Std 8 Mathematics · Chapter 8

8. Quadrilateral: Constructions and Types

Solutions to Practice Sets 8.1, 8.2 and 8.3, with construction steps and figures drawn to scale. Maharashtra State Board (Balbharati), Standard Eight Mathematics.

Practice Set 8.1

1Construct the following quadrilaterals of given measures.

In each case, first draw a rough figure and mark the given measures on it. It shows which triangle or which angles to draw first.

(1) In □MORE, l(MO) = 5.8 cm, l(OR) = 4.4 cm, m∠M = 58°, m∠O = 105°, m∠R = 90°.

Two adjacent sides and three angles are given.

Step 1: Draw seg MO of length 5.8 cm.

Step 2: At O, draw a ray making an angle of 105° with seg MO. On it, mark point R such that OR = 4.4 cm.

Step 3: At M, draw a ray making an angle of 58° with seg MO, on the same side as R.

Step 4: At R, draw a ray making an angle of 90° with seg RO.

Step 5: Name the point where the rays from M and R meet as E. □MORE is the required quadrilateral.

Figure drawn to scale. The fourth angle, ∠E, measures 360° − (58° + 105° + 90°) = 107°.

(2) Construct □DEFG such that l(DE) = 4.5 cm, l(EF) = 6.5 cm, l(DG) = 5.5 cm, l(DF) = 7.2 cm, l(EG) = 7.8 cm.

Three sides and both diagonals are given. The sides of △DEF and △DEG are all known.

Step 1: Draw seg DE of length 4.5 cm.

Step 2: With centre D and radius 7.2 cm, draw an arc above DE. With centre E and radius 6.5 cm, draw another arc cutting it. Name the point F.

Step 3: With centre D and radius 5.5 cm, draw an arc on the same side. With centre E and radius 7.8 cm, draw an arc cutting it. Name the point G.

Step 4: Draw seg EF, seg FG and seg GD. □DEFG is the required quadrilateral. (DF and EG are its diagonals.)

Figure drawn to scale. The small grey arcs show where F and G are located.

(3) In □ABCD, l(AB) = 6.4 cm, l(BC) = 4.8 cm, m∠A = 70°, m∠B = 50°, m∠C = 140°.

Two adjacent sides and three angles are given.

Step 1: Draw seg AB of length 6.4 cm.

Step 2: At B, draw a ray making an angle of 50° with seg BA. On it, mark point C such that BC = 4.8 cm.

Step 3: At A, draw a ray making an angle of 70° with seg AB, on the same side as C.

Step 4: At C, draw a ray making an angle of 140° with seg CB.

Step 5: Name the point where the rays from A and C meet as D. □ABCD is the required quadrilateral.

Figure drawn to scale. ∠D measures 360° − (70° + 50° + 140°) = 100°.

(4) Construct □LMNO such that l(LM) = l(LO) = 6 cm, l(ON) = l(NM) = 4.5 cm, l(OM) = 7.5 cm.

Four sides and the diagonal OM are given. The sides of △LOM and △NOM are known.

Step 1: Draw seg OM of length 7.5 cm.

Step 2: With centres O and M and radius 6 cm each, draw arcs on one side of OM cutting each other. Name the point L.

Step 3: With centres O and M and radius 4.5 cm each, draw arcs on the other side of OM cutting each other. Name the point N.

Step 4: Draw seg LM, seg LO, seg ON and seg NM. □LMNO is the required quadrilateral.

Figure drawn to scale. Two pairs of adjacent sides are equal, so □LMNO is a kite.

Practice Set 8.2

1Draw a rectangle ABCD such that l(AB) = 6.0 cm and l(BC) = 4.5 cm.

Step 1: Draw seg AB of length 6.0 cm.

Step 2: At A and at B, draw perpendiculars to seg AB on the same side.

Step 3: On them, mark D and C such that AD = BC = 4.5 cm.

Step 4: Draw seg DC. □ABCD is the required rectangle.

Figure drawn to scale.

2Draw a square WXYZ with side 5.2 cm.

Step 1: Draw seg WX of length 5.2 cm.

Step 2: At W and at X, draw perpendiculars to seg WX on the same side.

Step 3: On them, mark Z and Y such that WZ = XY = 5.2 cm.

Step 4: Draw seg ZY. □WXYZ is the required square.

Figure drawn to scale.

3Draw a rhombus KLMN such that its side is 4 cm and m∠K = 75°.

Step 1: Draw seg KL of length 4 cm.

Step 2: At K, draw a ray making an angle of 75° with seg KL. On it, mark N such that KN = 4 cm.

Step 3: With centre L and radius 4 cm, draw an arc. With centre N and radius 4 cm, draw an arc cutting it. Name the point M.

Step 4: Draw seg LM and seg NM. □KLMN is the required rhombus.

Figure drawn to scale. Opposite angles are equal (75°, 75°) and adjacent angles are supplementary (75° + 105° = 180°).

4If the diagonal of a rectangle is 26 cm and one side is 24 cm, find the other side.

Let □ABCD be the rectangle with diagonal AC = 26 cm and side AB = 24 cm. △ABC is a right angled triangle with ∠B = 90°.

By Pythagoras theorem, l(BC)² = l(AC)² − l(AB)²
= 26² − 24² = 676 − 576 = 100
∴ l(BC) = √100 = 10 cm

Answer: The other side is 10 cm.

5Lengths of diagonals of a rhombus ABCD are 16 cm and 12 cm. Find the side and perimeter of the rhombus.

Let the diagonals AC = 16 cm and BD = 12 cm meet at P. The diagonals of a rhombus are perpendicular bisectors of each other.

∴ l(AP) = 16 ÷ 2 = 8 cm, l(BP) = 12 ÷ 2 = 6 cm and ∠APB = 90°.

In △APB, by Pythagoras theorem, l(AB)² = 8² + 6² = 64 + 36 = 100 ∴ l(AB) = 10 cm

Perimeter = 4 × side = 4 × 10 = 40 cm

Answer: Side = 10 cm, perimeter = 40 cm

6Find the length of the diagonal of a square with side 8 cm.

Let □PQRS be the square with side 8 cm and diagonal PR. In △PQR, ∠Q = 90°.

By Pythagoras theorem, l(PR)² = l(PQ)² + l(QR)² = 8² + 8² = 64 + 64 = 128
∴ l(PR) = √128 cm

Answer: √128 cm

√128 = √(64 × 2) = 8√2 cm, which is about 11.31 cm.

7Measure of one angle of a rhombus is 50°. Find the measures of the remaining three angles.

Let □ABCD be the rhombus with m∠A = 50°.

Opposite angles of a rhombus are congruent, so m∠C = m∠A = 50°.
Adjacent angles are supplementary (a rhombus is a parallelogram), so m∠B = 180° − 50° = 130°.
m∠D = m∠B = 130° (opposite angles).

Answer: The remaining angles are 130°, 50° and 130°.

Practice Set 8.3

1Measures of opposite angles of a parallelogram are (3x − 2)° and (50 − x)°. Find the measure of each angle.

Opposite angles of a parallelogram are congruent.

∴ 3x − 2 = 50 − x
∴ 3x + x = 50 + 2 ∴ 4x = 52 ∴ x = 13

One angle = (3x − 2)° = 3 × 13 − 2 = 37°. Its opposite angle = (50 − x)° = 50 − 13 = 37°.

Adjacent angles are supplementary, so each of the other two angles = 180° − 37° = 143°.

Answer: 37°, 143°, 37°, 143°

2Referring to the adjacent figure of a parallelogram, write the answers of the questions given below.
(1) If l(WZ) = 4.5 cm then l(XY) = ? (2) If l(YZ) = 8.2 cm then l(XW) = ? (3) If l(OX) = 2.5 cm then l(OZ) = ? (4) If l(WO) = 3.3 cm then l(WY) = ? (5) If m∠WZY = 120° then m∠WXY = ? and m∠XWZ = ?

Parallelogram WXYZ; its diagonals WY and XZ meet at O.

(1) l(XY) = l(WZ) = 4.5 cm … opposite sides are congruent.
(2) l(XW) = l(YZ) = 8.2 cm … opposite sides are congruent.
(3) l(OZ) = l(OX) = 2.5 cm … diagonals bisect each other.
(4) l(WY) = 2 × l(WO) = 2 × 3.3 = 6.6 cm … diagonals bisect each other.
(5) m∠WXY = m∠WZY = 120° … opposite angles are congruent.
m∠XWZ = 180° − 120° = 60° … adjacent angles are supplementary.

Answer: (1) 4.5 cm (2) 8.2 cm (3) 2.5 cm (4) 6.6 cm (5) 120°, 60°

3Construct a parallelogram ABCD such that l(BC) = 7 cm, m∠ABC = 40°, l(AB) = 3 cm.

Step 1: Draw seg BC of length 7 cm.

Step 2: At B, draw a ray making an angle of 40° with seg BC. On it, mark A such that BA = 3 cm.

Step 3: With centre A and radius 7 cm, draw an arc. With centre C and radius 3 cm, draw an arc cutting it. Name the point D. (Opposite sides of a parallelogram are equal.)

Step 4: Draw seg AD and seg CD. □ABCD is the required parallelogram.

Figure drawn to scale. ∠A = ∠C = 140° and ∠D = 40°.

4Ratio of consecutive angles of a quadrilateral is 1 : 2 : 3 : 4. Find the measure of each angle. Write, with reason, what type of quadrilateral it is.

Let the consecutive angles be ∠A, ∠B, ∠C, ∠D, with measures x, 2x, 3x and 4x.

Sum of the angles of a quadrilateral = 360°
∴ x + 2x + 3x + 4x = 360° ∴ 10x = 360° ∴ x = 36°

∴ m∠A = 36°, m∠B = 72°, m∠C = 108°, m∠D = 144°

Type: m∠A + m∠D = 36° + 144° = 180°, so side AB ∥ side DC (interior angles are supplementary).
m∠A + m∠B = 36° + 72° = 108° ≠ 180°, so side AD is not parallel to side BC.

Only one pair of opposite sides is parallel.

Answer: 36°, 72°, 108°, 144°; the quadrilateral is a trapezium.

5Construct □BARC such that l(BA) = l(BC) = 4.2 cm, l(AC) = 6.0 cm, l(AR) = l(CR) = 5.6 cm.

Step 1: Draw seg AC of length 6.0 cm (this is a diagonal).

Step 2: With centres A and C and radius 4.2 cm each, draw arcs on one side of AC cutting each other. Name the point B.

Step 3: With centres A and C and radius 5.6 cm each, draw arcs on the other side of AC cutting each other. Name the point R.

Step 4: Draw seg BA, seg AR, seg RC and seg CB. □BARC is the required quadrilateral.

Figure drawn to scale. BA = BC and RA = RC, so □BARC is a kite.

6★Construct □PQRS such that l(PQ) = 3.5 cm, l(QR) = 5.6 cm, l(RS) = 3.5 cm, m∠Q = 110°, m∠R = 70°. If it is given that □PQRS is a parallelogram, which of the given information is unnecessary?

Step 1: Draw seg QR of length 5.6 cm.

Step 2: At Q, draw a ray making an angle of 110° with seg QR. On it, mark P such that QP = 3.5 cm.

Step 3: At R, draw a ray making an angle of 70° with seg RQ, on the same side. On it, mark S such that RS = 3.5 cm.

Step 4: Draw seg PS. □PQRS is the required quadrilateral.

Figure drawn to scale.

Unnecessary information: m∠Q + m∠R = 110° + 70° = 180°, so side PQ ∥ side RS, and PQ = RS = 3.5 cm. So □PQRS is a parallelogram.

If we already know that □PQRS is a parallelogram, then two adjacent sides and the angle between them are enough to draw it: l(PQ) = 3.5 cm, l(QR) = 5.6 cm and m∠Q = 110°.

Answer: l(RS) = 3.5 cm and m∠R = 70° are unnecessary.

They follow from the properties of a parallelogram: opposite sides are congruent (RS = PQ), and adjacent angles are supplementary (∠R = 180° − 110°).

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