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Eclipse geometry for CBSE Class 10

A ready-to-teach lesson on the geometry of eclipses: why the Moon can only just cover the Sun. A 5-minute starter, a 35-minute main activity on similar triangles, heights and distances and a pinhole viewer, an extension and full worked answers. No calculator.

Level
CBSE Class 10
Time
40 minutes, plus a 10-minute extension
Topics
Triangles: similarity; Some applications of trigonometry: heights and distances; Exponents and standard form
Equipment
No calculator. Every answer works out exactly.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A12 minWhy the Moon just covers the Sun
Main: task B12 minShadows and angles
Main: task C11 minA pinhole viewer
Extension10 minFast finishers or homework

Starter (5 minutes)

Quick-fire mental maths.

  1. Write 1 400 000 in standard form.
  2. Simplify 3500/1 400 000.
  3. A 1.8 m person casts a 2.4 m shadow. At the same time a tree casts a 12 m shadow. How tall is the tree?
  4. Write down tan 60°.

Main activity (35 minutes)

Task A: Why the Moon just covers the Sun (12 min)

Take these rounded values: Sun diameter 1 400 000 km at 150 000 000 km from us; Moon diameter 3500 km at 380 000 km.

  1. How many times wider is the Sun than the Moon?
  2. Use similar triangles to find the distance at which the Moon would exactly cover the Sun.
  3. When the Moon is 360 000 km away, what is the smallest diameter a disc there would need to cover the Sun? Can the 3500 km Moon cover it?

Task B: Shadows and angles (12 min)

Use the standard values tan 30° = 1/√3 and tan 60° = √3.

  1. A pole is 6√3 m tall. Find the length of its shadow when the Sun’s angle of elevation is 60°.
  2. Find the length of the shadow when the angle of elevation is 30°.
  3. By how much does the shadow grow as the angle of elevation falls from 60° to 30°?

Task C: A pinhole viewer (11 min)

Never look at the Sun. A pinhole in a card makes a safe image on a screen behind it; the image and the Sun make similar triangles with the pinhole.

  1. The screen is 1.5 m behind the pinhole. Find the width of the Sun’s image in millimetres.
  2. How far behind the pinhole must the screen be for an image 28 mm wide?
  3. A 1.5 m tall girl stands 4 m from a 6 m tall lamp post. Find the length of her shadow.

Extension (10 minutes)

For fast finishers, or as homework.

  1. Light travels about 3 × 105 km each second. Using a Sun distance of 1.5 × 108 km, how long does sunlight take to reach us? Give minutes and seconds.

For teachers

Teacher notes and full worked answers
  • Safety first: never look at the Sun directly, even during an eclipse. A pinhole viewer (task C) is a safe way to see it.
  • Every physical value is a rounded value given in the question; the real values vary.
  • Task B uses the standard angles from the Heights and distances chapter.

Starter

  1. 1.4 × 106
    • 1 400 000 = 1.4 × 1 000 000
  2. 1/400
    • 1 400 000 ÷ 3500 = 400
  3. 9 m
    • 12 × 1.8 ÷ 2.4 = 9
  4. √3
    • Standard value.

Task A: Why the Moon just covers the Sun

  1. 400
    • 1 400 000 ÷ 3500 = 400
  2. 375 000 km
    • d/3500 = 150 000 000/1 400 000
    • d = 3500 × 150 000 000/1 400 000 = 375 000
  3. 3360 km; yes
    • Diameter = 1 400 000 × 360 000/150 000 000 = 3360 km.
    • 3500 > 3360, so the Moon covers the Sun completely: a total eclipse.

Task B: Shadows and angles

  1. 6 m
    • tan 60° = 6√3/s, so s = 6√3/√3 = 6.
  2. 18 m
    • s = 6√3 ÷ (1/√3) = 6 × 3 = 18
  3. 12 m
    • 18 − 6 = 12

Task C: A pinhole viewer

  1. 14 mm
    • Image/1500 mm = 1 400 000/150 000 000 = 7/750
    • Image = 1500 × 7/750 = 14 mm
  2. 3 m
    • Twice the width needs twice the distance: 3 m.
  3. 4/3 m
    • Similar triangles: s/1.5 = (s + 4)/6.
    • 6s = 1.5s + 6, so 4.5s = 6 and s = 4/3.

Extension

  1. 500 s = 8 minutes 20 seconds
    • (1.5 × 108) ÷ (3 × 105) = 0.5 × 103 = 500 s
    • 500 = 8 × 60 + 20

The same answers are in the answers PDF. Every answer was recomputed by computer before publishing.

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