Question

the ratio of height of the two triangles which are drawn by suja and i is 3:4 and the ratio of their area is 4:3 let is write by calculating what will be the ratio of two bases​

Answer 1

GIVEN :-

• Ratio of height of two triangles is 3:4.
• Ratio of two triangles is 4:3.

TO FIND :-

• Ratio of their bases.

TO KNOW :-

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

Here ,

• b → base of triangle
• h → height of triangle

SOLUTION :-

Let the height of first triangle be 3h and second triangle be 4h as ratio is 3:4.

Also , let the area of first triangle be 4A and second triangle be 3A as ratio is 4:3.

Let the base of first triangle be b{1} and that of second triangle be b{2}.

So,

$\sf \: area \: of \: 1st \: triangle = \dfrac{1}{2} \times b_{1} \times3h \\ \\ \\ \sf \: area \: of \: 2nd \: triangle = \dfrac{1}{2} \times b_{2} \times 4h$

We know ,

Ratio of their area is 4:3.

$\\ \implies \sf \: \dfrac{area(1)}{area(2)} = \dfrac{ \cancel \dfrac{1}{2} \times b_{1} \times 3 \cancel h }{ \cancel\dfrac{1}{2} \times b_{2} \times 4 \cancel{h}} \\ \\ \\ \implies \sf \: \dfrac{4}{3} = \dfrac{3b_{1}}{4b_{2}} \\ \\ \\ \implies \boxed{\sf \: \dfrac{b_{1}}{b_{2}} = \dfrac{16}{9} }$

Hence , ratio of their bases is 16:9.

MORE TO KNOW :-

★ Area of rectangle = l × b

★ Area of square = side²

★ Area of parallelogram = l × h

★ Area of circle = πr²

★ Area of semi-circle = πr²/2