Question

The bases of two triangles with equal heights are 6cm and 9cm respectively. Find the ratio of areas of these two triangles​

Answer 1

Answer :-

The ratio of their areas are 4:9.

Step-by-step Explanation

:: To Find :-

• The ratio of areas of triangle.

:: Solution :-

Given that,

• The bases of two triangles with equal heights are 6cm and 9cm respectively.

As we know that,

$\underline{ \boxed{\sf{Area \: of \: triangle = \dfrac{1}{2} \times b \times h}}}$

Where,

b = Base ; h = Height.

Figure: Refer the attachment.

ㅤㅤㅤㅤㅤ Assumption

Let us assume the base and height of one triangle as $\sf{{b}_{1}}$ and $\sf{{h}_{1}}$ and the other as $\sf{{b}_{2}}$ and $\sf{{h}_{2}}$.

According the question,

• $\underline{ \boxed{\sf{ \dfrac{ \dfrac{1}{2} \times {b}_{1} \times {h}_{1}}{\dfrac{1}{2} \times {b}_{2} \times {h}_{2}}}}}$

$\:$

On solving,

$\longrightarrow \: \sf{ \dfrac{ \dfrac{1}{2} \times {b}_{1} \times {h}_{1}}{\dfrac{1}{2} \times {b}_{2} \times {h}_{2}}}$

$\\ \longrightarrow \: \sf{ \dfrac{ \dfrac{1}{2} \times 4 \times {h}_{1}}{\dfrac{1}{2} \times 9 \times {h}_{2}}}$

⇝ Cancelling 1/2 from both numerator and denominator.

$\\ \longrightarrow \: \sf{ \dfrac{ \cancel{ \dfrac{1}{2}} \times 4 \times {h}_{1}}{ \cancel{\dfrac{1}{2}} \times 9 \times {h}_{2}}}$

⇝Canceling $\sf{{h}_{1}}$ and $\sf{{h}_{2}}$.

$\\ \longrightarrow \: \sf{ \dfrac{4 \times \cancel{{h}}_{1}}{9 \times \cancel{{h}_{2}}}}$

$\\ \longrightarrow \: \sf{ \dfrac{4}{9}}$

$\underline{\textsf{Hence, The ratio of their areas of triangle is} \: \textbf{4:9}}$