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ABCD is a trapezium with parallel sides AD = 3 cm and BC = 5 cm. E and F are the midpoints of the non-parallel sides. Find the ratio of the areas of the 4-gons AEFD and EBCF.
By the midline theorem for a trapezium, segment EF joining the midpoints of the non-parallel sides is parallel to both bases with length EF = (AD + BC)/2 = (3 + 5)/2 = 4 cm. Since E and F bisect the legs, both resulting trapeziums AEFD and EBCF share an identical vertical height h. Evaluating theirRead more
By the midline theorem for a trapezium, segment EF joining the midpoints of the non-parallel sides is parallel to both bases with length EF = (AD + BC)/2 = (3 + 5)/2 = 4 cm. Since E and F bisect the legs, both resulting trapeziums AEFD and EBCF share an identical vertical height h. Evaluating their area formulas gives (1/2)(3 + 4)h and (1/2)(4 + 5)h, producing a final area ratio of 7:9.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/
See lessThe diagonals AC and BD of a parallelogram ABCD intersect at O. A line through O meets AB and CD at points P and Q respectively. Show that O is the midpoint of PQ. (Multiple proofs are possible. Which is the simplest?)
Diagonals of parallelogram ABCD bisect each other, giving OB = OD. Since AB ∥ CD, alternate interior angles yield ∠PBO = ∠QDO. Furthermore, ∠POB = ∠QOD as vertically opposite angles. Therefore, triangle POB ≅ triangle QOD by ASA congruence. Corresponding sides give OP = OQ, meaning O is the midpointRead more
Diagonals of parallelogram ABCD bisect each other, giving OB = OD. Since AB ∥ CD, alternate interior angles yield ∠PBO = ∠QDO. Furthermore, ∠POB = ∠QOD as vertically opposite angles. Therefore, triangle POB ≅ triangle QOD by ASA congruence. Corresponding sides give OP = OQ, meaning O is the midpoint of PQ. This standard ASA congruence proof is the simplest approach.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/
See lessA more general midpoint theorem and its converse. In a quadrilateral ABCD, suppose AB ∥ DC. Recall that such ABCD is called a trapezium. Let E be the midpoint of AD. A line drawn through E intersects side BC at F. (i) If EF ∥ AB, then show that F is the midpoint of BC. Conclude that EF = (AB + CD)/2. (ii) If F is the midpoint of BC, then show that EF ∥ AB. There are at least two ways to solve (ii). Do it directly by using the midpoint M of BD and showing that EM and FM are the same lines. Or use (i) along with the same idea used in the second proof of the converse of the Midpoint Theorem. Which way do you think is simpler?
(i) Draw diagonal BD meeting line EF at M. In triangle DAB, line EM through midpoint E parallel to AB bisects BD at M. In triangle BDC, MF ∥ DC bisects BC at F, yielding EF = EM + MF = AB/2 + CD/2 = (AB + CD)/2. (ii) Midpoints of AD and BC join to form EF. Using the uniqueness of a parallel line thrRead more
(i) Draw diagonal BD meeting line EF at M. In triangle DAB, line EM through midpoint E parallel to AB bisects BD at M. In triangle BDC, MF ∥ DC bisects BC at F, yielding EF = EM + MF = AB/2 + CD/2 = (AB + CD)/2. (ii) Midpoints of AD and BC join to form EF. Using the uniqueness of a parallel line through E combined with part (i) is simpler than collinearity proofs.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/
See lessA right-triangle shaped cutout of a paper is folded such that point A touches point B (Fig. 12.35). Show that the crease line can be used to find the midpoint of not only AB but also that of AC.
Folding point A onto point B makes the fold crease the perpendicular bisector of segment AB, directly marking the midpoint of AB. Because ∠B = 90° and the crease is perpendicular to AB, the crease line is parallel to base BC. By the Converse of the Midpoint Theorem, this line drawn through the midpoRead more
Folding point A onto point B makes the fold crease the perpendicular bisector of segment AB, directly marking the midpoint of AB. Because ∠B = 90° and the crease is perpendicular to AB, the crease line is parallel to base BC. By the Converse of the Midpoint Theorem, this line drawn through the midpoint of AB parallel to BC bisects side AC.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/
See lessIn a parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 12.34). Show that APCQ is a parallelogram.
Let O be the intersection point of the diagonals AC and BD of parallelogram ABCD. Since the diagonals of a parallelogram bisect each other, OA = OC and OB = OD. Given that DP = BQ, we have OP = OD - DP = OB - BQ = OQ. The diagonals AC and PQ of quadrilateral APCQ bisect each other at O, which provesRead more
Let O be the intersection point of the diagonals AC and BD of parallelogram ABCD. Since the diagonals of a parallelogram bisect each other, OA = OC and OB = OD. Given that DP = BQ, we have OP = OD – DP = OB – BQ = OQ. The diagonals AC and PQ of quadrilateral APCQ bisect each other at O, which proves APCQ is a parallelogram.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/
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