Question

The area of an acute-angled triangle is $336$ $m^2$, and the sides are $26$ m and $30$ m. Find its base.

• A. $35$ m
• B. $28$ m
• C. $20$ m
• D. $46$ m

Answer 1

The correct option is B $28$ m

Area of given $△ABC=336m^2$

Let base of $△ABC=bm$

Now, $S=\frac{26+30+b}{2}=\frac{56+b}{2}$

ar $△=\sqrt{s(s-a)(s-b)(s-c)}$

$336=\sqrt{\frac{56+b}{2}(\frac{56+b}{2}-2b)(\frac{56+b}{2}-30)(\frac{56+b}{2}-b)}$

$336=\sqrt{\left(\frac{56+b}{2}\right)\left(\frac{b+4}{2}\right)\left(\frac{b-4}{2}\right)\left(\frac{56-b}{2}\right)}=\sqrt{\frac{1}{16}({56}^2-b^2)(b^2-4^2)}$

${\left(336\right)}^2=\frac{1}{16}({56}^2-b^2)(b^2-4^2)$

$16\times 112896=(3136-b^2)(b^2-16)$

$3136b^2-b^4-50176+16b^2=1806336$

$\Rightarrow {-b}^4+3152b^2-50176-1806336=0$

$\Rightarrow {-b}^4+3152b^2-1856512=0\Rightarrow b^4-3152b^2+1856512=0$

After solving, we get $b=28$m

Hence base $=28$m