Exercise 8.1 — Congruency and Similarity

Congruent objects are identical in shape and size — one can be placed exactly on top of the other. Look around your home and classroom for pairs of objects made from the same mould or manufactured identically.

Part (a) — Are congruent figures also similar?

Yes — congruent figures are always similar. When two figures are congruent, their corresponding sides are equal, which means the ratio of every pair of corresponding sides is 1 : 1. Since all ratios are equal (= 1) and all corresponding angles are equal, the conditions for similarity are fully satisfied. Congruence is therefore a special case of similarity where the scale factor equals 1.

Part (b) — Does similarity survive slide, rotation, or flip?

Yes — similarity is preserved under all three transformations. Consider △ABC ~ △PQR with a ratio of corresponding sides 5 : 7. If you rotate △PQR through any angle, the side lengths of △PQR do not change — so the ratio 5 : 7 remains the same and the triangles are still similar. The same reasoning applies to sliding (translation) and flipping (reflection): none of these transformations alter side lengths or angle measures, so similarity is always preserved.

When we write △ABC ≅ △NMO, the order of vertices tells us exactly which parts correspond to each other: A↔N, B↔M, C↔O.

This activity reinforces the idea that a congruent figure can appear in any orientation — upright, tilted, or flipped — and still be congruent to the original. Below is a pentagon drawn on a dot grid, shown in its original position, slid (translated), rotated, and flipped.

Each copy in the diagram above has the same side lengths and angles as the original, confirming congruence. The position and direction change, but shape and size do not.

Draw rectangle ABCD with length 3 units and breadth 2 units. Then draw a similar rectangle EFGH with length 9 units and breadth 6 units (scale factor = 3).

Calculations and Ratio Comparison

This problem uses the property of similar triangles: when a slant line (the girder) crosses a series of equally-spaced vertical lines (the pillars), each pillar and the horizontal distance from O form a triangle similar to the largest triangle OMN. Because the horizontal distances increase uniformly, the pillar heights also increase proportionally.

Solution Using Similar Triangles

Since equal intervals are marked along the base (OA = AC = CE = EG = GI = IK = KM = 1 m), the total base OM = 7 m. Each small triangle formed (e.g. △OAB) is similar to the largest triangle △OMN. Using the property of similar triangles:

This is a classic indirect measurement problem that uses the properties of similar triangles. When two objects are aligned from a single viewpoint, the observer, the nearer object, and the farther object form two similar triangles.

Solution

Let Sudha's position be A, the pole base B, pole top C, building base M, building top N.

• AB = 5 m (Sudha to pole), BC = 3 m (pole height)
• BM = 10 m (pole to building), so AM = AB + BM = 5 + 10 = 15 m
• MN = ? (building height)

Since △ABC ~ △AMN (AA similarity — common angle at A, both have 90° at B and M):

Draw rectangle ABCD with sides AB = DC = 3 units and AD = BC = 2 units. Using a centre of dilation O at the origin, draw rays from O through each vertex and mark each new vertex at 3 times the distance from O. The result is rectangle A'B'C'D' with sides 9 units × 6 units.

Verifying Similarity through Side Ratios