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
