JEE Mathematics Areas of Parallelograms and Triangles MCQs Set 02

Answer: (b) equal in area

Explanation:
1. Both parallelograms have the same base and the same height (the gap between the parallels).
2. Area = base × height, so the areas are equal.

Teacher's Note:
1. This is Theorem 1 of the chapter and the starting point for all area comparisons.
2. Equal area does not mean equal perimeter.

Answer: (a) 60 cm²

Explanation:
1. Area = base × height = 12 × 5 = 60 cm².

Teacher's Note:
1. Do not halve the product: the half applies to triangles only.
2. Always use the height perpendicular to the chosen base.

Answer: (d) 24 cm²

Explanation:
1. The triangle is half of the parallelogram on the same base between the same parallels.
2. Area = 48 ÷ 2 = 24 cm².

Teacher's Note:
1. Picture the diagonal splitting the parallelogram into two equal triangles.
2. Both triangles on that base and between the same parallels have this area.

Answer: (c) equal in area

Explanation:
1. The two triangles have equal bases (the median bisects the side).
2. They have the same altitude from the opposite vertex, so their areas are equal.

Teacher's Note:
1. Equal in area is not the same as congruent.
2. The medians of a triangle meet at the centroid, which they divide in the ratio 2 : 1, not the areas.

Answer: (b) 9 cm²

Explanation:
1. AD is a median, so Area(ABD) = 36 ÷ 2 = 18 cm².
2. BE is a median of triangle ABD, so Area(BED) = 18 ÷ 2 = 9 cm².

Teacher's Note:
1. Apply the median result twice.
2. Each time you use a median, the area halves.

Answer: (a) AD is parallel to BC

Explanation:
1. Equal areas on the same base give equal altitudes from A and D to BC.
2. So A and D are at the same distance from BC on the same side, hence AD || BC.

Teacher's Note:
1. This is the converse of Theorem 2.
2. The condition 'same side' is essential.

Answer: (d) equal areas

Explanation:
1. The diagonals bisect each other, so each diagonal is cut into two equal parts.
2. Two adjacent triangles share an altitude and have equal bases, so all four areas are equal.

Teacher's Note:
1. Each of the four triangles is one quarter of the parallelogram.
2. They are equal in area but are not in general congruent.

Answer: (c) 1 : 2

Explanation:
1. Triangle APB and the parallelogram ABCD are on the same base AB and between the same parallels AB and CD.
2. So the triangle is half of the parallelogram, ratio 1 : 2.

Teacher's Note:
1. The position of P on CD does not matter.
2. Look for the same base and the same parallels first.

Answer: (b) 8 cm

Explanation:
1. Area = \( \frac{1}{2} \) × base × altitude, so 40 = \( \frac{1}{2} \) × 10 × h.
2. h = 80 ÷ 10 = 8 cm.

Teacher's Note:
1. Remember to double the area before dividing by the base.
2. Check: half of 10 × 8 is 40.

Answer: (a) 12 cm

Explanation:
1. First triangle: 54 = \( \frac{1}{2} \) × 9 × h, so h = 12 cm.
2. Equal areas and equal bases imply equal altitudes, so the second altitude is also 12 cm.

Teacher's Note:
1. This is Theorem 3 of the chapter.
2. You only need to compute one altitude.

Answer: (d) 30 cm²

Explanation:
1. Triangles ABD and ADC have the same altitude from A, so their areas are in the ratio BD : DC = 2 : 3.
2. Area(ABD) = 75 × 2/5 = 30 cm².

Teacher's Note:
1. Areas of triangles with a common altitude are in the ratio of their bases.
2. Do not use 2/3 directly; the whole is 2 + 3 = 5 parts.

Answer: (c) 16 cm²

Explanation:
1. EF is parallel to BC and equal to half of BC (midpoint theorem).
2. Triangle AEF has half the base and half the height of the triangle ABC in matching positions, so its area is 1/4 of 64 = 16 cm².

Teacher's Note:
1. The small triangle is similar to the big one with ratio 1 : 2, so areas are in the ratio 1 : 4.
2. The rest of the triangle splits into three triangles of equal area.

Answer: (b) 42 cm²

Explanation:
1. Triangles AOX and COY are congruent (ASA), so they have equal area.
2. Hence Area(AXYD) = Area(ACD) = half of the parallelogram = 42 cm².

Teacher's Note:
1. The line through O always halves the parallelogram.
2. O is the centre of symmetry of the parallelogram.

Answer: (a) 12 cm

Explanation:
1. The area is the same whichever side is taken as base: 120 = 10 × h.
2. h = 12 cm.

Teacher's Note:
1. The altitude to the longer side is the shorter one (8 cm here).
2. Area stays fixed; base and height change inversely.

Answer: (d) triangle AED

Explanation:
1. Triangles DBC and DBE are on base DB between parallels DB and CE, so they have equal area.
2. Area(ABCD) = Area(ABD) + Area(DBC) = Area(ABD) + Area(DBE) = Area(AED).

Teacher's Note:
1. This is the standard construction of a triangle equal to a quadrilateral.
2. The vertex D is kept fixed.

Answer: (c) 40 cm²

Explanation:
1. Triangles ABC and PBC are on the same base BC and between the parallels BC and the line through A.
2. They have equal areas, so Area(PBC) = 40 cm².

Teacher's Note:
1. Sliding the apex along a parallel line never changes the area.
2. The base stays fixed and the height stays fixed.

Answer: (b) Triangles on the same base and between the same parallels are equal in area.

Explanation:
1. Only the first statement is Theorem 2.
2. Equal area allows many different shapes (for example a base of 6 with height 4, or a base of 3 with height 8).

Teacher's Note:
1. Area alone does not determine the shape.
2. Watch for statements that reverse a theorem.

Answer: (a) 36 cm²

Explanation:
1. Area(ABEF) = Area(ABCD) = 72 cm².
2. The diagonal AE of parallelogram ABEF divides it into two equal triangles, so Area(ABE) = 36 cm².

Teacher's Note:
1. First transfer the area to the second parallelogram, then halve it.
2. A diagonal always bisects the area of a parallelogram.

Answer: (d) triangles AOD and BOC

Explanation:
1. Triangles ADC and BDC are on the base DC between parallels AB and DC, so they have equal area.
2. Subtracting the common triangle DOC from both leaves Area(AOD) = Area(BOC).

Teacher's Note:
1. The triangles beside the parallel sides (AOB and COD) are not equal in general.
2. Subtracting a common part is a standard trick.

Answer: (c) 63 cm²

Explanation:
1. Area = \( \frac{1}{2} \) × 14 × 9 = 63 cm².

Teacher's Note:
1. The product 14 × 9 = 126 is the area of a parallelogram on the same base and height.
2. Divide by 2 for a triangle.