Model Test Paper – Solved ഉത്തരങ്ങൾ
Every question of the model paper, with a full solution, an animated figure where a figure or construction is involved, and a marking hint so you can score your own paper honestly.
📖 How marks are usually given
1-mark questions: the correct answer only. 2- and 3-mark questions: about one mark for identifying the similar triangles or writing the correct ratio, the rest for the calculation and the answer with units. 5-mark questions: marks for each part as shown. A correct method with an arithmetic slip usually loses only one mark.
SECTION A8 × 1 = 8 marks
1.
Two triangles have the same angles 40°, 60° and 80°. The side opposite the 80° angle is 6 cm in the first triangle and 9 cm in the second. The scale factor from the first triangle to the second is ______.
(1)Answer: 1½ (= 1.5 = 3/2)
Scale factor = new length ÷ old length = 9 ÷ 6 = 1½. (Every side of the second triangle is 1½ times its partner.)Review: Page 5 · Scale factor
2.
In ΔPQR, ∠P = 70°, ∠Q = 60° and ∠R = 50°. The longest side of the triangle is (a) PQ (b) QR (c) PR (d) all sides are equal.
(1)Answer: (b) QR
The longest side faces the largest angle. ∠P = 70° is the largest, and the side that does not touch P is QR.Review: Page 1 · Page 5, section 2
3.
If every side of a triangle is multiplied by 3, its area is multiplied by ______.
(1)Answer: 9
Area changes by the square of the scale factor: 3² = 9 (Key idea 4).Review: Page 7 · Area
4.
Which pair of triangles need not be similar? (a) two equilateral triangles (b) two right triangles each having an angle of 30° (c) two isosceles triangles (d) a triangle with sides 3, 4, 5 cm and one with sides 6, 8, 10 cm.
(1)Answer: (c) two isosceles triangles
(a) both have angles 60°, 60°, 60° – similar. (b) both have angles 30°, 60°, 90° – similar. (d) all sides in the ratio 1 : 2 – similar. (c) An isosceles triangle with angles 20°, 80°, 80° and one with angles 50°, 50°, 80° are both isosceles but have different angles – not similar.Review: Page 9 · The three conditions
5.
The perpendicular from the right-angle vertex of a right triangle divides the hypotenuse into parts of 4 cm and 9 cm. The length of the perpendicular is ______ cm.
(1)Answer: 6 cm
h² = ab = 4 × 9 = 36 ⇒ h = 6 cm. (The full reasoning is in question 18.)Review: Page 6 · Exercise 1, Q2 & Q5
6.
In the figure, DE is parallel to BC. AD = 2 cm, DB = 4 cm and AE = 3 cm. Then EC = ______ cm.
(1)🎬 Solution
Answer: 6 cm
A line parallel to one side of a triangle divides the other two sides in the same ratio: AD/DB = AE/EC ⇒ 2/4 = 3/EC ⇒ EC = 6 cm.Review: Page 3 · Dividing a triangle
7.
Fill in the blank: "Triangles with the same angles have their sides, taken in order of size, in the same ______."
(1)Answer: ratio (അംശബന്ധം)
This is Key idea 1 from page 102 of the textbook.Review: Page 5
8.
In ΔABC and ΔPQR, ∠A = ∠P and ∠B = ∠Q. If AB = 5 cm, PQ = 10 cm and BC = 7 cm, then QR = (a) 3.5 cm (b) 7 cm (c) 14 cm (d) 12 cm.
(1)Answer: (c) 14 cm
Two equal angles ⇒ the third is equal too ⇒ similar. AB faces ∠C and PQ faces ∠R (equal angles), so the scale factor is PQ ÷ AB = 10 ÷ 5 = 2. BC faces ∠A and QR faces ∠P ⇒ QR = 2 × 7 = 14 cm.Review: Page 5 · Worked example B
SECTION B5 × 2 = 10 marks
9.
The sides of ΔABC are AB = 6 cm, BC = 9 cm and CA = 12 cm. ΔPQR has the same angles as ΔABC, and its shortest side is 4 cm. Find the other two sides of ΔPQR.
(2)Answer: 6 cm and 8 cm
Same angles ⇒ sides in order of size are in the same ratio. Shortest ↔ shortest: 4 ÷ 6 = ⅔ is the scale factor.Middle side: ⅔ × 9 = 6 cm. Longest side: ⅔ × 12 = 8 cm.
Marks: 1 for the scale factor ⅔ (or the correct pairing of sides), 1 for both answers.
Review: Page 5
10.
The sides of one triangle are 5 cm, 7 cm and 9 cm; the sides of another are 7.5 cm, 10.5 cm and 13.5 cm. Are the two triangles similar? Give the reason.
(2)Answer: Yes
7.5 ÷ 5 = 1.5, 10.5 ÷ 7 = 1.5, 13.5 ÷ 9 = 1.5. All three sides change by the same scale factor 1½, so by Key idea 3 (വശങ്ങളും കോണുകളും) the triangles have the same angles – they are similar.Marks: 1 for computing all three ratios, 1 for the conclusion with the reason (all sides in the same ratio ⇒ same angles).
Review: Page 7
11.
Two poles of heights 6 m and 3 m stand vertically on level ground. A rope is tied from the top of each pole to the foot of the other. At what height above the ground do the two ropes cross?
(2)🎬 Solution
Answer: 2 m
Let the ropes cross at X, at height h, with M the point directly below X. ΔXMB and ΔPAB have a right angle and the common angle at B ⇒ h/6 = MB/AB. ΔXMA and ΔQBA give h/3 = AM/AB. Adding: h/6 + h/3 = (MB + AM)/AB = 1 ⇒ h/2 = 1 ⇒ h = 2 m. (Formula: h = ab/(a + b) = 18/9 = 2.) The distance between the poles cancels out – it does not matter.Marks: 1 for the two pairs of similar triangles (or the formula 1/h = 1/a + 1/b), 1 for h = 2 m.
Review: Page 6 · Exercise 1, Q6
12.
In ΔABC, AB = 8 cm, AC = 6 cm and BC = 7 cm. The bisector of ∠A meets BC at P. Find BP and PC.
(2)🎬 Solution
Answer: BP = 4 cm, PC = 3 cm
The bisector of an angle divides the opposite side in the ratio of the two sides containing the angle: BP/PC = AB/AC = 8/6 = 4/3. So BC = 7 cm is split into 4 + 3 = 7 parts of 1 cm: BP = 4 cm, PC = 3 cm.Marks: 1 for BP : PC = 8 : 6 (= 4 : 3), 1 for the two lengths.
Review: Page 6 · Exercise 1, Q7
13.
The area of a triangle is 24 sq cm.
(2)- What is the area of the triangle whose sides are half the sides of this triangle?
- What is the area if the sides are made 1½ times?
Answer: (i) 6 sq cm (ii) 54 sq cm
Area scales by the square of the side scale factor. (i) (½)² = ¼ ⇒ 24 × ¼ = 6 sq cm. (ii) (1½)² = 9/4 ⇒ 24 × 9/4 = 54 sq cm.Marks: 1 each. Writing "12" or "36" (forgetting to square) gets no mark for that part.
Review: Page 7 · Area · Page 8, Q3
SECTION C4 × 3 = 12 marks
14.
A vertical stick of height 1.5 m casts a shadow 2 m long. At the same time a coconut tree casts a shadow 16 m long.
(3)- Draw a rough figure and explain why the two triangles have the same angles.
- Find the height of the tree.
🎬 Solution
Answer: (ii) 12 m
(i) The stick and the tree are vertical, so both triangles have a right angle at the base. The Sun's rays are parallel, so the angle at the tip of each shadow is the same. Two equal angles ⇒ the third is equal ⇒ same angles.(ii) Same angles ⇒ height ÷ shadow is the same: H/16 = 1.5/2 ⇒ H = 1.5 × 16 ÷ 2 = 12 m.
Marks: 1 for the figure, 1 for the reason (right angles + parallel rays ⇒ equal angles), 1 for 12 m.
Review: Page 11 · Shadows
15.
In the figure, a square is drawn inside a right triangle with legs 6 cm and 3 cm: one vertex of the square is at the right-angle corner and the other three vertices lie on the sides. Find the length of a side of the square.
(3)🎬 Solution
Answer: 2 cm
Let the side be s. The small triangle above the square has vertical leg 6 − s and horizontal leg s. Its sides are parallel to the big triangle's sides, so it has the same angles ⇒ same ratio of vertical to horizontal: (6 − s)/s = 6/3 = 2 ⇒ 6 − s = 2s ⇒ s = 2 cm. Check with the small triangle on the right: s/(3 − s) = 2/1 = 2 ✓. (Formula: s = ab/(a + b) = 18/9 = 2.)Marks: 1 for identifying a small triangle similar to the big one, 1 for the equation (6 − s)/s = 2, 1 for s = 2 cm.
Review: Page 6 · Exercise 1, Q4
16.
Two sides of a triangle are 4 cm and 6 cm and the angle between them is 50°. In another triangle two sides are 6 cm and 9 cm and the angle between them is also 50°.
(3)- Are the triangles similar? Which rule tells you so?
- The third side of the first triangle is 4.6 cm. Find the third side of the second triangle.
🎬 Solution
Answer: (i) Yes – two sides in the same ratio and equal included angles (Key idea 5, the "third way"/SAS) (ii) 6.9 cm
(i) 6 ÷ 4 = 1.5 and 9 ÷ 6 = 1.5; the 50° angles lie between these sides ⇒ the triangles are similar with scale factor 1½.(ii) Third side = 1.5 × 4.6 = 6.9 cm.
Marks: 1 for the equal ratios, 1 for naming the rule (two sides in the same ratio + equal included angle), 1 for 6.9 cm.
Review: Page 9
17.
A rectangle has sides 4 cm and 6 cm. A parallelogram has sides 6 cm and 9 cm and angles 60° and 120°.
(3)- Are the sides of the two figures in the same ratio?
- Are the two figures similar?
- What does this show about quadrilaterals compared with triangles?
🎬 Solution
Answer: (i) Yes (ii) No (iii) Proportional sides do not guarantee similarity for quadrilaterals
(i) 6 ÷ 4 = 1.5 and 9 ÷ 6 = 1.5 – every side of the parallelogram is 1½ times a side of the rectangle.(ii) No. The rectangle's angles are all 90°, the parallelogram's are 60° and 120°. Similar figures must have the same angles, so these are not similar (one is not an enlargement of the other).
(iii) For triangles, sides in the same ratio force the angles to be equal (Key idea 3). For quadrilaterals this fails: the same four side lengths can be "pushed" into different shapes. That is why the results of this chapter are about triangles (ത്രികോണവിശേഷം).
Marks: 1 each part.
Review: Page 8 · Exercise 2, Q2 · Page 5, section 6
SECTION D2 × 5 = 10 marks
18.
In the figure, ΔABC is right-angled at C and CD is perpendicular to AB. AD = 4 cm and DB = 9 cm.
(5)- Prove that ΔADC and ΔCDB have the same angles.
- Find CD.
- Find AC and BC.
- Find the area of ΔABC.
🎬 Solution
Answer: (ii) CD = 6 cm (iii) AC = √52 ≈ 7.2 cm, BC = √117 ≈ 10.8 cm (iv) 39 sq cm
(i) Let ∠A = x°. In ΔADC, ∠ADC = 90° so ∠ACD = 90° − x°. Since ∠ACB = 90°, ∠BCD = 90° − (90° − x°) = x°. In ΔCDB, ∠CDB = 90° so ∠B = 90° − x°. Both triangles have the angles x°, 90° − x°, 90°.(ii) Sides opposite equal angles are in the same ratio. Opposite x°: CD (in ΔADC) and DB (in ΔCDB). Opposite 90° − x°: AD and CD. So CD/AD = DB/CD ⇒ CD² = AD × DB = 4 × 9 = 36 ⇒ CD = 6 cm.
(iii) Pythagoras in ΔADC: AC² = 4² + 6² = 52 ⇒ AC = √52 = 2√13 ≈ 7.2 cm. In ΔCDB: BC² = 9² + 6² = 117 ⇒ BC = √117 = 3√13 ≈ 10.8 cm. (Check: 52 + 117 = 169 = 13² = AB² ✓.)
(iv) Area = ½ × AB × CD = ½ × 13 × 6 = 39 sq cm.
Marks: (i) 1 (ii) 1 (iii) 2 (iv) 1. Leaving the answers as √52 and √117 is fine.
Review: Page 6 · Exercise 1, Q2 & Q5
19.(5)
- Describe how to draw ΔABC with AB = 6 cm, BC = 5 cm and CA = 4 cm.
- Describe how to draw a triangle whose sides are 1¼ times the sides of ΔABC by measuring its sides.
- Describe how to draw the same enlarged triangle using the circumcircle of ΔABC and a second circle, without measuring any side. Explain why the sides of the triangle obtained are 1¼ times the sides of ΔABC.
🎬 Solution – the construction step by step
Answer
(i) Draw AB = 6 cm. With centre A and radius 4 cm draw an arc; with centre B and radius 5 cm draw another arc; they meet at C. Join AC and BC.(ii) 1¼ × 6 = 7.5 cm, 1¼ × 5 = 6.25 cm, 1¼ × 4 = 5 cm. Draw the triangle with these three sides in the same way (compass). Because all sides change by the same scale factor, the angles are the same as in ΔABC (Key idea 3).
(iii) Draw the perpendicular bisectors of two sides; they meet at the circumcentre O. Draw the circumcircle (radius OA = R). With the same centre O draw a circle of radius 1¼R. Extend OA, OB, OC to meet the bigger circle at A′, B′, C′. Join A′B′C′.
Why: in ΔOAB and ΔOA′B′ the angle at O is common, and OA′ = 1¼ OA, OB′ = 1¼ OB. Two sides in the same ratio with equal included angles ⇒ (Key idea 5) the third sides are in the same ratio: A′B′ = 1¼ AB. The same argument at the other two angles at O gives B′C′ = 1¼ BC and C′A′ = 1¼ CA.
Marks: (i) 1 (ii) 1 – with the three lengths 7.5, 6.25, 5 (iii) 2 for the construction (circumcentre, second circle, extending the radii), 1 for the explanation using the common angle at O.
Review: Page 9 · Enlarging with circles · Page 8, Q1
Marks summary
| Section | Questions | Marks | What is tested |
|---|---|---|---|
| A | 1 – 8 | 8 | Scale factor, side order, area ratio, recognising similar triangles, h² = ab, parallel-line ratio |
| B | 9 – 13 | 10 | Finding sides, SSS check, ropes on poles, angle bisector, area scaling |
| C | 14 – 17 | 12 | Shadow problem, inscribed square, SAS, quadrilateral counter-example |
| D | 18 – 19 | 10 | Altitude on the hypotenuse (proof + calculation), constructions and the circumcircle enlargement |
| Total | 40 | ||
⏱️ Time management (1½ hours = 90 minutes)
- Section A: about 10 minutes – do not get stuck; move on and come back.
- Section B: about 20 minutes (4 minutes a question). Always write the ratio before the arithmetic.
- Section C: about 30 minutes. Draw the figure first; a good figure earns a mark by itself in question 14.
- Section D: about 25 minutes. Question 19 is descriptive – use numbered steps.
- Keep the last 5 minutes to check units (cm, m, sq cm) and to make sure every blank is filled.
🎉 Finished the chapter?
Scored 32+? Excellent. Between 24 and 31? Revisit the pages named under the questions you lost marks on. Below 24? Go back to page 5 and work through the three sections again with the animations – then retake the test in a week.