Advertisements
Advertisements
प्रश्न
In the given Figure, if AB = 2, BC = 6, AE = 6, BF = 8, CE = 7, and CF = 7, compute the ratio of the area of quadrilateral ABDE to the area of ΔCDF. (Use congruent property of triangles)
Advertisements
उत्तर
Given: AB = 2 cm, BC = 6 cm, AE = 6 cm, BF = 8 cm, CE = 7 cm and CF = 7 cm
Consider ΔAEC and ΔBCF.
In ΔAEC, AE = 6 cm, EC = 7 cm and AC = 8 cm (2 + 6 = 8)
In ΔBCF, BC = 6 cm, CF = 7 cm and BF = 8 cm
∴ ΔAEC ≅ ΔBCF
∴ Area of ΔAEC = Area of ΔBCF ...(Two triangles are similar areas are equal)
Subtract area of ΔBDC on both sides we get,
Area of ΔAEC – Area of ΔBDC = Area of ΔBCF – Area of ΔBDC
Area of quadrilateral ABDE = Area of ΔCDF
APPEARS IN
संबंधित प्रश्न
In Fig., below, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠B meet at P, prove that AD = DP, PC = BC and DC = 2AD.

The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.
In a ΔABC median AD is produced to X such that AD = DX. Prove that ABXC is a
parallelogram.
In a parallelogram ABCD, if ∠A = (3x − 20)°, ∠B = (y + 15)°, ∠C = (x + 40)°, then find the values of xand y.
We get a rhombus by joining the mid-points of the sides of a
The figure formed by joining the mid-points of the adjacent sides of a rhombus is a
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
ABCD is a parallelogram in which diagonal AC bisects ∠BAD. If ∠BAC = 35°, then ∠ABC =
P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. If AD = 10 cm, then CD =
In the given figure, ∠A = 64°, ∠ABC = 58°. If BO and CO are the bisectors of ∠ABC and ∠ACB respectively of ΔABC, find x° and y°
