Grade 10 Notes of Triangles and Quadrilaterals|| Compulsory Mathematics

1.Introduction & Review

Compare figures by overlapping, area, and congruency.

Questions to discuss:

Do pairs of figures match when overlapped?

Are their areas equal?

Which figures are congruent and which are not?

Do all figures with equal area have to be congruent?

Observation: Congruency ≠ equal area necessarily, but equal area ≠ congruency.

2.Area of Parallelograms and Triangles

2.1 Theorems

Theorem 1: Parallelograms on the same base and between the same parallels have equal area  [Fig1]

Given: Parallelograms ABCD and ABEF on same base AB and between parallel lines AB and CF.

To Prove: Area of ABCD = Area of ABEF

Proof Idea:

1.Triangles formed by diagonals are congruent (AAS).

2.Adding the same trapezium to both sides → Areas equal.

3.Hence, parallelograms on the same base and between the same parallels are equal in area.

Formula for Area of Parallelogram: Area = base × height

Example 1: Rectangle ABPQ and parallelogram ABCD on same base AB and parallel lines → Area of ABCD = Area of ABPQ

Theorem 2: Area of a triangle = ½ area of parallelogram on the same base and between same parallels[FIG 2]

Given: Triangle DABE and parallelogram ABCD on same base AB and parallel lines.

To Prove: Area of DABE = ½ Area of ABCD

Proof Idea:

Draw a parallelogram using the triangle as half → ABFE.

Triangle is half of this parallelogram.

Both parallelograms on same base → Area of triangle = ½ Area of parallelogram.

Formula for Area of Triangle: Area = ½ × base × height

Example 2: Triangle with base 8 cm and height 6 cm → Area = 24 cm²; parallelogram on same base and height → Area = 48 cm²

Theorem 3: Triangles on the same base and between same parallels are equal in area [Fig 3]

Given: Triangles ABE and ABC on base AB and parallel lines.

To Prove: Area of ABE = Area of ABC

Proof Idea:

Construct parallelogram ABCD with triangles sharing the same base.

Diagonal divides parallelogram into two triangles of equal area.

Hence, triangles with same base and between same parallels → equal area.

3.Midpoint Theorems

Line joining midpoints of two sides of a triangle is parallel to third side and half its length.

Joining midpoints of a quadrilateral forms a parallelogram (Varignon Theorem).

Area of this midpoint parallelogram = ½ area of original

Example 3: Parallelogram ABCD, midpoints E, F, G, H → Area of EFGH = ½ Area of ABCD

4.Relations Between Figures

5.Examples & Problems

5.1 Parallelograms

ABCD and ABEF: Base = 7 cm, height = 3 cm → Area = 21 cm² each

Rectangle vs Parallelogram: Same base & parallel lines → Areas equal

5.2 Triangles

Triangle base = 8 cm, height = 6 cm → Area = 24 cm²

Parallelogram same base & height → Area = 48 cm² → Triangle = ½ parallelogram

5.3 Triangles inside Parallelogram

ABCD parallelogram, triangles ΔAPB + ΔPCD → ½ Area of ABCD

PQRS parallelogram, triangles ΔPQY = ΔQRX → Equal area

5.4 Midpoints

ABCD parallelogram, midpoints M, N of BC, AD → MN divides ABCD into 2 equal parallelograms

Triangles standing on same base & between same parallels → Equal area

6.Key Formulas

Area of Parallelogram = base × height

Area of Triangle = ½ × base × height

Trapezium = ½ × (sum of parallel sides) × height

Midpoint parallelogram = ½ area of original

7.Important Notes for Students

Equal area ≠ congruency

Base and perpendicular height are crucial for calculating area

Diagonals divide parallelogram into equal triangles → shortcut in proofs

Use midpoints to form smaller parallelograms → Varignon theorem

Triangles between same parallels & on same base are always equal in area

Parallelograms on same base & between same parallels → equal area

Triangle = ½ parallelogram on same base & between same parallels

Important Questions – Area of Parallelograms, Triangles, and Quadrilaterals

1.Parallelograms on Same Base and Between Same Parallels

Q1: Parallelograms ABCD and ABEF have the same base AB and lie between the same parallels AB and CF. Find the area of both and prove they are equal.

Solution:

Base AB = 7 cm, Height = 3 cm

Area of ABCD = base × height = 7 × 3 = 21 cm²

Area of ABEF = base × height = 7 × 3 = 21 cm²

Hence, Area of ABCD = Area of ABEF.

Concept: Parallelograms on same base and between same parallels are always equal in area.

Q2: ABPQ is a rectangle and ABCD is a parallelogram on the same base AB and between the same parallels. Prove their areas are equal.

Solution:

Area of ABCD = AB × height = base × height

Area of ABPQ = length × breadth = AB × height

Hence, Area of ABCD = Area of ABPQ.

Concept: Same base and same height → areas equal.

2.Triangle and Parallelogram on Same Base

Q3: Triangle DABE has base AB = 8 cm and height AE = 6 cm. Parallelogram ABCD has same base and height. Find the area of triangle and parallelogram, and relation between them.

Solution:

Area of triangle DABE = ½ × base × height = ½ × 8 × 6 = 24 cm²

Area of parallelogram ABCD = base × height = 8 × 6 = 48 cm²

Relation: Area of triangle = ½ area of parallelogram

Q4: Prove that triangle standing on same base and between same parallels = ½ parallelogram area.

Solution:

Construct a parallelogram using triangle → ABFE

Triangle = ½ parallelogram → ABFE

Parallelogram ABFE = ABCD → Triangle = ½ Area of ABCD

Hence proved.

3.Triangles on Same Base and Between Same Parallels

Q5: Prove that triangles ABE and ABC on base AB and between same parallels are equal in area.

Solution:

Construct parallelogram ABCD using triangles

Diagonal AC divides ABCD into 2 equal triangles → Area of ΔABE = Area of ΔABC

Hence, triangles on same base & between same parallels = equal area

Q6: In parallelogram ABCD, prove that ΔAPB + ΔPCD = ½ area of ABCD.

Solution:

Draw line MN through point P parallel to CD

ABNM and MNCD are parallelograms

Area of ΔAPB = ½ area of ABNM

Area of ΔPCD = ½ area of MNCD

Add → Area of ΔAPB + ΔPCD = ½ area of ABCD

4.Midpoint Theorems

Q7: ABCD is a parallelogram. E, F, G, H are midpoints of sides. Prove area of EFGH = ½ area of ABCD.

Solution:

Join midpoints → forms parallelogram EFGH (Varignon Theorem)

Each small triangle = ½ of its respective half-parallelogram

Area of EFGH = ½ (area of ABCD)

Q8: ABCD parallelogram, M and N are midpoints of BC and AD. Prove that MN divides ABCD into 2 equal parallelograms.

Solution:

MN joins midpoints → parallel to base → forms 2 equal parallelograms

Area of ABMN = Area of CDNM → proved

5.Parallelogram and Triangle Examples

Q9: PQRS parallelogram. Triangles ΔPQY and ΔQRX formed by points on PS and SR. Prove they are equal in area.

Solution:

Triangles share same base and parallel height

Area of ΔPQY = ½ area of PQRS

Area of ΔQRX = ½ area of PQRS

Hence, Area ΔPQY = Area ΔQRX

Q10: Parallelogram PQRS, SR extended to T, PT intersects QR at O. Prove Area ΔPQO = Area ΔRTO.

Solution:

Draw diagonal PR → divides into triangles with equal area on same base & height

Use whole-part axiom to subtract common areas

Area ΔPQO = Area ΔRTO → proved

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