What is the area of quadrilateral ABCD, if AB = 5 cm, BC = 13 cm, CD = 9 cm, DA = 15 cm, AC = 12 cm.

There are a few "Pythagorean triples" that show up on algebra problems regularly. One is (5, 12, 13). Another is (3, 4, 5). It can be useful to remember these.

Here, diagonal AC divides the figure into two right triangles.
ΔBAC has side lengths 5, 12, and 13.
ΔDCA has side lengths 9, 12, and 15, which are multiples of 3, 4, and 5.

The area of the quadrilateral will be the sum of the areas of the triangles.
A = (1/2)*AB*AC +(1/2)*CD*AC
A = (1/2)*AC*(AB +CD) . . . . . . . . . removing common factors
A = (1/2)*(12 cm)*(5 cm +9 cm)
A = 84 cm²

The area of the quadrilateral is 84 cm².

Blacklash Blacklash · Answer 2 · 2019-07-03T20:08:58+02:00

Answer:

Area of quadrilateral ABCD = 84 cm²

Step-by-step explanation:

From figure given

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

Area of triangle ABC

a = 5 cm, b= 13 cm, c = 12 cm

[tex]s=\frac{a+b+c}{2}=\frac{5+13+12}{2}=15[/tex]

[tex]\texttt{Area of ABC}=\sqrt{15\times (15-5)\times (15-13)\times (15-12)}=30cm^2[/tex]

Area of triangle ACD

a = 9 cm, b= 15 cm, c = 12 cm

[tex]s=\frac{a+b+c}{2}=\frac{9+15+12}{2}=18[/tex]

[tex]\texttt{Area of ACD}=\sqrt{18\times (18-9)\times (18-15)\times (18-12)}=54cm^2[/tex]

Area of quadrilateral ABCD = 30 + 54 = 84 cm²