HOTS Questions on Chapter 6 Class 10 Triangles

HOTS Questions on Class 10 Maths Chapter 6 Triangles help students understand similarity, ratios, and triangle properties in a simple way. They improve knowledge of the basic proportionality theorem, Pythagoras theorem, and areas of similar triangles. This chapter builds a strong base for solving geometry problems and improving math skills. A downloadable PDF is also available for easy revision and practice.

HOTS Questions Based on Similarity of Triangles

Q1: In triangle ABC and triangle DEF, angle A = angle D and AB/DE = AC/DF. Are the triangles similar? Name the criterion.

Answer: Yes. Two sides are in the same ratio and the included angle between those two sides is equal (angle A = angle D). This satisfies the SAS similarity criterion. Therefore triangle ABC is similar to triangle DEF.

Q2: Two triangles have sides in the ratio 3:5. What is the ratio of their perimeters and the ratio of their areas?

Answer: Since the triangles are similar with corresponding sides in ratio 3:5:

Ratio of perimeters = 3 : 5 (perimeter scales linearly with sides).

Ratio of areas = 3² : 5² = 9 : 25 (area scales as the square of the linear scale factor).

Q3: In triangle PQR, a line XY is drawn parallel to QR such that XY divides PQ in ratio 2:3. Find the ratio PX:XQ and also the ratio of triangle PXY to triangle PQR.

Answer: PX:XQ = 2:3 (given by the parallel line division).

Since XY is parallel to QR, triangle PXY is similar to triangle PQR (AA criterion — angle P is common, and corresponding angles are equal due to parallel lines).

Scale factor = PX/PQ = 2/(2+3) = 2/5.

Ratio of areas = (2/5)² = 4/25.

So area of triangle PXY : area of triangle PQR = 4 : 25.

Q4: In similar triangles ABC and DEF, the ratio AB:DE = 4:7. If the area of triangle ABC is 48 cm², find the area of triangle DEF.

Answer: Ratio of areas = (AB/DE)² = (4/7)² = 16/49.

48/Area(DEF) = 16/49

Area(DEF) = 48 × 49/16 = 3 × 49 = 147 cm²

Q5: The corresponding medians of two similar triangles are in ratio 5:8. What is the ratio of their areas?

Answer: In similar triangles, medians are in the same ratio as corresponding sides.

So the ratio of corresponding sides = 5:8.

Ratio of areas = 5² : 8² = 25 : 64.

Q6: In triangle ABC, D is a point on BC such that angle ADB = angle ABC. Prove that triangle ADB is similar to triangle ABC.

A: In triangle ADB and triangle ABC:

angle ABD = angle ABC (common angle at B).

angle ADB = angle ABC (given).

angle ADB = angle BAC (both given equal to angle ABC means angle ADB = angle BAC... let us re-read).

Re-stating: angle ADB = angle BAC (given).

In triangle ADB and triangle ABC: angle ABD = angle ABC, common angle B. angle ADB = angle BAC given.

Therefore by AA criterion, triangle ADB is similar to triangle ABC.

HOTS Questions Based on Basic Proportionality Theorem

Q1: In triangle ABC, DE is parallel to BC. If AD/DB = 3/5 and AE = 6 cm, find EC.

Answer: By BPT, since DE ∥ BC:

AD/DB = AE/EC

3/5 = 6/EC

EC = 6 × 5/3 = 10 cm

Q2: In triangle PQR, a line parallel to QR meets PQ at S and PR at T. If PS = 4 cm, SQ = 6 cm, and PT = 3 cm, find TR.

Answer:: By BPT: PS/SQ = PT/TR

4/6 = 3/TR

TR = 3 × 6/4 = 4.5 cm

Q3: In triangle ABC, D and E are points on AB and AC respectively. AD = 4, DB = 6, AE = 5, EC = 7.5. Is DE parallel to BC? Justify.

Answer: Check if AD/DB = AE/EC:

AD/DB = 4/6 = 2/3

AE/EC = 5/7.5 = 2/3

Since AD/DB = AE/EC = 2/3, by the converse of BPT, DE ∥ BC.

Q4: Two lines l and m are parallel to each other. A transversal cuts them and also cuts a triangle. Using BPT, explain why the intercepted segments on the two sides of the triangle are always proportional.

Answer: The parallel line inside the triangle acts like a line parallel to the third side. By BPT, any line drawn parallel to one side of a triangle divides the other two sides proportionally. Since l ∥ m and both cut the two sides of the triangle, the ratios of the intercepted segments on both sides are equal. This is a direct application of BPT, which states that the ratio of the segments on one side equals the ratio of the segments on the other side.

Q5: Prove that if a line divides two sides of a triangle in equal ratio, it must be parallel to the third side.

Answer: This is the Converse of BPT.

Given: In triangle ABC, D is on AB and E is on AC such that AD/DB = AE/EC.

To prove: DE ∥ BC.

Proof: Assume DE is not parallel to BC. Then draw DE' parallel to BC where E' is on AC.

By BPT (since DE' ∥ BC): AD/DB = AE'/E'C.

But we are given: AD/DB = AE/EC.

Therefore: AE/EC = AE'/E'C.

Adding 1 to both sides: (AE + EC)/EC = (AE' + E'C)/E'C

AC/EC = AC/E'C

EC = E'C.

So E and E' are the same point, meaning DE ∥ BC.

HOTS Questions Based on Proving Two Triangles Similar

Q1: In triangles ABC and PQR: angle A = 50°, angle B = 70°, angle Q = 70°, angle R = 60°. Are the triangles similar? Which criterion?

Answer: In triangle ABC: angle C = 180 − 50 − 70 = 60°.

Comparing: angle B = angle Q = 70° and angle C = angle R = 60°.

By AA criterion, triangle ABC is similar to triangle PQR (with B↔Q and C↔R).

Q2: Two triangles have sides: Triangle 1: 4, 6, 8. Triangle 2: 6, 9, 12. Are they similar?

Answer: Check if all ratios are equal: 4/6 = 2/3, 6/9 = 2/3, 8/12 = 2/3.

All three ratios are equal (2/3). By SSS similarity criterion, the triangles are similar.

Q3: Triangle ABC is similar to triangle DEF. AB = 6 cm, BC = 9 cm, AC = 12 cm, and DE = 4 cm. Find EF and DF.

Answer: Scale factor = DE/AB = 4/6 = 2/3.

EF = BC × 2/3 = 9 × 2/3 = 6 cm

DF = AC × 2/3 = 12 × 2/3 = 8 cm

Q4: In triangle ABC, angle A = 90°. A perpendicular AD is drawn from A to BC. Prove that triangle ABD is similar to triangle CAD.

Answer: In triangle ABD and triangle CAD:

angle ADB = angle CDA = 90° (AD is perpendicular to BC).

angle ABD = angle CAD (both equal to 90° − angle ACD, since in triangle ACD: angle CAD + angle ACD = 90°, and in triangle ABD: angle ABD + angle BAD = 90°, and angle BAD + angle CAD = 90°, so angle ABD = angle CAD).

By AA criterion, triangle ABD is similar to triangle CAD.

Q5: In a right triangle ABC (right angle at C), if CD is the altitude to the hypotenuse AB, prove that CD² = AD × DB.

Answer: From Q4, triangle ACD is similar to triangle CBD (both share angle D = 90°, and angle ACD = angle DBC).

So CD/DB = AD/CD.

Cross multiplying: CD² = AD × DB

This is the geometric mean relation, one of the most important results from altitude-on-hypotenuse problems.

HOTS Questions Based on Areas of Similar Triangles

Q1: The areas of two similar triangles are 25 cm² and 64 cm². If one side of the first triangle is 5 cm, find the corresponding side of the second triangle.

Answer: Ratio of areas = 25/64.

Ratio of sides = √(25/64) = 5/8.

If one side of the first triangle = 5 cm: 5/x = 5/8

x = 8 cm.

Q2: Triangle ABC is similar to triangle DEF. If AB:DE = 3:4, find the ratio of the area of triangle ABC to the area of triangle DEF.

Answer: Ratio of areas = (AB/DE)² = (3/4)² = 9 : 16.

Q3: Two similar triangles have areas 36 cm² and 81 cm². The perimeter of the smaller triangle is 24 cm. Find the perimeter of the larger triangle.

Answer: Ratio of areas = 36/81 = 4/9.

Ratio of sides = √(4/9) = 2/3.

Ratio of perimeters = 2/3 (perimeter scales with sides).

24/Perimeter₂ = 2/3

Perimeter₂ = 24 × 3/2 = 36 cm.

Q4: A triangle has sides 6, 8, 10 cm. Another similar triangle has its longest side as 15 cm. What is the ratio of their areas?

Answer: Longest side of first triangle = 10 cm. Longest side of second = 15 cm.

Scale factor = 15/10 = 3/2.

Ratio of areas = (3/2)² = 9/4 or 9 : 4.

Q5: The diagonal of a square is 6√2 cm. Another similar square (all squares are similar) has a diagonal of 9√2 cm. Compare their areas.

Answer: For squares, since all squares are similar, sides are proportional to diagonals.

Ratio of diagonals = 6√2 / 9√2 = 6/9 = 2/3.

Ratio of areas = (2/3)² = 4/9.

The first square's area is 4/9 of the second square's area.

HOTS Questions Based on Real-Life Applications of Triangles

Q1: A tree casts a shadow of 15 m when a vertical pole of 2 m casts a shadow of 3 m. Find the height of the tree.

Answer: The tree and pole are vertical. Their shadows are horizontal. The sun's rays are parallel, making the angles of elevation equal.

Triangle formed by tree and its shadow is similar to the triangle formed by pole and its shadow (AA criterion).

Height of tree / Shadow of tree = Height of pole / Shadow of pole

h / 15 = 2 / 3

h = 15 × 2/3 = 10 metres

Q2: A person 1.6 m tall stands 8 m from a lamp post. His shadow cast by the lamp is 2 m long. Find the height of the lamp post.

Answer: Let the height of the lamp = H.

The tip of the shadow, the tip of the lamp, and the tip of the person's head all form a triangle.

Using similar triangles: H/(8+2) = 1.6/2

H/10 = 0.8

H = 8 metres

Q3: A flagpole and a tree stand vertically. The flagpole is 10 m tall and casts a shadow of 8 m. If the tree's shadow at the same time is 12 m, how tall is the tree?

Answer: Similar triangles (same sun angle):

Height of tree / 12 = 10 / 8

Height of tree = 12 × 10/8 = 15 metres

Q4: In a photograph, the length of a room appears as 6 cm. The scale of the photograph is 1:50. Find the actual length of the room.

Answer: Actual length = 6 × 50 = 300 cm = 3 metres.

The photograph creates similar figures corresponding lengths are in the same ratio as the scale, which is the direct application of similarity.

Q5: A ramp is designed so that for every 12 m of horizontal distance, the height rises 5 m. A similar ramp is designed with a height of 7.5 m. Find the horizontal distance of the new ramp.

Answer: The two ramps form similar right triangles.

Horizontal/Height: First ramp = 12/5. Second ramp = x/7.5.

Since the ramps are similar: x/7.5 = 12/5

x = 7.5 × 12/5 = 18 metres

Download PDF - HOTS Questions on Chapter 6 Class 10 Triangles pdf

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