Q1 of 24 Page 247

In ∆ABC, ∠ABC = 90°, AD = DC, AB = 12 cm and BC = 6.5 cm. Find the area of ∆ADB.

Given: ∠ABC = 90°


AD = DC


AB = 12 cm and BC = 6.5 cm


Area of ∆ABC = 1/2 × BC × AB


= 1/2 × 6.5 × 12 …(given)


= 6 × 6.5


= 39 sq. cm


Area of ∆ABC = 39 sq. cm …(i)


AD = DC which means BD is the median


Median divides area of triangle in two equal parts


Therefore area(∆ABD) = area(∆CDB) …(ii)


From figure area(∆ABC) = area(∆ABD) + area(∆CDB)


Using equation (i) and (ii) we can write


39 = area(∆ABD) + area(∆ABD)


39 = 2 area(∆ABD)


Therefore area(∆ABD) = 19.5 sq. cm


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