5. Explain why a rectangle is a convex quadrilateral.
D
6. ABC is a right-angled triangle and O is the mid point of the side
opposite to the right angle. Explain why O is equidistant from A,
B and C. (The dotted lines are drawn additionally to help you).
B
C
Answers
Answer:
ABC is a right-angled triangle and O is the midpoint of the side opposite to the right angle. Explain why O is equidistant from A, B, and C. (The dotted lines are drawn additionally to help you).
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ANSWER
Between Δ AOD & Δ BOC we have AO=CO (given),BO=OD (by Construction).
∠ AOD =∠ BOC....Vertically opposite angle
∴ By SAS test Δ AOD & Δ BOC are congruent.
So AD=BC....(i)
similarly, between Δ AOB & Δ DOC we have AO=CO (given), BO=OD (by Construction)
∠ AOB =∠ DOC
∴ By SAS test Δ AOB & Δ DOC are congruent.
So AB=DC.....(ii)
Also ∠ ABC=90
o
....(iii)
∴ From (i) & (ii) & (iii) we conclude that ABCD is a rectangle.
So the diagonals AC & BD are equal and bisect each other at O.
∴ OA=OB=OC=OD.
i.e O is equidistant from A, B & C.
Step-by-step explanation:
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ANSWER
ABC is a right-angled triangle and O is the midpoint of the side opposite to the right angle. Explain why O is equidistant from A, B, and C. (The dotted lines are drawn additionally to help you).
8415
ANSWER
Between Δ AOD & Δ BOC we have AO=CO (given),BO=OD (by Construction).
etween Δ AOD & Δ BOC we have AO=CO (given),BO=OD (by Construction).∠ AOD =∠ BOC....Vertically opposite angle
etween Δ AOD & Δ BOC we have AO=CO (given),BO=OD (by Construction).∠ AOD =∠ BOC....Vertically opposite angle∴ By SAS test Δ AOD & Δ BOC are congruent.
etween Δ AOD & Δ BOC we have AO=CO (given),BO=OD (by Construction).∠ AOD =∠ BOC....Vertically opposite angle∴ By SAS test Δ AOD & Δ BOC are congruent.So AD=BC....(i)
similarly, between Δ AOB & Δ DOC we have AO=CO (given), BO=OD (by Construction)
AO=CO (given), BO=OD (by Construction)∠ AOB =∠ DOC
AO=CO (given), BO=OD (by Construction)∠ AOB =∠ DOC∴ By SAS test Δ AOB & Δ DOC are congruent.
AO=CO (given), BO=OD (by Construction)∠ AOB =∠ DOC∴ By SAS test Δ AOB & Δ DOC are congruent.So AB=DC.....(ii)
AO=CO (given), BO=OD (by Construction)∠ AOB =∠ DOC∴ By SAS test Δ AOB & Δ DOC are congruent.So AB=DC.....(ii)Also ∠ ABC=90
AO=CO (given), BO=OD (by Construction)∠ AOB =∠ DOC∴ By SAS test Δ AOB & Δ DOC are congruent.So AB=DC.....(ii)Also ∠ ABC=90 o
AO=CO (given), BO=OD (by Construction)∠ AOB =∠ DOC∴ By SAS test Δ AOB & Δ DOC are congruent.So AB=DC.....(ii)Also ∠ ABC=90 o ....(iii)
AO=CO (given), BO=OD (by Construction)∠ AOB =∠ DOC∴ By SAS test Δ AOB & Δ DOC are congruent.So AB=DC.....(ii)Also ∠ ABC=90 o ....(iii)∴ From (i) & (ii) & (iii) we conclude that ABCD is a rectangle.
AO=CO (given), BO=OD (by Construction)∠ AOB =∠ DOC∴ By SAS test Δ AOB & Δ DOC are congruent.So AB=DC.....(ii)Also ∠ ABC=90 o ....(iii)∴ From (i) & (ii) & (iii) we conclude that ABCD is a rectangle.So the diagonals AC & BD are equal and bisect each other at O.
AO=CO (given), BO=OD (by Construction)∠ AOB =∠ DOC∴ By SAS test Δ AOB & Δ DOC are congruent.So AB=DC.....(ii)Also ∠ ABC=90 o ....(iii)∴ From (i) & (ii) & (iii) we conclude that ABCD is a rectangle.So the diagonals AC & BD are equal and bisect each other at O.∴ OA=OB=OC=OD.
AO=CO (given), BO=OD (by Construction)∠ AOB =∠ DOC∴ By SAS test Δ AOB & Δ DOC are congruent.So AB=DC.....(ii)Also ∠ ABC=90 o ....(iii)∴ From (i) & (ii) & (iii) we conclude that ABCD is a rectangle.So the diagonals AC & BD are equal and bisect each other at O.∴ OA=OB=OC=OD. i.e O is equidistant from A, B & C.
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