find the area.
.
.
.
.
.
need help

Answers
Answers:
Area of the quadrilateral = area of the triangle ABD + area of the triangle DBC.
From ΔDBC, BC 2 =DC 2 −BD 2
[Using Pythagoras theorem]
⟹BC 2 =13 2−12 2=25 , ⟹BC= 25=5 cm
Area of the ΔDBC= 21 ×b×h= 21 ×5×12=30 cm 2Area of the ΔABD= s(s−a)(s−b)(s−c)
Here, s= 210+12+8=15 cm ∴ Area of the ΔABD= 15(15−10)(15−12)(15−8)=15 7 cm 2
∴ area of the quadrilateral=15+15 7 =15(1+ 7) cm 2
Answer:
since BC=10cm, BD=5 cm
In triangle ADB,
By Pythagoras theorem,
AB²= AD²+BD²
SO,
AD²=AB²-BD²
= 13²-5²
= 169-25
AD² =144
therefore AD=12CM
Area of triangle ABC=1/2×BASE×HEIGHT
=1/2×10×12
=5×12
=60cm²
area of triangle ABC=60cm²
please mark as brainliest