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NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry provides a downloadable FREE PDF. The foremost objective is to help students understand and crack these problems. We at Swiflearn have prepared the NCERT Solutions for Class 7 Maths wherein problems are solved step by step with complete descriptions. There are 5 exercises present in Chapter 10 – Practical Geometry of NCERT Solutions for Class 7 Maths.

NCERT Solutions for Class 7 Maths Chapter 10 by Swiflearn are by far the best and most reliable NCERT Solutions that you can find on the internet. These NCERT Solutions for Class 7 Maths Chapter 10 are designed as per the CBSE Class 7 Maths Syllabus. These NCERT Solutions will surely make your learning convenient & fun. Students can also Download FREE PDF of NCERT Solutions Class 7 Chapter 10. NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry: EX 10.1 PDF

Question 1.
Draw a line, say AB, take a point C outside it. Through C, draw a line
parallel to AB using ruler and compasses only.

Solution:- Steps for construction of the figure,
Step 1. Draw a straight line AB.
Step 2. Take a point Q on AB and a point P outside AB and join points PQ.
Step 3. With Q as centre and any radius draw an arc that cut AB at E and PQ at F.
Step 4. With P as centre and same radius draw an arc IJ that cut QP at G.
Step 5. Placing the pointed tip of the compass at E and adjust the opening so that the pencil tip is at F.
Step 6. The with the same opening as in last step(5) and with G as centre, draw an arc cutting the arc IJ at H.
Step 7. At last, join PH to draw a line CD.

Question 2.Draw a line L. Draw a perpendicular to L at any point on L. On thisperpendicular choose a point X, 4 cm away from l. Through X, draw a linem parallel to L.

Solution:- Steps for construction,
Step 1. Draw a line L.
Step 2. Take a point P on the line L.
Step 3. At point P, draw a perpendicular line N.
Step 4. Using a compass, place the pointed tip of the compass at point P and adjust the compass to a length of 4 cm, draw an arc cutting the perpendicular at point X.
Step 5. At last. At the point X, again draw a perpendicular line M.

Question 3.Let L be a line and P be a point not on L. Through P, draw a line mparallel to L. Now join P to any point Q on L. Choose any other point R onm. Through R, draw a line parallel to PQ. Let this meet L at S. What shapedo the two sets of parallel lines enclose?

Solution:- Steps for construction,
Step 1. Draw a line L.
Step 2. Take a point Q on the line L and a point P outside L and join points PQ.
Step 3. Make sure that angles are equal at point P and point Q i.e. ∠Q = ∠P
Step 4. At point P extend the line to get line M which is parallel line L.
Step 5. Then take a point R on line M.
Step 6. At point R draw an angle such that ∠P = ∠R
Step 7. At last, on point R extend the line which intersects line L at S and draw a line RS.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry: EX 10.2 PDF

Question 1.Construct ΔXYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm

Solution:- Steps for construction of the figure:
Step 1. Draw a straight line segment i.e. YZ = 5 cm.
Step 2. With taking Z as a centre and a radius of 6 cm, draw an arc.
Step 3. With taking Y as a centre and a radius of 4.5 cm, draw another arc, cutting the previous arc at X.
Step 4. Join the points XY and points XZ.
ΔXYZ is formed and is the required triangle.

Question 2.Construct an equilateral triangle of side 5.5 cm.

Solution:- Steps for construction of the figure,
Step 1. Draw a straight line segment i.e. AB = 5.5 cm.
Step 2. With taking A as a centre and radius of 5.5 cm, draw an arc.
Step 3. With taking B as a centre and radius of 5.5 cm, draw another arc, cutting the previous arc at C.
Step 4. Join the points CA and points CB.
Then, ΔABC is formed and is the required triangle.

Question 3.
Draw ΔPQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of
triangle is this?

Solution:- Steps for construction of the figure,
Step 1. Draw a straight line segment i.e. QR = 3.5 cm.
Step 2. With taking Q as a centre and radius of 4 cm, draw an arc.
Step 3. With taking R as a centre and radius of 4 cm, draw another arc, cutting the previous arc at P.
Step 4. Join points PQ and points PR.
Then, ΔPQR is formed and is the required triangle.

Question 4.Construct ΔABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm.Measure ∠B.

Solution:- Step 1. Draw a line segment BC = 6 cm.
Step 2. With taking B as a centre and of radius 2.5 cm, draw an arc.
Step 3. With taking C as a centre and of radius 6.5 cm, draw another arc, cutting the previous arc at A.
Step 4. Join points AB and points AC.
Then, ΔABC is formed and is the required triangle.
Step 5. Measuring the angle B of triangle by protractor, the angle comes out and is equal to ∠B = 80o

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry: EX 10.3 PDF

Question 1.Construct ΔDEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90o

.
Solution:- Steps for construction of the figure,
1. Draw a straight line segment i.e. DF = 3 cm.
2. At the point D, draw a straight ray DX making an angle of 90o
i.e. ∠XDF = 90o.
3. Then, Along DX, mark DE at. 5cm.
4. Join points EF.
Then, ΔEDF is formed and is the required triangle.

Question 2.Construct an isosceles triangle in which the lengths of each of its equalsides is 6.5 cm and the angle between them is 110o

.
Solution:- Steps for construction of the figure,
1. Draw a straight line segment AB i.e. 6.5 cm.
2. At the point A, draw a straight ray AX making an angle of 110o
i.e. ∠XAB = 110o.
3. Along AX, mark AC at 6.5cm.
4. Join CB.
Then, ΔABC is the required isosceles triangle formed.

Question 3.Construct ΔABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.

Solution:- Steps for construction of the figure,
1. Draw a straight line segment BC i.e. 7.5 cm.
2. At the point C, draw a straight ray CX to making an angle of 60o
i.e. ∠XCB = 60o .
3. Along CX, mark AC at 5cm.
4. Join points AB.
Then, ΔABC is formed and is the required triangle.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry: EX 10.4 PDF

Question 1.Construct ΔABC, given m ∠A =60o, m ∠B = 30o and AB = 5.8 cm.

Solution:- Steps for construction of the figure,
1. Draw a straight line segment AB of 5.8 cm.
2. At the point A, draw a straight ray P making an angle of 60o
i.e. ∠PAB = 60o.
3. At point B, draw a straight ray Q making an angle of 30o
i.e. ∠QBA = 30o.
4. Now mark the intersecting point of the two rays AP and BQ as point C.
Then, ΔABC is formed and is the required triangle.

Question 2.Construct ΔPQR if PQ = 5 cm, m∠PQR = 105o and m∠QRP = 40o.(Hint: Recall angle-sum property of a triangle).

Solution:- From the previous lessons we know that the sum all of the angles inside of a triangle is 180o.
∴ ∠PQR + ∠QRP + ∠RPQ = 180o
= 105o+ 40o+ ∠RPQ = 180o
= 145o + ∠RPQ = 180o
= ∠RPQ = 180o
– 1450
= ∠RPQ = 35o
Therefore, the measures of ∠RPQ comes out to be = 35o.
Steps for construction of the figure,
1. Draw a straight line segment PQ of 5 cm.
2. At the point P, draw a straight ray L that makes an angle of 35o
i.e. ∠LPQ = 35o.
3. At the point Q, draw a straight ray M that makes an angle of 105o
i.e. ∠PQM = 105o.
4. Now mark the intersecting point of the two rays PL and QM as point R.
Then, ΔPQR is formed and is the required triangle.

Question 3.Examine whether you can construct ΔDEF such that EF = 7.2 cm, m∠E =110° and m∠F = 80°. Justify your answer.

Solution:-
According to the question,
EF = 7.2 cm
∠E = 110o
∠F = 80o
Verifying if it is possible to construct ΔDEF from the given values.
Then,
∠D + ∠E + ∠F = 180o
∠D + 110o+ 80o= 180o
∠D + 190o = 180o
∠D = 180o– 1900
∠D = -10o
It can be seen that the sum of two angles is 190o
and is greater than 180o . So, the triangle is not possible.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry: EX 10.5 PDF

Question 1.Construct the Construct the right angled ΔPQR, where m∠Q = 90°, QR =8cm and PR = 10 cm.

Solution:- Steps for construction of the figure,
1. Draw a straight line segment QR of 8 cm.
2. At the point Q, draw a straight ray QY that makes an angle of 90o
i.e. ∠YQR = 90o.
3. With R as the centre and radius of 10 cm, draw an arc that cuts the straight ray QY at point P.
4. Join point PR.
Then, ΔPQR is formed and is the required triangle.

Question 2.Construct a right-angled triangle whose hypotenuse is 6 cm long and one ofthe legs is 4 cm long

Solution:-
Assuming ΔABC to be right angled triangle at ∠B = 90o
So, AC becomes hypotenuse = 6 cm
BC = 4 cm
Now, constructing the right angled triangle from the above values. Steps for construction of the figure,
1. Draw a straight line segment BC of 4 cm.
2. At the point B, draw a straight ray BX that makes an angle of 90o
i.e. ∠XBC = 90o.
3. With C as the centre and radius of 6 cm, draw an arc that cuts the ray BX at point A.
4. Join point AC.
Then, ΔABC is formed and is the required triangle.

Question 3.Construct an isosceles right-angled triangle ABC, where m ∠ACB = 90°and AC = 6 cm.

Solution:- Steps for construction of the figure,
1. Draw a straight line segment BC of 6 cm.
2. At the point C, draw a ray CX that makes an angle of 90o
i.e. ∠XCB = 90o .
3. With C as the centre and radius of 6 cm, draw an arc that cuts the ray CX at point A.
4. Join points AB.
Then, ΔABC is formed and is the required triangle. 