Question

N △ABC, point M is the midpoint of AB

point D is the midpoint of CM and ABMD=3 cm2.
Find ACDB, AAMC, AABD, AADC, and AABC.

Answer 1

ACDB = 3 cm²

AAMC = 6 cm²

AABD = 6 cm²

AADC = 3 cm²

AABC = 12 cm²

Explanation:

We're given that ABMD = 3 cm², so ACDB is also 3 cm². The sum of those two areas is ABMC, which is also AAMC, 6 cm².

Median AD also divides ΔAMC in half, so AADC = 3 cm² and AADM = 3 cm². AABD is the sum of ABMD (3 cm²) and AADM (3 cm²) so is 6 cm².

AABC is the sum of the smaller areas, so is 12 cm².

Answer 2

• ACDB = 3 cm²
• AAMC = 6 cm²
• AABD = 6 cm²
• AADC = 3 cm²
• AABC = 12 cm²

Explanation:

A median divides the area of a triangle into two equal parts. The first median, CM, divides triangle ABC into two equal areas, AMC and BMC. The second median, BD, divides triangle BMC into two equal areas, BMD and CDB.

We're given that ABMD = 3 cm², so ACDB is also 3 cm². The sum of those two areas is ABMC, which is also AAMC, 6 cm².

Median AD also divides ΔAMC in half, so AADC = 3 cm² and AADM = 3 cm². AABD is the sum of ABMD (3 cm²) and AADM (3 cm²) so is 6 cm².

AABC is the sum of the smaller areas, so is 12 cm².