NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 12 Quadrilaterals Think and Reflect

Think and Reflect

Think and Reflect (p. 51). Can we use any given quadrilateral to tile the plane? If not, which quadrilaterals can be used and which cannot?
Answer:
Yes. Every quadrilateral, convex or non-convex, tiles the plane. (Only planar, non-self-intersecting quadrilaterals are meant.)
The reason, proved step by step in Section 12.4:
1. The four internal angles of any quadrilateral add to 360°. Call them 1, 2, 3, 4.
2. Turn a copy of the quadrilateral through 180° about the midpoint of any side. The copy fits exactly along that side, because the side is carried onto itself (its ends just swap).
3. Doing this at every side of every copy gives copies that meet only along complete edges, and at every vertex the angles 1, 2, 3, 4 meet once each, filling 360° with no gap or overlap.

So there is no quadrilateral that cannot be used: the plane can be tiled with copies of any given 4-gon.

Think and Reflect (p. 52). Informally, a quadrilateral is a figure with four straight sides, as in the first figure ABCD in Fig. 12.2 below. But consider the other six figures in Fig. 12.2: the five plane figures NOPE, SILY, DART, CUTS, OPENS and the non-planar BENT. Should we call all these figures quadrilaterals? If you answer 'no' for any of them, how will you define a quadrilateral so that such a figure is excluded? As you can see, some care is needed to precisely define what we think of as a quadrilateral.

Answer:
Checking each figure:
1. ABCD: four vertices, four sides meeting only at their ends. Quadrilateral.
2. NOPE: E, P, O are on one line, so EP and PO lie along one straight line and the figure looks like triangle NOE. No.
3. SILY: S, L, Y are collinear and side YS overlaps side LY. No.
4. DART: it has a dent at D, but the sides meet only at the vertices. Yes (a non-convex quadrilateral).
5. CUTS: sides UT and SC cross at a point that is not a vertex. Not an ordinary quadrilateral (self-intersecting).
6. OPENS: the segments OP, PE, EN, NS do not close up; the last one does not return to the start. No.
7. BENT: the four vertices are not in one plane. Not a (planar) quadrilateral.
Requirements for a definition of quadrilateral ABCD:
1. A, B, C, D are four distinct points in one plane (rules out BENT);
2. no three of them are collinear (rules out NOPE and SILY);
3. the sides are exactly AB, BC, CD and DA, so the figure closes up (rules out OPENS);
4. every point of the sides other than A, B, C, D lies on exactly one side, so sides do not cross or overlap (rules out CUTS and SILY).
These are the conditions that Definition 1 of the chapter puts together.

Think and Reflect (p. 53). Can we similarly define a quadrilateral ABCD?
Answer:
A triangle ABC was defined as the points on the three segments AB, BC, CA, provided A, B, C are not collinear.
For a quadrilateral, 'the points on AB, BC, CD, DA' is not enough (Fig. 12.2): the segments may overlap (SILY), cross (CUTS), not lie in one plane (BENT) or form a triangle-like shape (NOPE). So we define it in the same way but with extra conditions:
Let A, B, C, D be four distinct points in a plane, no three of them collinear. The points on the four segments AB, BC, CD and DA form quadrilateral ABCD if every such point other than A, B, C and D lies on exactly one of these four segments.
This is Definition 1 of the chapter, together with the condition (in the Chapter Summary) that no three vertices are collinear.

Think and Reflect (p. 59). The angles of a quadrilateral add up to 360°. Therefore, if the opposite angles are equal, what can we say about adjacent angles? Is the converse of your answer true? Conclude that Theorem 3 can also be stated as follows. "If each pair of adjacent angles in a quadrilateral ABCD … then ABCD is a parallelogram." Fill in the blank.
Answer:
Opposite angles equal ⇒ adjacent angles supplementary. If ∠A = ∠C = x and ∠B = ∠D = y, then 2x + 2y = 360°, so x + y = 180°. Every pair of adjacent angles adds to 180°.
Converse: adjacent angles supplementary ⇒ opposite angles equal. If ∠A + ∠B = 180° and ∠B + ∠C = 180°, subtracting gives ∠A = ∠C. Likewise ∠B + ∠C = 180° and ∠C + ∠D = 180° give ∠B = ∠D. So the converse is also true and the two conditions are equivalent.
Hence Theorem 3 can be stated as: "If each pair of adjacent angles in a quadrilateral ABCD adds up to 180° (is supplementary), then ABCD is a parallelogram."

Think and Reflect (p. 62). Try to prove directly that the four smaller triangles are congruent. You will see that no congruence test applies, because many quantities are unknown.

Answer:
Compare, for example, △APQ and △PBR in Fig. 12.12.
1. We know AP = PB (P is the midpoint of AB).
2. But we do not know how AQ compares with PR, or PQ with BR, and we know none of the angles at P, Q, R.
So there is only one pair of equal sides and no equal angles, and none of SSS, SAS, ASA, AAS or RHS can be applied. The same happens for every pair of the four triangles.
The trouble is that the lengths PQ, QR, RP and the angles at the new points are unknown. The chapter gets round this by first proving the Midpoint Theorem (PQ ∥ BC and PQ = \( \frac{1}{2} \)BC) for each new segment; after that the congruence follows by SSS (Exercise Set 12.3, Q1).

Think and Reflect (p. 63). Read the statement of the Midpoint Theorem again. The following questions seem natural to ask. What can we say about (1) a segment PQ (with P on side AB and Q on side AC) that is parallel to the third side and has half the length, and (2) the line that passes through the midpoint of one side and is parallel to another side? We will address question (1) in Exercise 22 at the end of the chapter. For now, experiment and see if you can make a guess about question (2). It appears that this line must bisect the remaining side! Let us try to justify this.

Answer:
Question (2), guess. Draw several triangles. Through the midpoint P of AB draw the line parallel to BC, meeting AC at Q. Measuring AQ and QC we always find AQ = QC.
So the guess is: the line through the midpoint of one side, parallel to a second side, bisects the third side (and PQ is half of BC). This is Theorem 7, proved in the chapter and again in End-of-Chapter Exercise 15(ii).
Question (1). If PQ ∥ BC and PQ = \( \frac{1}{2} \)BC (P on AB, Q on AC), then P and Q must be the midpoints of AB and AC. This is proved in End-of-Chapter Exercise 22(i).

Think and Reflect (p. 65). In Grade 10, we will generalise Theorem 7 by dropping the condition that P is the midpoint of AB and assuming only that PQ ∥ BC. Can you guess the conclusion in this case? (Hint: the name of the more general theorem includes the word proportionality.)

Answer:
If P is any point on AB and the line through P parallel to BC meets AC at Q, then P and Q divide the two sides in the same ratio:
\( \frac{AP}{PB} \) = \( \frac{AQ}{QC} \)
This is the Basic Proportionality Theorem (also called Thales' theorem). When P is the midpoint, \( \frac{AP}{PB} \) = 1, so \( \frac{AQ}{QC} \) = 1 and Q is the midpoint, which is Theorem 7. (The case when \( \frac{AP}{PB} \) is rational is proved in End-of-Chapter Exercise 11(iv).)

Think and Reflect (p. 66). Give another proof of the Centroid Theorem using a clever construction that we will describe only partly and ask you to complete. Once again let M be the intersection point of the two medians CP and BQ. We want to show that the intersection point of line AM with side BC is the midpoint of BC. For this, extend AM up to a carefully chosen point S such that BSCM becomes a parallelogram. (Hint: To arrange MC ∥ BS, first note that PM and MC are along the same line. Now where should M be on segment AS? Does your placement of S also prove MB ∥ CS?)

Answer:
Construction. Extend AM beyond M to S with MS = AM, so M is the midpoint of AS. Let AS meet BC at X.
Step 1: MC ∥ BS. In △ABS, P and M are midpoints of AB and AS, so PM ∥ BS (Midpoint Theorem). P, M, C lie on the median CP, so line MC is line PM. Hence MC ∥ BS.
Step 2: MB ∥ CS. In △ACS, Q and M are midpoints of AC and AS, so QM ∥ CS. Q, M, B lie on the median BQ, so line MB is line QM. Hence MB ∥ CS. So yes, the same S also gives MB ∥ CS.
Step 3. Both pairs of opposite sides of BSCM are parallel, so BSCM is a parallelogram. Its diagonals BC and MS bisect each other (Theorem 1(c)). MS lies along line AM, so the diagonals meet at X and BX = XC.
So line AM passes through the midpoint X of BC: AX is the third median, and all three medians pass through M.
The 2 : 1 ratio. The diagonals bisect each other, so MX = XS = \( \frac{1}{2} \)MS = \( \frac{1}{2} \)AM. Hence AM : MX = 2 : 1. Starting with another pair of medians gives the ratio for the other two.

Teacher's Note:
1. Placing M at the midpoint of AS is what lets the Midpoint Theorem work in two triangles at once.
2. The same parallelogram gives both the concurrency and the 2 : 1 ratio.

Think and Reflect (p. 70). Suppose we have a tiling of the plane. Consider any vertex. As we go around this vertex and consider the angles made by consecutive lines, the total of these angles must be 360°. This suggests an idea. What if we take 4 copies of SOME and fit them together around a common point so that each angle is used once as we go around?
Answer:
The four internal angles 1, 2, 3, 4 of SOME add to 360°: ∠1 + ∠2 + ∠3 + ∠4 = 360°.
So if four copies of SOME are placed around a common point, using each of the angles 1, 2, 3, 4 exactly once there, the four angles together make a full turn of 360°. The copies fit around the point with no gap and no overlap.

This is the key idea behind tiling the plane with any 4-gon: at every vertex of the tiling, the four different angles of SOME meet once each.