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NCERT Practical Geometry class 7 - The word geometry comprises of two greek words geo and metron. Geo means earth and metron means measurements. Geometry is applied in everyday life from designing simple things like your furniture, doors and windows, the wall clock to the design of complex structures like the designing of a new house. To design all the above structures you need to know about geometry. In CBSE NCERT solutions for Class 7 Maths chapter 10 Practical Geometry, you will get questions related to applications of geometry in real life. You will study a few rulers and compass constructions in this chapter. You will learn how to draw a parallel line to a given line through a point outside the line, how to construct triangles if three sides of triangle are given (SSS), how to construct a triangle if measure of two sides and the angle between them are given (SAS), how to draw a triangle if value of two angles and the length of side included between them are given (ASA), how to draw a triangle if length of hypotenuse and the measure of one leg of a right-angled triangle are given (RHS). There are practice questions given on all the above topics in the NCERT .
You should try to solve all these questions by yourself. You can take help from NCERT Solutions for Class 7 . While solving the problems learn the steps and draw the figures. Just only going through the steps of drawings may not help in the exams. If you have tried to draw all the figures mentioned in the solutions of NCERT for Class 7 Maths chapter 10 Practical Geometry it will be easy for you solve all the questions in the exam related to this chapter. You can get NCERT Solutions from Classes 6 to 12 for Science and Maths by clicking on the above link. Here you will get solutions to five exercises of this chapter.
Criteria for Congruence of Triangles:
SSS Congruence: For triangles ABC and DEF, If AB = DE, BC = EF, and AC = DF, Then △ABC ≅ △DEF.
SAS Congruence: For triangles ABC and DEF, If AB = DE, ∠BAC = ∠EDF, and BC = EF, Then △ABC ≅ △DEF.
ASA Congruence: For triangles ABC and DEF, If ∠BAC = ∠EDF, ∠ABC = ∠DEF, and AC = DF, Then △ABC ≅ △DEF.
AAS Congruence: For triangles ABC and DEF, If ∠BAC = ∠EDF, ∠ACB = ∠EFD, and BC = EF, Then △ABC ≅ △DEF.
RHS Congruence: For right-angled triangles ABC and DEF, If ∠CAB = ∠FDE, AC = DF, and BC = EF, Then △ABC ≅ △DEF.
For more, Download Ebook - NCERT Class 7 Maths: Chapterwise Important Formulas And Points
Five Measurements Determine a Quadrilateral:
A quadrilateral can be uniquely determined when five measurements related to its sides and angles are known.
Construction with Four Sides and a Diagonal:
If the lengths of all four sides and one diagonal of a quadrilateral are given, the quadrilateral can be constructed uniquely.
Construction with Diagonals and Three Sides:
When the lengths of both diagonals and three sides of a quadrilateral are known, the quadrilateral can be constructed uniquely.
Construction with Adjacent Sides and Angles:
If the lengths of two adjacent sides and the measures of three angles of a quadrilateral are provided, the quadrilateral can be constructed uniquely.
Construction with Three Sides and Included Angles:
If the lengths of three sides and the measures of two angles included between those sides are given, the quadrilateral can be constructed uniquely.
Free download NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry PDF for CBSE Exam.
Answer:
1. Draw the line AB
2. Mark any point C outside AB and a point D on AB
3. Join CD
4. Draw an arc with convenient radius with D as centre cutting AB at E and DC at F
5. With the same radius as in step 4 draw an arc IGH with C as a centre and cutting line AC at G.
6 Place one tip of the compass on E and adjust the opening so that pencil tip touches at F.
7. With the same opening in step 6 and G as centre cut an arc at J on IGH.
8. Join CJ and extend the line. This line will be parallel to AB
Answer:
1. Draw a line l. Mark a point P on the line
2. With p as a centre and with any radius draw an arc which touches the line at A and B
3. Take a length more than AP with A as centre draw an arc. with the same length on compass and B as centre draw an arc to cut the previous arc at Q.
4. Join PQ. PQ is perpendicular to the line l
5. Take 4 cm on the compass and with P as centre cut an arc on line PQ at C. The length PC=4 cm
6. with P as a centre and any radius, draw an arc to cut line l at D and PQ at E.
7. with the same radius as in step 6 draw an arc GF with centre C so that it cuts the line PQ at G.
8. Take the length DE on the compass and with G as centre draw an arc to cut FG at J
9. Join CJ and extend the line to name it as line m. Now the line l is parallel to line m
Answer:
1. Draw the line PQ parallel to l using the steps to draw parallel lines
2. Draw the line RS parallel to PQ using the steps to draw parallel lines
the two parallel lines make the shape of a parallelogram PQRS.
Question:1 Construct in which XY= 4.5 cm, YZ = 5 cm and ZX = 6 cm.
Answer:
in which XY= 4.5 cm, YZ = 5 cm and ZX = 6 cm :
1. Draw a line XY = 4.5 cm
2. with x as centre and length= 6cm draw an arc
3. with y as centre and length = 5cm draw another arc to cut the arc drawn in step 2 at Z
4. Join XZ and YZ. XYZ is the required triangle
Question:2 Construct an equilateral triangle of side 5.5 cm.
Answer:
an equilateral triangle of side 5.5 cm :
1. Draw a line AB = 5.5 cm
2. with A as centre and length= 5.5cm draw an arc
3. with B as centre and length = 5.5cm draw another arc to cut the arc drawn in step 2 at C
4. Join AC and BC.ABC is the required triangle
Question:3 Draw ?PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?
Answer:
PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm:
1.Draw a line QR = 3.5 cm
2. With Q as centre and length = 4 cm draw an arc
3. With R as centre and length = 4cm draw another arc to cut the previous arc at P
4. Join QP and QR. Then PQR is the required triangle
As we can see two sides of the triangle are same so The triangle is an isosceles triangle.
Question:4 Construct such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure .
Answer:
such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure :
1. Draw a line AB = 2.5 cm
2. with A as centre and length= 6.5cm draw an arc
3. with B as centre and length = 6cm draw another arc to cut the arc drawn in step 2 at C
4. Join AC and BC. ABC is the required triangle
On measuring angle B comes out to be 90 degrees.
Question:1 Construct such that DE = 5 cm, DF = 3 cm and .
Answer:
such that DE = 5 cm, DF = 3 cm and :
1. Draw a line DE = 5cm
2. Draw a line DX making 90 degrees with DE.
3. cut the length DF =3cm using the compass on the line DX
4. Join E and F
DEF is the required triangle
Answer:
1. Draw a line DE = 6.5 cm
2. Draw a line DX making 110 degrees with DE.
3. cut the length DF =6.5 cm using the compass on the line DX
4. Join E and F
DEF is the required triangle
Question:3 Construct with BC = 7.5 cm, AC = 5 cm and .
Answer:
with BC = 7.5 cm, AC = 5 cm and :
1. Draw a line AC = 5 cm
2. Draw a line CX making 60 degrees with AC from C.
3. cut the length BC =7.5 cm using the compass on line CX
4. Join A and B
ABC is the required triangle
Question:1 Construct , given , and AB = 5.8 cm.
Answer:
, given , and AB = 5.8 cm :
1. Draw the line AB=5.8 cm
2. At A making an angle 60 degrees with AB draw a line AX
3. At B making an angle 30 degrees with AB draw a line BY
4. The lines AX and BY intersect at C ( in case the drawn line AX or BY is small extend it to intersect each other)
ABC is the requires triangle
Question:2 Construct if PQ = 5 cm, and .
Answer:
if PQ = 5 cm, and :
the sum of angles of a triangle is 180 degrees. Given angles PQR and QRP, so the angle QPR= 180-(105+40)=35 degrees
1. Draw the line PQ=5 cm
2. At Q making an angle 105 degrees with PQ draw a line QY
3. At P making an angle 35 degrees with PQ draw a line PX
4. The lines PX and QY intersect at R ( in case the drawn line PX or QY is small extend it to intersect each other)
PQR is the required triangle
Question:3 Examine whether you can construct such that EF = 7.2 cm, and . Justify your answer.
Answer:
No, we cannot construct such that EF = 7.2 cm, and . This is because of the property of the triangle of having a sum of internal angles equal to 180 degrees. if we have , then we cannot have because then the sum would exceed 180 degree which is impossible.
Question:1 Construct the right angled , where , QR = 8cm and PR = 10 cm.
Answer:
the right-angled , where , QR = 8cm and PR = 10 cm:
1. Draw QR = 8 cm
2. Draw a perpendicular QX to QR at Q
3. From R and with length = 10 cm cut an arc on QX. The arc meet QX at P
4. Join PR
PQR is the required triangle
Question:2 Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
Answer:
a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long:
1. Draw QR = 4 cm
2. Draw a perpendicular QX to QR at Q
3. From R and with length = 6 cm cut an arc on QX. The arc meet QX at P
4. Join PR
Triangle PQR is the required triangle
Question:3 Construct an isosceles right-angled triangle ABC, where and AC = 6 cm.
Answer:
1. Draw CA = 6 cm
2. Draw a perpendicular CX to CA at C
3. From C and with length = 6 cm cut an arc on CX. The arc meet CX at B
4. Join BA
Triangle ABC is the required triangle
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | Practical Geometry |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
In CBSE NCERT solutions for Class 7 Maths chapter 10 Practical Geometry, you have learnt some fundamental of geometry for simple mathematical geometry. There are many other applications of geometry in the field of designing, civil engineering, CAD(computer-aided design), mapping, GPS and many more which you will learn in higher classes.
Happy Reading!!!
Also Check NCERT Books and NCERT Syllabus here:
Practical geometry Class 7th is a branch of geometry in which we learn the method of constructing various geometrical shapes like line segments, angles, triangles, and different polygons. We use various geometrical tools to construct the required shapes with the exact measurements given.
There are 5 exercises in NCERT chapter 10 maths class 7th.
Practical Geometry Class 7 NCERT Exercise 10.1- 3 questions
Practical Geometry Class 7th NCERT Exercise 10.2- 4 questions
Practical Geometry Class 7th NCERT Exercise Exercise 10.3 - 3 questions
Practical Geometry Class 7 NCERT Exercise 10.4 - 3 questions
Practical Geometry Class 7th NCERT Exercise Exercise 10.5 - 3 questions
The Father of Geometry is Euclid. He was a Greek Mathematician. He discussed many concepts related to Geometry, students can get brief idea from class 7 chapter 10 maths about these concepts. Also you can practice practical geometry class 7 pdf to get deeper understanding of concepts.
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