Question

5. Find the area of a pentagon ABCDE in which each one of BF, CH, and EG is perpendicular to
AD such that AF = 9 cm, AG = 13 cm, AH = 19 cm, AD = 24 cm, BF = 6 cm, CH = 8 cm, and EG - 9 cm. ( Hint : answer is 225cm square )​

Answer 1

Answer:

$225 {cm}^{2}$

Step-by-step explanation:

explanation is in the image

Answer 2

Given:

AF=9cm, AG=13 cm, AH=19 cm, AD=24 cm, BF=6 cm, CH=8 cm, EG=9 cm

To find:

The area of the pentagon ABCDE

Solution:

The area of the pentagon ABCDE is 225 $cm^{2}$.

We will obtain the required area by adding the area of the triangles ABF, AGE, EGD, CHD, and trapezium BFHC.

Since the lines BF, CH, and EG are at a perpendicular distance from AD, the triangles ABF, AGE, EGD, CHD are right-angled.

So, the area of triangles= 1/2×base×height

Also, from the given information, GD=GH+HD=11cm, FH=FG+GH=10cm.

Area of ΔABF=1/2×AF×BF

=1/2×9×6

=27 $cm^{2}$

Area of ΔAGE=1/2×AG×EG

=1/2×13×9

=58.5 $cm^{2}$

Area of ΔEGD=1/2×GD×EG

=1/2×11×9

=49.5 $cm^{2}$

Area of ΔCHD=1/2×HD×CH

=1/2×5×8

=20 $cm^{2}$

Area of trapezium BFHC=1/2×FH×(BF+CH)

=1/2×10×(6+8)

=5×14

=70 $cm^{2}$

The total area of ABCDE=27+58.5+49.5+20+70

=225 $cm^{2}$

Thus, the required area of the pentagon ABCDE is 225 $cm^{2}$.