The Rajasthan Board of Secondary Education (RBSE) for Class 7 Maths, which aligns with the NCERT curriculum, features Chapter 10 on Practical Geometry (प्रायोगिक ज्यामिति). This chapter focuses on the basic constructions using a ruler and compass, helping students develop fundamental geometric skills.
Table of Contents
Here are the key topics and an overview of the solutions for the exercises in this chapter:
📝 Chapter 10: Practical Geometry Overview
Practical Geometry involves constructing various geometric figures accurately. The exercises cover the construction of:
- A line parallel to a given line through a point not on the line.
- Triangles given different combinations of measurements:
- SSS (Side-Side-Side): When the lengths of all three sides are known.
- SAS (Side-Angle-Side): When the lengths of two sides and the measure of the angle between them are known.
- ASA (Angle-Side-Angle): When the measures of two angles and the length of the included side are known.
- RHS (Right angle-Hypotenuse-Side): When the length of the hypotenuse and one side of a right-angled triangle are known.
🔑 Solutions for Key Exercises
Solutions typically involve detailed, step-by-step construction processes and often include the final drawing or figure.
| Exercise | Key Construction Topics | Example Problem Type |
| Ex. 10.1 | Construction of a line parallel to a given line. | Draw a line $l$, take a point $P$ outside it, and construct a line $m$ parallel to $l$ through $P$. |
| Ex. 10.2 | Construction of a triangle (SSS Criterion). | Construct $\triangle XYZ$ where $XY = 4.5 \text{ cm}$, $YZ = 5 \text{ cm}$, and $ZX = 6 \text{ cm}$. |
| Ex. 10.3 | Construction of a triangle (SAS Criterion). | Construct $\triangle DEF$ such that $DE = 5 \text{ cm}$, $DF = 3 \text{ cm}$, and $\angle EDF = 90^\circ$. |
| Ex. 10.4 | Construction of a triangle (ASA Criterion). | Construct $\triangle ABC$ given $\angle A = 60^\circ$, $\angle B = 30^\circ$, and $AB = 5.8 \text{ cm}$. |
| Ex. 10.5 | Construction of a right-angled triangle (RHS Criterion). | Construct a right-angled triangle $PQR$, where $\angle Q = 90^\circ$, $QR = 8 \text{ cm}$, and $PR = 10 \text{ cm}$. |
NCERT Class 7 Maths Chapter 10 – Practical Geometry Exercise 10.1
1. Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using a ruler and compasses only.
Solution:-
Steps for construction,
1. Draw a line AB.
2. Take any point Q on AB and a point P outside AB and join PQ.
3. With Q as the centre and any radius, draw an arc to cut AB at E and PQ at F.
4. With P as the centre and the same radius, draw an arc IJ to cut QP at G.
5. Place the pointed tip of the compass at E and adjust the opening so that the pencil tip is at F.
6. With the same opening as in step 5 and with G as the centre, draw an arc cutting the arc IJ at H.
7. Now, join PH to draw a line CD.
2. Draw a line L. Draw a perpendicular to L at any point on L. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to L.
Solution:-
Steps for construction,
1. Draw a line L.
2. Take any point P on line L.
3. At point P, draw a perpendicular line N.
4. Place the pointed tip of the compass at P and adjust the compass up to the length of 4 cm. Draw an arc to cut this perpendicular at point X.
5. At point X, again draw a perpendicular line M.
3. Let L be a line and P be a point not on L. Through P, draw a line m parallel to L. Now join P to any point Q on L. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet L at S. What shape do the two sets of parallel lines enclose?
Solution:-
Steps for construction,
1. Draw a line L.
2. Take any point Q on L and a point P outside L and join PQ.
3. Make sure that angles at point P and point Q are equal, i.e. ∠Q = ∠P
4. At point P, extend the line to get line M which is parallel to L.
5. Then take any point R on line M.
6. At point R, draw an angle such that ∠P = ∠R
7. At point R, extend the line which intersects line L at S and draw a line RS.
NCERT Class 7 Maths Chapter 10 – Practical Geometry Exercise 10.2
1. Construct ΔXYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.
Solution:-
Steps of construction
1. Draw a line segment YZ = 5 cm.
2. With Z as a centre and radius 6 cm, draw an arc.
3. With Y as a centre and radius 4.5 cm, draw another arc, cutting the previous arc at X.
4. Join XY and XZ.
Then, ΔXYZ is the required triangle.
2. Construct an equilateral triangle of side 5.5 cm.
Solution:-
Steps of construction
1. Draw a line segment AB = 5.5 cm.
2. With A as a centre and radius 5.5 cm, draw an arc.
3. With B as a centre and radius 5.5 cm, draw another arc, cutting the previous arc at C.
4. Join CA and CB.
Then, ΔABC is the required equilateral triangle.
3. Draw ΔPQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?
Solution:-
Steps of construction:
1. Draw a line segment QR = 3.5 cm.
2. With Q as a centre and radius 4 cm, draw an arc.
3. With R as a centre and radius 4 cm, draw another arc, cutting the previous arc at P.
4. Join PQ and PR.
Then, ΔPQR is the required isosceles triangle.
4. Construct ΔABC, such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
Solution:-
1. Draw a line segment BC = 6 cm.
2. With B as a centre and radius 2.5 cm, draw an arc.
3. With C as a centre and radius 6.5 cm, draw another arc, cutting the previous arc at A.
4. Join AB and AC.
Then, ΔABC is the required triangle.
5. When we measure the angle B of the triangle with a protractor, then the angle is equal to ∠B = 90o
NCERT Class 7 Maths Chapter 10 – Practical Geometry Exercise 10.3
1. Construct ΔDEF, such that DE = 5 cm, DF = 3 cm and m∠EDF = 90o.
Solution:-
Steps of construction:
1. Draw a line segment DF = 3 cm.
2. At point D, draw a ray DX to make an angle of 90o, i.e., ∠XDF = 90o
3. Along DX, set off DE = 5cm
4. Join EF.
Then, ΔEDF is the required right-angled triangle.
2. Construct an isosceles triangle in which the lengths of each of its equal sides are 6.5 cm, and the angle between them is 110o.
Solution:-
Steps of construction:
1. Draw a line segment AB = 6.5 cm.
2. At point A, draw a ray AX to make an angle of 110o, i.e., ∠XAB = 110o.
3. Along AX, set off AC = 6.5cm.
4. Join CB.
Then, ΔABC is the required isosceles triangle.
3. Construct ΔABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.
Solution:-
Steps of construction:
1. Draw a line segment BC = 7.5 cm.
2. At point C, draw a ray CX to make an angle of 60o, i.e., ∠XCB = 60o.
3. Along CX, set off AC = 5cm.
4. Join AB.
Then, ΔABC is the required triangle.
NCERT Class 7 Maths Chapter 10 – Practical Geometry Exercise 10.4
1. Construct ΔABC, given m ∠A =60o, m ∠B = 30o and AB = 5.8 cm.
Solution:-
Steps of construction:
1. Draw a line segment AB = 5.8 cm.
2. At point A, draw a ray P to making an angle of 60o i.e. ∠PAB = 60o.
3. At point B, draw a ray Q to making an angle of 30o i.e. ∠QBA = 30o.
4. Now the two rays AP and BQ intersect at the point C.
Then, ΔABC is the required triangle.
2. Construct ΔPQR if PQ = 5 cm, m∠PQR = 105o and m∠QRP = 40o.
(Hint: Recall angle-sum property of a triangle).
Solution:-
We know that the sum of the angles of a triangle is 180o.
∴ ∠PQR + ∠QRP + ∠RPQ = 180o
= 105o+ 40o+ ∠RPQ = 180o
= 145o + ∠RPQ = 180o
= ∠RPQ = 180o– 1450
= ∠RPQ = 35o
Hence, the measures of ∠RPQ is 35o.
Steps of construction:
1. Draw a line segment PQ = 5 cm.
2. At point P, draw a ray L to making an angle of 105o i.e. ∠LPQ = 35o.
3. At point Q, draw a ray M to making an angle of 40o i.e. ∠MQP = 105o.
4. Now the two rays PL and QM intersect at the point R.
Then, ΔPQR is the required triangle.
3. Examine whether you can construct ΔDEF such that EF = 7.2 cm, m∠E = 110° and
m∠F = 80°. Justify your answer.
Solution:-
From the question it is given that,
EF = 7.2 cm
∠E = 110o
∠F = 80o
Now we have to check whether it is possible to construct ΔDEF from the given values.
We know that the sum of the angles of a triangle is 180o.
Then,
∠D + ∠E + ∠F = 180o
∠D + 110o+ 80o= 180o
∠D + 190o = 180o
∠D = 180o– 1900
∠D = -10o
We may observe that the sum of two angles is 190o is greater than 180o. So, it is not possible to construct a triangle.
NCERT Class 7 Maths Chapter 10 – Practical Geometry Exercise 10.5
1. Construct the right-angled ΔPQR, where m∠Q = 90°, QR = 8cm and PR = 10 cm.
Solution:-
Steps of construction
1. Draw a line segment QR = 8 cm.
2. At point Q, draw a ray QY to make an angle of 90o, i.e., ∠YQR = 90o.
3. With R as a centre and radius of 10 cm, draw an arc that cuts the ray QY at P.
4. Join PR.
Then, ΔPQR is the required right-angled triangle.
2. Construct a right-angled triangle whose hypotenuse is 6 cm long, and one of the legs is 4 cm long.
Solution:-
Let us consider ΔABC is a right-angled triangle at ∠B = 90o
Then,
AC is hypotenuse = 6 cm … [given in the question]
BC = 4 cm
Now, we have to construct the right-angled triangle by the above values.
Steps of construction
1. Draw a line segment BC = 4 cm.
2. At point B, draw a ray BX to make an angle of 90o, i.e., ∠XBC = 90o.
3. With C as a centre and radius of 6 cm, draw an arc that cuts the ray BX at A.
4. Join AC.
Then, ΔABC is the required right-angled triangle.
3. Construct an isosceles right-angled triangle ABC, where m∠ACB = 90° and AC = 6 cm.
Solution:-
Steps of construction
1. Draw a line segment BC = 6 cm.
2. At point C, draw a ray CX to make an angle of 90o, i.e., ∠XCB = 90o.
3. With C as a centre and radius of 6 cm, draw an arc that cuts the ray CX at A.
4. Join AB.
Then, ΔABC is the required right-angled triangle.