AP State Syllabus AP Board 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 8th Lesson Quadrilaterals Exercise 8.2

Question 1.

In the given figure ABCD is a parallelogram. ABEF is a rectangle. Show that ΔAFD ≅ ΔBEC

Solution:

Given that □ABCD is a parallelogram.

□ABEF is a rectangle.

In ΔAFD and ΔBEC

AF = BE ( ∵ opp. sides of rectangle □ABEF)

AD = BC (∵ opp. sides of //gm □ABCD)

DF = CE (∵ AB = DC = DE + EC , AB = EF = DE + DF)

∴ ΔAFD ≅ ΔBEC (SSS congruence)

Question 2.

Show that the diagonals of a rhombus divide it into four congruent triangles.

Solution:

□ABCD is a rhombus.

Let AC and BD meet at O’.

In ΔAOB and ΔCOD

∠OAB = ∠OCD (alt.int. angles)

AB = CD (def. of rhombus)

∠OBA = ∠ODC ………………….(1) (alt. int. angles)

∴ ΔAOB ≅ ΔCOD (ASA congruence)

Thus AO = OC (CPCT)

Also ΔAOD ≅ ΔCOD …………..(2)

[ ∵ AO = OC; AD = CD; OD = OD SSS congruence]

Similarly we can prove

ΔAOD ≅ ΔCOB ……………. (3)

From (1), (2) and (3) we have

ΔAOB ≅ ΔBOC ≅ ΔCOD ≅ ΔAOD

∴ Diagonals of a rhombus divide it into four congruent triangles.

Question 3.

In a quadrilateral ABCD, the bisector of ∠C and ∠D intersect at O. Prove that ∠COD = \(\frac{1}{2}\) (∠A + ∠B) .

(OR)

In a quadrilateral ABCD, the bisectors of ∠A and ∠B are intersects at ‘O’ then prove that ∠AOB = \(\frac{1}{2}\) (∠C + ∠D)

Solution:

In a quadrilateral □ABCD

∠A + ∠B + ∠C + ∠D = 360°

(angle sum property)

∠C + ∠D = 360° – (∠A + ∠B)

\(\frac{1}{2}\) (∠C + ∠D) = 180 – \(\frac{1}{2}\) (∠A + ∠B) ………….. (1)

(∵ dividing both sides by 2) .

But in ΔCOD

\(\frac{1}{2}\)∠C + \(\frac{1}{2}\)∠D + ∠COD = 180°

\(\frac{1}{2}\)∠C + \(\frac{1}{2}\)∠D = 180° – ∠COD

∴\(\frac{1}{2}\)(∠C +∠D) = 180° -∠COD………….(2)

From (1) and (2);

180° – ∠COD = 180° – \(\frac{1}{2}\) (∠A + ∠B)

∴ ∠COD = \(\frac{1}{2}\) (∠A + ∠B)

Hence proved.