A- B- C are any points on the circle with centre O- marc BC-140- and marc AB-100- Then marc ABC - -

By property of sum of measure of arcs, we must have m(arc ABC) = m(arc AB) + m(arc BC) nbsp; (0.5 marks) nbsp;m(arc ABC) = nbsp;100^∘ + 140^∘ nbsp; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VIII

Mathematics

Terms Related to Circles

A, B, C are a...

Question

A, B, C are any points on the circle with centre O. m(arc BC)=140 $^{\circ}$ and m(arc AB)=100 $^{\circ}$ . Then m(arc ABC) = .

Open in App

Solution

By property of sum of measure of arcs, we must have
m(arc ABC) = m(arc AB) + m(arc BC) (0.5 marks)
$⟹$ m(arc ABC) = 100 $^{\circ}$ + 140 $^{\circ}$
= 240 $^{\circ}$ (0.5 marks)

Suggest Corrections

Similar questions

Q. A, B, C are any points on the circle with centre O. m(arc BC)=120

^{\circ}

and m(arc AB)=130

^{\circ} .

Then, find the value of m(arc ABC) .

Q. A, B, C are any points on the circle with centre O. m(arc BC) = 120

^{\circ}

and m(arc AB) = 130

^{\circ} .

Then, find m(arc ABC), m (arc ACB), and m (arc BAC).

Q. A, B, C are any points on the circle with centre O. m(arc BC)=120

^{\circ}

and m(arc AB)=130

^{\circ} .

Then, find m(arc ABC), m (arc ACB), and m (arc BAC).
[3Marks ]

Q. Points

A, B, C

are on a circle,
such that

m (a r c A B) = m (arc

B C) = 120^{\circ}

. No point, except point B, is common to the arcs. Which is the type of

△ A B C

?
(A) Equilateral triangle
(B) Scalene triangle
(C) Right angled triangle
(D) Isosceles triangle

Given circle with centre

O

and

B C ∥ E D

m (a r c B C) = 94^{0}

m (a r c E D) = 86^{0}

∠ A D E = 8^{0}

The m

∠ E C B

is :

Join BYJU'S Learning Program

A, B, C are any points on the circle with centre O. m(arc BC)=140∘ and m(arc AB)=100∘. Then m(arc ABC) = .

By property of sum of measure of arcs, we must have m(arc ABC) = m(arc AB) + m(arc BC) (0.5 marks) ⟹ m(arc ABC) = 100∘ + 140∘ = 240∘ (0.5 marks)

A, B, C are any points on the circle with centre O. m(arc BC)=140 $^{\circ}$ and m(arc AB)=100 $^{\circ}$ . Then m(arc ABC) = .

By property of sum of measure of arcs, we must have
m(arc ABC) = m(arc AB) + m(arc BC) (0.5 marks)
$⟹$ m(arc ABC) = 100 $^{\circ}$ + 140 $^{\circ}$
= 240 $^{\circ}$ (0.5 marks)