CBSE Class 9 Maths Ganita Manjari Part 2 Ch 12 Quadrilaterals Very Short Answer Questions with Answers

Very Short Answer Questions

Question 1. State the chapter's definition of the quadrilateral ABCD for four distinct points A, B, C and D in a plane.
Answer:
Let A, B, C, D be four distinct points in a plane. The points of the segments AB, BC, CD and DA form the quadrilateral ABCD, provided every such point other than A, B, C, D lies on exactly one of the four segments.
The 'exactly one' condition rules out sides that overlap or cross.

Teacher's Note:
1. Underline 'in a plane' and 'exactly one'.
2. Each phrase excludes a kind of figure.

Question 2. A four-sided figure may fail to be a quadrilateral for three different reasons. State any two of them, and name the kind of figure that results in each case.
Answer:
Any two of:
1. Three vertices collinear: two sides lie on one line, giving a triangle-like shape or overlapping sides.
2. Two sides cross: a self-intersecting quadrilateral.
3. Vertices not in one plane: a skew (non-planar) quadrilateral.

Teacher's Note:
1. Give the name of each figure.
2. Two correct reasons are enough.

Question 3. Write down all the names, other than ABCD itself, by which the quadrilateral ABCD may be called.
Answer:
BCDA, CDAB, DABC, DCBA, ADCB, BADC, CBAD.
With ABCD, that makes 8 names: start at any of 4 vertices and go either way round.

Teacher's Note:
1. 4 starting points × 2 directions = 8.
2. Letters must follow the boundary.

Question 4. Define a convex quadrilateral, and state the test for convexity that uses the diagonals.
Answer:
A quadrilateral is convex when all four internal angles are less than 180° (it has no dent).
Diagonal test: it is convex exactly when its two diagonals intersect.

Teacher's Note:
1. Two equivalent tests: angles and diagonals.
2. Non-convex: diagonals do not meet.

Question 5. Three angles of a quadrilateral measure 72°, 108° and 96°. Find the fourth angle.
Answer:
72 + 108 + 96 = 276.
Fourth angle = 360° − 276° = 84°.

Teacher's Note:
1. Angle sum of a quadrilateral is 360°.
2. Check the total at the end.

Question 6. In quadrilateral PQRS, name the two sides adjacent to PQ and the side opposite to PQ.
Answer:
Adjacent sides: QR and SP (each shares a vertex with PQ).
Opposite side: RS (shares no vertex with PQ).

Teacher's Note:
1. Adjacent sides share a vertex.
2. PR and QS are diagonals, not sides.

Question 7. State the three properties of a parallelogram given by Theorem 1 of this chapter.
Answer:
In any parallelogram:
1. the opposite sides are equal;
2. the opposite angles are equal;
3. the diagonals bisect each other.
All three have true converses, which become tests for a parallelogram.

Teacher's Note:
1. Learn the three as a set.
2. Converses are Theorems 2, 3, 4.

Question 8. Write the converse of 'the opposite sides of a parallelogram are equal', and state whether the converse is true.
Answer:
Converse: if the opposite sides of a quadrilateral are equal, then it is a parallelogram.
It is true (Theorem 2), and it is one of the four tests for a parallelogram.

Teacher's Note:
1. Swap hypothesis and conclusion.
2. Say 'quadrilateral' in the hypothesis, not 'parallelogram'.

Question 9. Name the congruence test used in the proof that a quadrilateral with equal opposite sides is a parallelogram, and state why that test is the natural one to use there.
Answer:
The test is SSS.
Only sides are known: the two pairs of equal opposite sides and the common diagonal AC. No angle is known, so SSS is the only test that fits.

Teacher's Note:
1. Match the test to the given data.
2. Angles come out of the congruence, not into it.

Question 10. State the equal parallel sides test for a parallelogram.
Answer:
If one pair of opposite sides of a quadrilateral is both equal and parallel, then it is a parallelogram.
Both facts are needed; either one alone is not enough.

Teacher's Note:
1. Only one pair is checked.
2. An isosceles trapezium shows 'parallel only' is not enough.

Question 11. In △ABC, P and Q are the midpoints of AB and AC respectively. Given that PQ = 6.5 cm, find BC.
Answer:
By the Midpoint Theorem PQ = ½BC.
BC = 2 × 6.5 = 13 cm.

Teacher's Note:
1. Double to go back to the third side.
2. Check: 13 ÷ 2 = 6.5.

Question 12. State the Midpoint Theorem.
Answer:
The segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

Teacher's Note:
1. Two conclusions: direction and length.
2. Used later for centroid and Varignon results.

Question 13. State the converse of the Midpoint Theorem.
Answer:
The line through the midpoint of one side of a triangle, parallel to another side, bisects the third side.
The segment so formed is half the length of the parallel side.

Teacher's Note:
1. Given: one midpoint + parallel.
2. Proved: the other midpoint.

Question 14. State the Centroid Theorem, and give the name of the point at which the three medians of a triangle meet.
Answer:
The three medians of a triangle pass through one common point, which divides each median in the ratio 2:1, the longer part next to the vertex.
This point is the centroid.

Teacher's Note:
1. State the direction of the ratio.
2. Concurrent = meet at one point.

Question 15. A median of a triangle is 21 cm long. Find the lengths of the two parts into which the centroid divides it.
Answer:
The parts are ⅔ and ⅓ of the median.
\( \frac{2}{3} \) × 21 = 14 cm (next to the vertex) and \( \frac{1}{3} \) × 21 = 7 cm.

Teacher's Note:
1. Check 14 + 7 = 21.
2. Say which part is at the vertex.

Question 16. Explain what is meant by the Varignon parallelogram of a quadrilateral.
Answer:
Mark the midpoints of the four sides of a quadrilateral and join them in order.
The figure obtained is always a parallelogram, called the Varignon parallelogram of the quadrilateral.

Teacher's Note:
1. Midpoints of sides, joined in order.
2. It is a parallelogram for every quadrilateral.

Question 17. State the Midpoint Theorem for Quadrilaterals.
Answer:
The midpoints of the four sides of any quadrilateral are the vertices of a parallelogram.
It holds for convex, non-convex and even self-intersecting quadrilaterals.

Teacher's Note:
1. Also called Varignon's theorem.
2. No condition on the quadrilateral is needed.

Question 18. In a trapezium the two parallel sides measure 7 cm and 13 cm. Find the length of the segment joining the midpoints of the two non-parallel sides.
Answer:
The segment is the average of the parallel sides.
\( \frac{7 + 13}{2} \) = 10 cm.

Teacher's Note:
1. The answer must lie between 7 and 13.
2. If one side shrinks to 0, this becomes the Midpoint Theorem.

Question 19. Find the number of diagonals of a decagon.
Answer:
For n = 10: \( \frac{n(n-3)}{2} \) = \( \frac{10 \times 7}{2} \) = 35.

Teacher's Note:
1. n − 3 diagonals per vertex.
2. Halve because each is counted twice.

Question 20. Find the sum of the internal angles of a heptagon.
Answer:
For n = 7: (n − 2) × 180° = 5 × 180° = 900°.

Teacher's Note:
1. n − 2 triangles from one vertex.
2. Heptagon = 7 sides.

Question 21. One angle of a parallelogram measures 65°. Find the other three angles.
Answer:
The opposite angle is also 65°.
Each adjacent angle is 180° − 65° = 115°.
The angles are 65°, 115°, 65°, 115°.

Teacher's Note:
1. Opposite equal, adjacent add to 180°.
2. Check the total is 360°.

Question 22. Explain why a square can tile the plane while a regular pentagon cannot, referring to the angles at a vertex of the tiling.
Answer:
The angles at each vertex of a tiling must add to exactly 360°.
Square: 360 ÷ 90 = 4, a whole number, so four squares fit.
Pentagon: 360 ÷ 108 = 3⅓, not whole; three leave a 36° gap and four overlap.

Teacher's Note:
1. The test is whole-number division.
2. Mention both gap and overlap.

Question 23. Explain why the statement 'a rhombus with perpendicular diagonals is a square' is false.
Answer:
Every rhombus already has perpendicular diagonals, so this condition adds nothing.
Example: a rhombus of side 5 with diagonals 8 and 6 has perpendicular diagonals but no right angle, so it is not a square.

Teacher's Note:
1. Check whether the extra condition is already built in.
2. One counterexample settles it.

Question 24. Give a counterexample to the statement 'a quadrilateral whose diagonals are equal is a rectangle'.
Answer:
Take the isosceles trapezium A(0, 0), B(10, 0), C(7, 4), D(3, 4).
AC = \( \sqrt{7^2 + 4^2} = \sqrt{65} \) and BD = \( \sqrt{(-7)^2 + 4^2} = \sqrt{65} \), so the diagonals are equal.
It has no right angle, so it is not a rectangle.

Teacher's Note:
1. The isosceles trapezium is the standard counterexample.
2. Show both diagonal lengths.

Question 25. State what the Varignon parallelogram of ABCD becomes when AC = BD, and give the reason.
Answer:
It becomes a rhombus.
Its sides are ½AC and ½BD, so when AC = BD all four sides are equal.

Teacher's Note:
1. Diagonal lengths control side lengths.
2. Perpendicular diagonals would give a rectangle.

Question 26. Name the parallelogram-based device that makes exact scaled copies of a drawing, and name the mathematician whose parallel motion linkage formed part of his steam engine.
Answer:
The device is the pantograph.
The mathematician is James Watt.

Teacher's Note:
1. Both use the parallelogram's property of staying parallel as it moves.
2. Link maths to engineering.

Question 27. Name the shape, and the year of its discovery, that was the first single tile found to tile the plane without the pattern ever repeating.
Answer:
The 'hat', discovered in 2023 by Smith, Myers, Kaplan and Goodman-Strauss.
Such a never-repeating tiling is called aperiodic.

Teacher's Note:
1. Remember the word 'aperiodic'.
2. Recent mathematics is still being discovered.

Question 28. In parallelogram ABCD the diagonal AC bisects ∠A. State what kind of quadrilateral ABCD must be.
Answer:
ABCD must be a rhombus.
Since AB ∥ DC, ∠BAC = ∠DCA (alternate). AC bisects ∠A, so ∠BAC = ∠DAC. Hence ∠DAC = ∠DCA, so AD = DC.
Two adjacent sides are equal, so all four sides are equal.

Teacher's Note:
1. Use alternate angles, then an isosceles triangle.
2. Adjacent sides equal → rhombus.

Question 29. Explain why the angles meeting at any one vertex of a tiling of the plane must add up to 360°.
Answer:
Going once round a vertex is a full turn of 360°.
The tiles there fill this turn with no gap and no overlap, so their angles add to exactly 360°.
Less would leave a gap; more would cause an overlap.

Teacher's Note:
1. A full turn = 360°.
2. This one fact explains most tiling results.

Question 30. Three sides of a quadrilateral measure 2 cm, 4 cm and 15 cm. Find the range of possible lengths of the fourth side.
Answer:
Each side must be less than the sum of the other three.
d < 2 + 4 + 15 = 21, and 15 < 2 + 4 + d gives d > 9.
So 9 cm < d < 21 cm.

Teacher's Note:
1. Check the given longest side against the others.
2. Inequalities are strict.