Answer to Question #313826 in Analytic Geometry for Vick

Question #313826

A trapezoid has an area of 36 sq. m and an altitude of 2m. Its two bases have ratio of 4:5. Find the perimeter of the trapezoid. Find the percentage increase in the volume if its edge is increased by 1%


1
Expert's answer
2022-03-19T02:40:19-0400

S=36m2BK=2mBC:AD=4:5S=36m^2\\ BK=2m\\ BC:AD=4:5

BK, CM - an altitude

Lets

BC=4x,AD=5xS=BC+AD2BKS=4x+5x22=369x=36x=4BC=16m,AD=20mBC=4x, AD=5x\\ S=\frac{BC+AD}{2}\cdot BK\\ S=\frac{4x+5x}{2}\cdot2=36\\ 9x=36\\ x=4\\ BC=16m, AD=20m

AK=MD=ADBC2=20162=2(m)AK=MD=\frac{AD-BC}{2}=\frac{20-16}{2}=2(m)\\

ABK,K=900AB2=AK2+BK2AB2=22+22=4+4=8AB=8=22(m)AB=CD=22mP=AB+BC+CD+AD==22+16+22+20=36+42(m)\triangle ABK, \angle K=90^0\\ AB^2=AK^2+BK^2\\ AB^2=2^2+2^2=4+4=8\\ AB=\sqrt{8}=2\sqrt{2}(m)\\ AB=CD=2\sqrt{2}m\\ P=AB+BC+CD+AD=\\ =2\sqrt{2}+16+2\sqrt{2}+20=36+4\sqrt{2}(m)

If AB and CD is increased by 1%=0.01:

AB=1.01(22)=2.022(m)P=16+20+2.022+2.022=36+4.042(m)P100%P=(36+4.042)100%36+42100.14%AB'=1.01(2\sqrt{2})=2.02\sqrt{2}(m)\\ P'=16+20+2.02\sqrt{2}+2.02\sqrt{2}=36+4.04\sqrt{2}(m)\\ \frac{P'\cdot100\%}{P}=\frac{(36+4.04\sqrt{2})\cdot100\%}{36+4\sqrt{2}} \approx100.14\%



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment