GSEB Class 8 Maths Solutions Chapter 4 Practical Geometry InText Questions

Try These (Page 58)

Question 1. Arshad has five measurements of a quadrilateral ABCD. These are AB = 5 cm, \( \angle A = 50^\circ \), AC = 4 cm, BD = 5 cm and AD = 4 cm, BD = 5 cm, and AD = 6cm. Can he construct a unique quadrilateral? Give reasons for answer?
Answer: No, Arshad cannot build the quadrilateral ABCD with the given set of measurements. For this construction, knowing the length of BC or DC is essential.
In simple words: Arshad cannot draw the shape. He needs to know either how long side BC is or how long side DC is to complete it.

Exam Tip: A unique quadrilateral usually requires five independent measurements, but these five must be in a specific combination (e.g., three sides and two included angles) to fully define the shape.

Question 2.
1. We saw that 5 measurements of a quadrilateral can determine a quadrilateral uniquely. Do you think any five measurement of the quadrilateral can do this?
2. Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and AS = 6.5 cm? Why?
3. Can you draw a rhombus ZEAL where ZE = 3.5 cm, diagonal EL = 5 cm? Why?
4. Attempted to draw a quadrilateral PLAY where PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY = 6 cm, but could not draw it? What is the reason? [Hint: Discuss it using a rough sketch].
Answer:
1. No, any five measurements (elements) cannot always define a quadrilateral. To build a quadrilateral, we actually require a particular combination of measurements, such as:
(a) Four sides and one diagonal or
(b) Three sides and two diagonals or
(c) Two adjacent sides and three angles or
(d) Three sides and two included angles or
(e) Some special properties are given.
In simple words: Just any five measurements aren't enough to draw a quadrilateral perfectly. You need a specific mix, like knowing all four sides and one diagonal, or three sides and two angles that are next to each other.

Exam Tip: Remember the specific conditions (like SSS, SAS, ASA) for triangle construction and how they extend to quadrilaterals, which are composed of triangles.

Answer:
2. Let us create a rough drawing of BATS as provided.

We cannot find the precise locations of points T and B without knowing the lengths of measurements ST and SB. Therefore, we cannot draw a parallelogram with the given measurements.
In simple words: We can't draw the parallelogram because we don't know how long sides ST and SB are. We need more information to place points T and B correctly.

Exam Tip: For parallelogram construction, you typically need two adjacent sides and an included angle, or two sides and a diagonal. Just three sides (even with the "parallelogram" property) isn't enough.

Answer:
3. Let us make a rough sketch of the rhombus quadrilateral ZEAL and label the given measurements on it.

Yes, we can build the needed rhombus, as all its sides measure 3.5 cm and a diagonal EL is provided.
In simple words: Yes, this rhombus can be drawn because we know all its sides are 3.5 cm long, and we also know one of its diagonals is 5 cm. This gives us enough information.

Exam Tip: Remember that all sides of a rhombus are equal. Knowing one side and a diagonal is enough to construct it, as it forms two congruent triangles (SSS criterion).

Answer:
4. Let us draw a rough sketch of the quadrilateral PLAY and mark its measurements. No, this quadrilateral cannot be drawn. Point P cannot be precisely located. In triangle LPY, the sum of the lengths of PL and PY is less than LY. This means \( (2 \text{ cm} + 3 \text{ cm}) < 6 \text{ cm} \).
In simple words: This quadrilateral cannot be drawn. In the triangle LPY, the two shorter sides (PL and PY) add up to less than the longest side (LY). This means they can't meet to form a triangle, so point P can't be placed.

Exam Tip: Always check the triangle inequality theorem when given three side lengths. The sum of any two sides of a triangle must be greater than the third side.

Try These (Page 62)

Question 1. “Construct a quadrilateral ABCD, given that BC = 4.5 cm, AD = 5.5 cm, CD = 5 cm, the diagonal AC = 5.5 cm and diagonal BD = 7 cm.” Can we draw the quadrilateral by drawing \( \triangle ABD \) first and then find the fourth point C?
Answer: Since the measurement of AB is not provided, we cannot precisely locate point B. Thus, the quadrilateral cannot be drawn.

In simple words: We cannot draw this quadrilateral properly because the length of side AB is not given. Without AB, we cannot find the exact spot for point B.

Exam Tip: When constructing quadrilaterals, ensure you have sufficient and appropriate measurements. For example, knowing four sides and one diagonal, or three sides and two diagonals, or two adjacent sides and three angles, etc.

Question 2. Can you construct a quadrilateral PQRS with PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm and SQ = 4 cm? Justify your answer?
Answer: Yes, a quadrilateral PQRS is constructible using the specified dimensions: PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm, and SQ = 4 cm.
In simple words: Yes, you can draw this quadrilateral. All the necessary lengths for the sides and diagonals are provided.

Exam Tip: This case involves three sides (PQ, RS, PS) and two diagonals (PR, SQ), which is generally sufficient to construct a unique quadrilateral.

Try These (Page 64)

Question 1. Can you construct the following quadrilateral MIST if we have \( 100^\circ \) at M instead of \( 75^\circ \)?
Answer: Yes, the quadrilateral MIST is constructible with \( \angle M = 100^\circ \) instead of \( 75^\circ \).

In simple words: Yes, you can build the quadrilateral MIST even if the angle at M is \( 100^\circ \) instead of \( 75^\circ \).

Exam Tip: Constructing quadrilaterals usually involves creating triangles first. Knowing sides and an included angle (SAS) is a common starting point.

Question 2. Can you construct the quadrilateral PLAN if PL = 6 cm, LA = 9.5 cm, \( \angle P = 75^\circ \), \( \angle L = 150^\circ \) and \( \angle A = 140^\circ \)? (Hint: Recall angle-sum property)
Answer: Here, \( \angle P + \angle L + \angle A + \angle N = 75^\circ + 150^\circ + 140^\circ + \angle N = 365^\circ + \angle N \). But the total sum of all the angles of any quadrilateral is \( 360^\circ \). As a result, the construction of quadrilateral PLAN is not possible.
In simple words: No, you cannot draw this quadrilateral. When you add up the three given angles (\( 75^\circ + 150^\circ + 140^\circ \)), they already add up to \( 365^\circ \). Since all angles in a quadrilateral must sum to exactly \( 360^\circ \), it's impossible to make this one.

Exam Tip: Always remember that the sum of the interior angles of any quadrilateral is \( 360^\circ \). This is a fundamental property to check before attempting construction involving angles.

Question 3. In a parallelogram, the lengths of adjacent sides are known. Do we still need measures of the angles to construct as in the Q – 1 above?
Answer: No, the measures of three angles are not needed in the case of a parallelogram because its opposite sides are always parallel.
In simple words: No, you don't need all the angle measurements for a parallelogram if you know the lengths of its sides next to each other. Because opposite sides are parallel, the angles are set once the side lengths are known.

Exam Tip: For special quadrilaterals like parallelograms, rhombuses, and rectangles, fewer measurements are often needed due to their inherent properties (e.g., opposite sides parallel and equal, specific angles).

Try These (Page 67)

Question 1. How will you construct a rectangle PQRS if you know only the lengths PQ and QR?
Answer: A rectangle can be constructed by drawing PQ as the length and then making an angle of \( 90^\circ \) at Q. Cut off QR, which is the breadth, from the ray \( \overrightarrow{QY} \). The remaining two points R and S can be located by taking P and R as centers and radii as QR and PQ respectively to draw arcs that intersect at S. Thus, PQRS is the required rectangle.

In simple words: To build a rectangle, first draw the length PQ. Then, at point Q, draw a line straight up at a \( 90^\circ \) angle. Measure and mark the breadth QR on this line. To find the last point S, use a compass: from P, draw an arc with radius QR, and from R, draw an arc with radius PQ. Where these arcs cross is point S, completing your rectangle.

Exam Tip: Rectangles have four right angles and opposite sides equal. Knowing just two adjacent sides is enough because these properties define the other angles and sides.

Question 2. Construct the kite EASY if AY = 8 cm, EY = 4 cm and SY = 6 cm. Which properties of the kite did you use in the process?
Answer: The general characteristics used when building a kite include:
1. The diagonals always cross each other at right angles.
2. One of the diagonals cuts the other into two equal parts.
3. Adjacent sides come in pairs of equal length (for example, EA = EY and SA = SY).
Now, for constructing the kite EASY with AY = 8 cm, EY = 4 cm, and SY = 6 cm, follow these steps:
I. First, draw a line segment AY measuring 8 cm.
II. Next, draw the ray \( \overrightarrow{PQ} \), which serves as the perpendicular bisector of AY, making sure it intersects AY at point O.

III. With the given measurements, we cannot find point E on PQ such that it is 4 cm away from both Y and A. That is, having EY = 4 cm and EA = 4 cm simultaneously is not possible in a typical kite shape. This condition can only happen if E and O become the same point. In such a specific instance, a distinct kite, where E is separate from AY, would not be formed.

In simple words: First, list the properties of a kite: diagonals cross at \( 90^\circ \), one diagonal cuts the other in half, and adjacent sides are equal. Then, to construct this specific kite, draw line AY (8 cm) and its perpendicular bisector. However, a point E that is 4 cm from both A and Y cannot be found unless E is exactly on the midpoint of AY. This means a distinct kite shape wouldn't be formed with these exact measurements.

Exam Tip: Be mindful of side length relationships when constructing. For a kite, the non-equal adjacent sides should not make one pair equal to half the main diagonal, otherwise it might degenerate into a triangle.