Question

Find the altitude and area of an isoceles triangle whose perimeter is 64 cm and base is 24 cm.

Answer 1

Answer:

let the two sides of isosceles triangle be AB and CA and base = BC = 24cm

perimeter = 64

now AB + BC + CA = 64

AB + 24+ AB = 64 (AB = CA therefore IN place of CA there will be AB)

2AB + 24 = 64

2AB = 64-24

2AB = 40

AB = 40/2 = 20

now AB = 20cm

and since AB = CA therefore CA= 20

now deciding the triangle in two equal part such that AD bisects Angle A and AD also bisects BC

Now the two triangles will be right angled triangle :-

AD is perpendicular to BC

now using Pythagoras theorem :-

BC = 24

so BD = 24/2 = 12(AD bisects BC)

AD ^2+ BD ^2 = AB ^2

AD ^2 + 12^2 = 20^2

AD ^2 = 400-144

AD ^2 = 256

AD = √256

AD = 16 cm

area of first right angled triangle :-

1/2 * b*h

1/2 * BD * AD

1/2 * 12*16 = 96 cm

now the both right angled triangle are equal so their area will be also equal so :-

triangle ABD = triangle ADC

area of ABD = area of ADC

96 = 96

total area of isosceles triangle = area of ABD + area of ADC

area of isosceles triangle = 96 + 96 = 192 cm ^2

so the answers are

altitude = 16 cm

area of isosceles triangle = 192cm^2

by the way this question is of maths exam which was held on 17 th Feb ...am I right???

that was question is easy just divide the triangle into two equal parts.........

Answer 2

Answer:

area=240cm square

Step-by-step explanation:

altitude of the triangle let be 'h' cm

perimeter of the triangle=64cm

base of the triangle = 24cm

ATQ, 24+h+h=6

= 2h=64-24=40

=h=40/2=20cm

height=20cm

area=1/2×b×h

1/2×24×20

= 24×10

= 240cm square