Question

can anyone tell me the answer to this questions​

Answer 1

given: In rectangle ABCD, diagonals AC and

BD intersect at O such that angle

BOC = 70°

to find: angle ODA = ?

solution: diagonals of a rectangle are equal to and bisect each other

Hence, in ∆BOC

OB = OC

angle OBC = angle OCB

... (angles opposite to equal sides of a triangle are equal)

In ∆BOC,

angle BOC + angle OBC + angle OCB = 180°

70° + 2 angle OBC = 180°

2 angle OBC = 110°

angle OBC = 55°

angle ODA = 55°

... (alternate interior angles)

Question 2)

In ∆ABC

AC = AB = BC ... (i)

(sides of equilateral triangle)

In ∆ADC

AC = AD = CD ... (ii)

(sides of equilateral triangle)

Hence, from (i) and (ii)

AB = BC = CD = AD ... (iii)

Now, angle BAC = angle ACD = 60°

(angles of equilateral triangle)

Hence, AB || CD ... (iv)

(alternate interior angles are equal)

similarly, angle BCA = angle CAD = 60°

Hence, AD || BC ... (v)

Hence, from (iii), (iv) and (v)

[] ABCD is a rhombus

... (by definition)

angle ABC = angle ADC = 60°

(angles of equilateral triangle)

angle DAB = angle BAC + angle DAC

= 60 ° + 60° (angles of equi. ∆)

= 120°

angle BCD = angle BCA + angle DCA

= 60° + 60° (angles of equi. ∆)

= 120°

Hence, angles of the quadrilateral are 60°, 60°, 120° and 120°