Question

The perimeter of two similar triangles ABC and PQR are 32 cm and 24 cm respectively. If PQ=12 cm , find AB​

Answer 1

Answer:

• AB = 16 cm .

Step-by-step explanation:

According to the Question

It is given that,

• ∆ABC is similar to ∆PQR
• Length of PQ = 12cm
• Perimeter of ∆ABC = 32 cm
• Perimeter of ∆PQR = 24 cm

As it is given that triangle ABC is similar to Triangle PQR .

by similarity of triangles we get

• AB/PQ = BC/QR = AC/PR

Also ,

→ Perimeter of ∆ABC = AB + BC + AC = 32

And,

→ Perimeter of ∆PQR = PQ + QR + PR = 24

Ratio of Perimeter of these traingle is equal sum of all ratio of length of triangle.

→ ∆ABC/∆PQR = AB+BC+AC/PQ+QR+PR

by putting the value we get

(as ratio of length of similar triangles are equal )

↠ 32/24 = AB/12

↠ 4/3 = AB/12

↠ AB = 4×12/3

↠ AB = 4×4

↠ AB = 16

• Hence , the length of AB will be 16 cm .

Answer 2

Given :

• ➙ ABC = PQR
• ➙ Area of ABC = 32 cm
• ➙ Area of PQR = 24 cm
• ➙ PQ = 12 cm

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To Find :

• ➙ AB = ?

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Solution :

⚘ As these two triangles are similar :

${:\longmapsto{\sf{\dfrac{Triangle \: 1}{Triangle \: 2}}}}$

$\qquad{:\longmapsto{\sf{\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR}}}}$

⚘ Area of Triangles :

• ABC = 32 cm
• PQR = 24 cm

⚘ Ratio of perimeter of these triangles :

$\qquad{:\longmapsto{\sf{ \dfrac{ABC}{PQR} = \dfrac{AB + BC + AC}{PQ + QR + PR}}}}$

⚘ Applying the Values :

$\qquad{:\longmapsto{\sf{ \dfrac{Area{\small_{(Triangle \: 1)}}}{Area {\small_{(Triangle \: 2)}}} = \dfrac{PQ}{AB}}}}$

$\qquad{:\longmapsto{\sf{ \dfrac{32}{24} = \dfrac{12}{AB}}}}$

$\qquad{:\longmapsto{\sf{ \dfrac{32}{24} = \dfrac{12}{AB}}}}$

$\qquad{:\longmapsto{\sf{ \cancel\dfrac{32}{24} = \dfrac{12}{AB}}}}$

$\qquad{:\longmapsto{\sf{ \dfrac{4}{3} = \dfrac{12}{AB}}}}$

$\qquad{:\longmapsto{\sf{ AB = 4 \times \dfrac{12}{3}}}}$

$\qquad{:\longmapsto{\sf{ AB = 4 \times \cancel\dfrac{12}{3}}}}$

$\qquad{:\longmapsto{\sf{ AB = 4 \times 4}}}$

${\qquad{\sf{ AB \: = {\blue{\underline{\orange{\sf{16 \: cm}}}}}}}}$

Therefore :

❝ ${\sf{ Length \: of \: AB \: is \: {\blue{\underline{\red{\sf{ 16 \: cm}}}}}}}$ . ❞

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