RS Aggarwal Chapter 14 Class 9 Maths Exercise 14.1 Solutions: You may recall that the part of the plane enclosed by a simple closed figure is called a planar region corresponding to that figure. The magnitude or measure of this planar region is called its area. This magnitude or measure is always expressed with the help of a number (in some unit) such as 5 cm2 , 8 m2 , 3 hectares etc. So, we can say that area of a figure is a number (in some unit) associated with the part of the plane enclosed by the figure
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Important Definition for RS Aggarwal Chapter 14 Class 9 Maths Ex 14a Solutions
- AB and CD are parallel sides of trapezium ABCD. Diagonals AC and BD intersect at O. prove that ar(ΔAOD) = ar(ΔBOC).
- If D is the mid .point of side of side BC of a ΔABC, P and Q are two points lying respectively on the sides AB and BC such that DP is parallel to QA. Prove that ar(ΔCQP) = 1/2 ar(ΔABC).
- A rectangle is formed by joining the mid-points of the sides of a rhombus. Show that the area of rectangle is half the area of rhombus.
- In a parallelogram ABCD, AE is perpendicular to DC and CF is perpendicular to AD. If AB = 10 cm, AE = 6 cm and CF = 8 cm, then find AD.
- The adjacent sides of a rectangle are 16 cm and 8 cm. Find the area of the rectangle.
- PQRS is a square. T and U are the mid-points of sides PS and QR respectively. Find the area of ΔOTS, if PQ= 8 cm, where O is the point of intersection of TU and OS.
- If two sides of one triangle are equal to two sides of another triangle and the contained angles are supplementary, show that the two sides are equal in area.
- In a trapezium ABCD where AB is parallel to CD, E is the mid-point of BC, prove that ΔAED = 1/2 trapezium ABCD.
- The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD.
- The area of parallelogram PQRS is 88 cm sq. A perpendicular from S is drawn to intersect PQ at M. If SM = 8 cm, then find the length of PQ.
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