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

## AP State Syllabus 9th Class Maths Solutions 8th Lesson Quadrilaterals InText Questions

**Try This**

Question

Extend AB to E. Find ∠CBE. What do you notice? What kind of angles are ∠ABC and ∠CBE?

Solution:

Given that □ABCD is a parallelogram and∠A = 40°

∴ ∠ABC = 180°-40° = 140°

∠CBE = 40° ( ∵ ∠A and, ∠CBE are corresponding angles) And ∠CBE and ∠ABC are linear pair of angles.

**Do This**

Question

Cut out a parallelogram from a sheet of paper again and cut along one of its diagonal. What kind of shapes you obtain ? What can you say about these triangles 7 [Page No. 179]

Solution:

We get two triangles.

The two triangles are congruent to each other.

**Think, Discuss and Write**

Question 1.

Show that the diagonals of a square are equal and right bisectors of each other. (Page No. 185)

Solution:

Let □ABCD be a square.

Thus AB = BC = CD = DA

In ΔABC and ΔBAD

AB = AB (common base)

∠B =∠A (each 90°)

BC = AD (equal sides)

∴ ΔABC = ΔBAD (SAS congruence)

⇒ AC = BD (CPCT)

Also in ΔAOB and ΔCOD

∠OAB = ∠OCD [∵ alt. int. angles]

∠OBA = ∠ODC [∵ alt. int. angles]

AB = DC (sides of a square)

∴ ΔAOB = ΔCOD (ASA congruence)

Thus AO = OC (CPCT) ⇒ O is midpoint of AC

Also BO = OD (CPCT) ⇒ O is midpoint of BD

∴ O is midpoint of both AC and BD.

∴ Diagonals bisect each other.

In ΔAOB and ΔCOB

AB = BC (given)

OB = OB (common)

AO = OC (proved)

∴ ΔAOB ≅ ΔCOB (SSS congruence)

∠AOB = ∠COB (CPCT)

But ∠AOB + ∠COB = 180° (∵ linear pair of angles)

∴ ∠AOB = ∠COB = 180°/2 = 90°

Also ∠AOB = ∠COD (∵ vertically opposite angles)

∠BOC = ∠AOD (∵ vertically opposite angles)

∴ AC ⊥ BD

(i.e.,) In a square diagonals bisect at right angles.

Question 2.

Show that the diagonals of a rhombus divide it into four congruent triangles. (Page No. 185)

Solution:

□ABCD is a rhombus

Let AC and BD meet at ‘O’

In ΔAOB and ΔCOD

∠OAB = ∠ODC (alt.int.angles)

AB = CD (def. of rhombus)

∠OBA = ∠ODC (alt. mt, angles)

∴ ΔAOB ≅ΔCOD …………(1)

(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.

**Try This**

Question

Draw a triangle ABC and mark the mid points E and F of

two sides of triangle. \(\overline{\mathbf{A B}}\) and \(\overline{\mathbf{A C}}\) respectively. Join the point E and F as shown in the figure. Measure EF and the third side BC of triangle. Also measure ∠AEF and ∠ABC.

We find ∠AEF = ∠ABC and \(\overline{\mathrm{EF}}=\frac{1}{2} \overline{\mathrm{BC}}\)

As these are corresponding angles made by the transversal AB with lines EF and BC, we say EF//BC.

Repeat this activity with some more triangles. (Page No. 188)

Solution:

P, Q are mid points of XY and XZ

PQ // YZ

PQ = \(\frac{1}{2}\)YZ

X, Y are mid points of PQ and PR

XY // QR

XY = \(\frac{1}{2}\)QR

A, B are midpoints of DE and DF

AB // EF

AB = \(\frac{1}{2}\)EF

( ∵ In all cases the pairs of respective corresponding angles are equal.)