Class 10 Mathematics Unit 10 – Triangles and Quadrilaterals | Nepal eNotes

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CLASS 10 MATHEMATICS • UNIT 10

Unit 10: Triangles and Quadrilaterals

Area theorems, constructions and geometry proofs — WordPress-ready notes reconstructed from the supplied 22-page handwritten PDF.

Original Scanned PDF – View Notes

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Source note: The uploaded filename says “Unit 10 – Indices”, but page 1 of the handwritten PDF is headed “Lesson 10: Triangles and Quadrilateral”. The scanned pages themselves are therefore used as the source of truth. Several construction pages contain mainly compass-and-ruler drawings with little or no written procedure; those drawings are summarized without inventing missing construction steps.

Important Area Formulae

Area of a Triangle

A = ½ × b × h

Area of a Parallelogram

A = b × h

Area of a Square from Diagonal

A = ½d²

Isosceles Triangle Area

A = b/4 √(4a² − b²)

A recurring theorem in the notes is: parallelograms standing on the same base and between the same parallels are equal in area. Likewise, triangles standing on the same base and between the same parallels are equal in area.

PDF Page 1

1. Find the Area of the Given Figures

(a) Triangle

Base = 12 cm, height = 4 cm

Area = ½ × 12 × 4

∴ Area = 24 cm²

(b) Parallelogram ABCD

The source applies:

Area of parallelogram = base × height

The handwritten calculation gives:

∴ Area = 24 cm²

PDF Pages 2–6

2. Area Theorems and Related Problems

(a) Square WXYZ and an equal-area parallelogram

WXYZ is a square and its diagonal is given as 2√2 cm.

Area of square = ½d²

= ½(2√2)²

= ½ × 8

∴ Area of square = 4 cm²

The source then uses the same-base / same-parallels area theorem to conclude that the required parallelogram also has area:

∴ 4 cm²

(b) Find the area of parallelogram ABCD

The source uses BE = EC = 10 cm and AD = BC = 16 cm.

For isosceles triangle EBC:

Area = b/4 √(4a² − b²)

= 16/4 √[4(10)² − (16)²]

= 4√144

= 48 cm²

Triangle EBC is half of parallelogram ABCD in the source figure.

Area of ABCD = 2 × 48

∴ Area of parallelogram ABCD = 96 cm²

(c) Parallelogram-area problem

A perpendicular height of 8 cm is used in the source, and the related triangle has area 64 cm².

The source doubles the triangle area:

Area of parallelogram = 2 × 64 = 128 cm²

Now, A = b × h:

128 = b × 8

∴ b = 16 cm

(d) Equal-area proof inside a parallelogram

Pages 4–5 contain a statement-and-reason proof using a parallelogram PQRS and points inside/on its sides. The proof repeatedly uses:

A triangle on the same base and between the same parallels has half the area of the corresponding parallelogram.
Equal areas remain equal after adding or subtracting equal areas.

The exact labels in the question statement and the final pair of triangles are partially obscured in the scan. The source visibly concludes that the two required triangular regions are equal in area, but the labels are not rewritten here with guessed letters.

(e) Two equal-area parallelogram/triangle regions

The source states BC = CE, so AC acts as a median and divides the relevant triangle into two equal-area parts.

It also uses triangles standing on the same base and between the same parallels.

Two triangular parts are each shown as 24 cm².

Area of quadrilateral ABCD = 24 + 24

∴ Area = 48 cm²

PDF Page 7

3. Trapezium Problem

The handwritten figure is a trapezium PQRS with parallel sides. The source gives lengths 18 cm, 13 cm and 23 cm and drops a perpendicular to form a right triangle.

Difference of parallel-side segments:

23 − 18 = 5 cm

Using Pythagoras in the right triangle:

13² = p² + 5²

169 − 25 = p²

p² = 144

∴ p = 12 cm

The source then calculates the required triangle area as:

½ × 18 × 12 = 108 cm²

PDF Page 8

Construction Notes

Page 8 begins the construction section. The handwritten headings indicate constructions involving:

Parallelograms equal in area
Triangles equal in area
A triangle and a parallelogram equal in area
A triangle and a quadrilateral equal in area

The source primarily records the final compass-and-straightedge diagrams. It does not consistently provide numbered construction steps, so the prompts and the intended equal-area results are preserved without fabricating procedures.

PDF Pages 9, 12

Parallelogram Constructions

Construction 1

Construct parallelogram ABCD with:

AB = 4 cm
BC = 5.5 cm
∠ABC = 60°

Then construct another parallelogram whose area is equal to the area of parallelogram ABCD.

Construction 2

Page 12 shows another construction beginning from a parallelogram with given adjacent sides and a 60° included angle, followed by construction of a triangle equal in area to the parallelogram.

The measurement text on page 12 is partly faint. The diagram clearly shows conversion between a parallelogram and an equal-area triangle.

PDF Pages 10–11, 18–19

Triangle Constructions

Triangle and equal-area parallelogram

Pages 10–11 contain a construction based on a triangle ABC with:

AB = 9 cm
BC = 6.5 cm
∠ABC = 120°

The task then constructs a parallelogram with a specified side and area equal to the area of triangle ABC.

Triangle equal in area to a rectangle

Pages 18–19 state:

Construct a triangle whose one angle is 60°.
Its area is to be equal to the area of a rectangle.
Rectangle length = 6 cm
Rectangle breadth = 4.5 cm

PDF Pages 12–17

Quadrilateral Constructions and Conversion to Equal-Area Triangles

Quadrilateral PQRS

Pages 12–13 show a quadrilateral PQRS constructed from four sides and a diagonal, followed by construction of a triangle equal in area to the quadrilateral.

Question 5(a)

Page 14 constructs another quadrilateral PQRS from given side lengths and a diagonal, then converts it into a triangle equal in area to PQRS.

Question 5(b)

Page 15 constructs quadrilateral ABCD from given side lengths and a diagonal, then constructs triangle ADE equal in area to quadrilateral ABCD.

Question 5(c)

Pages 16–17 contain further equal-area conversion constructions involving quadrilateral PQRS and a triangle.

Several measurements in pages 12–17 are faint or partially overwritten. Because the source is primarily graphical, unclear values have not been guessed. The original PDF embedded above should be used when the exact construction measurements are required.

PDF Page 18

Rhombus and Equal-Area Construction

The source asks to construct rhombus PQRS using its diagonals:

PR = 6 cm
QS = 8 cm

Then construct a parallelogram equal in area to rhombus PQRS.

PDF Page 20

Theorem: The Central Angle is Double the Inscribed Angle on the Same Arc

To prove: ∠AOB = 2∠ACB

Statement	Reason
∠ACO = ∠CAO	OA = OC, so ΔAOC is isosceles.
Exterior angle at O = ∠ACO + ∠CAO = 2∠ACO	Exterior angle of a triangle equals the sum of the two opposite interior angles.
Similarly, the corresponding central part on arc BC = 2∠BCO.	Same argument in the second isosceles triangle.
∠AOB = 2(∠ACO + ∠BCO)	Add the two central-angle parts.
∠AOB = 2∠ACB	∠ACB = ∠ACO + ∠OCB.

PDF Page 21

Theorem: Angles on the Circumference Based on the Same Arc are Equal

To prove: ∠ADB = ∠ACB

Statement	Reason
∠AOB = 2∠ADB	Central angle is double the circumference angle standing on the same arc AB.
∠AOB = 2∠ACB	Same theorem on the same arc AB.
2∠ADB = 2∠ACB	From the first two statements.
∠ADB = ∠ACB	Dividing both sides by 2.

PDF Page 22

Theorem: Opposite Angles of a Cyclic Quadrilateral are Supplementary

To prove: ∠ADC + ∠ABC = 180°

Statement	Reason
Reflex ∠AOC = 2∠ABC	Central angle is double the angle at the circumference standing on the corresponding arc AC.
Minor ∠AOC = 2∠ADC	Central angle is double the angle at the circumference standing on the other arc AC.
2∠ABC + 2∠ADC = 360°	Minor angle AOC + reflex angle AOC = 360°.
∠ABC + ∠ADC = 180°	Divide both sides by 2.