Chapter 6 · Exercise 1Textbook pages 105–107

Exercise 1 – Animated Solutions

Seven questions, all built on one idea from Section 1: same angles ⇒ sides opposite equal angles are in the same ratio. For each question, read it, try it in your notebook (use the hint if you are stuck), then press Next ▶ to watch the solution grow step by step.

📖 Tools you will use

Question 1 · Draw a 1½-times copy

Q1One side of a triangle is 8 cm and two of its angles are 60° and 70°. Draw a triangle with the same angles and sides 1½ times as long.

(ഒരു ത്രികോണത്തിന്റെ ഒരു വശം 8 സെ.മീ. ഉം അതിലെ രണ്ട് കോണുകൾ 60° യും 70° യും ആണ്. കോണുകൾ മാറാതെ വശങ്ങൾ ഇതിന്റെ ഒന്നരമടങ്ങായ ത്രികോണം വരയ്ക്കുക.)

You only need to scale one side and copy the two angles. Which side do you know? What is 1½ × 8?

🎬 Solution animation
Answer: Draw a 12 cm line, a 60° angle at one end and a 70° angle at the other. The rays meet at the third vertex. Because the angles are the same, every side is 1½ times the original (Key idea 2) – no measuring of the other sides is needed.
🌍 Real life

This is how a tailor enlarges a paper pattern or an architect redraws a roof truss for a bigger house: keep the angles, scale one length, and everything else follows.

Question 2 · The perpendicular to the hypotenuse

Q2In a right triangle, the perpendicular from the right-angle vertex to the hypotenuse divides the hypotenuse into pieces of 2 cm and 3 cm.
  1. Prove that the two small right triangles cut by the perpendicular have the same angles.
  2. Taking the height of the perpendicular as h, prove that h2 = 3h.
  3. Calculate the perpendicular sides (legs) of the large right triangle.
  4. Prove that if the perpendicular from the right angle to the hypotenuse has length h and it divides the hypotenuse into pieces a and b, then h² = ab.

Call one angle of the big triangle x°. Chase the angles: every right triangle has angles x°, 90° − x° and 90°. Then pair the sides opposite equal angles.

🎬 Solution animation (parts i–iv)
Answer: (i) both small triangles have angles x°, 90° − x°, 90°. (ii) h/2 = 3/h. (iii) h = √6, legs √10 ≈ 3.16 cm and √15 ≈ 3.87 cm. (iv) h/a = b/h ⇒ h² = ab.
💡 A pattern worth remembering

The perpendicular to the hypotenuse cuts a right triangle into two smaller copies of itself (same angles). This one picture gives three famous results: h² = ab, AC² = AD·AB, BC² = DB·AB – and adding the last two even proves Pythagoras!

Question 3 · Dividing a line in a given ratio

Q3At the two ends of a horizontal line, angles of the same size are drawn, one above and one below the line. A line joins a point on one slanted line to a point on the other, crossing the horizontal line.
  1. Prove that the parts of the horizontal (blue) line and the parts of the slanted (red) line are in the same ratio.
  2. Prove that the two slanted (green) pieces at the ends of the horizontal line are also in this ratio.
  3. Using this, how can a 6 cm line be divided in the ratio 3 : 4?

The crossing point makes two triangles. Look for equal angles: one pair is given, and one pair is vertically opposite.

🎬 Solution animation (parts i–iii)
Answer: ΔAMP and ΔBMQ have the same angles, so AM/MB = PM/MQ = AP/BQ. To divide a 6 cm line AB in the ratio 3 : 4, draw equal angles at A (above) and B (below), mark AP = 3 cm and BQ = 4 cm, and join PQ; it cuts AB at M with AM : MB = 3 : 4 (AM ≈ 2.57 cm).
🌍 Real life

Draftsmen and carpenters use exactly this trick to split a plank or a line into parts with a given ratio without any calculation – only a ruler and two equal angles.

Question 4 · A square inside a right triangle

Q4In the figure, a square is drawn with the right-angle corner of a right triangle as one vertex and its other vertices on the three sides. The legs are 2 cm (vertical) and 1 cm (horizontal).
  1. Calculate the side of the square.
  2. What is the side of such a square drawn in a right triangle with sides 3, 4 and 5 cm?

Call the side of the square s. The little triangle above the square has legs 2 − s and s, and its sides are parallel to the big triangle's sides…

🎬 Solution animation (parts i and ii)
Answer: (i) (2 − s)/s = 2/1 ⇒ s = 2/3 cm. (ii) (3 − s)/s = 3/4 ⇒ s = 12/7 ≈ 1.71 cm. In general s = ab/(a + b).
🎚️ Play: change the legs and watch the square
📖 Side box: a square inside a semicircle (അർധവൃത്തവും സമചതുരവും)

The textbook shows a lovely construction: draw a square inside a semicircle of diameter 8 cm without any calculation. Watch how it works and why.

🎬 Square in a semicircle – construction and proof

Question 5 · Area of the large right triangle

Q5In the figure, the perpendicular from the right angle divides the hypotenuse into pieces of 9 cm and 4 cm. Calculate the area of the large right triangle.

Use the result of Q2(iv): h² = ab. Then take the hypotenuse as the base.

🎬 Solution animation
Answer: h² = 9 × 4 = 36 ⇒ h = 6 cm. Area = ½ × 13 × 6 = 39 sq cm.

Question 6 · Ropes between two poles

Q6Two poles of heights 3 m and 2 m stand vertically on the ground. A rope is tied from the top of each pole to the foot of the other.
  1. At what height above the ground do the ropes cross?
  2. Prove that this height does not change whatever the distance between the poles.
  3. Taking the pole heights as a, b and the crossing height as h, find the relation between a, b, h.

Drop a perpendicular from the crossing point to the ground. It creates two small right triangles, each similar to one of the big right triangles made by a rope and a pole.

🎬 Solution animation (parts i–iii)
Answer: (i) 1.2 m. (ii) the distance d cancels out. (iii) h/a + h/b = 1, i.e. 1h = 1a + 1b, or h = ab/(a + b).
🎚️ Play: move the poles apart – the crossing height never changes
🌍 Real life

This is the famous crossed ladders problem. Electric cables tied between two poles, two ladders leaning across an alley, or the ropes of a stage backdrop all cross at a height that depends only on the two heights – never on the gap between them. Notice that 1/h = 1/a + 1/b is the same formula physicists use for two resistors in parallel!

Question 7 · The angle bisector

Q7In the figure, AP is the bisector of ∠A of triangle ABC. AB = 5 cm, AC = 3 cm, and Q is the point on AP extended with CQ = 3 cm.
  1. Prove that triangles ABP and CPQ have the same angles.
  2. Calculate BPPC.
  3. Prove that in any triangle, the bisector of an angle divides the opposite side in the ratio of the sides containing that angle.

CA = CQ = 3 cm makes triangle ACQ isosceles. What does that tell you about the angles at A and Q? Also look for vertically opposite angles at P.

🎬 Solution animation (parts i–iii)
Answer: (i) ∠BAP = ∠CQP (isosceles ΔACQ and the bisector) and ∠APB = ∠CPQ (vertically opposite) ⇒ same angles. (ii) BP/PC = AB/QC = 5/3. (iii) In general BP/PC = AB/AC – the Angle Bisector Theorem.
💡 Looking ahead

You will meet this theorem again in Class 10. The trick of adding a point (Q) to create an isosceles triangle is a classic move in geometry – keep it in your toolbox.

Check yourself

🧭 What Exercise 1 taught us

Next: Section 2 – Sides and Angles ▶