Grade 10 Notes of Construction|| Compulsory Mathematics

Topic 1: Parallelograms on the same base and between the same parallels

Notes:
• If two parallelograms stand on the same base and lie between the same parallels, their areas are equal.
• This property allows constructing new parallelograms with the same area by shifting a vertex along a line parallel to the opposite side.
• Changing the angle or side of the parallelogram does not affect the area if the height from the same base remains unchanged.

Topic 2: Triangle and Parallelogram area relationship

Notes:
• A triangle standing on the same base and between the same parallels as a parallelogram has half the area of the parallelogram.
• If a parallelogram is made on half the base of a triangle between the same parallels, then the triangle area equals the parallelogram.
• This helps in constructing triangles equal in area to given parallelograms and vice versa.

Topic 3: Constructing equal-area parallelograms

Notes:
• Draw a line through a vertex parallel to the opposite side.
• Mark a required length on that parallel line (new side or angle).
• Join corresponding vertices to complete the parallelogram.
• Any parallelogram drawn on the same base and between the same parallels automatically has equal area.

Topic 4: Constructing equal-area triangles

Notes:
• Draw a line from a vertex parallel to the base of the given triangle.
• Choose any point on this parallel line for the new vertex to satisfy given conditions (side length or angle).
• Join with the endpoints of the base to get the new triangle.
• Any triangle on the same base and between the same parallels as the original has equal area.

Topic 5: Triangle and quadrilateral equal-area constructions

Notes:
• A quadrilateral can be split into two triangles using a diagonal.
• A triangle equal to the quadrilateral is obtained by shifting one of these triangles using parallel lines and combining them into a single triangle.
• Equality is based on triangles on the same base and same parallels.

Important Solved Constructions

1.Construct a parallelogram with one angle 45° equal in area to parallelogram AB = 4 cm, AD = 6 cm, ∠BAD = 60°
Solution:
• Draw original parallelogram ABCD.
• Extend side CD.
• At vertex A, draw a line making 45°.
• Mark point P on this line such that AP = AD.
• From P, draw a line parallel to AB to meet extended CD at Q.
• Join A to P and B to Q. Parallelogram ABPQ is equal in area to ABCD.

2.Construct a triangle with side 7 cm equal in area to triangle ABC (BC = 6.4 cm, AB = 5.6 cm, AC = 6 cm)
Solution:
• Construct triangle ABC.
• Draw a line from A parallel to BC.
• On this parallel, mark D such that AD = 7 cm.
• Join D to C. Triangle DBC has the same base BC and lies between the same parallels, so area is equal to triangle ABC.

3.Construct a parallelogram equal in area to triangle PQR (PQ = 6.5 cm, QR = 6 cm, PR = 5.5 cm) with angle 75°
Solution:
• Draw triangle PQR.
• Draw a line through Q parallel to PR.
• At R, draw 75° to meet this parallel at S.
• Draw a line from S parallel to QR meeting the extension of PR at T.
• Parallelogram RSTQ is equal in area to triangle PQR (triangle = ½ parallelogram on same base).

4.Construct a triangle equal in area to parallelogram EF = 5 cm, FG = 4 cm, ∠EFG = 120°
Solution:
• Construct parallelogram EFGH.
• Extend EF.
• Draw a line from G parallel to EF meeting the extension of EF at P.
• Join P to F. Triangle EFP has the same area as parallelogram EFGH.

5.Construct a triangle equal in area to quadrilateral PQRS (PQ = QR = 5.5 cm, RS = SP = 4.5 cm, ∠SPQ = 75°)
Solution:
• Construct quadrilateral PQRS.
• Draw diagonal QS.
• From R, draw a line parallel to QS.
• Extend PQ to meet this line at T.
• Triangle PST has area equal to quadrilateral PQRS.

6.onstruct parallelogram equal in area to a triangle ABC (AB = 4 cm, BC = 6.5 cm, ∠ABC = 120°), parallelogram side PB = 5 cm
Solution:
• Draw triangle ABC.
• Draw line ST parallel to BC from A.
• Bisect BC, mark midpoint D. Draw arc of radius 5 cm from B along ST to cut at P.
• Draw an arc of radius BP from D on line PT to mark Q.
• Join D and Q. Parallelogram PBDQ has area equal to triangle ABC.

7.Construct triangle equal in area to parallelogram ABCD (AB = 5 cm, BC = 4 cm, ∠ABC = 60°, triangle side PB = 6 cm)
Solution:
• Draw parallelogram ABCD.
• Extend BC to E such that CE = BC.
• Draw line ST through AD.
• Draw an arc of radius 6 cm from B along ST to mark P.
• Join B to P and P to E. Triangle BPE has area equal to parallelogram ABCD.

8.Construct triangle equal in area to quadrilateral PQRS (PQ = 5 cm, PS = 4 cm, QR = 4.4 cm, RS = 3.6 cm, diagonal PR = 6.4 cm)
Solution:
• Draw quadrilateral PQRS.
• Draw diagonal SQ.
• From R, draw a line parallel to SQ and extend PQ to meet it at H.
• Join S to G (intersection point). Triangle PSG has area equal to quadrilateral PQRS.

9.Construct rectangle equal in area to triangle ABC (AB = 2.6 cm, BC = 3.4 cm, CA = 4 cm)
Solution:
• Draw triangle ABC with given sides.
• Draw rectangle CDEF such that area = area of triangle ABC.
• Use base BC as rectangle length and calculate breadth using formula: rectangle area = triangle area × 2 / base.

10.Construct triangle equal in area to a parallelogram (EF = 5 cm, FG = 4 cm, ∠EFG = 120°)
Solution:
• Draw parallelogram EFGH.
• Extend EF.
• Draw line from G parallel to EF.
• Join intersection point to F to get triangle with area equal to parallelogram.

Key Area Theorems and Formulas

Formulas:
• Parallelogram = base × height
• Triangle = ½ × base × height
• Trapezium = ½ × (sum of parallel sides) × height
• Rhombus = ½ × d₁ × d₂

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