Question

The area of triangle is 90 cm². If it's base is 15 cm, find its altitude​

Answer 1

Step-by-step explanation:

Area of triangle=90cm²

Given,its base=15cm

1/2×base×height=area of triangle

1/2×15×h=90

h=90×1/15×2

h=12cm

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Answer 2

$\bf \pink{Given}\begin{cases}&\sf{Area\:of\:the\:triangle\:is\:\bf{90\:cm^2.}} \\ \\ &\sf{Base\:of\:the\:triangle\:is\:\bf{15\:cm.}}\end{cases}$

To FinD:-

The altitude of the triangle.

Solution:-

• Let the height or altitude be h.

Diagram:

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(-.2,2.5){\large\bf h\:cm}\put(2.8,.3){\large\bf 15\:cm}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\end{picture}$

We know that,

$\normalsize{\pink{\underline{\boxed{\bf{Area_{(triangle)}=\dfrac{1}{2}\times\:base\times\:height}}}}}$

where,

• Area = 90 cm²
• Base = 15 cm
• Height or altitude = h

Putting the values,

$\normalsize\implies{\sf{90=\dfrac{1}{2}\times15\times\:h}}$

$\normalsize\implies{\sf{\dfrac{90\times2}{15\times1}=h}}$

$\normalsize\implies{\sf{\dfrac{\cancel{90}\times2}{\cancel{15}\times1}=h}}$

$\normalsize\implies{\sf{\dfrac{6\times2}{1}=h}}$

$\normalsize\implies{\sf{\dfrac{12}{1}=h}}$

$\normalsize\implies{\sf{12=h}}$

$\normalsize\therefore\boxed{\mathfrak{\pink{Height=12\:cm.}}}$

Verification:-

$\normalsize\implies{\sf{Area_{(triangle)}=\dfrac{1}{2}\times\:base\times\:height}}$

where,

• Base = 15 cm
• Height = 12 cm
• Area = 90 cm²

Putting the values,

$\normalsize\implies{\sf{90=\dfrac{1}{2}\times15\times12}}$

$\normalsize\implies{\sf{90=\dfrac{1}{\cancel{2}}\times15\times\cancel{12}}}$

$\normalsize\implies{\sf{90=1\times15\times6}}$

$\normalsize\implies{\sf{90=90}}$

$\normalsize\therefore\boxed{\bf{LHS=RHS.}}$

• Hence verified.

The altitude of the triangle is 12 cm.