Advertisements
Advertisements
Question
P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. If AD = 10 cm, then CD =
Options
5 cm
6 cm
8 cm
10 cm
Advertisements
Solution
Parallelogram ABCD is given such that ∠BAP = ∠DAP
We haveAD = 10cm and `BC = 1/2 BC`
We need to find the measure of side CD.
Since BC || AD and AP as transversal
∠I = ∠DAP
But it is given that
∠BAP = ∠DAP
Therefore, we get:
∠I = ∠BAP
Also, sides opposite to equal angles are equal.
Then,
AB = BP …… (I)
Also, `BP = 1/2 BC`
Substituting`BP = 1/2 BC`in (I), We get:
`AB = 1/2 BC`
It is given that AD = 10cm, this means opposite side BC = 10cm,asABCD is a parallelogram. Therefore,
`AB = 1/2 (10cm)`
AB = 5cm
Or, CD = 5cm
Hence the correct choice is (a).
APPEARS IN
RELATED QUESTIONS
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 Fig. below, AB = AC and CP || BA and AP is the bisector of exterior ∠CAD of ΔABC.
Prove that (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram
In a parallelogram ABCD, if ∠A = (3x − 20)°, ∠B = (y + 15)°, ∠C = (x + 40)°, then find the values of xand y.
In a parallelogram ABCD, the bisector of ∠A also bisects BC at X. Find AB : AD.
PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following case, PQRS is a parallelogram?
∠P = 100°, ∠Q = 80°, ∠R = 95°
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
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°
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)
