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Standard IX

Mathematics

The Mid-Point Theorem

P is the mid ...

Question

P is the mid -point of side BC of a parallelogram ABCD such that $∠ B A P = ∠ D A P$ . If AD = 10 cm, then CD =

5 cm

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6 cm

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8 cm

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10 cm

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Solution

The correct option is A
5 cm

In || gm ABCD, P is mid -point of BC.

AD = 10 cm

$∠ B A P = ∠ D A P Through P, draw P Q | | A B ∴ A B P Q i s r h o m b u s ∴ A B = B P = A Q = \frac{1}{2} A B = \frac{1}{2} \times 10 = 5 c m B u t C D = A B (Opposite sides of | | g m) ∴ C D = 5 c m$

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Similar questions

Q. If

P

is mid-point of side

B C

of a parallegram

A B C D

such that

∠ B A P = ∠ D A P

A D = 10

cm then length of

C D

is:

Q. Question 10
In figure, P is the mid-point of side BC of a parallelogram ABCD such that

∠ B A P = ∠ D A P

. Prove that AD = 2CD .

Thinking process

Firstly, use the property that sum of cointeritor angles is $180^{\circ}$ .Secondly use the property that sum of all angles in a triangle Is $180^{\circ}$ and then prove the required result.

The given figure shows a parallelogram ABCD such that AB = 15 cm, BC = 12 cm, and CN = 10 cm.

What is the length of AM?

Q. Answer the following as per the exact requirement

(i)

A B C D

is a parallelogram in which

A B ∥ C D

and

A B = C D = 10 c m

. If the perpendicular distance between

A B

and

C D

8 c m

, find the area of the parallelogram

A B C D

(i i)

A B C D

is a parallelogram having area

240 c m^{2}

B C = A D = 20 c m

and

B C ∥ A D

. Find the distance between the parallel sides

B C

and

A D

Construct a quadrilateral ABCD given that AB=8 cm, BC= 8 cm, CD=10 cm, AD=10 cm and $∠ A = 45^{\circ}$

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P is the mid -point of side BC of a parallelogram ABCD such that ∠BAP=∠DAP. If AD = 10 cm, then CD =

The correct option is A 5 cm In || gm ABCD, P is mid -point of BC. AD = 10 cm ∠BAP=∠DAPThrough P, draw PQ||AB∴ABPQ is rhombus∴AB=BP=AQ=12AB=12×10=5cmBut CD=AB (Opposite sides of ||gm)∴CD=5cm

P is the mid -point of side BC of a parallelogram ABCD such that $∠ B A P = ∠ D A P$ . If AD = 10 cm, then CD =

The correct option is A
5 cm

In || gm ABCD, P is mid -point of BC.

AD = 10 cm

$∠ B A P = ∠ D A P Through P, draw P Q | | A B ∴ A B P Q i s r h o m b u s ∴ A B = B P = A Q = \frac{1}{2} A B = \frac{1}{2} \times 10 = 5 c m B u t C D = A B (Opposite sides of | | g m) ∴ C D = 5 c m$