Question

Find the area of the triangle whose sides are $50$ m, $78$ m, $112$ m respectively and also find the perpendicular from the opposite angle on the side $112$ m.

• A. $1680{\text{m}}^2,30$ m
• B. $1686{\text{m}}^2,20$ m
• C. $1780{\text{m}}^2,25$ m
• D. none

Answer 1

The correct option is A $1680{\text{m}}^2,30$ m

Here, $ABC$ is a triangle,

$AB=50m,BC=112m$ and $AC=78m$

Part $I:$

$S=\frac{50+112+78}{2}=\frac{240}{2}=120$

By Heron's formula,

$\Rightarrow$ Area of $△ABC=\sqrt{S(S-a)(S-b)(S-c)}$

$=\sqrt{120(120-50)(120-112)(120-78)}$

$=\sqrt{120\times 70\times 8\times 42}$

$=\sqrt{(2\times 2\times 2\times 3\times 5)\times (2\times 5\times 7)\times (2\times 2\times 2)\times (2\times 3\times 7)}$

$=\sqrt{2^8\times 5^2\times 3^2\times 7^2}$

$=2^4\times 5\times 3\times 7$

$=1680m^2$

Part $II:$

Area of $△ABC=\frac{1}{2}\times BC\times h$

$\Rightarrow$ $1680=\frac{1}{2}\times 112\times h$

$\Rightarrow$ $1680=56\times h$

$\Rightarrow$ $h=\frac{1680}{56}$

$\therefore$ $h=30m$