Question

the base of an isoceles triangle is 16cm.if the height on the base is 4cm shorter than the equal sides,find the sides,height and area of the triangle.​

Answer 1

Given :-

The base of an Isosceles triangle is 16 cm

The height on the base is 4 cm shorter than the equal sides.

To find:-

♦ The height of the triangle.

♦ The sides of the triangle.

♦ The area of the triangle.

Solution :-

Given that

The base of the Isosceles triangle (b) = 16 cm

Let the equal sides of the Isosceles triangle be

X cm each

The height on the base of the Isosceles triangle (h) = Equal Side - 4 cm

=> h = (X-4) cm

Let Consider an Isosceles triangle for the given data

We have,

AB = AC = X cm

BC = 16 cm

AD = (X-4) cm

We know that

The altitude of an Isosceles triangle bisects the angle of the vertex and bisects the base.

BD =DC = 16/2 = 8 cm

We have,

∆ ABD is a right angled triangle.

By Pythagoras Theorem

=> AB² = BD²+AD²

=> X² = (8)²+(X-4)²

=> X² = 64+X²-2(X)(4)+(4)²

Since, (a-b)² = a²-2ab+b²

Where, a = X and b = 4

=> X² = 64+X²-8X+16

=> X² = 80+X²-8X

=> X²-X²+8X = 80

=> 0+8X = 80

=> 8X = 80

=> X = 80/8

=> X = 10 cm

Therefore, X = 10 cm

The lengths of the equal sides = 10 cm each

The height of the triangle = X-4 = 10-4 = 6 cm

We know that

Area of a triangle = (1/2)bh sq.units

=> Area = (1/2)×16×6 cm²

=> Area = 8×6

=> Area = 48 cm²

(Or)

Area of an Isosceles triangle = (b/4)√(4a²-b²) sq.units

We have,

a = 10 cm , b = 16 cm

Area = (16/4)√[4(10)²-(16)²]

=> Area = 4√[4(100)-256]

=> Area = 4√(400-256)

=> Area = 4√144

=> Area = 4×12

Area = 48 cm²

Answer :-

♦ The sides of the Isosceles triangle are 10 cm, 10 cm and 16 cm

♦ The height of the Isosceles triangle is 6 cm

♦ The area of the Isosceles triangle is 48 cm²

Used formulae:-

♦ Area of a triangle = (1/2)bh sq.units

♦ Area of an Isosceles triangle = (b/4)√(4a²-b²)

Used Concept :-

♦ The altitude of an Isosceles triangle bisects the angle of the vertex and bisects the base.

Used Theorem :-

Pythagoras Theorem :-

" In a right angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides".

Answer 2

Given:-

• The base of an isosceles triangle = 16cm.
• Height = 4 cm shorter than the equal sides.
• Height = ?

To find:-

• Legth of sides.
• Length of height.
• Area of the triangle.

Solution:-

• Let the equal sides be x cm.
• Then, The height of the isosceles triangle = (x-4)cm.

We have:-

• AB=AC=x cm
• BC=16cm
• AD=(x-4) cm

Now we have sides of the isosceles triangle as a variable x cm and the base as 16 cm. We have also found the height as (x-4) cm. Now we will use Pythagorean theorem to find the side and then we will subtract 4 from side to get height. Are that we will use the formula for area of isosceles triangle.

We know that the height bisects the angle of vertex and bisects the base.

So, we have

BD=BC

16/2

=>8cm

Now we have two triangles as follows

ABD=ACD

We also have ∆ABD as a right a right angled triangle.

Now we can use Pythagoras Theorem to find the hypotenuse.

$\large \blue{\underline{ \green{\boxed{\sf{ \red{{H}^{2}= {P}^{2} + {B}^{2} }}}}}}$

where,

• H=Hypotenuse=AB
• P=Perpendicular=AD
• B=Base=BD

Now we will substitute the values in the formula.

$\large\underline{\sf{ \implies{x}^{2} = {(x - 4)}^{2} + {8}^{2} }}$

Now we will use identity (a-b)²=a²-2ab+b²

Where,

• a=x
• b=4

$\large\underline{\sf{ \implies {x}^{2} = {x}^{2} - 2 \times x\times 4 + {4}^{2} + {8}^{2} }}$

$\large\underline{\sf{ \implies {x}^{2} = {x}^{2} - 8x + 16 + {8}^{2} }}$

$\large\underline{\sf{ \implies {x}^{2} - {x}^{2} = - 8x + 16 + 64}}$

$\large\underline{\sf{ \implies0 = - 8x + 16 + 64}}$

$\large\underline{\sf{ \implies8x = 16 + 64}}$

$\large\underline{\sf{ \implies8x = 80}}$

$\large\underline{\sf{ \implies x = \frac{80}{8} }}$

$\large \green{\underline{ \blue{ \boxed{{\sf{ \red{ \implies x = 10cm}}}}}}}$

Now we have the equal sides as 10 cm.

Now, given height is 4 less than equal sides.

So,

$\large\underline{\sf{ \implies (x - 4)cm}}$

$\large\underline{\sf{ \implies(10 - 4)cm}}$

$\large \green{\underline{ \blue{ \boxed{\sf{ \red{ \implies6cm}}}}}}$

Area of isosceles triangle is:-

$\large \blue{\underline{ \green{ \boxed{\sf{ \red{ \frac{1}{2} \times b \times h}}}}}}$

On substituting values we get:-

$\large\underline{\sf{ \implies \frac{1}{2} \times 16cm \times 6cm }}$

$\large\underline{\sf{ \implies8cm \times 6cm}}$

$\large \blue{\underline{ \green {\boxed{\sf{ \red{\implies 48 {cm}^{2} }}}}}}$

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Basic concept!!

• An isosceles triangle is a 2-D figure.
• It has 2 equal sides and a base.
• If it is divides into two equal parts the two triangles formed are right-angled.

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Deriviation of formula for triangle:-

• We know that the area of a parallelogram is b×h.
• If we divide the parallelogram in two parts from one corner to the opposite corner we get a triangle.
• Hence it is the half of parallelogram area.

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