Question

for two triangles if the ratio of their bases is 1:2 and the ratio of their corresponding altitudes is also 1:2 then find the ratio of their areas​

Answer 1

See bro

Area of triangle is 1/2×base×(altitude or height)

Let the ratio be

area1:area2

(1/2*b1*H1):(1/2*b2*H2)

1*1:2*2

so the ratio of their area is 1:4

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Answer 2

Given:-

• Ratio of bases of two triangles is 1:2 .
• Ratio of their corresponding heights is 1:2 .

To Find :-

• The ratio of their areas .

Formula Used:-

$\large\purple{\underline{\boxed{\pink{\tt{\dag Area_{\triangle}\:\:=\:\:\dfrac{1}{2}\times(base)\times(height)\:\: }}}}}$

Answer :-

Given that the ratio of base if two triangles is 1 : 2 .

So , let the given ratio be 1x : 2x .

Also given that the ratio of altitudes is 1:2 .

So , let this ratio be 1y : 2y .

Now , we know that area of triangle is , ½ × base × height .

Let the area of first ∆ be $\tt A_1$ and that of second be $\tt A_2$

❒ Area of first ∆ :-

$\tt:\implies A_1=\dfrac{1}{2}\times(base)\times(height)$

$\tt:\implies A_1 =\dfrac{1}{2}\times1x\times1y$

$\underline{\boxed{\red{\tt{\longmapsto A_2\:\:=\:\:\dfrac{xy}{2}}}}}$

❒ Area of second ∆ :-

$\tt:\implies A_2=\dfrac{1}{2}\times(base)\times(height)$

$\tt:\implies A_2 =\dfrac{1}{2}\times2x\times2y$

$\tt:\implies A_2=\dfrac{\cancel{4xy}}{\cancel{2}}$

$\underline{\boxed{\red{\tt{\longmapsto A_2\:\:=\:\:2xy}}}}$

______________________________________

Now let's find the ratio :-

The required ratio will be $\tt A_1:A_2$

$\tt:\implies \dfrac{A_1}{A_2}=\dfrac{\dfrac{xy}{2}}{2xy}$

$\tt:\implies\dfrac{A_1}{A_2}=\dfrac{\cancel{xy}}{2\times2\cancel{xy}}$

$\tt:\implies \dfrac{A_1}{A_2}=\dfrac{1}{4}$

$\underline{\boxed{\red{\tt{\longmapsto A_1\::\: A_2\:\:=\:\:1\::\:4\:\:}}}}$

$\red{\boxed{\blue{\bf{\dag Hence\:the\: required\:ratio\:is\:1:4.}}}}$