Simplify Practical Geometry with Free PDF Downloads of NCERT Solutions for Class 7 Math Chapter 10, "Practical Geometry," designed to make the understanding of geometry problems accessible for students. At [Your School Name], we have meticulously crafted these solutions to facilitate students in cracking problems with ease. Our stepbystep explanations ensure students can grasp the concepts thoroughly. Our NCERT Solutions aim to empower students to excel in geometry and develop confidence in solving practical geometry problems.
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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.
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 center and any radius, draw an arc to cut AB at E and PQ at F.
4. With P as the center 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 center, draw an arc cutting the arc IJ at H.
7. Now, join PH to draw a line CD.
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.
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.
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?
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.
Construct ΔXYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.
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.
Construct an equilateral triangle of side 5.5 cm.
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.
Draw ΔPQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?
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.
Construct ΔABC, such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
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
Construct ΔDEF, such that DE = 5 cm, DF = 3 cm and m∠EDF = 90o.
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 rightangled triangle.
Construct ΔDEF, such that DE = 5 cm, DF = 3 cm and m∠EDF = 90o.
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 rightangled triangle.
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.
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.
Construct ΔABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.

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.
Construct ΔABC, given m ∠A =60o, m ∠B = 30o and AB = 5.8 cm.
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.
Construct ΔPQR if PQ = 5 cm, m∠PQR = 105o and m∠QRP = 40o.
(Hint: Recall anglesum property of a triangle).
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.
Examine whether you can construct ΔDEF such that EF = 7.2 cm, m∠E = 110° and
m∠F = 80°. Justify your answer.
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.
Construct the rightangled ΔPQR, where m∠Q = 90°, QR = 8cm and PR = 10 cm.
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 rightangled triangle.
Construct a rightangled triangle whose hypotenuse is 6 cm long, and one of the legs is 4 cm long.
Let us consider ΔABC is a rightangled triangle at ∠B = 90o
Then,
AC is hypotenuse = 6 cm … [given in the question]
BC = 4 cm
Now, we have to construct the rightangled 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 rightangled triangle.
Construct an isosceles rightangled triangle ABC, where m∠ACB = 90° and AC = 6 cm.
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 rightangled triangle.
The NCERT solution for Class 7 Chapter 10: Practical Geometry is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 7 Chapter 10: Practical Geometry is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. Practical Geometryally, they can solve the practice questions and exercises that allow them to get examready in no time.
You can get all the NCERT solutions for Class 7 Maths Chapter 10 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 7 Maths Chapter 10: Practical Geometry as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 7 Maths Chapter 10 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.
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