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25 questions

Unit 11 · Construction

Worksheet to print (PDF, 4 pages) Answer key for teachers

Harder worksheet (PDF, 3 pages) Answer key for teachers

Question 1: parallelograms

Construct a parallelogram with an angle of 45° and the same area as the parallelogram ABCD with AB = 4 cm, AD = 6 cm and ∠BAD = 60°.

Answer

Draw AB = 4 cm.
At A construct ∠BAD = 60°, and cut AD = 6 cm on its arm.
With center B and radius 6 cm, and center D and radius 4 cm, draw arcs that cut at C. Join BC and DC.
Produce DC both ways to make the line ST.
At A construct ∠BAP = 45°; its arm meets ST at P. Cut PQ = AB = 4 cm on PT and join BQ: ABQP is the parallelogram.

Figure 11.8

ABQP and ABCD stand on the base AB between $AB \parallel ST$ , so they are equal in area.

Construct a parallelogram with an angle of 75° and the same area as the parallelogram ABCD with AB = 5 cm, AD = 6 cm and the diagonal BD = 6 cm.

Answer

Draw AB = 5 cm.
With center A and radius 6 cm, and center B and radius 6 cm, draw arcs that cut at D. Join AD and BD.
With center B and radius 6 cm, and center D and radius 5 cm, draw arcs that cut at C. Join BC and DC: ABCD is the parallelogram.
Produce CD both ways to make the line ST.
At A construct ∠BAP = 75°; its arm meets ST at P. Cut PQ = AB = 5 cm along ST and join BQ: ABQP is the parallelogram.

Figure 11.9

The same base AB, between $AB \parallel ST$ : equal areas.

Construct a parallelogram with a diagonal of 7.2 cm and the same area as a parallelogram whose diagonals are 6 cm and 4.8 cm, with an angle of 30° between them.

Hint

The diagonals of a parallelogram bisect each other: start with one diagonal and its midpoint.

Answer

Draw the diagonal AC = 6 cm and mark its midpoint O by bisecting it.
At O construct an angle of 30° with OC and draw the line through O both ways. Cut OB = OD = 2.4 cm on it, on either side of O.
Join AB, BC, CD and DA: ABCD is the parallelogram.
Produce DC both ways to make the line ST, parallel to AB.
With center A and radius 7.2 cm, cut ST at X. With center X and radius AB, cut ST at Y between C and X. Join BX and AY: ABXY is the parallelogram, and its diagonal AX is 7.2 cm.

Figure 11.10

ABXY and ABCD stand on the base AB between $AB \parallel ST$ , so they are equal in area.

Construct a parallelogram ABXY equal in area to the parallelogram ABCD with AB = 5 cm, AD = 6 cm and the diagonal BD = 8 cm.

Answer

Draw AB = 5 cm.
With center A and radius 6 cm, and center B and radius 8 cm, draw arcs that cut at D. Join AD and BD.
With center B and radius 6 cm, and center D and radius 5 cm, draw arcs that cut at C. Join BC and DC: ABCD is the parallelogram.
Produce DC both ways to make the line ST.
Any parallelogram on AB with its top on ST will do. Here: at B construct ∠ABX = 60°, meeting ST at X; with center X and radius BA, cut ST at Y and join AY: ABXY is the parallelogram.

Figure 11.11

No side or angle is given for ABXY, so any point X on ST will do; the same base and the same parallels make the areas equal.

Construct a parallelogram with an angle of 30° and the same area as the parallelogram PQRS with PQ = 4.2 cm, QR = 6 cm and ∠PQR = 60°.

Answer

Draw QR = 6 cm.
At Q construct ∠RQP = 60°, and cut QP = 4.2 cm on its arm.
With center R and radius 4.2 cm, and center P and radius 6 cm, draw arcs that cut at S. Join RS and PS.
Produce PS both ways to make the line MN.
At Q construct ∠RQX = 30°; its arm meets MN at X. Cut XY = QR = 6 cm along it and join RY: QRYX is the parallelogram.

Figure 11.12

QRYX and PQRS stand on the base QR between $QR \parallel MN$ : equal areas.

Question 2: triangles

Construct △ABC with BC = 6.4 cm, AB = 5.6 cm and AC = 6 cm. Then construct a triangle with a side of 7 cm equal in area to △ABC.

Answer

Draw BC = 6.4 cm.
With center B and radius 5.6 cm, and center C and radius 6 cm, draw arcs that cut at A. Join AB and AC.
Through A draw ST parallel to BC, making ∠BAS = ∠ABC.
With center B and radius 7 cm, cut ST at D. Join DB and DC: △DBC, with DB = 7 cm, has the same area as △ABC.

Figure 11.13

The same base BC, between $BC \parallel ST$ : equal areas.

Construct △LMN with LM = 4.3 cm, ∠NLM = 30° and ∠LMN = 45°. Then construct △OLM with OM = 7.5 cm, equal in area to △LMN.

Answer

Draw LM = 4.3 cm.
At L construct ∠NLM = 30° and at M construct ∠LMN = 45°; the arms meet at N.
Through N draw ST parallel to LM, making ∠MNT = ∠NML.
With center M and radius 7.5 cm, cut the line at O. Join OL and OM: △OLM has the same area as △LMN.

Figure 11.14

The same base LM, between $LM \parallel ST$ : equal areas.

Construct △PQR with PQ = 4.5 cm, QR = 7 cm and PR = 6 cm. Then construct a triangle with a side of 8 cm equal in area to △PQR.

Answer

Draw QR = 7 cm.
With center Q and radius 4.5 cm, and center R and radius 6 cm, draw arcs that cut at P. Join PQ and PR.
Through P draw a line parallel to QR, making ∠QPX = ∠PQR.
With center Q and radius 8 cm, cut the line at S. Join SQ and SR: △SQR, with SQ = 8 cm, has the same area as △PQR.

Figure 11.15

The same base QR, between $QR \parallel XY$ : equal areas.

Construct a triangle equal in area to △ABC with AB = 4.2 cm, BC = 5.2 cm and CA = 3.5 cm.

Answer

Draw BC = 5.2 cm.
With center B and radius 4.2 cm, and center C and radius 3.5 cm, draw arcs that cut at A. Join AB and AC.
Through A draw ST parallel to BC, making ∠BAS = ∠ABC.
Any point of ST makes a triangle on BC of the same area. Here: at B construct ∠CBD = 90°, meeting ST at D. Join DC: △DBC is right-angled at B.

Figure 11.16

Nothing is given for the new triangle, so any point of ST other than A will do; here a right angle at B.

Construct a triangle MBC with one side 8 cm, equal in area to △ABC with AB = 7.2 cm, BC = 5.9 cm and CA = 6.1 cm.

Answer

Draw BC = 5.9 cm.
With center B and radius 7.2 cm, and center C and radius 6.1 cm, draw arcs that cut at A. Join AB and AC.
Through A draw ST parallel to BC, making ∠BAS = ∠ABC.
With center B and radius 8 cm, cut ST at M. Join MB and MC: △MBC, with MB = 8 cm, has the same area as △ABC.

Figure 11.17

The same base BC, between $BC \parallel ST$ : equal areas.

Question 3: a parallelogram equal to a triangle

Construct △PQR with PQ = 6.5 cm, QR = 6 cm and PR = 5.5 cm. Then construct a parallelogram RSTI with ∠TSR = 75°, equal in area to △PQR.

Answer

Draw QR = 6 cm.
With center Q and radius 6.5 cm, and center R and radius 5.5 cm, draw arcs that cut at P. Join PQ and PR.
Through P draw a line parallel to QR, making alternate angles equal.
Bisect QR at S.
At S construct ∠TSR = 75°; its arm meets the line at T.
With center T and radius SR, cut the line at I; join RI: RSTI is the parallelogram.

Figure 11.18

S is the midpoint of QR, so △PSR is half of △PQR and also half of RSTI (the base SR, between QR and the parallel): area of RSTI = area of △PQR.

Construct △ABC with AC = 5 cm, BC = 4.8 cm and ∠ABC = 45°. Then construct a parallelogram CDEF with CD = 7.5 cm, equal in area to △ABC.

Answer

Draw BC = 4.8 cm.
At B construct ∠ABC = 45°. With center C and radius 5 cm, cut its arm at A. Join AC.
Through A draw ST parallel to BC, making ∠BAS = ∠ABC.
Bisect BC at F.
With center C and radius 7.5 cm, cut ST at D; with center D and radius CF, cut ST at E on the side of A. Join CD and FE: CDEF is the parallelogram.

Figure 11.19

CDEF stands on half the base, CF, between $BC \parallel ST$ : its area is twice △AFC, which is △ABC.

Construct △ABC with AB = 4 cm, BC = 3.2 cm and AC = 3.5 cm. Then construct a parallelogram BXYE with BE = 5 cm, equal in area to △ABC.

Answer

Draw BC = 3.2 cm.
With center B and radius 4 cm, and center C and radius 3.5 cm, draw arcs that cut at A. Join AB and AC.
Through A draw ST parallel to BC, making ∠BAS = ∠ABC.
Bisect BC at X.
With center B and radius 5 cm, cut ST at E; with center E and radius BX, cut ST at Y on the side of T. Join BE and XY: BXYE is the parallelogram.

Figure 11.20

Half the base, BX, between $BC \parallel ST$ : area of BXYE = 2 × area of △ABX = area of △ABC.

Construct a rectangle equal in area to △XYZ with XY = 4 cm, YZ = 6.8 cm and ZX = 6.5 cm.

Answer

Draw YZ = 6.8 cm.
With center Y and radius 4 cm, and center Z and radius 6.5 cm, draw arcs that cut at X. Join XY and XZ.
Through X draw ST parallel to YZ, making ∠YXS = ∠XYZ.
Bisect YZ at M; the bisector, perpendicular to YZ, meets ST at N.
With center N and radius MZ, cut ST at K on the side of Z. Join ZK: MZKN is the rectangle.

Figure 11.21

MZKN has right angles at M (the perpendicular bisector) and stands on half the base, MZ, between $YZ \parallel ST$ : its area is that of △XYZ.

Construct △PQR with PQ = 7.1 cm, ∠RPQ = 60° and PR = 5.7 cm. Then construct a parallelogram with one side 7.5 cm, equal in area to △PQR.

Answer

Draw PQ = 7.1 cm.
At P construct ∠QPR = 60°; cut PR = 5.7 cm on its arm. Join RQ.
Through R draw a line parallel to PQ, making alternate angles equal.
Bisect PQ at M.
With center P and radius 7.5 cm, cut the line at S; with center S and radius PM, cut it at T beyond S. Join PS and MT: PMTS is the parallelogram, with PS = 7.5 cm.

Figure 11.22

Half the base, PM, between PQ and the parallel through R: area of PMTS = area of △PQR.

Question 4: a triangle equal to a parallelogram

Construct a parallelogram ABCD with AB = 6 cm, BC = 4.5 cm and ∠DAB = 60°. Then construct △AEF with FE = 7.5 cm, equal in area to the parallelogram.

Answer

Draw AB = 6 cm.
At A construct ∠BAD = 60°, and cut AD = 4.5 cm on its arm.
With center B and radius 4.5 cm, and center D and radius 6 cm, draw arcs that cut at C. Join BC and DC.
Produce AB to E with BE = AB = 6 cm (an arc from B). Produce DC to T.
With center E and radius 7.5 cm, cut DT at F. Join AF and EF: △AEF is the triangle.

Figure 11.23

△AEF stands on twice the base, AE = 2AB, between $AE \parallel DT$ : area of △AEF = ½ × 2 × AB × height = area of ABCD.

Construct a parallelogram ABCD with AB = 6 cm, BC = 4 cm and ∠BAD = 45°. Then construct △APQ with ∠APQ = 60°, equal in area to ABCD.

Answer

Draw AB = 6 cm.
At A construct ∠BAD = 45°, and cut AD = 4 cm on its arm.
With center B and radius 4 cm, and center D and radius 6 cm, draw arcs that cut at C. Join BC and DC.
Produce AB to P with BP = AB = 6 cm. Produce DC to T.
At P construct ∠APQ = 60°; its arm meets DT at Q. Join AQ: △APQ is the triangle.

Figure 11.24

Twice the base, AP = 2AB, between $AP \parallel DT$ : equal areas.

Construct a triangle equal in area to the parallelogram PQRS with PQ = 5 cm and the diagonals PR = 6 cm and QS = 8 cm.

Hint

The diagonals bisect each other, so △PQO has sides 5 cm, 3 cm and 4 cm.

Answer

Draw PQ = 5 cm. The diagonals bisect each other, so they meet at O with PO = 3 cm and QO = 4 cm: draw arcs of those radii from P and Q to cut at O.
Produce PO to R and QO to S so that OR = PO and OS = QO. Join QR, RS and SP: PQRS is the parallelogram.
Produce PQ to E with QE = PQ = 5 cm. Join SE: △PES, on twice the base and between the same parallels, has the same area as PQRS.

Figure 11.25

△PES stands on twice the base, PE = 2PQ, between $PE \parallel SR$ : its area is ½ × 2PQ × height = area of PQRS.
(Since 3² + 4² = 5², the diagonals cross at a right angle and PQRS is a rhombus.)

Construct a triangle equal in area to the parallelogram EFGH with EF = 5 cm, FG = 4 cm and ∠EFG = 120°.

Answer

Draw FG = 4 cm.
At F construct ∠GFE = 120°, and cut FE = 5 cm on its arm.
With center G and radius 5 cm, and center E and radius 4 cm, draw arcs that cut at H. Join GH and EH.
Produce FG to K with GK = FG = 4 cm. Join EK: △FKE, on twice the base FG and between the same parallels, has the same area as EFGH.

Figure 11.26

△FKE stands on twice the base, FK = 2FG, between $FK \parallel EH$ : equal areas.

Construct a triangle equal in area to the parallelogram IJKL with IJ = 5 cm, IK = 6 cm and JL = 8 cm.

Answer

Draw IJ = 5 cm. With center I and radius 3 cm (half of IK) and center J and radius 4 cm (half of JL), draw arcs that cut at O.
Produce IO to K and JO to L with OK = IO and OL = JO. Join JK, KL and LI: IJKL is the parallelogram.
Produce LK to N with KN = LK, and join IN: △ILN, with twice the side LK for its base and its corner I on the line IJ, has the same area as IJKL.

Figure 11.27

IK and JL are the diagonals; as in c, they cross at right angles.

Question 5: a triangle equal to a quadrilateral

Construct quadrilateral PQRS with PQ = QR = 5.5 cm, RS = SP = 4.5 cm and ∠SPQ = 75°. Then construct △PST equal in area to PQRS.

Answer

Draw PQ = 5.5 cm.
At P construct ∠SPQ = 75° and cut PS = 4.5 cm on its arm.
With center S and radius 4.5 cm, and center Q and radius 5.5 cm, draw arcs that cut at R. Join SR and QR: PQRS is the quadrilateral.
Draw the diagonal SQ. Through R draw a line parallel to SQ, making ∠QRT = ∠SQR; produce PQ to meet it at T.
Join ST: △PST has the same area as PQRS.

Figure 11.28

△QRS and △QTS stand on the base SQ, between SQ and the line $RT \parallel SQ$ , so they are equal; adding △PQS gives area of △PST = area of PQRS.

Construct quadrilateral ABCD with AB = 4.5 cm, BC = 5.5 cm, CD = 5.7 cm, DA = 4.9 cm and the diagonal BD = 5.9 cm. Then construct △DAE equal in area to ABCD.

Answer

Draw AB = 4.5 cm.
With center A and radius 4.9 cm, and center B and radius 5.9 cm, draw arcs that cut at D. Join AD and BD.
With center B and radius 5.5 cm, and center D and radius 5.7 cm, draw arcs that cut at C. Join BC and DC: ABCD is the quadrilateral.
Through C draw a line parallel to DB, making ∠BCE = ∠DBC; produce AB to meet it at E.
Join DE: △DAE has the same area as ABCD.

Figure 11.29

△BCD = △BED (the base BD, between $BD \parallel CE$ ); add △ABD.

Construct △QRT equal in area to the quadrilateral PQRS with PQ = 5 cm, QR = 7 cm, RS = 4.5 cm, SP = 5.4 cm and the diagonal QS = 6.5 cm.

Answer

Draw QR = 7 cm.
With center Q and radius 6.5 cm, and center R and radius 4.5 cm, draw arcs that cut at S. Join QS and RS.
With center Q and radius 5 cm, and center S and radius 5.4 cm, draw arcs that cut at P. Join QP and SP: PQRS is the quadrilateral.
Through P draw a line parallel to QS, making alternate angles equal; produce RS to meet it at T.
Join QT: △QRT has the same area as PQRS.

Figure 11.30

△QPS = △QTS (the base QS, between $QS \parallel PT$ ); add △QRS.

Construct a rhombus PQRS with the diagonals PR = 6 cm and QS = 8 cm. Then construct △PSA equal in area to the rhombus.

Hint

The diagonals of a rhombus bisect each other at right angles.

Answer

Draw the diagonal PR = 6 cm and its perpendicular bisector, which crosses it at O.
Cut OQ = OS = 4 cm on the bisector. Join PQ, QR, RS and SP: PQRS is the rhombus.
Through Q draw a line parallel to PR, making alternate angles equal; produce SR to meet it at A.
Join PA: △PSA has the same area as the rhombus.

Figure 11.31

△PQR = △PAR (the base PR, between $PR \parallel QA$ ); add △PRS. The rhombus’s area is ½ × 6 × 8 = 24 cm², and so is the triangle’s.

Construct a triangle with one angle of 60° and the same area as a rectangle 6 cm long and 4.5 cm broad.

Answer

Draw AB = 6 cm. At A and at B construct right angles; cut AD = BC = 4.5 cm on them and join DC: ABCD is the rectangle.
Produce AB to E with BE = AB = 6 cm. Produce DC to T.
At E construct ∠AEF = 60°; its arm meets DT at F. Join AF: △AEF has an angle of 60° and the rectangle’s area.

Figure 11.32

Twice the base, AE = 2AB, between $AE \parallel DT$ : area of △AEF = ½ × 12 × 4.5 = 27 cm² = 6 × 4.5, the rectangle’s area.