Question

6. dn isosceles triangle has perimeter 30 .cm each of the
equal sides is 12 cm. Find the area of the triangle.​

Answer 1

Answer:

PERIMETER = SIDE 1+SIDE 2+SIDE 3

30= SIDE 1+12+12

SIDE 1= 30-24= 6 CM=BC

DRAW AD⊥BC

IN ΔABD AND ΔACD

AB=AC (GIVEN)

∠ADB=∠ADC (90°)

AD=AD (COMMON)

ΔABD≅ΔACD (RHS)

BD=CD   (CPCT)

BC=BD+CD

6=2CD

CD=3CM

IN ΔACD,

AC²=AD²+CD²

12²=AD²+3²

144-9=AD²

135=AD²

AD=11.6 CM

AREA= 1/2 *BASE *HEIGHT

AREA= 1/2 *6 *11.6

AREA= 3*11.6

AREA=34.8 CM²

Step-by-step explanation:

Answer 2

Answer:

here given an isosceles triangle hence the three sides are 12,12,6

here area

$\sqrt{p(p - a)(p - b)(p - c)}$

here p =(a+b+c)/2

so

$\sqrt{15(15 - 12)(15 - 12)(15 - 6)}$

=34.8 sq cm