Question

Find the altitude and area of an isoscles triangle whose perimeter is 32cm and whose base is 12cm

Answer 1

★Given:-

• For an isosceles triangle,
• Perimeter = 32cm
• Base = 12cm

★To find:-

• Altitude
• Area

★Solution:-

Let the congruent sides of the isosceles triangle =a

The base =b  

We know,

✦Perimeter of the triangle = Sum of all sides

Putting values,

→a + a + b =32

→2a+12=32

→2a = 20

→a = 20/2

→a=10cm

Therefore,length of the congruent sides of the triangle is 10cm.

The altitude of the triangle divides it into two right angled triangles,

For each equal triangle,

• Hypotenuse = 10 cm

Base = (12/2)

• Base = 6 cm

By pythagoras theorem,

Height = √hypotenuse² - base²

              = √(10)² - (6)²

              = √100 - 36

              = √64

             = 8 cm

Therefore,altitude = 8 cm.

Using the formula,

✦Area = 1/2(Base × Height)

Putting values,

→Area = (12 ×8)/2  cm²

→(96/2) cm²

→48 cm²

∴The area of the triangle is 48 cm².

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Answer 2

AnswEr-:

• $\boxed {\mathrm {Area\:of\:\triangle = 48cm^{2}}}$
• $\boxed {\mathrm {Altitude \:of\:\triangle = 8cm}}$

Explanation-:

• $\sf{Given-:}$

• Perimeter of an isosceles triangle is 32cm .
• Base of an isosceles triangle is 12 cm.

• $\sf{To\:Find-:}$

• The altitude and area of an isosceles triangle .

$\dag{\mathrm{Solution \:of\:Question \:-:\:}}$

$\sf{Let's\:Assume-:}$

• The same side of an isosceles triangle be x

As , We know that ,

• $\underline{\boxed{\mathrm {\dag{\red{ Perimeter _{(Isosceles\:Triangle)}-: Side + Side + Base }}}}}$

• Here ,

• Side = x cm

• Base = 12 cm

• Perimeter = 32 cm

Now , By Putting Known Values-:

• $\longrightarrow {\mathrm {x + x + 12 = 32}}$

• $\longrightarrow {\mathrm {2x + 12= 32}}$

• $\longrightarrow {\mathrm {2x= 32-12}}$

• $\longrightarrow {\mathrm {2x = 20}}$

• $\longrightarrow {\mathrm {x = \dfrac{\cancel {20}}{\cancel {2}}}}$

• $\longrightarrow {\mathrm {x = 10}}$

Therefore,

• $\longrightarrow {\mathrm {The\:length \;of\:same\:side\:of\:triangle\:is\:= 10}}$

The altitude of triangle devices the isosceles Triangle in two equal parts -:

• Then ,

• For Each Equal Triangle-:

• Hypotenuse = 10 [ It will remain same in both triangles ]

• Base = 12 /2 = 6 cm [ The base will also devided in two Equal parts . ]

As , We know that -:

• $\underline{\boxed{\mathrm {\dag{\red{ By\:Pythagoras\:Theorem\:-: Height = \sqrt{(Hypotenuse)^{2}- (Base)^{2}}}}}}}$

• Here ,

• Hypotenuse = 10 cm
• Base = 6cm

Now , By Putting known Values-:

• $\longrightarrow {\mathrm {Height= \sqrt{(10)^{2}-(6)^{2}}}}$

• ☆ 10² = 100 , 6² = 36 .

• $\longrightarrow {\mathrm {Height= \sqrt{100-36}}}$

• $\longrightarrow {\mathrm {Height= \sqrt{64}}}$

• ☆$\sf{\sqrt{64}= 8}$

• $\longrightarrow {\mathrm {Height= 8cm}}$

Therefore,

• Height of Triangle is 8 cm .

As we know that ,

• $\underline{\boxed{\mathrm {\dag{\red{ Area _{(Triangle)}-: \dfrac{1}{2} \times Base \times Height }}}}}$

• Here ,

• Base of Triangle is = 12 cm
• Height of Triangle = 8cm

Now, By Putting known Values-:

• $\longrightarrow {\mathrm {Area\:of\:\triangle = \dfrac{1}{2} \times 12 \times 8}}$

• $\longrightarrow {\mathrm {Area\:of\:\triangle = \dfrac{1}{\cancel{2}} \times \cancel{12} \times 8}}$

• $\longrightarrow {\mathrm {Area\:of\:\triangle = 6 \times 8}}$

• $\longrightarrow {\mathrm {Area\:of\:\triangle = 48cm^{2}}}$

Therefore ,

• $\boxed {\mathrm {Area\:of\:\triangle = 48cm^{2}}}$

___________________________________________

Hence ,

• $\boxed {\mathrm {Area\:of\:\triangle = 48cm^{2}}}$

• $\boxed {\mathrm {Altitude \:of\:\triangle = 8cm}}$

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