English

ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium. - Mathematics

Advertisements
Advertisements

Question

ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.

Short Note

Solution

If in cyclic quadrilateral `angle B = 70°` , then we have to find the other three angles. 

Since, AD is parallel to BC, So,
 `angleB + angleA = 180 ` (Alternate interior angles)

`70 + angleA = 180`

`⇒ angle A = 180 - 70 = 110°`

Now, since ABCD is cyclic quadrilateral, so

`angle A + angle C = 180`

`⇒ 110 + angle C = 180`

`⇒ angle C = 180 - 110 = 70°`

And,

`angle B + angleD = 180`\

`⇒ 70 + angleD = 180`

`⇒ angleD = 180 - 70 = 110°`

\[Hence, \angle A = 110°, \angle C = 70° \text{ and }  \angle D = 110°\] 

                                    

shaalaa.com
  Is there an error in this question or solution?
Chapter 15: Circles - Exercise 15.5 [Page 103]

APPEARS IN

RD Sharma Mathematics [English] Class 9
Chapter 15 Circles
Exercise 15.5 | Q 22 | Page 103

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.


ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.


Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.


Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.


In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°

 

 

In the given figure, ABCD is a cyclic quadrilateral. Find the value of x.

 


Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals. 


Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).


ABCD is a cyclic quadrilateral. M (arc ABC) = 230°. Find ∠ABC, ∠CDA, and ∠CBE.


If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×