks]
AD/DB = 2/3. AE/EC = 3/4.5 = 2/3.
Since AD/DB = AE/EC, by the converse of BPT, DE ∥ BC.
SA 6
The sides of a triangle are 7 cm, 24 cm and 25 cm. Show that it is a right triangle.
[2 marks]
Longest side = 25 cm. Check: 7² + 24² = 49 + 576 = 625 = 25².
Since a² + b² = c², by the converse of Pythagoras the triangle is right-angled (right angle opposite the 25 cm side).
SA 7
Two similar triangles have corresponding medians in the ratio 4:9. Find the ratio of their areas.
[2 marks]
Medians of similar triangles are in the same ratio as the sides.
Ratio of areas = (4/9)² = 16/81.
SA 8
State whether the following are always similar: (i) two circles (ii) two squares (iii) two right triangles.
[3 marks]
(i) Two circles — always similar ✓
(ii) Two squares — always similar ✓ (all angles 90°, sides proportional)
(iii) Two right triangles — NOT always similar ✗ (the other two angles may differ).
SA 9
A vertical pole 6 m high casts a shadow 4 m long. At the same time a tower casts a shadow 28 m long. Find the height of the tower.
[3 marks]
The pole and tower form similar triangles (same sun angle).
height/shadow constant: 6/4 = h/28
h = (6 × 28)/4 = 42 m.
SA 10
In a trapezium ABCD, AB ∥ DC and the diagonals intersect at O. If AO = 3, OC = 6 and BO = 4, find OD.
[2 marks]
Diagonals of a trapezium divide each other proportionally: AO/OC = BO/OD.
3/6 = 4/OD → OD = (4×6)/3 = 8 units.
SA 11
Find the length of the diagonal of a rectangle whose sides are 12 cm and 5 cm.
[2 marks]
Diagonal² = 12² + 5² = 144 + 25 = 169.
Diagonal = √169 = 13 cm (a 5-12-13 triplet).
SA 12
In ΔABC, the perpendicular from A meets BC at D. If AD² = BD·DC, prove ∠BAC = 90°.
[3 marks]
In right triangles ABD and ADC, AD² = BD·DC is given.
This makes ΔABD ~ ΔCAD (SAS-type, with the shared perpendicular), so ∠BAD = ∠ACD and ∠DAC = ∠ABD.
Then ∠BAC = ∠BAD + ∠DAC = ∠ACD + ∠ABD.
Since angles of ΔABC sum to 180°: ∠BAC + (∠ABD + ∠ACD) = 180° → 2∠BAC = 180° → ∠BAC = 90°.
Section D — Long Answer Questions (5 Marks Each)
LA 1
State and prove the Basic Proportionality Theorem (Thales Theorem).
[5 marks]
Statement: 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.
Given: ΔABC with DE ∥ BC meeting AB at D and AC at E. To prove: AD/DB = AE/EC. Construction: Join BE and CD; draw EM ⊥ AB and DN ⊥ AC.
Proof: ar(ADE) = ½·AD·EM and ar(BDE) = ½·DB·EM.
∴ ar(ADE)/ar(BDE) = AD/DB … (i)
Similarly ar(ADE)/ar(CDE) = AE/EC … (ii)
ΔBDE and ΔCDE are on the same base DE and between the same parallels DE, BC → equal areas.
So RHS of (i) and (ii) are equal → AD/DB = AE/EC. ∎
LA 2
Prove that in a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides (Pythagoras Theorem).
[5 marks]
Given: Right ΔABC, right-angled at B. To prove: AC² = AB² + BC². Construction: Draw BD ⊥ AC.
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
[5 marks]
Given: ΔABC ~ ΔPQR. To prove: ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)². Construction: Draw altitudes AM ⊥ BC and PN ⊥ QR.
Proof: ar(ABC)/ar(PQR) = (½·BC·AM)/(½·QR·PN) = (BC/QR)(AM/PN).
In ΔABM and ΔPQN: ∠B = ∠Q (similar Δ) and ∠M = ∠N = 90° → ΔABM ~ ΔPQN (AA).
So AM/PN = AB/PQ. Also, since ΔABC ~ ΔPQR, BC/QR = AB/PQ.
∴ ar(ABC)/ar(PQR) = (AB/PQ)(AB/PQ) = (AB/PQ)². ∎
LA 4
Two poles of heights 6 m and 11 m stand on level ground. If the distance between their feet is 12 m, find the distance between their tops.
[5 marks]
Let the poles be AB = 6 m and CD = 11 m, with feet B and D on the ground, BD = 12 m.
Draw AE ∥ BD from the top of the shorter pole to meet CD at E.
Then AE = BD = 12 m and CE = 11 − 6 = 5 m.
In right ΔAEC: AC² = AE² + CE² = 12² + 5² = 144 + 25 = 169.
AC = √169 = 13 m. The distance between the tops is 13 m.
LA 5
In ΔABC, D is a point on BC such that ∠ADC = ∠BAC. Prove that CA² = CB·CD.
[5 marks]
In ΔADC and ΔBAC:
∠ADC = ∠BAC (given)
∠C = ∠C (common angle)
∴ ΔADC ~ ΔBAC (AA similarity).
From similarity, corresponding sides are proportional:
CA/CB = CD/CA
Cross-multiplying: CA² = CB·CD. ∎