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 is used to construct 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 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 its area.
• If a parallelogram is made on half the base of a triangle between the same parallels, then triangle area = area of parallelogram.
• This helps construct triangles equal in area to given parallelograms and vice-versa.
Topic 3: Constructing equal-area parallelograms
Notes:
• To construct a parallelogram equal in area to a given one, draw a line through a vertex parallel to the opposite side.
• Mark a required length on that parallel line (example: new side or angle).
• Join corresponding vertices to complete the parallelogram.
• Any parallelogram drawn on same base and between same parallels will automatically have equal area.
Topic 4: Constructing equal-area triangles
Notes:
• To make a triangle equal in area to a given triangle, draw a line from a vertex parallel to the base.
• Choose any point on this parallel line for the new vertex such that the given condition (side length or angle) is satisfied.
• Join with the endpoints of the base to get the new triangle.
• Any triangle on the same base and between 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 a quadrilateral is obtained by shifting one of these triangles using parallel lines and combining them into a single triangle.
• The equality is based on triangles on the same base and same parallels.
Five important solved questions
Construct a parallelogram with angle 45° equal in area to parallelogram AB = 4 cm, AD = 6 cm, angle BAD = 60°.
Solution:
• Construct original parallelogram ABCD.
• Extend CD.
• At A, draw a line making 45°.
• Mark point P on that line such that AP = AD.
• From P draw a line parallel to AB to meet extended CD at Q.
• ABPQ is the required equal-area parallelogram because both stand on base AB and lie between parallels.
Construct a triangle with a 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 line, mark point D so that AD = 7 cm.
• Join D to C.
• Triangle DBC has same base BC and lies between same parallels, so area is equal.
Construct a parallelogram equal in area to triangle PQR (PQ = 6.5 cm, QR = 6 cm, PR = 5.5 cm) with angle TSR = 75°.
Solution:
• Construct triangle PQR.
• Use base QR.
• Draw a line through Q parallel to PR.
• At R draw 75° and meet this parallel at S.
• From S draw a line parallel to QR meeting extension of PR at T.
• RSTQ is the required equal-area parallelogram since triangle area is half of parallelogram area on same base.
Construct a triangle equal in area to parallelogram EF = 5 cm, FG = 4 cm, angle EFG = 120°.
Solution:
• Construct the given parallelogram EFGH.
• Use base EF.
• Extend EF.
• At G draw a line parallel to EF meeting extension of EF at P.
• Triangle EFP has same height as parallelogram, but base is double, so their areas match if PF = FG + FH.
• Join P to F to get required triangle.
Construct triangle equal in area to quadrilateral PQRS (PQ = QR = 5.5 cm, RS = SP = 4.5 cm, angle SPQ = 75°).
Solution:
• Construct PQRS.
• Draw diagonal QS.
• From R draw line parallel to QS.
• Extend PQ to meet this line at T.
• Triangle PST has area equal to quadrilateral PQRS because triangle SQR shifts to triangle SQT.
Visit this link for further practice!!
https://besidedegree.com/exam/s/academic