Practical Geometry — Questions and Answers
Welcome to HSLC Guru. This lesson gives the complete, step-by-step solutions with labelled figures for every worked example, Exercise 4.1, Exercise 4.2 and the activity boxes of Chapter 4 Practical Geometry of ASSEB (Assam State School Education Board) Class 8 General Mathematics.
Summary
In Class VII you learnt to construct triangles, for which three suitable measurements are enough. In this chapter you learn to construct a four-sided closed figure — a quadrilateral.
A quadrilateral has four sides, four angles and two diagonals — ten measurements in all. However, only five suitable measurements are needed to draw a quadrilateral uniquely. Four sides alone are not enough, because the same four sides can form many differently shaped quadrilaterals (for example, both a rectangle and a parallelogram).
The five measurements can be given in several ways — (a) four sides and one diagonal, (b) three sides and two diagonals, (c) four sides and one angle, (d) two adjacent sides and three angles, (e) three sides and two included angles, and (f) special properties (square, rectangle, rhombus, parallelogram). A ruler, a compass and a protractor are used for the constructions.
Summary: This chapter of ASSEB Class 8 General Mathematics Chapter 4 (Practical Geometry) shows how to construct a unique quadrilateral. A quadrilateral has ten measurements (four sides, four angles and two diagonals), but only five suitable measurements are needed to draw it uniquely — four sides and a diagonal, three sides and two diagonals, four sides and an angle, two adjacent sides and three angles, three sides and two included angles, or its special properties. Every worked example, Exercise 4.1, Exercise 4.2 and the activity boxes are solved here with step-by-step construction and labelled figures.
Textbook Questions and Answers
Idea of quadrilateral construction
Question: How many measurements does a quadrilateral have in all, and at least how many are needed to construct a unique quadrilateral?
Answer: A quadrilateral has four sides, four angles and two diagonals — $4 + 4 + 2 = 10$ measurements in all, and its shape depends on these ten. Still, to construct a quadrilateral uniquely only five suitable measurements are needed.
Question: Can a unique quadrilateral be drawn from four sides only? Why?
Answer: No. Suppose the four sides are $3\ \text{cm}, 5\ \text{cm}, 6.5\ \text{cm}$ and $7\ \text{cm}$. Many differently shaped quadrilaterals (with different angles and diagonal lengths) can be drawn from the same four sides. For example, the sides $5\ \text{cm}, 3\ \text{cm}, 5\ \text{cm}, 3\ \text{cm}$ can form a rectangle as well as a parallelogram. Only after one diagonal or one angle is also given does the shape become fixed. Hence four sides alone cannot fix a unique quadrilateral — at least five measurements are needed.
4.1.1 When four sides and one diagonal are given
Example: Construct a quadrilateral ABCD where $\overline{AB} = 3.5\ \text{cm}$, $\overline{BC} = 4\ \text{cm}$, $\overline{CD} = 5\ \text{cm}$, $\overline{DA} = 4.5\ \text{cm}$ and diagonal $\overline{BD} = 6.5\ \text{cm}$.
Answer: Steps of construction—
- First draw a rough sketch marking the measurements (optional).
- With a ruler, draw the diagonal $\overline{BD} = 6.5\ \text{cm}$.
- With B as centre draw an arc of radius $3.5\ \text{cm}$ and with D as centre draw an arc of radius $4.5\ \text{cm}$; their point of intersection is A. Join $\overline{BA}$ and $\overline{DA}$.
- On the opposite side of A with respect to BD, with B as centre draw an arc of radius $4\ \text{cm}$ and with D as centre draw an arc of radius $5\ \text{cm}$; their point of intersection is C. Join $\overline{BC}$ and $\overline{CD}$.
Thus ABCD is the required quadrilateral. (In the figure the red dashed line is the diagonal $\overline{BD}$, and the two dashed arcs meet at A.)
4.1.2 When two diagonals and three sides are given
Example (i): Construct a quadrilateral PQRS where $\overline{QR} = 7.5\ \text{cm}$, $\overline{PS} = 6\ \text{cm}$, $\overline{RS} = 5\ \text{cm}$, diagonal $\overline{PR} = 6\ \text{cm}$ and diagonal $\overline{QS} = 10\ \text{cm}$.
Answer: Steps of construction—
- Draw the diagonal $\overline{PR} = 6\ \text{cm}$.
- With P as centre draw an arc of radius $6\ \text{cm}$ ($\overline{PS}$) and with R as centre an arc of radius $5\ \text{cm}$ ($\overline{RS}$) to fix S above PR.
- On the side opposite to S (below PR), with S as centre draw an arc of radius $10\ \text{cm}$ ($\overline{QS}$) and with R as centre an arc of radius $7.5\ \text{cm}$ ($\overline{QR}$); their intersection is Q.
- Join $\overline{PQ}, \overline{QR}, \overline{RS}, \overline{SP}$ to complete PQRS.
Thus PQRS is the required quadrilateral (the two red dashed lines are the diagonals $\overline{PR}$ and $\overline{QS}$).
Example (ii): Construct a parallelogram ABCD where $\overline{AB} = 4\ \text{cm}$, $\overline{BC} = 2.8\ \text{cm}$ and diagonal $\overline{AC} = 5\ \text{cm}$.
Answer: Opposite sides of a parallelogram are equal and parallel, so $\overline{AD} = \overline{BC} = 2.8\ \text{cm}$ and $\overline{DC} = \overline{AB} = 4\ \text{cm}$; $\overline{AC} = 5\ \text{cm}$ is the diagonal. Steps—
- Draw $\overline{AB} = 4\ \text{cm}$.
- With A as centre draw an arc of radius $5\ \text{cm}$ (diagonal $\overline{AC}$) and with B as centre an arc of radius $2.8\ \text{cm}$ ($\overline{BC}$) to fix C. Join $\overline{AC}$ and $\overline{BC}$.
- With A as centre draw an arc of radius $2.8\ \text{cm}$ ($\overline{AD}$) and with C as centre an arc of radius $4\ \text{cm}$ ($\overline{DC}$) to fix D.
- Join $\overline{AD}$ and $\overline{CD}$ to complete parallelogram ABCD.
Example (iii): Construct a rhombus ABCD where diagonal $\overline{AC} = 6\ \text{cm}$ and diagonal $\overline{BD} = 8\ \text{cm}$.
Answer: The two diagonals of a rhombus bisect each other at right angles. Steps—
- Draw $\overline{AC} = 6\ \text{cm}$.
- Draw the perpendicular bisector of $\overline{AC}$; it cuts AC at O.
- With O as centre draw arcs of radius $4\ \text{cm}$ (half of $\overline{BD}$) above and below AC; they cut the perpendicular bisector at B and D.
- Join $\overline{AB}, \overline{BC}, \overline{CD}, \overline{DA}$ to complete rhombus ABCD. Each side measures $\sqrt{3^2 + 4^2} = 5\ \text{cm}$.
Example (iv): Construct a rhombus ABCD where side $\overline{AB} = 4\ \text{cm}$ and diagonal $\overline{AC} = 6\ \text{cm}$.
Answer: All four sides of a rhombus are equal, so $\overline{AB} = \overline{BC} = \overline{CD} = \overline{DA} = 4\ \text{cm}$. Steps—
- Draw $\overline{AC} = 6\ \text{cm}$.
- With A and C as centres draw arcs of radius $4\ \text{cm}$ above and below AC; the upper intersection is B and the lower intersection is D.
- Join $\overline{AB}, \overline{BC}, \overline{CD}, \overline{DA}$ to complete rhombus ABCD.
4.1.3 When four sides and one angle are given
Example: Construct a quadrilateral ABCD where $\overline{AB} = 4.5\ \text{cm}$, $\overline{BC} = 3.5\ \text{cm}$, $\overline{CD} = 4.8\ \text{cm}$, $\overline{AD} = 4\ \text{cm}$ and $\angle B = 120^\circ$.
Answer: Steps of construction—
- Draw $\overline{AB} = 4.5\ \text{cm}$.
- At B, taking $\overline{AB}$ as one arm, construct $\angle B = 120^\circ$.
- With B as centre draw an arc of radius $3.5\ \text{cm}$ cutting the other arm of $\angle B$ at C.
- With A as centre draw an arc of radius $4\ \text{cm}$ ($\overline{AD}$).
- With C as centre draw an arc of radius $4.8\ \text{cm}$ ($\overline{CD}$) cutting the previous arc; the point of intersection is D.
- Join $\overline{AD}$ and $\overline{CD}$ to complete ABCD.
4.1.4 When two adjacent sides and three angles are given
Example: Construct a quadrilateral ABCD where $\angle A = 60^\circ$, $\angle B = 105^\circ$, $\angle C = 105^\circ$, $\overline{AB} = 6\ \text{cm}$ and $\overline{BC} = 4.5\ \text{cm}$.
Answer: Steps of construction—
- Draw $\overline{AB} = 6\ \text{cm}$.
- At A construct $\angle A = 60^\circ$ and draw ray AZ.
- At B construct $\angle B = 105^\circ$ and draw ray BX.
- With B as centre draw an arc of radius $4.5\ \text{cm}$ cutting ray BX at C.
- At C construct $\angle C = 105^\circ$ and draw ray CY.
- The point of intersection of rays AZ and CY is the fourth vertex D. Then $\angle D = 360^\circ – (60^\circ + 105^\circ + 105^\circ) = 90^\circ$, and ABCD is the required quadrilateral.
4.1.5 When three sides and two included angles are given
Example: Construct a quadrilateral PQRS where $\overline{PQ} = 3.5\ \text{cm}$, $\overline{QR} = 3\ \text{cm}$, $\overline{RS} = 4\ \text{cm}$, $\angle Q = 75^\circ$ and $\angle R = 120^\circ$.
Answer: Steps of construction—
- Draw $\overline{QR} = 3\ \text{cm}$.
- At Q construct $\angle Q = 75^\circ$ and draw ray QX.
- At R construct $\angle R = 120^\circ$ and draw ray RY.
- With Q as centre draw an arc of radius $3.5\ \text{cm}$ cutting QX at P.
- With R as centre draw an arc of radius $4\ \text{cm}$ cutting RY at S.
- Join $\overline{PS}$ to complete PQRS.
4.1.6 Construction using special properties
Example: Construct a square of side $5\ \text{cm}$.
Answer: Properties of a square — all four sides equal and each angle $90^\circ$. Steps—
- Draw a ray AX slightly longer than $5\ \text{cm}$. With A as centre draw an arc of radius $5\ \text{cm}$ cutting AX at B; $\overline{AB}$ is one side of the square.
- At A construct an angle of $90^\circ$ (ray AY). With A as centre draw an arc of radius $5\ \text{cm}$ cutting AY at D; so $\overline{AD} = 5\ \text{cm}$.
- With D and B as centres draw two arcs of radius $5\ \text{cm}$ intersecting at C.
- Join $\overline{DC}$ and $\overline{BC}$ to complete the square ABCD ($\overline{AB} = \overline{AD} = \overline{BC} = \overline{CD} = 5\ \text{cm}$).
Exercise 4.1
1. Construct the following quadrilaterals.
(i) Quadrilateral ABCD where $\overline{AB} = 4\ \text{cm}$, $\overline{BC} = 6\ \text{cm}$, $\overline{CD} = 5\ \text{cm}$, $\overline{DA} = 5.5\ \text{cm}$ and diagonal $\overline{AC} = 7\ \text{cm}$.
Answer: This is the four-sides-and-a-diagonal case. Steps—
- Draw the diagonal $\overline{AC} = 7\ \text{cm}$.
- With A as centre draw an arc of radius $4\ \text{cm}$ ($\overline{AB}$) and with C as centre an arc of radius $6\ \text{cm}$ ($\overline{BC}$) to fix B on one side.
- On the other side of AC, with A as centre draw an arc of radius $5.5\ \text{cm}$ ($\overline{DA}$) and with C an arc of radius $5\ \text{cm}$ ($\overline{CD}$) to fix D.
- Join $\overline{AB}, \overline{BC}, \overline{CD}, \overline{DA}$.
(ii) Quadrilateral ABCD where $\overline{AB} = 4\ \text{cm}$, $\overline{BC} = 3\ \text{cm}$, $\overline{DA} = 2.8\ \text{cm}$, diagonal $\overline{AC} = 5\ \text{cm}$ and diagonal $\overline{BD} = 4.5\ \text{cm}$.
Answer: This is the three-sides-and-two-diagonals case. Steps—
- Draw the diagonal $\overline{AC} = 5\ \text{cm}$.
- With A as centre draw an arc of radius $4\ \text{cm}$ ($\overline{AB}$) and with C an arc of radius $3\ \text{cm}$ ($\overline{BC}$) to fix B on one side (this is a $3$-$4$-$5$ right triangle).
- On the other side of AC, with A as centre draw an arc of radius $2.8\ \text{cm}$ ($\overline{DA}$) and with B an arc of radius $4.5\ \text{cm}$ (diagonal $\overline{BD}$) to fix D.
- Join $\overline{AB}, \overline{BC}, \overline{CD}, \overline{DA}$ (the fourth side $\overline{CD}$ is fixed automatically).
(iii) Quadrilateral PQRS where $\overline{QR} = 4.5\ \text{cm}$, $\overline{PS} = 5.5\ \text{cm}$, $\overline{RS} = 5\ \text{cm}$, diagonal $\overline{PR} = 5.5\ \text{cm}$ and diagonal $\overline{QS} = 7\ \text{cm}$.
Answer: The three-sides-and-two-diagonals case. Steps—
- Draw the diagonal $\overline{PR} = 5.5\ \text{cm}$.
- With P as centre draw an arc of radius $5.5\ \text{cm}$ ($\overline{PS}$) and with R an arc of radius $5\ \text{cm}$ ($\overline{RS}$) to fix S above PR.
- Below PR, with S as centre draw an arc of radius $7\ \text{cm}$ ($\overline{QS}$) and with R an arc of radius $4.5\ \text{cm}$ ($\overline{QR}$) to fix Q.
- Join $\overline{PQ}, \overline{QR}, \overline{RS}, \overline{SP}$.
(iv) Parallelogram EFGH where $\overline{FG} = 7\ \text{cm}$, $\overline{GH} = 5.5\ \text{cm}$ and diagonal $\overline{HF} = 8.5\ \text{cm}$.
Answer: Opposite sides of a parallelogram are equal, so $\overline{EF} = \overline{GH} = 5.5\ \text{cm}$ and $\overline{HE} = \overline{FG} = 7\ \text{cm}$. Steps—
- Draw $\overline{FG} = 7\ \text{cm}$.
- With F as centre draw an arc of radius $8.5\ \text{cm}$ (diagonal $\overline{HF}$) and with G an arc of radius $5.5\ \text{cm}$ ($\overline{GH}$) to fix H.
- With F as centre draw an arc of radius $5.5\ \text{cm}$ ($\overline{EF}$) and with H an arc of radius $7\ \text{cm}$ ($\overline{HE}$) to fix E.
- Join $\overline{EF}, \overline{FG}, \overline{GH}, \overline{HE}$ to complete the parallelogram.
(v) Rhombus DEFG where $\overline{DE} = 5\ \text{cm}$ and diagonal $\overline{EG} = 6.5\ \text{cm}$.
Answer: All four sides of a rhombus are equal ($= 5\ \text{cm}$). Steps—
- Draw the diagonal $\overline{EG} = 6.5\ \text{cm}$.
- With E and G as centres draw arcs of radius $5\ \text{cm}$ above and below EG; the upper intersection is D and the lower one is F.
- Join $\overline{DE}, \overline{EF}, \overline{FG}, \overline{GD}$ to complete rhombus DEFG.
(vi) Rhombus LMNO where diagonal $\overline{LN} = 6\ \text{cm}$ and diagonal $\overline{MO} = 7\ \text{cm}$.
Answer: The two diagonals of a rhombus bisect each other at right angles. Steps—
- Draw the diagonal $\overline{LN} = 6\ \text{cm}$ and draw its perpendicular bisector; the point of intersection is the centre.
- From the centre mark M and O on the perpendicular bisector at $3.5\ \text{cm}$ (half of $\overline{MO}$) above and below.
- Join $\overline{LM}, \overline{MN}, \overline{NO}, \overline{OL}$. Each side $= \sqrt{3^2 + 3.5^2} = \sqrt{21.25} \approx 4.6\ \text{cm}$.
Activity and Group Activity
Activity (i): Try to construct the $5\ \text{cm}$ square by another method.
Answer: One easy alternative — (1) draw $\overline{AB} = 5\ \text{cm}$; (2) at A and B erect perpendiculars ($90^\circ$) to AB; (3) cut off $\overline{AD} = 5\ \text{cm}$ and $\overline{BC} = 5\ \text{cm}$ on these perpendiculars; (4) join $\overline{DC}$. This gives the square ABCD.
Activity (ii): Construct a rectangle with adjacent sides of length $4\ \text{cm}$ and $5\ \text{cm}$.
Answer: Each angle of a rectangle is $90^\circ$ and opposite sides are equal. Steps — (1) draw $\overline{AB} = 5\ \text{cm}$; (2) construct $90^\circ$ at A and B; (3) cut off $\overline{AD} = 4\ \text{cm}$ and $\overline{BC} = 4\ \text{cm}$ on these perpendiculars; (4) join $\overline{DC}$ to complete rectangle ABCD.
Group Activity (1): Can a quadrilateral with one side $7\ \text{cm}$ and four angles $75^\circ, 85^\circ, 110^\circ$ and $90^\circ$ be drawn?
Answer: The four angles add up to $75^\circ + 85^\circ + 110^\circ + 90^\circ = 360^\circ$, so the angle condition of a quadrilateral is satisfied. But of the four angles only three are independent (their sum is fixed at $360^\circ$); together with one side that gives only four independent measurements, one short of the five needed. Hence infinitely many quadrilaterals with the same angles but different sizes are possible — it can be drawn, but not uniquely.
Group Activity (2): Can a kite ABCD be constructed where $\overline{AD} = 4\ \text{cm}$, diagonal $\overline{AC} = 8\ \text{cm}$ and $\overline{CD} = 6\ \text{cm}$? (Use the property of a kite.)
Answer: Yes, it can be drawn. By the kite property, two pairs of adjacent sides are equal — $\overline{AB} = \overline{AD} = 4\ \text{cm}$ and $\overline{CB} = \overline{CD} = 6\ \text{cm}$. In triangle ACD, $\overline{AC} = 8, \overline{AD} = 4, \overline{CD} = 6$ (since $4 + 6 = 10 > 8$, the triangle is possible). Steps — (1) draw the diagonal $\overline{AC} = 8\ \text{cm}$; (2) with A as centre draw arcs of radius $4\ \text{cm}$ and with C arcs of radius $6\ \text{cm}$ on both sides of AC; the upper intersection is D and the lower one is B; (3) join $\overline{AB}, \overline{BC}, \overline{CD}, \overline{DA}$.
Group Activity (3): A quadrilateral cannot be constructed if its four angles and one side are given. Justify.
Answer: A unique quadrilateral needs at least five independent measurements. Since the four angles always sum to $360^\circ$, only three of them are independent; together with one side this gives $3 + 1 = 4$ independent measurements, which is less than five. Therefore the sides can take any length — infinitely many different quadrilaterals are possible — and no unique quadrilateral can be constructed.
Exercise 4.2
1. Draw a quadrilateral ABCD where $\overline{AB} = 6\ \text{cm}$, $\overline{BC} = 7\ \text{cm}$, $\overline{CD} = 6.5\ \text{cm}$, $\overline{DA} = 5.5\ \text{cm}$ and $\angle B = 105^\circ$.
Answer: Four sides and one angle. Steps — (1) draw $\overline{AB} = 6\ \text{cm}$; (2) at B construct $\angle B = 105^\circ$; (3) with B as centre draw an arc of radius $7\ \text{cm}$ cutting the other arm at C; (4) with A as centre draw an arc of radius $5.5\ \text{cm}$ ($\overline{DA}$) and with C an arc of radius $6.5\ \text{cm}$ ($\overline{CD}$) to fix D; (5) join $\overline{AD}$ and $\overline{CD}$.
2. Draw a quadrilateral ABCD where $\overline{AB} = 5\ \text{cm}$, $\overline{BC} = 4\ \text{cm}$, $\overline{CD} = 3.5\ \text{cm}$, $\overline{DA} = 4.5\ \text{cm}$ and $\angle C = 75^\circ$.
Answer: Four sides and one angle. Steps — (1) draw $\overline{BC} = 4\ \text{cm}$; (2) at C construct $\angle C = 75^\circ$; (3) with C as centre draw an arc of radius $3.5\ \text{cm}$ cutting the other arm at D; (4) with B as centre draw an arc of radius $5\ \text{cm}$ ($\overline{AB}$) and with D an arc of radius $4.5\ \text{cm}$ ($\overline{DA}$) to fix A; (5) join $\overline{AB}$ and $\overline{AD}$.
3. Draw a quadrilateral ABCD where $\overline{AB} = 4\ \text{cm}$, $\overline{BC} = 7\ \text{cm}$, $\angle A = 105^\circ$, $\angle B = 75^\circ$ and $\angle C = 120^\circ$.
Answer: Two adjacent sides and three angles ($\angle D = 360^\circ – (105^\circ + 75^\circ + 120^\circ) = 60^\circ$). Steps — (1) draw $\overline{AB} = 4\ \text{cm}$; (2) at A construct $\angle A = 105^\circ$, ray AZ; (3) at B construct $\angle B = 75^\circ$, ray BX; (4) with B as centre draw an arc of radius $7\ \text{cm}$ cutting BX at C; (5) at C construct $\angle C = 120^\circ$, ray CY; (6) the intersection of AZ and CY is D.
4. Draw a quadrilateral EFGH where $\overline{EF} = 5\ \text{cm}$, $\overline{FG} = 7.5\ \text{cm}$, $\angle E = 90^\circ$, $\angle G = 105^\circ$ and $\angle H = 80^\circ$.
Answer: The four angles sum to $360^\circ$, so $\angle F = 360^\circ – (90^\circ + 105^\circ + 80^\circ) = 85^\circ$. Steps — (1) draw $\overline{EF} = 5\ \text{cm}$; (2) at F construct $\angle F = 85^\circ$ and with F as centre draw an arc of radius $7.5\ \text{cm}$ to fix G; (3) at E construct $\angle E = 90^\circ$, ray EH; (4) at G construct $\angle G = 105^\circ$, ray GH; (5) the intersection of the two rays is H.
5. Draw a parallelogram PQRS where $\overline{PQ} = 6\ \text{cm}$, $\overline{QR} = 7\ \text{cm}$ and $\angle S = 85^\circ$.
Answer: Opposite angles of a parallelogram are equal, so $\angle Q = \angle S = 85^\circ$; also $\overline{RS} = \overline{PQ} = 6\ \text{cm}$ and $\overline{SP} = \overline{QR} = 7\ \text{cm}$. Steps — (1) draw $\overline{PQ} = 6\ \text{cm}$; (2) at Q construct $\angle Q = 85^\circ$ and with Q as centre draw an arc of radius $7\ \text{cm}$ to fix R; (3) with P as centre draw an arc of radius $7\ \text{cm}$ and with R an arc of radius $6\ \text{cm}$ to fix S; (4) join $\overline{RS}$ and $\overline{SP}$.
6. Draw a rectangle LMNO where $\overline{LM} = 6\ \text{cm}$ and $\overline{MN} = 4\ \text{cm}$.
Answer: Each angle of a rectangle is $90^\circ$. Steps — (1) draw $\overline{LM} = 6\ \text{cm}$; (2) construct $90^\circ$ at L and M; (3) cut off $\overline{LO} = 4\ \text{cm}$ and $\overline{MN} = 4\ \text{cm}$ on these perpendiculars; (4) join $\overline{NO}$ to complete rectangle LMNO.
7. Draw a quadrilateral PQRS where $\overline{PQ} = 6\ \text{cm}$, $\overline{QR} = 7\ \text{cm}$, $\overline{RS} = 7.5\ \text{cm}$, $\angle Q = 105^\circ$ and $\angle R = 80^\circ$.
Answer: Three sides and two included angles. Steps — (1) draw $\overline{QR} = 7\ \text{cm}$; (2) at Q construct $\angle Q = 105^\circ$ and with Q as centre draw an arc of radius $6\ \text{cm}$ to fix P; (3) at R construct $\angle R = 80^\circ$ and with R as centre draw an arc of radius $7.5\ \text{cm}$ to fix S; (4) join $\overline{PS}$.
8. Draw a quadrilateral ABCD where $\overline{AB} = 4.5\ \text{cm}$, $\overline{BC} = 5.5\ \text{cm}$, $\overline{CD} = 5\ \text{cm}$, $\angle B = 68^\circ$ and $\angle C = 90^\circ$.
Answer: Three sides and two included angles. Steps — (1) draw $\overline{BC} = 5.5\ \text{cm}$; (2) at B construct $\angle B = 68^\circ$ and with B as centre draw an arc of radius $4.5\ \text{cm}$ to fix A; (3) at C construct $\angle C = 90^\circ$ and with C as centre draw an arc of radius $5\ \text{cm}$ to fix D; (4) join $\overline{AD}$.
9. Draw a rectangle having adjacent sides of length $5\ \text{cm}$ and $7\ \text{cm}$. (A protractor may be used where necessary.)
Answer: Each angle of a rectangle is $90^\circ$. Steps — (1) draw $\overline{AB} = 7\ \text{cm}$; (2) at A and B construct $90^\circ$ using a protractor; (3) cut off $\overline{AD} = 5\ \text{cm}$ and $\overline{BC} = 5\ \text{cm}$ on these perpendiculars; (4) join $\overline{DC}$ to complete rectangle ABCD.
Additional Questions and Answers
Multiple Choice Questions (MCQ)
1. At least how many measurements are needed to construct a unique quadrilateral? (a) 3 (b) 4 (c) 5 (d) 6
Answer: (c) 5.
2. How many measurements (sides, angles and diagonals) does a quadrilateral have in all? (a) 8 (b) 9 (c) 10 (d) 12
Answer: (c) 10 (4 sides + 4 angles + 2 diagonals).
3. The two diagonals of a rhombus bisect each other at what angle? (a) 30° (b) 45° (c) 60° (d) 90°
Answer: (d) 90°.
4. What is the sum of the four angles of a quadrilateral? (a) 180° (b) 270° (c) 360° (d) 540°
Answer: (c) 360°.
5. Given only four sides, one can construct— (a) a unique quadrilateral (b) infinitely many quadrilaterals (c) no quadrilateral (d) only a square
Answer: (b) infinitely many quadrilaterals.
6. At least how many independent measurements are needed to construct a square? (a) 1 (b) 2 (c) 4 (d) 5
Answer: (a) 1 (only the side length, using its properties).
7. Besides two adjacent sides, what more is needed to construct a parallelogram? (a) one diagonal or one angle (b) two diagonals (c) nothing (d) three angles
Answer: (a) one diagonal or one angle.
8. Three sides and ___ angles are enough to construct a quadrilateral. (a) one included (b) two included (c) three (d) four
Answer: (b) two included angles.
9. Which instrument is used to measure angles during a construction? (a) ruler (b) compass (c) protractor (d) eraser
Answer: (c) protractor.
10. If a rhombus ABCD has diagonals $\overline{AC} = 6\ \text{cm}$ and $\overline{BD} = 8\ \text{cm}$, what is the length of each side? (a) 4 cm (b) 5 cm (c) 7 cm (d) 10 cm
Answer: (b) 5 cm ($\sqrt{3^2 + 4^2} = 5$).
Fill in the Blanks
1. To construct a quadrilateral uniquely, at least ___ measurements are needed.
Answer: five.
2. The two diagonals of a rhombus bisect each other at ___.
Answer: right angles (90°).
3. Each angle of a rectangle is ___.
Answer: a right angle (90°).
4. A ___ is used to draw angles.
Answer: protractor.
5. A quadrilateral has ___ diagonals.
Answer: two.
True or False
1. A unique quadrilateral can be drawn from four sides only.
Answer: False.
2. Opposite sides of a parallelogram are equal and parallel.
Answer: True.
3. To construct a square, knowing only the side length is enough.
Answer: True.
4. The sum of the four angles of a quadrilateral is 180°.
Answer: False (the sum is 360°).
5. All sides of a rhombus are equal.
Answer: True.
Short Answer Questions
1. At least how many measurements are needed to construct a unique quadrilateral, and in what ways can they be given?
Answer: At least five measurements are needed. They can be given as — (a) four sides and one diagonal, (b) three sides and two diagonals, (c) four sides and one angle, (d) two adjacent sides and three angles, (e) three sides and two included angles, and (f) through special properties (square, rectangle, rhombus, parallelogram).
2. Why can a unique quadrilateral not be drawn from four sides only?
Answer: The same four sides can form many quadrilaterals with different angles and diagonals (for example a rectangle and a parallelogram); the shape is not fixed. Only after one diagonal or one angle is added does the shape become fixed. Hence at least five measurements are needed for a unique quadrilateral.
3. Which properties are used while constructing a rhombus and a parallelogram?
Answer: Rhombus — all four sides are equal and the diagonals bisect each other at right angles. Parallelogram — opposite sides are equal and parallel and opposite angles are equal. These properties let us construct them from fewer measurements.
4. Why can a quadrilateral not be constructed if its four angles and one side are given?
Answer: Since the four angles always sum to $360^\circ$, only three of them are independent. With one side that gives $3 + 1 = 4$ independent measurements, which is less than the five required. Hence a unique quadrilateral cannot be constructed.
Key Terms
| Term | Meaning |
|---|---|
| Quadrilateral | A closed figure bounded by four sides |
| Diagonal | A line segment joining two opposite vertices |
| Included angle | The angle between two given sides |
| Adjacent sides | Sides meeting at the same vertex |
| Radius | The distance from the centre of a circle to its circumference |
| Arc | A part of a circle |
| Perpendicular bisector | A line that divides a segment into two equal parts at right angles |
| Parallelogram | A quadrilateral whose opposite sides are parallel |
| Rhombus | A quadrilateral with all four sides equal |
| Kite | A quadrilateral with two pairs of equal adjacent sides |
| Protractor | An instrument used to measure and draw angles |
| Vertex | A corner point of a quadrilateral |