Chapter 11 · Kerala SSLC Class 10 Maths
Geometry and Algebra
Two ways to learn — a crisp formula reference, or a story that makes the geometry feel obvious.
Three scenes that reveal the maths naturally.
How this works
Each scene is a real-world situation that secretly uses the maths from Chapter 11. Read the story, then open "The Maths" box to see the formula behind it.
The Map That Lied
Arjun is planning a school trip. He opens a map and measures the distance between two towns — it reads 5 cm on paper. The map says the scale is 1 cm = 20 km. Easy enough.
So the real distance is 100 km. But wait — if I zoom in and print this on a bigger sheet, does the real distance change?
Of course not. The real towns don't move. What changes is the drawing — the picture is scaled up, but all proportions stay the same. This is exactly what similar triangles do.
Two triangles are similar when one is a scaled version of the other. Every angle is identical. Every side is longer (or shorter) by the same scale factor. The shape is preserved; only the size changes.
The Shadow Trick
Diya wants to find the height of a tall tree without climbing it. She drives a stick 1 m tall into the ground and measures its shadow: 1.5 m. Then she measures the tree's shadow: 9 m.
Same sun, same time of day. The shadows form the same angle with the ground. So the triangles — stick-and-shadow vs tree-and-shadow — must be similar.
stick height / tree height = stick shadow / tree shadow
1 / h = 1.5 / 9
h = 9 / 1.5 = 6 m
The tree is 6 m tall. But there's a deeper idea hidden here. Draw a line across the two triangles at the height of the stick. That horizontal line is parallel to the ground (parallel to the base of the big triangle). And when a line parallel to the base cuts the two other sides, it divides them in equal ratios. That is BPT.
Dividing the Field Fairly
A farmer owns a triangular plot ABC. He wants to divide it into two parts by building a straight fence from vertex A to a point D on BC. He asks: "Where exactly should D be so that the fence bisects angle A — splitting the field into two triangles with the same top angle?"
I want AD to cut angle A exactly in half. Where does D land on BC?
The Angle Bisector Theorem answers this precisely. If AD bisects ∠A, then D divides BC in the ratio of the two sides adjacent to angle A — that is AB : AC.
AB = 12 m, AC = 8 m, BC = 10 m
BD/DC = AB/AC = 12/8 = 3/2
BD = 10 × 3/5 = 6 m
DC = 10 × 2/5 = 4 m
D is 6 m from B and 4 m from C. The fence that bisects angle A automatically lands at this exact point — no guessing needed.
The Big Picture
Chapter 11 has three core tools. Each one handles a different situation:
Line parallel to base cuts two sides?
BPT: AD/DB = AE/EC
Triangles are scaled copies?
Similarity: AB/XY = BC/YZ = CA/ZX
Areas of two similar triangles?
Area ratio = (side ratio)²
Angle bisector hits opposite side?
Angle bisector: BD/DC = AB/AC
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