Two triangles have the same base lengths. One Triangle has the height that is 4 times the height of the other triangle. Are the heights and the areas of the two triangles proportional?

YES

Base=b

Height= h

Area 1=A1=1/2*b*h

Base=b

Height= 4h

Area of triangle 2=A2=1/2*b*4h

A2/A1= (1/2*b*4h)/(1/2*b*h)

= (1/2*b*4h)/(1/2*b*h)

=4/1

Ratio of heights =h2/h1= 4/1

Ratio of areas =A2/A1= 4/1

They are proportional.

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

using pemdas, (order of operations) you'll see the answer is 24.