Question

each of equal sides of an isosceles traingle is 4cm greater than its height .if the base of the traingle is 24cm.calculate the perimeter and area of the triangle.​

Answer 1

GIVEN :-

base of the isosceles triangle = AC = 24cm

➡ AD = AC/2 = 24/2 = 12cm

ATQ,

the two equal sides of the isosceles triangle is 4m greater than it's height.

let the height of the triangle be x.

therefore it's two equal sides are x + 4 each

by Pythagoras theorem, we get

➡ AB² = BD² + AD²

➡ (x + 4)² = (12)² + (x)²

using identity (a + b)² = a² + 2ab + b²

➡ (x)² + 2(x)(4) + (4)² = 144 + x²

➡ x² + 8x + 16 = 144 + x²

➡ x² - x² + 8x = 144 - 16

➡ 8x = 128

➡ x = 128/8

➡ x = 16cm

therefore

» height of the triangle = x = 16cm

» length of the two equal sides = x + 4 = 20cm each

hence, it's perimeter = sum of all three sides

= 24 + 20 + 20

= 64cm

And area of the triangle = 1/2 × b × h

= 1/2 × 24 × 16

= 12 × 16

= 192cm²

Answer 2

• Let ABC a isoscele triangle in which BC = base, AD = height and AB & AC are sides.

__________ [ ASSUME ]

Base (BC) = 24cm and each of equal sides of an isoscele triangle i.e. AB & AC are 4 cm greater than it's height.

i.e.

AB = AD + 4

AC = AD + 4

__________ [ GIVEN ]

• We have to find the perimeter and area of traingle.

______________________________

In $\triangle$ADB

By Pythagoras Theorem

=> AB² = AD² + BD²

=> BD² = AB² - AD² _____ (eq 1)

Similarly,

In $\triangle$ADC

By Pythagoras Theorem

=> AC² = AD² + DC²

=> DC² = AC² - AD² ______ (eq 2)

_____________________________

Now,

Add both the equation (1) and (2)

=> BD² + DC² = AB² - AD² + AC² - AD²

=> BC² = AB² + AC² - (AD² + AD²)

=> BC² = AB² + AC² - 2AD²

=> BC² = AB² + AB² - 2AD²

(As AB = AC)

=> BC² = 2AB² - 2AD²

=> BC² = 2(AB² - AD²)

=> BC = 2$( \sqrt{ {AB}^{2} \: - \: {AD}^{2} }$

On squaring both sides we get,

=> BC² = 4(AB² - AD²)

Put the known values above

=> (24)² = 4[(AD + 4)² - (AD)²]

=> 576 = 4[AD² + 16 + 8AD - AD²]

=> 576 = 4(16 + 8AD)

=> 144 = 16 + 8AD

=> 144 - 16 = 8AD

=> 128 = 8AD

=> AD = 16 (height)

Now..

AB = AD + 4

AB = 16 + 4

AB = 20

Similarly, AC = 20

_____________________________

Area of traingle = 1/2 × b × h

=> 1/2 × 24 × 16

=> 192 cm²

Perimeter of triangle = AB + BC + AC

=> 20 + 24 + 20

=> 64 cm

______________________________

Area of traingle = 192 cm² and Perimeter of triangle = 64 cm.

_______________ [ ANSWER ]

______________________________