Mathematics
In the adjoining figure, ABCD is a square. E and F are mid-points of sides BC and CD respectively. If R is mid-point of EF, prove that:
area of ∆AER = area of ∆AFR
Theorems on Area
3 Likes
Answer
In ∆ABE and ∆ADF,
AB = AD (Sides of a square)
∠B = ∠D (Each angle of a square = 90°)
BE = DF (E is mid-point of BC and F is mid-point of DC)
∴ ∆ABE ≅ ∆ADF (SAS axiom)
∴ AE = AF (c.p.c.t.)
Again in ∆AER and ∆AFR
AE = AF (Proved above)
AR = AR (Common)
ER = FR (R is mid-point of EF)
∴ ∆AER ≅ ∆AFR (SSS axiom)
∴ Area of ∆AER = Area of ∆AFR
Hence, proved that area of ∆AER = area of ∆AFR.
Answered By
1 Like
Related Questions
In the adjoining figure, ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area of ∆DFB = 3 cm2, find the area of parallelogram ABCD.
In the adjoining figure, X and Y are mid-points of the sides AC and AB respectively of ∆ABC. QP || BC and CYQ and BXP are straight lines. Prove that area of ∆ABP = area of ∆ACQ.
Perpendiculars are drawn from a point within an equilateral triangle to the three sides. Prove that the sum of the three perpendiculars is equal to the altitude of the triangle.
If each diagonal of a quadrilateral divides it into two triangles of equal areas, then prove that the quadrilateral is a parallelogram.