CBSE ADDA

Sample questions are 1. The angle between two altitudes of parallelogram through vertex of an obtuse angle of parallelogram is 70. Find all angles of parallelogram. 2. ABCD is a parallelogram. If AB = 2AD and P is the mid-point of AB, then find <CPD 3. ABCD is a trapezium, AB||CD, X and Y are the mid points of AD and BC respectively. If AB = 30 cm and CD = 50 cm, then show that ar (DCYX) = ar (XYBA). 4. In a parallelogram ABCD the bisector of angle A also bisects BC at X . Prove that AD = 2AB 5. In a trapezium AB||CD, E and F are midpoint of non-parallel sides AD and BC. Show that EF =1/2 (AB +CD)  For full list of questions download  9th Area of parallelogram and quadrilateral guess questions for CBSE Exam 2018 9th Area of parallelogram and quadrilateral guess questions 2018 -1 Download File 9th Area of parallelogram and quadrilateral guess questions 2018 -2 Download File For more study help visit  website jsuniltutorial.in 

Quadrilaterals  A quadrilateral is a closed plane figure bounded by four line segments. E.g. The figure ABCD shown here is a quadrilateral. A line segment drawn from one vertex of a quadrilateral to the opposite vertex is called a diagonal of the quadrilateral. For example, AC is a diagonal of quadrilateral ABCD. Types of Quadrilaterals There are six basic types of quadrilaterals: 1. Rectangle: Opposite sides of a rectangle are parallel and equal. All angles are 90º. 2. Square Opposite sides of a square are parallel and all sides are equal. All angles are 90º. 3. Parallelogram Opposite sides of a parallelogram are parallel and equal. Opposite angles are equal. 4. Rhombus All sides of a rhombus are equal and opposite sides are parallel. Opposite angles of a rhombus are equal. The diagonals of a rhombus bisect each other at right angles. 5. Trapezium A trapezium has one pair of opposite sides parallel. A regular trapezium has non-parallel sides equal and its ba

CBSE ADDA 9th Constructions Construct a triangle ABC in which BC = 5 cm, < B = 60° and the sum of the other two sides is 7 cm Step of Constructions: (i) Draw BC = 5 cm.  (ii) Draw < XBC = 60° (iii) On the ray BX, mark off point D such that BD = 7 cm.  (iv) Join D to C. (v) Draw perpendicular bisector EF of CD. Let it intersects BD at A.  (vi) Join A to C.  ABC is the required triangle Construct a Triangle ABC in which < B = 45°, <C = 60° and AB + BC + AC = 13 cm. Step of constructions: (i) Draw XY = 13 cm. (ii) Draw < MXY = 45° and < NYX = 60° (iii) Draw angle bisectors of < MXY and < NYX, meeting at a point, say A. (iv) Draw perpendicular bisector of XA and YA, meeting XY at B and C respectively. (v) Join A to B and A to C. ABC is the required triangle Construct a Triangle ABC in which BC = 5.5 cm, < B = 30° and AB – AC = 2 cm. Step of constructions: (i) Draw BC = 5.5 cm. (ii) Draw < XB

CBSE ADDA Area of Parallelogram and Triangle Prove that followings: Parallelograms on the same base and between the same parallels are equal in area.  Two triangles on the same base (or equal base) and between the same parallels are equal in area.  Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.  If a triangles and a parallelogram are on the same base and between the same parallels, then prove that the area of the triangle is equal to half the area of the parallelogram.  In ABCD is parallelogram and EFCD is a rectangle. Also, AL ┴ DC. Prove that(i) ar (ABCD) = (EFCD)(ii) ar (ABCD) = DCxAL. ABCD is a parallelogram, AE DC and CF AD. If AB =16cm, AE=8cm and CF=10cm, find AD.  If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) =1/2 ar (ABCD).  P and Q are any two points lying on the sides DC and AD respectively of parallelogram ABCD. Show that ar (APB) =ar (BQC).  P is