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To prove:
∠P +∠Q+ ∠ R+∠S+∠T = 2(right angle) = 180°.
Proof:
In ΔBTR,
∠T+∠R + ∠BTR =180° (angle sum property of Δ) ------- (i)
In ΔAQS
∠Q+∠S + ∠QAS =180°(angle sum property of Δ) ---(ii)
In ΔPER
∠P+∠R +∠PER = 180°(angle sum property of Δ) ---(iii)
In ΔTQD
∠T +∠Q +∠TQD =180°(angle sum property of Δ) ----(iv)
In ΔPSC
∠P +∠S +∠PSC = 180°(angle sum property of Δ) ----(v)
Adding all 5 equations,
2(∠P +∠Q+ ∠ R+∠S+∠T ) + ∠BTR+∠QAS+∠PER+∠TQD+∠PSC= 5(180°) = 900°
You might be familiar that the sum of all interior angles on Pentagon equals 180°.
(If not try searching a 7th or 8th standard maths text books)
2(∠P +∠Q+ ∠ R+∠S+∠T ) + 540° = 900°
2(∠P +∠Q+ ∠ R+∠S+∠T ) = 900°- 540° = 360°
∠P +∠Q+ ∠ R+∠S+∠T = 360°/2 = 180°.
Hence proved.
∠P +∠Q+ ∠ R+∠S+∠T = 2(right angle) = 180°.
Proof:
In ΔBTR,
∠T+∠R + ∠BTR =180° (angle sum property of Δ) ------- (i)
In ΔAQS
∠Q+∠S + ∠QAS =180°(angle sum property of Δ) ---(ii)
In ΔPER
∠P+∠R +∠PER = 180°(angle sum property of Δ) ---(iii)
In ΔTQD
∠T +∠Q +∠TQD =180°(angle sum property of Δ) ----(iv)
In ΔPSC
∠P +∠S +∠PSC = 180°(angle sum property of Δ) ----(v)
Adding all 5 equations,
2(∠P +∠Q+ ∠ R+∠S+∠T ) + ∠BTR+∠QAS+∠PER+∠TQD+∠PSC= 5(180°) = 900°
You might be familiar that the sum of all interior angles on Pentagon equals 180°.
(If not try searching a 7th or 8th standard maths text books)
2(∠P +∠Q+ ∠ R+∠S+∠T ) + 540° = 900°
2(∠P +∠Q+ ∠ R+∠S+∠T ) = 900°- 540° = 360°
∠P +∠Q+ ∠ R+∠S+∠T = 360°/2 = 180°.
Hence proved.
allysia:
Good job girl. ^_^
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