Question

4. In a A ABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of A ABC andhence its altitude on AC.
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Answer 1

Step-by-step explanation:

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

By Heron's formula ,

S = ( A + B + C ) / 2

S = ( 15 + 13 + 14 ) / 2

S = 42 / 2

S = 21

Now, area of triangle

= √S (S–A) (S–B) (S–C)

= √21 (21–15) (21–13) (21–14)

= √21 (6) (8) (7)

= √7056

= 84 cm2

Again,area of triangle = 1/2×base×altitude=84 cm2

1/2×14×altitude = 84

7 × altitude=84

altitude = 84/7

Altitude = 12cm

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

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Answer 2

Correct Question :

In a ∆ ABC , AB = 15 cm , BC = 13 and AC = 14 cm. Find the area of the ∆ ABC and the it's altitude on AC.

Given :

• Length of AB = 15 cm
• Length of BC = 13 cm
• Length of AC = 14 cm

To Find :

• The area of the ∆ ABC
• Height of the ∆ ABC on AC

Solution :

⠀⠀⠀⠀⠀⠀To find the area of the triangle :

According to the Question, the triangle has three different sides.

Hence, the ∆ ABC is an Scalene triangle.

Now, we know the formula for area of an Scalene triangle i.e,

$\underline{\boxed{\bf{A = \sqrt{s(s - a)(s - b)(s - c)}}}}$

Where :-

$\bf{Sides\:of\:the\: triangle}\begin{cases} \bf{a} \\ \bf{b} \\ \bf{c} \end{cases}$

$\bf{s = Semi-perimeter}$

Here , Semi-perimeter is :

$\underline{\boxed{\bf{s = \dfrac{a + b + c}{2}}}}$

Now , using the formula and substituting the values in it, we get :-

$\underline{\bf{A = \sqrt{s(s - a)(s - b)(s - c)}}}$

$:\implies \bf{A = \sqrt{\bigg(\dfrac{a + b + c}{2}\bigg)\bigg\{\bigg(\dfrac{a + b + c}{2}\bigg) - a\bigg\}\bigg\{\bigg(\dfrac{a + b + c}{2}\bigg) - b\bigg\}\bigg\{\bigg(\dfrac{a + b + c}{2}\bigg) - c\bigg\}}}$$\:\:\:\bigg[\because s = \dfrac{a + b + c}{2}\bigg]$

$:\implies \bf{A = \sqrt{\bigg(\dfrac{13 + 14 + 15}{2}\bigg)\bigg\{\bigg(\dfrac{13 + 14 + 15}{2}\bigg) - 13\bigg\}\bigg\{\bigg(\dfrac{13 + 14 + 15}{2}\bigg) - 13\bigg\}}$$\bf{\bigg\{\bigg(\dfrac{13 + 14 + 15}{2}\bigg) - 15\bigg\}}}$$(\because s = a = 13 ; b = 14 ; c = 15)$

$:\implies \bf{A = \sqrt{\bigg(\dfrac{42}{2}\bigg)\bigg\{\bigg(\dfrac{42}{2}\bigg) - 13\bigg\}\bigg\{\bigg(\dfrac{42}{2}\bigg) - 13\bigg\}\bigg\{\bigg(\dfrac{42}{2}\bigg) - 15\bigg\}}}$

$:\implies \bf{A = \sqrt{21(21 - 13)(21 - 14)(21 - 15)}}$

$:\implies \bf{A = \sqrt{21 \times 8 \times 7 \times 6}}$

$:\implies \bf{A = \sqrt{21 \times 336}}$

$:\implies \bf{A = \sqrt{7056}}$

$:\implies \bf{A = 84}$

$\therefore \bf{Area = 84\:cm^{2}}$

Hence, the Area of the Triangle is 84 cm²

Now ,

⠀⠀⠀⠀⠀⠀⠀The height on AC :

We know the formula for area of an right-triangle :

$\underline{\boxed{\bf{A = \dfrac{1}{2} \times base \times height}}}$

Now using the above formula and substituting the values in it, we get :

Let the height be h cm.

$:\implies \bf{A = \dfrac{1}{2} \times base \times h}$

$:\implies \bf{84 = \dfrac{1}{2} \times 14 \times h}$

$:\implies \bf{84 \times 2 = 14 \times h}$

$:\implies \bf{84 \times 2 = 14 \times h}$

$:\implies \bf{6 \times 2 = h}$

$:\implies \bf{12 = h}$

$\therefore \bf{h = 12\:cm}$

Hence, the height of the triangle ∆ ABC on the base 14 cm is 12 cm.