Question

solve for 100 points​

Answer 1

Answer:

$\bf\huge\underline\green{hello,nmste}$Given : A rectangle ABCD with AC and BD are its diagonals.

Prove that : AC = BD

Statements Reasons

1) ABCD is a rectangle. 1) Given

2) AD = BC 2) Property of rectangle (opposite sides are equal)

3) AB = AB 3) Reflexive (common side)

4) ∠A = ∠B 4) Each right angle.(property of rectangle)

5) Δ DAB ≅ Δ CBA 5) SAS Postulate

6) AC = BD 6) CPCTC

Example :

1) The diagonals of a rectangle ABCD meet at ‘O’. If ∠BOC = 44 0 , find ∠OAD.

Solution :

∠ BOC + ∠BOA = 180 [ Linear pair angles are supplementary]

⇒ 44 + ∠ BOA = 180

⇒ ∠BOA = 180 – 44

⇒ ∠ BOA = 136

As diagonals of a rectangle are equal and bisect each other.

So, OA = OB

⇒ ∠1 = ∠2

∠1 + ∠2 + ∠BOA = 180

2∠1 + 136 = 180

2∠1 = 180 -136

2∠1 = 44

∴ ∠1 = 22

0

As ∠A = 90 0

∠A = ∠1 + ∠3

90 = 22 + ∠3

So, ∠3 = 90 – 22

∠3 = 68

So, ∠OAD = 68 0

Quadrilateral

• Introduction to Quadrilateral

• Types of Quadrilateral

• Properties of Quadrilateral

• Parallelogram and its Theorems

• Rectangle and its Theorems

• Square and its Theorems

• Rhombus and its Theorems

• Trapezoid (Trapezium)and its Theorems

• Kite and its Theorems

• Mid Point Theorem

Answer 2

Step-by-step explanation:

To Proue:-ar( triangle CPD)=ar( Delta AQD) (Proue)/(Delta*cPDondp)

501: parallelogram ABCD are on the same base and b/w same parallel lines DC and A B therefore ar( Delta CPD)= 1 2 ar(AB CD) (

Similarly

ar(Delta*AQD) = 1/2 * a * r * (ABcD) * Q

N 0 omega , from xi q^ n odot and overline 2 we

dr( Delta CPD) 2 dr( Delta AQD).

I HOPE IT HELPS YOU.