4exam

In the given figure, P is a point in the interior of a parallelogram ABCD. Show that (i) ar (APB) + ar (PCD) =   ar (ABCD) (ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD) [Hint: Through. P, draw a line parallel to AB] Consider ΔACD. Line-segment CD is bisected by AB at O. Therefore, AO is the median of ΔACD. ∴ Area (ΔACO) = Area (ΔADO) ... (1) Considering ΔBCD, BO is the median. ∴ Area (ΔBCO) = Area (ΔBDO) ... (2) Adding equations (1) and (2), we obtain Area (ΔACO) + Area (ΔBCO) = Area (ΔADO) + Area (ΔBDO) ⇒ Area (ΔABC) = Area (ΔABD)

In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE) ans) AD is the median of ΔABC. Therefore, it will divide ΔABC into two triangles of equal areas.  ∴ Area (ΔABD) = Area (ΔACD) ... (1) ED is the median of ΔEBC. ∴ Area (ΔEBD) = Area (ΔECD) ... (2) On subtracting equation (2) from equation (1), we obtain Area (ΔABD) − Area (EBD) = Area (ΔACD) − Area (ΔECD) Area (ΔABE) = Area (ΔACE)

We know that diagonals of parallelogram bisect each other. Therefore, O is the mid-point of AC and BD. BO is the median in ΔABC. Therefore, it will divide it into two triangles of equal areas. ∴ Area (ΔAOB) = Area (ΔBOC) ... (1) In ΔBCD, CO is the median. ∴ Area (ΔBOC) = Area (ΔCOD) ... (2) Similarly, Area (ΔCOD) = Area (ΔAOD) ... (3) From equations (1), (2), and (3), we obtain Area (ΔAOB) = Area (ΔBOC) = Area (ΔCOD) = Area (ΔAOD) Therefore, it is evident that the diagonals of a parallelogram divide it into four triangles of equal area.

AD is the median of ΔABC. Therefore, it will divide ΔABC into two triangles of equal areas. ∴ Area (ΔABD) = Area (ΔACD) ⇒  ... (1) In ΔABD, E is the mid-point of AD. Therefore, BE is the median. ∴ Area (ΔBED) = Area (ΔABE) ⇒ Area (ΔBED) =  Area (ΔABD) ⇒ Area (ΔBED) =  Area (ΔABC) [From equation (1)] ⇒ Area (ΔBED) =  Area (ΔABC)

Question 1   ( 1.0 marks) A certain plant  ‘X’ shows the presence of vascular tissue, but does not produce seeds. To which division does plant ‘X’ belong? Solution: Pteridophytes Question 2   ( 1.0 marks) Fill in the blanks . Reptiles have __________-chambered heart, while fishes have __________-chambered heart. Solution: Reptiles have  three -chambered heart, while fishes have  two -chambered heart. Question 3   ( 1.0 marks) Give two examples of cartilaginous fishes. Solution: Scoliodon  and shark Question 4   ( 2.0 marks) Differentiate between bilateral symmetry and radial symmetry. Solution: - Radial Symmetry - Bilateral  Symmetry ( i) If the body of an animal is divided into two equal parts by cutting it in any plane passing through the centre of its body, then the body of the animal is said to have radial symmetry. (i) If the body of an animal is divided into left and right halves by one plane only, then the b

Question 1   ( 1.0 marks) Carl Woese further subdivided kingdom monera into two categories. What are these? Solution: Eubacteria and archaebacteria Question 2   ( 1.0 marks) Name the group of plants that do not contain well-differentiated body components. Solution: Thallophyta (algae) Question 3   ( 1.0 marks) Which fundamental characteristic separates animals from plants? Solution: Locomotion is present in animals, and not in plants. Question 4   ( 2.0 marks) Draw a branch diagram to represent the five-kingdom classification. Solution: Question 5   ( 2.0 marks) Mention the main features of kingdom monera. Solution: Main features of kingdom monera: (i) Absence of a well-defined nucleus and membrane-bound organelles (ii) Absence of multicellular design; all of them are unicellular Question 6   ( 2.0 marks) State any two general features of division gymnospermae. Solution: Two features of division gymno

Question 1   ( 1.0 marks) Calculate the mass  (in grams) of the iron sulphide formed when 5.6 g of iron combines with 3.2 g of sulphur. (Atomic mass of Fe = 56 u and Atomic mass of S = 32 u) Solution: Fe + S → FeS 56 g of Fe + 32 g of S = 88 g of FeS ∴ 5.6 g of Fe + 3.2 g of S = 8.8 g of FeS Question 2   ( 1.0 marks) What is the mass of 0.5 mole of Cl 2  gas? Solution: The molecular mass of chlorine is 35.5  u. Mass = Molar mass × N umber of moles = 35.5 × 2 = 71 g Question 3   ( 1.0 marks) Define one mole of a substance. Solution: One mole of a substance is the quantity of the substance containing 6.022 × 10 23  molecules of particles. Question 4   ( 2.0 marks) Cal culate the molecular mass of Ca (OH) 2 . Solution: M olecular mass of Ca (OH)  2  = (1 × 40 + 2 × 16 + 2 × 1) g = (40 + 32 + 2) g = 74 g Question 5   ( 2.0 marks) What is the significance  of the symbols of elements? Solution: S ignificance of t

Mass can neither be created nor destroyed in a chemical reaction. This is known as the law of conservation of mass. It means that the sum of the masses of the reactants and the products remains the same during a reaction. The law of constant proportion states that a chemical substance always contains the same elements in a fixed proportion by mass, irrespective of the source of compound. Matter is made up of very tiny particles and these particles are called atoms. Atoms cannot be divided further i.e., atoms are indivisible. An atom can be defined as the smallest particle of matter that can neither be created nor destroyed by chemical means. The symbol of the element is made from one or two letters of the English or the Latin name of the element. The mass of an atom is known as the atomic mass. The atomic mass of an atom of an element is also known as its relative atomic mass, since it is determined relative to the mass of C-12 isotope. A molecule is formed when two or

Grade 9 Why Do We Fall Ill Health  Health - A state of physical, mental, and social well-being, which includes a unity and harmony within the mind, body, and soul of an organism Diseases  Disease - Any condition that can lead to discomfort, distress, health problems, and even death of the affected person  Symptoms - Indications of disease, such as headache, stomach pain, nausea, etc.; can only be felt by the patient  Signs of a disease include fever, vomiting, diarrhoea, etc.; can be observed by a physician  Incubation period - The time interval between infection and appearance of symptoms Causes of diseases  On the basis of its duration - Acute and Chronic Acute - Lasts for a short period of time, eg. cold, cough, influenza, etc. Chronic - Lasts for long periods of time, eg. diabetes, kidney stones, etc.  On the basis of causative agents - Infectious and Non-infectious Infectious - Diseases such as influenza, cold, etc., which are caused due to infectious

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Lung Cancer    pop Lung cancer is the most frequently lethal cancer in the United States. Among all causes of death, lung cancer ranks second after heart disease for males, and third after heart and cerebrovascular disease for females. In 2002 alone, over 150,000 people in the United States died of lung cancer. http://e-lungcancer.com (Added: Fri Apr 07 2006 Hits: 2296 Downloads: 0 Rating: 0.00 Votes: 0)    Rate It     Review It     Details     LinkBack   Lung Cancer in Non-Smokers The incidence of lung cancer in non-smokers has increased over the years.Smokers and non-smokers alike are vulnerable to a disease which is largely incurable. .In the face of a disease that seems to have neither rhyme nor reason, what can we do to protect ourselves? http://www.GreatBodyat50.com (Added: Tue Apr 18 2006 Hits: 1148 Downloads: 0 Rating: 0.00 Votes: 0)    Rate It     Review It     Details     LinkBack   Lung Cancer Stages Cancers are staged depending on how far they have spread. Staging a

             Class IX, NCERT Mathematics  (CBSE Board)   Chapter NCERT Textbook Exercise Solution NUMBER SYSTEM Exercise 1.1 (Page 5) Read Exercise 1.2 (Page 8) Read Exercise 1.3 (Page 14) Read Exercise 1.4 (Page 18) Read Exercise 1.5 (Page 24) Read Exercise 1.6 (Page 26) Read POLYNOMIALS Exercise 2.1 (Page 32) Read Exercise 2.2 (Page 34, 35) Read Exercise 2.3 (Page 40) Read Exercise 2.4 (Page 43, 44) Read Exercise 2.5 (Page 48, 49, 50) Read COORDINATE GEOMETRY Exercise 3.1 (Page 53) Read Exercise 3.2 (Page 60, 61) Read Exercise 3.3 (Page 65) Read LINEAR EQUATIONS IN TWO VARIABLES Exercise 4.1 (Page 68) Read Exercise 4.2 (Page 70) Read Exercise 4.3 (Page 74, 75) Read Exercise 4.4 (Page 77) Read INTRODUCTION TO EUCLID’S GEOMETRY Exercise 5.1 (Page 85, 86) Read Exercise 5.2 (Page 88) Read LINES AND ANGLES Exercise 6.1 (Page 96, 97) Read Exercise 6.2 (Page 103