Question

Q. Find the area of ∆ having sides 7cm, 9cm and 14cm using heron's formula.​

Answer 1

Answer:

Step-by-step explanation:

In triangle ABC

AB=7cm

BC=9cm

CA=14cm

Perimeter = AB+BC+CA

                =7+9+14

                = 30cm

semi perimeter=AB+BC+CA/2

                       =30/2

                       =15cm

Using Heron's formula,

Area of triangle ABC=√s(s-a)(s-b)(s-c)

                                 =√15(15-7)(15-9)(15-14)

                                 =√15(8)(6)(1)

                                 =√15×48

                                 =√720

                                 =26.84cm²

hence,Area of triangle ABC = 26.84 cm²

i hope you understood

Answer 2

AnswEr:

⋆ DIAGRAM:

$\setlength{\unitlength}{1.3cm}\begin{picture}(6,8)\linethickness{0.3mm}\qbezier(1,.5)(2,1)(4,2)\qbezier(4,2)(2,3)(2,3)\qbezier(2,3)(2,3)(1,0.5)\put(.7, .3){ C }\put(4.05, 1.9){ B }\put(1.7, 2.95){ A }\put(3.2, 2.5){\sf{7 cm}}\put(0.5,1.7){\sf{9 cm}}\put(2.7, 1.05){\sf{14 cm}}\end{picture}$

${\underline{\underline{\bf{GivEn\;:}}}}\\\\ \;\;\;\bullet\;\;\sf AB = \bf{7\;cm}\\\\ \;\;\;\bullet\;\;\sf BC = \bf{14\;cm}\\\\ \;\;\;\bullet\;\;\sf AC = \bf{9\;cm}$

$\rule{150}2$

${\underline{\bf{\bigstar\; Using\;Heron's\;Formula\;:}}}\\\\ \sf The\;area\;of\;triangle\;a,\;b,\;c\;and\;s\;as\;semi - perimeter\;is\;given\;by\;:\\\\ \maltese\;{\boxed{\sf{A = \sqrt{s(s - a)(s - b)(s - c)}}}}\\\\ \sf where,\\\\ \sf s = \dfrac{a + b + c}{2}$

$\sf Therefore,\;area\;of\;\triangle\;ABC\;is\;given\;by\;:\\\\ \sf here,\\\\ \;\;\;\bullet\;\;\sf a = 7\;cm\\\\ \;\;\;\bullet\;\;\sf b = 9\;cm\\\\ \;\;\;\bullet\;\;\sf c = 14\;cm\\\\ \sf Now,\\\\ :\implies\sf s = \dfrac{a + b + c}{2}\\\\ :\implies\sf s = \dfrac{7 + 9 + 14}{2}\\\\ :\implies\sf s = \cancel{ \dfrac{30}{2}}\\\\ :\implies{\underline{\boxed{\sf{\pink{s = 15\;cm}}}}}\\\\ \sf Therefore,$

$:\implies\sf A = \sqrt{s(s - a)(s - b)(s - c)}\\\\ \;\;\dag\;\sf \underline{Put\;the\;given\;values\;:}\\\\ :\implies\sf A = \sqrt{15(15 - 7)(15 - 9)(15 - 14)}\\\\ :\implies\sf A = \sqrt{15 \times 8 \times 6 \times 1}\\\\:\implies\sf A = \sqrt{720}\\\\ :\implies{\underline{\boxed{\sf{\purple{A = 12 \sqrt{5}\;cm^2}}}}}\\\\ \therefore\;\sf \underline{Area\;of\;\triangle\;ABC\;is\; \bf{12 \sqrt{5}\;cm^2}}.$