RowQ
The Vault
RowQ
The Vault
CBSE Class 10 Maths · 11 questions · 26 marks
Two triangles are similar when they have the same shape but not necessarily the same size, and that single idea lets you measure a tower using its shadow. The chapter builds from the Basic Proportionality Theorem to the AA, SSS and SAS similarity criteria, and then to the neat result that areas of similar triangles are in the ratio of the squares of corresponding sides. Draw a clean figure and mark equal angles first — the proportion you need usually becomes obvious.
In ΔABC, D lies on AB and E lies on AC with DE ∥ BC. If AD = 3 cm, DB = 5 cm and AE = 4.5 cm, then EC equals:
Answer
EC = 7.5 cm. Since DE ∥ BC, the Basic Proportionality Theorem gives AD/DB = AE/EC. So 3/5 = 4.5/EC. Cross-multiplying, 3 × EC = 5 × 4.5 = 22.5, so EC = 22.5/3 = 7.5 cm. The option 2.7 cm comes from wrongly writing the proportion upside down.
The areas of two similar triangles are in the ratio 25 : 64. The ratio of their corresponding sides is:
Answer
The sides are in the ratio 5 : 8. For similar triangles, ar(Δ₁)/ar(Δ₂) = (corresponding side ratio)². So (side ratio)² = 25/64, and taking positive square roots, side ratio = 5/8. Squaring back, (5/8)² = 25/64, which confirms the answer.
If ΔABC ~ ΔPQR with AB/PQ = 3/5 and the perimeter of ΔPQR is 40 cm, then the perimeter of ΔABC is:
Answer
The perimeter of ΔABC is 24 cm. In similar triangles the ratio of the perimeters equals the ratio of corresponding sides. So perimeter(ΔABC)/perimeter(ΔPQR) = 3/5. perimeter(ΔABC) = (3/5) × 40 = 24 cm.
If ΔDEF ~ ΔMNP and ∠D = 55° and ∠E = 65°, then ∠P equals:
Answer
∠P = 60°. In ΔDEF the angles add to 180°, so ∠F = 180° - 55° - 65° = 60°. Because ΔDEF ~ ΔMNP, corresponding vertices are D↔M, E↔N and F↔P, so ∠P = ∠F. Hence ∠P = 60°.
Assertion (A): In ΔABC, D lies on AB and E lies on AC such that AD/DB = 2/3 and AE/EC = 2/3; then DE is parallel to BC. Reason (R): If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
Answer
Both A and R are true and R is the correct explanation of A. Reason R is the converse of the Basic Proportionality Theorem, a true statement. In the Assertion, D and E divide AB and AC in the same ratio 2 : 3. Applying the converse of BPT directly gives DE ∥ BC, so Assertion A is true and follows from R.
In ΔPQR, S lies on PQ and T lies on PR with ST ∥ QR. If PS = 4 cm, SQ = 6 cm and PT = 5 cm, find TR.
Answer
Since ST ∥ QR, by the Basic Proportionality Theorem PS/SQ = PT/TR. 4/6 = 5/TR Cross-multiplying, 4 × TR = 6 × 5 = 30. TR = 30/4 = 7.5 cm.
The areas of two similar triangles are 81 cm² and 144 cm². If a side of the smaller triangle is 6 cm, find the length of the corresponding side of the larger triangle.
Answer
Let the required side be x cm. For similar triangles, ratio of areas = (ratio of corresponding sides)². 81/144 = (6/x)² Taking square roots of both sides: 9/12 = 6/x, i.e. 3/4 = 6/x. Cross-multiplying, 3x = 24, so x = 8. The corresponding side of the larger triangle is 8 cm. Check: (6/8)² = 36/64 = 9/16 = 81/144 ✔
The diagonals of a trapezium ABCD, in which AB ∥ DC, intersect at O. If AO = 3 cm, OC = 6 cm and BO = 4 cm, find OD.
Answer
In triangles AOB and COD, ∠AOB = ∠COD (vertically opposite angles) and ∠OAB = ∠OCD (alternate angles, since AB ∥ DC). So ΔAOB ~ ΔCOD by the AA criterion, and therefore AO/CO = BO/DO. 3/6 = 4/OD Cross-multiplying, 3 × OD = 24, so OD = 8 cm.
State the Basic Proportionality Theorem and prove it. Use the theorem to find x if, in ΔABC, D and E lie on AB and AC with DE ∥ BC, AD = x cm, DB = (x - 2) cm, AE = (x + 2) cm and EC = (x - 1) cm.
Answer
Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then it divides those two sides in the same ratio. Proof: In ΔABC let DE ∥ BC, with D on AB and E on AC. Join BE and CD, and draw EM ⊥ AB and DN ⊥ AC. ar(ΔADE) = (1/2) × AD × EM and ar(ΔBDE) = (1/2) × DB × EM, so ar(ΔADE)/ar(ΔBDE) = AD/DB. ...(i) Similarly, using DN as the common height, ar(ΔADE)/ar(ΔCDE) = AE/EC. ...(ii) Now ΔBDE and ΔCDE stand on the same base DE and lie between the same parallels DE and BC, so they have equal areas: ar(ΔBDE) = ar(ΔCDE). ...(iii) From (i), (ii) and (iii), AD/DB = AE/EC. Hence proved. Application: since DE ∥ BC, AD/DB = AE/EC. x/(x - 2) = (x + 2)/(x - 1) Cross-multiplying: x(x - 1) = (x - 2)(x + 2) x² - x = x² - 4 -x = -4, so x = 4. Then AD = 4 cm, DB = 2 cm, AE = 6 cm and EC = 3 cm, all positive lengths, and AD/DB = 4/2 = 2 = 6/3 = AE/EC. ✔
In ΔABC, right-angled at B, BD is drawn perpendicular to the hypotenuse AC. If AB = 9 cm and BC = 12 cm, find AC, BD, AD and DC, justifying the similarity you use.
Answer
Step 1 — Find AC by Pythagoras Theorem. AC² = AB² + BC² = 9² + 12² = 81 + 144 = 225, so AC = 15 cm. Step 2 — Establish similarity. In ΔADB and ΔABC, ∠ADB = ∠ABC = 90° and ∠A is common, so ΔADB ~ ΔABC by AA. Likewise ΔBDC ~ ΔABC by AA, since ∠BDC = ∠ABC = 90° and ∠C is common. Step 3 — From ΔADB ~ ΔABC: AD/AB = AB/AC, so AD = AB²/AC = 81/15 = 5.4 cm. Step 4 — From ΔBDC ~ ΔABC: DC/BC = BC/AC, so DC = BC²/AC = 144/15 = 9.6 cm. Check: AD + DC = 5.4 + 9.6 = 15 = AC ✔ Step 5 — Find BD using the area of ΔABC in two ways. (1/2) × AB × BC = (1/2) × AC × BD, so 9 × 12 = 15 × BD. BD = 108/15 = 7.2 cm. So AC = 15 cm, BD = 7.2 cm, AD = 5.4 cm and DC = 9.6 cm.
Read the following and answer the questions that follow: On a sunny afternoon, a student who is 1.6 m tall stands upright on level ground and casts a shadow 2 m long. At the very same moment, a nearby communication tower casts a shadow 25 m long on the same ground. The sun's rays reaching the two objects may be treated as parallel. (a) Explain why the triangle formed by the student and her shadow is similar to the triangle formed by the tower and its shadow. (b) Find the height of the tower. (c) Later in the day the tower's shadow shortens to 15 m. How long is the student's shadow at that moment? (d) Find the ratio of the areas of the two triangles in part (b).
Answer
(a) Both the student and the tower stand vertically on level ground, so each makes a 90° angle with the ground. The sun's rays are parallel, so the angle of elevation of the sun is the same at both places, giving a second pair of equal angles. By the AA criterion the two triangles are similar. (b) Let the height of the tower be h metres. Corresponding sides of similar triangles are proportional: 1.6/2 = h/25 So h = 25 × (1.6/2) = 25 × 0.8 = 20 metres. (c) Let the student's shadow be s metres. Using the same proportionality at the new instant: 1.6/s = 20/15 20s = 1.6 × 15 = 24, so s = 24/20 = 1.2 metres. (d) The ratio of corresponding sides is 1.6/20 = 2/25 (equivalently 2/25 from the shadows). Ratio of areas = (2/25)² = 4/625.
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