Telangana SCERT Class 9 Math Solution Chapter 8 Quadrilaterals Exercise 8.2

Telanagana SCERT Solution Class IX (9) Math Chapter 8 Quadrilaterals Exercise 8.2

(1) In the adjacent figure ABCD is a parallelogram and ABEF is a rectangle show that ΔAFD ΔBEC

Solution: Given,

ABCD is a parallelogram

ABEF is a rectangle

∴ In △AFD & △BEC

(i) ∠AFD = 90o [angle of rectangle ABEF]

∠BEF = 90o [angle of a rectangle ABEF]

∴ ∠BEF + ∠BEC = 180o

= 90o

∴ ∠AFD = ∠BEC = 90o

(ii) AF = BE [opposite side of rectangle ABEF]

(iii) AD = BC [opposite side of parallelogram ABCD]

Now,

∴ △AFD is a right angled triangle at ∠F = 90o

△BEC is aright angled triangle at ∠E = 90o with BC as hypotenuse

∴ △AFD ≅ △BEC by RHS

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

Solution: ABCD is a rhombus

AC and BD are its diagonals

In △AOD & △BOC

(i) ∠AOD = ∠BOC [vertically opposite angles]

(ii) AO = OC

(iii) DO = DB [∵ diagonals of a rhombus bisects each other]

∴ △AOD ≅ △BOC by SAS

Similarly,

△AOB ≅ △DOC by SAS

Now,

In △AOD and △DOC

(i) AD = DC [∵ all sides of a rhombus are equal]

(ii) ∠AOD = ∠DOC = 90o [∵ diagonals of a rhombus intersects each other at right angles]

(iii) DO is the common side

Hence,

△AOD & △DOC are right angled triangle with AD and DC being their hypotenuses respectively.

∴ △AOD ≅ △DOC by RHS

Similarly,

△BOC ≅ △AOB by RHS

∴ △AOD ≅ △AOB ≅ △BOC ≅ △DOC

∴ We can conclude that diagonals of a rhombus divided it into four congruent triangles

(3) In a quadrilateral ABCD, the bisector of C and D intersect at O

Solution: Given,

ABCD is a quadrilateral

DO and CO are bisects of ∠D and ∠C respectively

Now,

In △DOC

△COD = 180o – (∠DOC + ∠OCD)

= 180o – ( +(∠C)/2) [DO and CO are bisects of D and C]

= 180o – ½(∠C+ ∠D)

Or, COD = ½ 360o – (∠C+ ∠D)

Or, 2COD = 360o – (∠C+ ∠D)

Or, ∠C+ ∠D = 360o – 2∠COD – (i)

Now, We know

Sum of angles of a quadrilateral is 360o

∠A + ∠B + ∠C + ∠D = 360o

Putting the value of ∠C + ∠D from equation (i) we get

∠A + ∠B + 360o – 2∠COD = 360o

Or, ∠A + ∠B – 2∠COD = 0

Or, 2∠COD = ∠A + ∠B

Or, ∠COD = ½(∠A + ∠B) Hence proved


Updated: October 21, 2020 — 1:18 pm

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