Mathematics
In the figure (2) given below, PQRS and ABRS are parallelograms, and X is any point on the side BR. Show that
area of ∆AXS = area of || gm PQRS.
Theorems on Area
9 Likes
Answer
From figure,
Since, PB is a straight line and PQ || SR so,
PB || SR.
|| gm PQRS and ABRS are on the same base SR and between the same parallel lines PB and SR.
So, Area of ||gm PQRS = Area of ||gm ABRS ………(i)
∆AXS and || gm ABRS are on the same base AS and between the same parallel lines AS and BR.
So, Area of ∆AXS = Area of ||gm ABRS
= area of ||gm PQRS [From (i)]
Hence, proved that Area of ∆AXS = Area of || gm PQRS.
Answered By
7 Likes
Related Questions
In figure (1) given below, ABCD is a parallelogram. P, Q are any two points on the sides AB and BC respectively. Prove that
area of ∆CPD = area of ∆AQD.
In the adjoining figure, D, E and F are midpoints of the sides BC, CA and AB respectively of ∆ABC. Prove that BCEF is a trapezium and area of the trap. BCEF = area of ∆ABC.
D, E and F are mid-points of the sides BC, CA and AB respectively of a ∆ABC. Prove that
(i) FDCE is a parallelogram
(ii) area of ∆DEF = area of ∆ABC
(iii) area of || gm FDCE = area of ∆ABC
If E, F, G and H are mid-points of the sides AB, BC, CD and DA, respectively of a parallelogram ABCD, prove that area of the quad. EFGH = area of || gm ABCD.