plz solve these three que 28,29,30
if not so plz only one ques plz solve it
Answers
Answer:
Step-by-step explanation:
28) Let us consider, a line segment AB.
Assume that it has two midpoints say C and D
Recall that the midpoint of a line segment divides it into two equal parts
That is AC = BC and AD = DB
Since C is midpoint of AB, we have A, C and B are collinear
∴ AC + BC = AB → (1)
Similarly, we get AD + DB = AB → (2)
From (1) and (2), we get
AC + BC = AD + DB
2 AC = 2AD
∴ AC = AD
This is a contradiction unless C and D coincide.
Therefore our assumption that a line segment AB has two midpoints is incorrect.
Thus every line segment has one and only one midpoint.
29) ⇒ Given :- ABCD is a rhombus
AC and BD are diagonals of rhombus intersecting at O.
⇒ To prove :- ∠BOC = ∠DOC = ∠AOD = ∠AOB = 90°
⇒ Proof :- All Rhombus are parallelogram, Since all of its sides are equal.
AB = BC = CD = DA ────(1)
The diagonal of a parallelogram bisect each other
Therefore, OB = OD and OA = OC ────(2)
In ∆ BOC and ∆ DOC
BO = OD [ From 2 ]
BC = DC [ From 1 ]
OC = OC [ Common side ]
∆ BOC ≅ ∆ DOC [ By SS congruency criteria ]
∠BOC = ∠DOC [ C.P.C.T ]
∠BOC + ∠DOC = 180° [ Linear pair ]
2∠BOC = 180° [ ∠BOC = ∠DOC ]
∠BOC = 180°/2
∠BOC = 90°
∠BOC = ∠DOC = 90°
Similarly, ∠AOB = ∠AOD = 90°
Hence, ∠BOC = ∠DOC = ∠AOD = ∠AOB = 90°
we proved
30) given : ABCD is a parallelogram
diagonals AC and BD of parallelogram ABCD are equal in length .
proof : Consider triangles ABD and ACD.
AC = BD [Given]
AB = DC [opposite sides of a parallelogram]
AD = AD [Common side]
∴ ΔABD ≅ ΔDCA [SSS congruence criterion]
∠BAD = ∠CDA [CPCT]
∠BAD + ∠CDA = 180° [Adjacent angles of a parallelogram are supplementary.]
So, ∠BAD and ∠CDA are right angles as they are congruent and supplementary.
Therefore, parallelogram ABCD is a rectangle since a parallelogram with one right interior angle is a rectangle.