Find the area of the quadrilateral.

Step-by-step explanation:

the area of a triangle is baseline × height / 2.

we have 2 major triangles that define this quadrilateral :

A-D-B and C-D-B.

both triangles have the same baseline : M_D + M_B = 14 cm

what we need to do is get the 2 heights M_C and N_A.

N_A we know already : 8 cm

so,

area A-D-B = 14×8/2 = 14×4 = 56 cm²

M_C we get via Pythagoras of the right-angled triangle D-M-C :

D_C² = M_C² + M_D²

10² = M_C² + 6²

100 = M_C² + 36

64 = M_C²

M_C = 8 cm

so,

area C-D-B = 14×8/2 = 56 cm²

so, the total area of the quadrilateral is

56 + 56 = 112 cm²

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

the answer is d. 75%