find the value of angle LQT and angle LQO, if angle OST is a rectangle
princess please reply me fast
Answers
Answer:
(i) Given : ABCD is rectangle
To prove : Each angle of rectangle =90
o
Proof:
In a rectangle opposite angles of a rectangle are equal
So, ∠A=∠C and ∠B=∠C
But, ∠A+∠B+∠C+∠D=360
o
[Sum of angles of a quadrilateral]
∠A+∠B+∠A+∠B=360
o
2(∠A+∠B)=360
o
(∠A+∠B)=360
o
/2
∠A+∠B=180
o
But, ∠A=∠B
So, ∠A=∠B=90
o
Thus,
∠A=∠B=∠C=∠D=90
o
Hence, each angle of a rectangle is 90
o
.
(ii) Given : In quadrilateral ABCD, we have
∠A=∠B=∠C=∠D
To prove: ABCD is a rectangle
Proof:
∠A=∠B=∠C=∠D
But these are opposite angle of the quadrilateral.
So, ABCD is parallelogram
And, as angleA=∠B=∠C=∠D
Therefore, ABCD is a rectangle.
(iii) Given : ABCD is a rhombus in which AC=BD
To prove : ABCD is a square
Proof:
Join, AC and BD.
Now, in ΔABC and ΔDCB we have
∠AB=DC [Sides of a rhombus]
∠BC=∠BC [Common]
∠AC=∠BD [Given]
So, ΔABC≅ΔDCB by S.S.S. axiom of of congruency
Thus,
∠ABC=∠DBC [By C.P.P.T.]
But these are made by transversal BC on the same side of parallel lines
AB and DC
So, angleABC+∠DBC=180
o
∠ABC=90
o
Hence, ABCD is a square.
(iv) Given : ABCD is rhombus
To prove : Diagonals AC and BD bisects ∠A,∠C,∠B and ∠D respectively
Proof:
In ΔAOD and ΔCOD, we have
AD=CD [sides of a rhombus are all equal]
OD=OD [Common]
AO=OC [Diagonal of rhombus bisect each other]
So, ΔAOD≅ΔCOD by S.S.S. axiom of congruency
Thus,
∠AOD=∠COD [By C.P.P.T.]
So, ∠AOD+∠COD=180
o
[Linear pair]
∠AOD=180
o
∠AOD=90
o
And, ∠COD=90
o
Thus,
OD⊥AC⟹BD⊥AC
Also, ∠ADO=∠CDO [By C.P.C.T.]
So,
OD bisect ∠DDB bisect ∠D
Similarly, we can prove that BD bisect ∠B and AC bisect the ∠A and ∠C.
l hope it will help u ☺️
Explanation:
no buddy l didn't think any thing l jst tld l don't want to fall in lv tk that's all l jst want to be as a bst frds
ab kya tum samajh sakate ho
Answer:
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