NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry (EX 4.2) Exercise 4.2

NCERT Solutions for Class 8 Maths Chapter 4 – Practical Geometry Ex 4.2 are given here in simple, step-by-step form. NCERT stands for National Council of Educational Research and Training, the body that makes your textbook. Practical Geometry is about drawing shapes accurately with a ruler and compass, not about calculating with them.

Exercise 4.2 covers one case only: constructing a quadrilateral when two diagonals and three sides are given. There is 1 question with 3 parts. Every construction below is written as numbered steps you can follow with a compass, with a diagram showing the finished shape and the arcs that locate each vertex. Students can use this page for homework and revision, parents can use it to check the work at home, and teachers can use it in class. A free printable PDF of these solutions is also available for offline study.

The Idea Behind Exercise 4.2

A quadrilateral needs five independent measurements to be fixed. Here you are given two diagonals and three sides, which is five.

The trick is always the same:
1. Find a triangle among the given measurements whose three lengths you know. A diagonal plus two sides usually makes one.
2. Draw that triangle first. Its three corners are three of your four vertices.
3. Use the remaining two lengths as radii to swing two arcs. Where they cross is the fourth vertex.
4. Join up the sides.

A compass is the tool for "a point at this exact distance". Two arcs crossing means "the one point that is this far from here and that far from there". That is the whole method.

Note. Draw a rough sketch first and write the given lengths on it. It takes half a minute and it shows you which triangle to start with.

Q 1. Construct the following quadrilaterals.

(i) Quadrilateral LIFT: LI = 4 cm  IF = 3 cm  TL = 2.5 cm  LF = 4.5 cm  IT = 4 cm
(ii) Quadrilateral GOLD: OL = 7.5 cm  GL = 6 cm  GD = 6 cm  LD = 5 cm  OD = 10 cm
(iii) Rhombus BEND: BN = 5.6 cm  DE = 6.5 cm

Answer:

(i) Quadrilateral LIFT
Given: LI = 4 cm  IF = 3 cm  TL = 2.5 cm  LF = 4.5 cm  IT = 4 cm
LF and IT are the two diagonals. Triangle LIT has all three lengths known (LI = 4, TL = 2.5, IT = 4), so start there.

Steps of construction

1. Draw LI = 4 cm with a ruler.
2. With L as centre, draw an arc of radius 2.5 cm (this is TL).
3. With I as centre, draw an arc of radius 4 cm (this is the diagonal IT). Where the two arcs cross is T. Join TL and TI.
4. With L as centre, draw an arc of radius 4.5 cm (this is the diagonal LF).
5. With I as centre, draw an arc of radius 3 cm (this is IF), on the same side as T. Where these two arcs cross is F.
6. Join IF, FT and LF. LIFT is the required quadrilateral.

Note. The fourth side FT is not given — it comes out of the construction, at about 2.7 cm.

(ii) Quadrilateral GOLD
Given: OL = 7.5 cm  GL = 6 cm  GD = 6 cm  LD = 5 cm  OD = 10 cm
GL and OD are the two diagonals. Triangle GLD has all three lengths known (GL = 6, GD = 6, LD = 5), so start there.

Steps of construction

1. Draw LD = 5 cm with a ruler.
2. With L as centre, draw an arc of radius 6 cm (this is the diagonal GL).
3. With D as centre, draw an arc of radius 6 cm (this is GD). Where the two arcs cross is G. Join GL and GD.
4. With L as centre, draw an arc of radius 7.5 cm (this is OL).
5. With D as centre, draw an arc of radius 10 cm (this is the diagonal OD), on the same side as G. Where these two arcs cross is O.
6. Join OL, OD and OG. GOLD is the required quadrilateral.

Note. Watch the order of the letters. GOLD means the sides are GO, OL, LD and DG, so GL and OD are the diagonals, not sides. The fourth side GO is not given and comes out at about 4.7 cm.

(iii) Rhombus BEND
Given: BN = 5.6 cm  DE = 6.5 cm
Only two measurements are given, and both are diagonals. That is enough because it is a rhombus: the diagonals of a rhombus bisect each other at right angles, which supplies the three missing facts.

Steps of construction

1. Draw BN = 5.6 cm with a ruler.
2. Construct the perpendicular bisector of BN. Call the point where it cuts BN O. So BO = ON = 5.6 ÷ 2 = 2.8 cm.
3. Because the diagonals bisect each other, O is also the midpoint of DE. So OE = OD = 6.5 ÷ 2 = 3.25 cm.
4. With O as centre, draw arcs of radius 3.25 cm to cut the perpendicular bisector above and below BN. Call these points E and D.
5. Join BE, EN, ND and DB. BEND is the required rhombus.

Note. Check your drawing: every side should measure about 4.3 cm, because each side is the hypotenuse of a right triangle with legs 2.8 cm and 3.25 cm.

Frequently Asked Questions (FAQs)

1. Why are five measurements needed to construct a quadrilateral?
Four sides alone do not fix a quadrilateral — it can flex like a hinge. A fifth measurement, such as a diagonal or an angle, locks it into one shape.

2. What is different about Exercise 4.2?
It deals with only one case: when two diagonals and three sides are given. Exercise 4.1 gives four sides and one diagonal instead.

3. How do I decide which triangle to draw first?
Look for three given lengths that form a closed triangle. A diagonal together with two sides that meet it usually works. Draw that triangle, then swing arcs for the last vertex.

4. Why is a rhombus constructible from just two diagonals?
Because the diagonals of a rhombus bisect each other at right angles. That property supplies the extra information, so two measurements are enough.

5. What if the two arcs never cross?
Then the quadrilateral cannot be drawn with those measurements. It means one length is too short to reach, just as a triangle is impossible when two sides together are shorter than the third.

6. How many questions are there in Exercise 4.2?
There is 1 question with 3 parts, so students complete 3 constructions.

7. Can I download these solutions in PDF?
Yes. A free printable PDF with all 3 constructions and their diagrams is available on this page.