Question

Base of triangle is 10 and height is 7. Base of another triangle is 9.and height is 6. Find ratio of area of these triangles.

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Answer 1

Given:

• We have been given two triangles with their dimensions
• Base and Height of first triangle is 10 and 7 unit , respectively
• Base and Height of second triangle is 9 and 6 unit , respectively

To Find:

• We have to find the ratio of area of two given triangles

Requirement:

We need to know that the area of Triangle is given by following formula

$\large{\boxed{\sf{\green{Area = \dfrac{1}{2} \times B \times H}}}}$

Where the variable stands for

• B = Length of Base of Triangle
• H = Height of Triangle

Solution:

Firstly we will find the area of first Triangle

Let the area of first Triangle be = $\sf{A_{1}}$

$\longrightarrow \sf{A_{1} = \dfrac{1}{2} \times B \times H}$

$\longrightarrow \sf{A_{1} = \dfrac{1}{2} \times 10 \times 7}$

$\longrightarrow \sf{A_{1} = \dfrac{70}{2}}$

$\longrightarrow \underline{\boxed{\sf{A_{1} = 35 \: unit^{2}}}}$

Similarly , we will find the area of second Triangle

Let the area of first Triangle be = $\sf{A_{2}}$

$\longrightarrow \sf{A_{2} = \dfrac{1}{2} \times B \times H}$

$\longrightarrow \sf{A_{2} = \dfrac{1}{2} \times 9 \times 6}$

$\longrightarrow \sf{A_{2} = \dfrac{54}{2}}$

$\longrightarrow \underline{\boxed{\sf{A_{2} = 27 \: unit^{2}}}}$

Now , Ratio of areas of triangle can easily be determined by dividing the area of both triangles.

$\hookrightarrow \sf{Ratio = \dfrac{Area \: of \: first \: triangle }{Area \: of \: second \: triangle}}$

$\hookrightarrow \sf{Ratio = \dfrac{A_{1} }{A_2}}$

$\hookrightarrow \underline{\boxed{\sf{Ratio = \dfrac{35 }{27}}}}$

The ratio of area of two triangle is 35:27

$[/tex]</p><p></p><p>[tex]\large{\underline{\boxed{\sf{\blue{Ratio \: of \: Area = 35 : 27} }}} }$

Answer 2

Given : Base of Triangle is 10 cm and height is 7 cm and Base of another Triangle is 9 cm and Height is 6 cm.

To Find : The ratio of area of these Triangles ?

__________________

❍ Let's consider Base of one Triangle as b¹, Height of one Triangle as h² & Base of another Triangle as b² & Height of another Triangle as h².

$~$

$\underline{\frak{As ~we ~know ~that~:}}$

• $\boxed{\sf\pink{Area_{(Triangle)}~=~\dfrac{1}{2}~×~b~×~h}}$★

$~$

Here b is the Base of Triangle in cm & h is the Height of Triangle in cm.

$~$

Then,

• Their ratios of their Area are :

• $\boxed{\sf\pink{ \dfrac{ Area_{(Triangle)}1 }{ Area_{(Triangle)}2 } = \dfrac{ \dfrac{1}{2} \times b_{1} \times h_{1} }{ \dfrac{1}{2} \times b_{2} \times h_{2} }}}$★

$~$

$\underline{\bf{Now~ By ~Substituting ~the ~Given ~Values~:}}$

$~$

$~~~~~~~~~:\implies{\sf{\dfrac{Area~of~\triangle~1}{Area~of~\triangle~2}~=~\dfrac{ \dfrac{1}{2} \times 10 \times 7}{ \dfrac{1}{2} \times 9 \times 6 } }}$

$~~~~~~~~~:\implies{\sf{\dfrac{Area~of~\triangle~1}{Area~of~\triangle~2}~=~\dfrac{\dfrac{1}{\cancel{2}}~×~\cancel{10}~×~7}{\dfrac{1}{\cancel{2}}~×~9~×~\cancel{6}}}}$

$~~~~~~~~~:\implies{\sf{\dfrac{Area~of~\triangle~1}{Area~of~\triangle~2}~=~\dfrac{5~×~7}{9~×~3}}}$

$~~~~~~~~~:\implies{\underline{\boxed{\frak{\pink{\dfrac{Area~of~\triangle~1}{Area~of~\triangle~2}~=~\dfrac{35}{27}}}}}}$★

$~$

Hence,

$\therefore\underline{\sf{The~Ratio~of~Area~of~Both~Triangles~are~\bf{35~:~27}}}$