Question

Find the missing lengths and angle measures in kite ABCD

Answer 1

Answer:

Part 1) $AC=40\ units$

Part 2) $DC=29\ units$

Part 3) $m\angle ABE=39^o$

Part 4) $m\angle BCE=51^o$

Step-by-step explanation:

we know that

A kite has two pairs of consecutive, congruent sides the diagonals are perpendicular and the non-vertex angles are congruent

Part 1) Find AC

we know that

BD is the axis of symmetry, bisects the diagonal AC

so

$AE=EC$

we have

$EC=20\ units$

$AC=AE+EC$ ----> by segment addition postulate

therefore

$AC=20+20=40\ units$

Part 2) Find CD

we know that

CDE is a right triangle (the diagonals are perpendicular)

so

Applying Pythagorean Theorem

$DC^2=EC^2+ED^2$

substitute the values

$DC^2=20^2+21^2$

$DC^2=841\\DC=29\ units$

Part 3) Find m∠ABE

we know that

In the right triangle ABE

$51^o+m\angle ABE=90^o$ ----> by complementary angles

$m\angle ABE=90^o-51^o=39^o$

Part 4) Find m∠BCE

we know that

$m\angle BCE=m\angle BAE$ ----> diagonal BD is the axis of symmetry

we have

$m\angle BAE=51^o$

therefore

$m\angle BCE=51^o$