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

Short Answer Questions

Question 1. With reference to the chapter's definition, explain why a four-sided figure is excluded from being a quadrilateral (i) when three of its vertices are collinear, and (ii) when two of its sides cross each other.
Answer:
The definition says every point of the four segments, apart from the vertices, must lie on exactly one segment.
(i) Three collinear vertices: If A, B, C lie on one line, sides AB and BC lie along the same line. Points of the shorter side also lie on the longer one, so they belong to two segments. The figure looks like a triangle or has a side folded back on another.
(ii) Sides crossing: The crossing point lies on two segments at once, so 'exactly one' fails. Such a figure is called a self-intersecting quadrilateral.
In simple words: In both cases some point sits on two sides, which the definition does not allow.

Teacher's Note:
1. Quote the 'exactly one' condition.
2. Draw a small sketch for each case.

Question 2. Prove that the opposite sides of a parallelogram are equal.
Answer:
Let ABCD be a parallelogram (AB ∥ DC, AD ∥ BC). Draw diagonal AC.
AC is a transversal of AB ∥ DC, so ∠BAC = ∠DCA (alternate angles).
AC is a transversal of AD ∥ BC, so ∠DAC = ∠BCA (alternate angles).
In △ACD and △CAB: ∠DCA = ∠BAC, AC = CA (common), ∠DAC = ∠BCA.
So △ACD ≅ △CAB (ASA), and by CPCT, AB = DC and AD = BC.
In simple words: The diagonal splits the parallelogram into two congruent triangles, so opposite sides match.

Teacher's Note:
1. Name each angle reason.
2. The common side lies between the two angles.

Question 3. Prove that if the opposite sides of a quadrilateral are equal, then it is a parallelogram.
Answer:
Let ABCD have AB = DC and AD = BC. Draw diagonal AC.
In △ACD and △CAB: CD = AB, AD = CB (given), AC = CA (common).
So △ACD ≅ △CAB (SSS). Hence ∠DCA = ∠BAC and ∠DAC = ∠BCA.
With AC as transversal, these equal alternate angles give AB ∥ DC and AD ∥ BC.
Both pairs of opposite sides are parallel, so ABCD is a parallelogram.
In simple words: Equal sides make two congruent triangles, and their equal angles show the sides are parallel.

Teacher's Note:
1. This is the previous proof run backwards.
2. SSS replaces ASA here.

Question 4. The diagonals of a quadrilateral ABCD meet at a point E which is the midpoint of each of them. Prove that ABCD is a parallelogram, naming the congruence test you use.
Answer:
Given: EA = EC and EB = ED.
In △AED and △CEB: EA = EC, ∠AED = ∠CEB (vertically opposite), ED = EB.
So △AED ≅ △CEB (SAS), giving ∠DAE = ∠BCE.
These are alternate angles on transversal AC, so AD ∥ BC.
Similarly △AEB ≅ △CED gives AB ∥ DC. Hence ABCD is a parallelogram.
In simple words: Bisecting diagonals make matching triangles, which give parallel opposite sides.

Teacher's Note:
1. The included angle is the vertically opposite angle.
2. Repeat for the second pair of sides.

Question 5. Prove the equal parallel sides test: a quadrilateral with one pair of opposite sides both equal and parallel is a parallelogram.
Answer:
Let ABCD have AB ∥ DC and AB = DC. Let the diagonals meet at E.
With BD as transversal, ∠ABE = ∠CDE; with AC as transversal, ∠BAE = ∠DCE (alternate angles).
In △EAB and △ECD: ∠BAE = ∠DCE, AB = CD, ∠ABE = ∠CDE, so △EAB ≅ △ECD (ASA).
Hence EA = EC and EB = ED: the diagonals bisect each other.
By the bisecting-diagonals test, ABCD is a parallelogram.
In simple words: Equal parallel sides make congruent triangles, so the diagonals bisect each other.

Teacher's Note:
1. This proof reuses the previous result.
2. Both 'equal' and 'parallel' are used.

Question 6. In parallelogram ABCD, ∠A = (3x − 10)° and ∠C = (x + 50)°. Find the value of x and hence all four angles of the parallelogram.
Answer:
Opposite angles are equal: 3x − 10 = x + 50, so 2x = 60 and x = 30.
∠A = 3(30) − 10 = 80° and ∠C = 30 + 50 = 80°.
∠B = ∠D = 180° − 80° = 100°.
Angles: 80°, 100°, 80°, 100° (total 360°).
In simple words: Set opposite angles equal to find x, then use 180° for the neighbours.

Teacher's Note:
1. Check ∠A and ∠C both give 80°.
2. Adjacent angles add to 180°.

Question 7. Prove that a parallelogram in which one angle is a right angle is a rectangle.
Answer:
Let ABCD be a parallelogram with ∠A = 90°.
AD ∥ BC with transversal AB, so ∠A + ∠B = 180° (co-interior). Hence ∠B = 90°.
Opposite angles are equal, so ∠C = ∠A = 90° and ∠D = ∠B = 90°.
All four angles are right angles, so ABCD is a rectangle.
In simple words: One right angle spreads to all four corners.

Teacher's Note:
1. Use co-interior angles first, then opposite angles.
2. Rectangle = parallelogram with right angles.

Question 8. Prove that if the diagonals of a parallelogram are equal in length, then the parallelogram is a rectangle.
Answer:
Let ABCD be a parallelogram with AC = BD.
In △ABC and △BAD: AB = BA (common), BC = AD (opposite sides), AC = BD (given).
So △ABC ≅ △BAD (SSS), hence ∠ABC = ∠BAD.
These are co-interior angles, so they add to 180°. Two equal angles adding to 180° are each 90°.
By the previous result, ABCD is a rectangle.
In simple words: Equal diagonals make two neighbouring angles equal, and since they add to 180°, both are right angles.

Teacher's Note:
1. BC = AD comes from being a parallelogram.
2. Equal + supplementary ⇒ 90° each.

Question 9. In △ABC, AB = 8 cm, BC = 10 cm and CA = 12 cm. The points P, Q and R are the midpoints of AB, AC and BC respectively. Find the three sides of △PQR and its perimeter, justifying each length.
Answer:
By the Midpoint Theorem each side of △PQR is half the third side of △ABC.
PQ joins midpoints of AB and AC: PQ = ½BC = 5 cm.
QR joins midpoints of AC and BC: QR = ½AB = 4 cm.
RP joins midpoints of BC and AB: RP = ½CA = 6 cm.
Perimeter = 5 + 4 + 6 = 15 cm, half of 30 cm.
In simple words: Each side of the middle triangle is half the side it does not touch.

Teacher's Note:
1. Match each segment with the side it is parallel to.
2. Perimeter is always halved.

Question 10. In △ABC, P and Q are the midpoints of AB and AC. The line through C parallel to BA meets the line PQ at R.
Using this construction, prove that PQ ∥ BC and PQ = ½BC.
Answer:
In △APQ and △CRQ: AQ = CQ (Q is the midpoint), ∠AQP = ∠CQR (vertically opposite), ∠APQ = ∠CRQ (alternate angles, AP ∥ RC).
So △APQ ≅ △CRQ (AAS), giving PQ = QR and CR = AP.
AP = BP (P is the midpoint), so CR = BP. Also CR ∥ BP.
By the equal parallel sides test, BCRP is a parallelogram, so PR ∥ BC and PR = BC.
PQ lies on PR, so PQ ∥ BC; and PQ = QR, so PQ = ½PR = ½BC.
In simple words: The construction builds a parallelogram whose side PR equals BC, and PQ is exactly half of PR.

Teacher's Note:
1. The key step is proving BCRP is a parallelogram.
2. State AAS clearly, not ASA.

Question 11. Prove the converse of the Midpoint Theorem by using the fact that through a given point there is exactly one line parallel to a given line.
Answer:
In △ABC let P be the midpoint of AB, and let the line through P parallel to BC meet AC at Q. We show Q is the midpoint of AC.
Let M be the midpoint of AC. By the Midpoint Theorem, PM ∥ BC.
So PM and PQ both pass through P and are parallel to BC.
Only one such line exists, so PM and PQ are the same line, and they meet AC at the same point.
Hence Q = M, the midpoint of AC.
In simple words: The true midpoint gives a parallel line through P, and there is only one such line, so Q must be that midpoint.

Teacher's Note:
1. This is not circular: a new fact about parallels is used.
2. Name M before comparing it with Q.

Question 12. In △ABC the medians BQ and CP meet at M. Given that CM = 10 cm and MQ = 4 cm, find the lengths of the medians CP and BQ.
Answer:
The centroid M divides each median 2:1 from the vertex.
On CP: CM:MP = 2:1, so MP = 10 ÷ 2 = 5 cm and CP = 15 cm.
On BQ: BM:MQ = 2:1, so BM = 2 × 4 = 8 cm and BQ = 12 cm.
In simple words: The vertex part is double the side part on each median.

Teacher's Note:
1. Identify which piece is given on each median.
2. Halve one, double the other.

Question 13. The points P, Q, R and S are the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD.
Prove that PQRS is a parallelogram.
Answer:
Draw diagonal AC.
In △ABC, P and Q are midpoints of AB and BC, so PQ ∥ AC and PQ = ½AC (Midpoint Theorem).
In △ADC, S and R are midpoints of DA and CD, so SR ∥ AC and SR = ½AC.
Hence PQ ∥ SR and PQ = SR.
One pair of opposite sides is equal and parallel, so PQRS is a parallelogram.
In simple words: Both PQ and SR are half of AC and parallel to it, so PQRS is a parallelogram.

Teacher's Note:
1. One diagonal is enough for the proof.
2. The other diagonal gives the same for QR and PS.

Question 14. In a quadrilateral ABCD the diagonals measure AC = 16 cm and BD = 12 cm. Find the lengths of the sides of its Varignon parallelogram and its perimeter.
Answer:
Each side of the Varignon parallelogram is half a diagonal.
PQ = SR = ½AC = 8 cm; QR = PS = ½BD = 6 cm.
Perimeter = 2(8 + 6) = 28 cm, which equals AC + BD.
In simple words: The perimeter of the Varignon parallelogram equals the sum of the two diagonals.

Teacher's Note:
1. Useful shortcut: perimeter = AC + BD.
2. Opposite sides equal in pairs.

Question 15. Prove that the Varignon parallelogram of a quadrilateral ABCD is a rhombus if and only if AC = BD.
Answer:
The sides of PQRS are PQ = SR = ½AC and QR = PS = ½BD.
If AC = BD: then ½AC = ½BD, so all four sides are equal and PQRS is a rhombus.
If PQRS is a rhombus: adjacent sides are equal, so ½AC = ½BD, giving AC = BD.
Both directions hold, so PQRS is a rhombus exactly when AC = BD.
In simple words: The sides are halves of the diagonals, so equal diagonals and equal sides go together.

Teacher's Note:
1. 'If and only if' needs both directions.
2. Show each direction in its own line.

Question 16. ABCD is a trapezium in which AB ∥ DC, with AB = 6 cm and DC = 14 cm. The points E and F are the midpoints of the non-parallel sides AD and BC. Find EF and prove that EF ∥ AB.
Answer:
Draw diagonal AC meeting EF at G.
In △ADC: E is the midpoint of AD and EG ∥ DC, so G is the midpoint of AC (converse of the Midpoint Theorem) and EG = ½DC = 7 cm.
In △ABC: G and F are midpoints of AC and BC, so GF ∥ AB and GF = ½AB = 3 cm.
EF = 7 + 3 = 10 cm = \( \frac{6 + 14}{2} \).
E, G, F lie on one line and GF ∥ AB, so EF ∥ AB.
In simple words: Split EF at the diagonal into two midpoint segments, 7 cm and 3 cm.

Teacher's Note:
1. The diagonal turns the trapezium into two triangles.
2. Result: EF = average of parallel sides.

Question 17. Show that a regular pentagon cannot be used to tile the plane, and name a regular polygon with more than four sides that can.
Answer:
At a vertex of a tiling the angles must total 360°.
Interior angle of a regular pentagon = \( \frac{(5-2) \times 180^\circ}{5} \) = 108°.
360 ÷ 108 = 3⅓, not whole: three leave a gap of 36°, four need 432° and overlap. So it cannot tile.
The regular hexagon can: its angle is 120° and 360 ÷ 120 = 3, as in a honeycomb.
In simple words: 108° does not fit into 360° a whole number of times, but 120° does.

Teacher's Note:
1. Show the gap and the overlap.
2. Honeycomb is a real-life example.

Question 18. Describe the first of the chapter's methods for tiling the plane with copies of any given quadrilateral, and explain why the four angles fit together exactly at each vertex of the tiling.
Answer:
Method: Place one copy. Pick an edge and mark its midpoint. Place a second copy on the first and rotate it 180° about that midpoint; it lands alongside, sharing the edge. Repeat at every edge of every copy.
Why it fits: A half-turn about an edge's midpoint maps the edge onto itself, so neighbouring copies share the whole edge.
Going round each vertex, all four angles of the quadrilateral appear once. They total 360°, so they close up with no gap or overlap.
In simple words: Half-turns about edge midpoints make the edges match, and the 360° angle sum makes the corners fit.

Teacher's Note:
1. Demonstrate with a paper cut-out.
2. Works for convex and non-convex shapes.

Question 19. Find a formula for the number of diagonals of an n-gon, and verify it for a pentagon and for a hexagon.
Answer:
From one vertex a diagonal can go to every vertex except itself and its two neighbours: n − 3 diagonals.
n vertices give n(n − 3), but each diagonal is counted twice, so the number is \( \frac{n(n-3)}{2} \).
Pentagon: \( \frac{5 \times 2}{2} \) = 5 (the five-pointed star).
Hexagon: \( \frac{6 \times 3}{2} \) = 9 (3 long through the centre + 6 short).
In simple words: Count n − 3 from each vertex, multiply by n, and halve.

Teacher's Note:
1. Explain the halving clearly.
2. Verify by drawing.

Question 20. Prove that the sum of the internal angles of a quadrilateral is 360° by drawing one diagonal.
Answer:
In quadrilateral ABCD draw diagonal AC, making △ABC and △ACD.
∠BAC + ∠ABC + ∠BCA = 180° and ∠DAC + ∠ACD + ∠CDA = 180°.
Adding: the six angles total 360°.
∠BAC + ∠DAC = ∠A, ∠BCA + ∠ACD = ∠C, ∠ABC = ∠B, ∠CDA = ∠D.
So ∠A + ∠B + ∠C + ∠D = 360°.
In simple words: A diagonal makes two triangles, and 2 × 180° = 360°.

Teacher's Note:
1. The diagonal must lie inside the figure.
2. Show how the six angles regroup into four.