Skip to main content

10th Maths Similar triangle Guess Paper For CBSE Board Exam SA-1




10th Triangle (Similarity) Practice Questions For SA-1 By JSUNIL

Similar figures: “Two similar figures have the same shape but not necessarily the same sizes are called similar figures. “ This verifies that congruent figures are similar but the similar figures need not be congruent.

Conditions for similarity of polygon: Two polygons of the same number of sides are similar, if

(i) Their corresponding angles are equal and

(ii) Their corresponding sides are in the same ratio (or proportion).

Note: The same ratio of the corresponding sides is referred to as the scale factor (or the Representative Fraction) for the polygons.

Equiangular triangles: If corresponding angles of two triangles are equal, then they are known as equiangular triangles.

A famous Greek mathematician Thales gave an important truth relating to two equiangular triangles which is as follows: “The ratio of any two corresponding sides in two equiangular triangles is always the same.”

Q. The Basic Proportionality Theorem (now known as the Thales Theorem) : “If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. “ [Prove it.]

Q. The converse of The Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. [Prove it by contradiction methods]

Q. In a triangle ABC, E and F are point on AB and AC and EF || BC. Prove that AB/AE = AC/AF

Q. Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Q. Prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Q. In a triangle ABC, E and F are point on AB and AC Such that AE/EB = AF/FC and <AEF =<ACB. Prove that ABC is an isosceles Triangle.

Q. In a trapezium ABCD , AB || DC and E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB .Show that AE/ ED = BF /FC [join AC to intersect EF at G]

Q. In a trapezium ABCD , AB || DC and its diagonals intersect each other at the point O. Show that AO/ BO = CO/DO

Q. If the diagonals of a quadrilateral divide each other proportionally, then it is a trapezium.

Q. In ΔABC, DE || BC
(a) IF AD /DB = 2/3 and AC = 18cm, find AE.
(b) IF AD = x, DB = x – 2 , AE = x + 2, EC = x -1, find x.
(c) If AD = 8cm, AB = 12cm, AE = 12cm, find CE.

Q. In the given figure, AB || DC. If EA = 3x - 19, EB = x - 4, EC = x - 3 and ED = 4, find x.

Download Full Pdf File : DOWNLOAD


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 dissolved in it

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

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 cm. Therefor

Electricity numerical for class 10 CBSE Trend Setter 50 Problems

1. The current passing through a room heater has been halved. What will happen to the heat produced by it? 2. An electric iron of resistance 20 ohm draws a current of 5 amperes. Calculate the heat produced in 30 seconds. 3. An electric heater of resistance 8 ohm takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater. 4. A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination in half a minute. 5. A resistance of 25 ohm is connected to a 12 V battery. Calculate the heat energy in joules generated per minute. 6. 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor? 7. An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Ca