NCERT Solutions for Class 9 Maths Chapter 12 — Exercise Set 12.1#
Definition (from the book): for four distinct points $A, B, C, D$ in a plane, with no three collinear, the points on segments $AB$, $BC$, $CD$, $DA$ form quadrilateral ABCD if every such point other than $A, B, C, D$ lies on exactly one of these segments. A quadrilateral is convex when all its internal angles are less than 180° — equivalently, when its diagonals intersect.
1. Let ABCD be a quadrilateral. #
(i) List all sides of ABCD adjacent to side AB. List all sides opposite to AB.
Sides sharing an endpoint with $AB$ are adjacent: $BC$ and $DA$.
The side with no common endpoint with $AB$ is opposite: $CD$.
(ii) List all angles of ABCD adjacent to ∠A. List all angles opposite to ∠A.
Adjacent angles are at the ends of a common side: $\angle B$ (side $AB$) and $\angle D$ (side $DA$).
Opposite angle: $\angle C$.
(iii) Define a pair of opposite sides and a pair of opposite angles without using the names of the vertices.
- Opposite sides: two sides of a quadrilateral that have no vertex in common.
- Opposite angles: the angles at two vertices that are not joined by a side (that is, the two ends of a diagonal).
*2. You have used internal angles of quadrilaterals, but they too require an exact definition, just like how we gave one for a quadrilateral. Precisely define the internal angle of a quadrilateral at a given vertex. Your answer should work for a non-convex quadrilateral too. (Hint: use the opposite vertex as well.) #
Definition
At vertex $A$ of quadrilateral $ABCD$, the two sides $AB$ and $AD$ form an angle $\angle BAD$ less than 180°. The internal angle at A is:
- $\angle BAD$, if $A$ does not lie inside the triangle $BCD$ formed by the other three vertices;
- $360° - \angle BAD$ (a reflex angle), if $A$ lies inside triangle $BCD$.
Why this works: if $A$ is inside triangle $BCD$, the quadrilateral is that triangle with a “bite” $\triangle BAD$ taken out at $A$, so the quadrilateral wraps around $A$ on the outside of $\angle BAD$. This is exactly the dent in DART: the vertex $D$ lies inside triangle $ART$, and its internal angle is more than 180°. At every other vertex, the quadrilateral lies inside the ordinary angle.
3. In a quadrilateral ABCD, suppose AB ∥ DC. Can ABCD be non-convex? What if we instead assume AB = CD? What if we instead assume ∠A = ∠C? #
AB ∥ DC
No, it must be convex. A non-convex quadrilateral has a vertex inside the triangle of the other three (Q2). But $D$ lies on the line through $C$ parallel to $AB$, and this line touches triangle $ABC$ only at $C$ — so $D$ cannot be inside triangle $ABC$. In the same way, $C$ cannot be inside triangle $ABD$, and neither $A$ nor $B$ can be inside the triangle of the other three (each lies on a line parallel to $DC$ through a vertex). So every internal angle is less than 180°.
AB = CD
Yes, it can be non-convex. Example: $A(0, 0)$, $B(6, 0)$, $C(3, 8)$, $D(3, 2)$. Then
$$AB = 6, \qquad CD = 8 - 2 = 6$$but $D$ lies inside triangle $ABC$, so the internal angle at $D$ is reflex.
∠A = ∠C
Yes, it can be non-convex, with the reflex angle at $B$ or $D$. Example: an arrowhead symmetric about $BD$: $A(-2, 0)$, $B(0, 3)$, $C(2, 0)$, $D(0, 1)$. By symmetry $\angle A = \angle C$, and $D$ is inside triangle $ABC$, so the angle at $D$ is reflex.
(The reflex angle cannot be at $A$ or $C$: then both would be more than 180°, and the angles would add up to more than 360°.)
4. Consider three non-collinear points A, B, C and draw the lines AB, BC, CA. For every possible location of point D in the plane outside these lines, decide if ABCD is self-intersecting, non-convex, or convex. (Hint: the three lines divide the plane into 7 regions.) #
Solution
The three lines divide the plane into: the inside of triangle $ABC$; three edge regions, each across one side; and three corner regions, each beyond one vertex (between the two lines through it). The quadrilateral $ABCD$ uses sides $AB$, $BC$, $CD$, $DA$.
| Where D is | ABCD is | Reason |
|---|---|---|
| Inside triangle $ABC$ | non-convex | $D$ is inside the triangle of the other three, so its angle is reflex |
| Across side $AC$ | convex | $D$ and $B$ are on opposite sides of $AC$, and the diagonals $AC$ and $BD$ cross |
| Across side $AB$ | self-intersecting | side $CD$ crosses side $AB$ |
| Across side $BC$ | self-intersecting | side $DA$ crosses side $BC$ |
| Corner region beyond $A$ | non-convex | $A$ lies inside triangle $BCD$ |
| Corner region beyond $B$ | non-convex | $B$ lies inside triangle $ACD$ |
| Corner region beyond $C$ | non-convex | $C$ lies inside triangle $ABD$ |
So: convex in 1 region, self-intersecting in 2 regions, non-convex in 4 regions.
5. Can a quadrilateral be both self-intersecting and non-planar? #
Solution
No. If $ABCD$ is self-intersecting, two opposite sides cross, say $AB$ and $CD$ at a point $E$. Two lines that meet lie in one plane, so the lines $AB$ and $CD$ — and with them all four points $A, B, C, D$ — lie in a single plane. So the quadrilateral is planar.