Skip to main content

CBSE 8tm maths:Construction of quadrilateral (Ch:Practical Geometry)

Point to Remember
1. .Five measurement can determine a quadrilateral uniquely.

2. A quadrilateral can be constructed uniquely if

  • Its four sides and one diagonal is given.
  • Its two adjacent sides and three angles are known.
  • Its three sides and two included angles are given.
  • Its other special properties are known (rectangle, square, rhombus, parallelogram).
  • Its three sides and two diagonals given.

For more study

Construction of quadrilateral Practice Questions
Construction of quadrilateral  CBSE Test Paper
Construction of quadrilateral  CBSE Test Paper

1. How many measurement s can determine a quadrilateral uniquely?

Ans: 5 

2. How many measurements can determine a square? 

Ans: 1 

3. How many measurements can determine a parallelogram uniquely? 

Ans: 3 

4. How many measurement can determine a rhombus? 

Ans: 2 

5. Which property is used to construct a parallelogram. If its one side and two diagonals are given.

Ans: Diagonals are bisects to each other. 

6. What property is used to construct a rhombus. If its two diagonals are given. 

Ans: Diagonals of a rhombus bisect each other at right angle.

Practice Sample paper

Q.1. Construct a quadrilateral BDEF where DE = 4.5 cm, EF = 3.5cm, FB = 6.5cm, <F = 50 and <E = 100.

Q.2. Construct a quadrilateral ABCD give that AB = 4cm, BC = 3cm, <A = 75, <B=80  and <C=120.

Q.3. Construct a quadrilateral PQRS where PQ = 3.5 cm, QR = 6.5 cm, <P = <R = 105 and <S = 75

Q.4. Construct a quadrilateral ABCD, where BC = 5.5 cm, CD = 4.1 cm, <A = 70 ,<B = 110 and <D=85

Q.5. Construct a rhombus ABCD where AB = 6cm, <B = 75.

Q.6. Construct a rectangle WXYZ in which WX = 7cm and XY = 4cm.

Q.7. Construct a rhombus ABCD where AB = 4cm and AC = 6cm.

Q.8.Construct a square ABCD where AB = 6cm.

Q.9. Construct a parallelogram PQRS in which PQ = 4.5 cm, QR = 5.1cm and <Q=70.

Q.10. Construct a square of diagonal 6 cm long.
                                                                       OR,
Construct a parallelogram ABCD in which AC = 3.8 cm, BD = 4.6 cm and the angle between AC and BD is 60

11. Construct a parallelogram ABCD in which AB=6.5 AC=3.8 and the altitude AL from A is 2.5cm draw an altitude from C

[ Hint : Follow the given steps to construct the parallelogram ABCD.

Step 1. Draw AB = 6.5 cm.

Step 2. At A, draw AE ⊥ AB. With A as centre and radius 2.5 cm, mark an arc which intersects AE in F.

Step 3. Draw a line parallel to AB and passing through F.

Step 4. With A as centre and radius 3.4 cm, mark an arc which intersects the line parallel to AB in C.

Step 5. With C as centre and radius 6.5 cm, mark an arc which intersects the line CF in D. (Opposite sides of parallelogram are equal)

Step 6. Join BC and DA.

Thus, ABCD is the required parallelogram. 


12. Construct a quadrilateral PQRS m which PQ = 5 cm, QR = 6.5 cm, ∠P = ∠R = 100° and ∠S = 75°.

13.Construct a quadrilateral ABCD in which AB = 3.6 cm, BC = 3.3 cm, AD = 2.7 cm, diagonal AC = 4.6 cm and diagonal BD = 4 cm.

14. Construct a trapezium ABCD in which AB = 6 cm, BC = 4 cm, CD = 3.2 cm, ∠B = 75° and DC ∥ AB.

15. Construct a parallelogram ABCD, in which diagonal AC = 3.8 cm, diagonal BD = 4.6 cm and the angle between AC and BD is 60°.

Comments

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink   Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get d...

CBSE I NCERT 10th Numerical Problem solved Reflection and reflection of light

Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?  Solution: V= -50 cm F= -20cm From mirror formula 1/u = 1/f – 1/v = -1/20+ 1/50 = - 3/100  U = - 33.3 cm Therefore, the distance of the object from the wall x =  50 – u X = 50 – 33.3 = 16.7 cm. Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced? Answer: Here f = - 15 cm, u = - 40 cm Now 1/f = 1/u + 1/v Then 1/v = 1/f – 1/u Or V= uf/u-f =( - 40 x -15)/25 = -24 cm Then object is displaced towards the mirror let u1 be the distance object from the Mirror in its new position. Then u1 = -(40-20) = -20cm If the image is formed at a distance u1 from the mirror then v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 ...

Living science ratna sagar class 6 solutions

Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page ...

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination(Solved)

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination Law of conservation of mass Law of constant proportion Empirical formula 1. If 10 grams of CaCO 3 on heating gave 4.4g of CO 2 and 5.6g of CaO, show that these observations are in agreement with the law of conservation of mass.(Based on Law of conservation of mass) Solution:  Mass of the reactants = 10g ;  Mass of the products = 4.4 + 6.6g = 10g Since the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass. 2. 1.375 g of cupric oxide was reduced by heating and the weight of copper that remained was 1.098g.  In another experiment 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate was converted into cupric oxide by ignition . The weight of cupric oxide formed was 1.476 g. which law of chemical combinations does this data state? Solution: in first experiment: Copper oxide = 1....