Question

in a triangle, AB=15 cm,BC =13 cm,AC= 14 cm. find the area of triangle ABC and hence its altitude on AC​

Answer 1

Answer:

HeyaHeya ‼

In ∆ABC, AB = 15 cm , BC = 13 cm and AC = 14 cm.

↪ By Heron's formula ,

S = ( A + B + C ) / 2S=(A+B+C)/2

S = ( 15 + 13 + 14 ) / 2S=(15+13+14)/2

S = 42 / 2S=42/2

S = 21S=21

↪ Now, area\:of\:triangle:-Now,areaoftriangle:−

= √S (S–A) (S–B) (S–C)

= √21 (21–15) (21–13) (21–14)

= √21 (6) (8) (7)

= √7056

= 84 cm^2=84cm

2

↪ Again,area\:of\:triangle,Again,areaoftriangle,

1/2×base×altitude=84 cm^21/2×base×altitude=84cm

2

1/2×14×altitude=841/2×14×altitude=84

7×altitude=847×altitude=84

altitude=84/7altitude=84/7

Altitude=12cmAltitude=12cm

Therefore, the area of ∆ABC is 84cm² and its altitude on AC is 12 cm.

Answer 2

AnswEr:

$\bf Given = \begin{cases} & \text{AB = \bf{15 cm}} \\ & \text{BC = \bf{13 cm} } \\ & \text{AC = \bf{14 cm}}\end{cases}$

$\star\; {\underline{\bf{\blue{As\;we\;know\;that\;:}}}}\\\\ \sf \bigstar\;Area\;of\;\triangle\;can\;be\;find\;by\;using\; following\;formulas\;:\\\\ \maltese\;{\boxed{\sf{Heron's\;formula\;:\;\sqrt{s(s - a)(s - b)(s - c)}}}}\\\\ \maltese\;{\boxed{\sf{Area\;:\; \dfrac{1}{2} \times base \times Altitude}}}$

$\rule{150}{2}$

$\underline{\bigstar\:\boldsymbol{As\:per\:given\: Question\::}}\\\\ \normalsize\;\;\bullet\;\sf s = \bf{semi - perimeter}\\\\ \normalsize\;\;\bullet\;\sf a = AB = \bf{15\;cm}\\\\ \normalsize\;\;\bullet\;\sf b = BC = \bf{13\;cm}\\\\ \normalsize\;\;\bullet\;\sf c = AC = \bf{14\;cm}\\\\ \bf \underline{\bigstar\;Using\; Heron's\;Formula\;:}\\\\ \dashrightarrow\sf s = \dfrac{a + b + c}{2}\\\\ \dashrightarrow\sf s = \dfrac{15 + 13 + 14}{2}\\\\ \dashrightarrow\sf s = \cancel{ \dfrac{42}{2}}\\\\ \dashrightarrow\bf \red{s = 21\;cm}$

$\rule{150}{2}$

$\bf \underline{\bigstar\;Area\;of\;triangle\;:}\\\\ \dashrightarrow\sf \sqrt{s(s - a)(s - b)(s - c)}\\\\ \dashrightarrow\sf \sqrt{21(21 - 15)(21 - 13)(21 - 14)}\\\\ \dashrightarrow\sf \sqrt{ 21 \times 6 \times 8 \times 7}\\\\ \dashrightarrow\sf \sqrt{7056}\\\\ \dashrightarrow{\underline{\boxed{\sf{\purple{ Area_{( \triangle ABC)} = 84\;cm^2}}}}}\;\bigstar$

$\bf \underline{\bigstar\;Again,\;Area\;of\;triangle\;:}\\\\ \dashrightarrow\sf 84 = \dfrac{1}{ \cancel{2}} \times \cancel{14} \times Altitude\\\\ \dashrightarrow\sf 84 = 7 \times Altitude\\\\ \dashrightarrow\sf Altitude = \cancel{ \dfrac{84}{7}}\\\\ \dashrightarrow{\underline{\boxed{\sf{\pink{Altitude (BD) = 12\;cm}}}}}\;\bigstar\\\\ \sf\therefore\; \underline{Hence,\;Area\;of\;\triangle\;ABC\;is\; 84\;cm^2 \;and\; Altitude\;on\;AC\;is\; 12\;cm.}$