Question

in ∆BCD,

angle C = 90°

BC = 12m, CD = 5m and BD = 13m

if we simply find it's area by basic method '1/2 × b × h' then answer is 30m² but if we find it's area by heron's formula, then answer is different.. why so???

plz explain..

no spam answers..

Answer 1

the answer would be the same if u do it in any method.it may just come in different form.

if u find the height of the triangle with the given values it would 11.7

then the are would be 1/2*5*11.7=29.25

Answer 2

Nope , it will come the same thing .

Δ BCD has ∠C = 90° .

Area = 1/2 ×b × h

⇒ Area = 1/2 × 12 m × 5 m

⇒ Area = 6 m × 5 cm

⇒ Area = 30 m² .

By Heron's formula we will find the semi perimeter first :-

s = ( 12 m + 13 m + 5 m )/2

⇒ 30 m / 2

⇒ 15 m

$\bf{A=\sqrt{s(s-a)(s-b)(s-c)}}\\\\\implies A=\sqrt{15\times (15-12)(15-13)(15-5)}\\\\\implies A=\sqrt{15\times 3\times 2\times 10}\\\\\implies A=\sqrt{900}\\\\\implies A=30$

Hence the area will always come 30 m²