(1) In ∆ABC, join C to midpoint E of AB. CE is the perpendicular bisector of AB because ∆ABC is an isosceles triangle. BE=AE=20. CE=√(BC²-BE²)=√(625-400)=15. ∆s BCD and ACD are right triangles because CD is perpendicular to the base triangle ABC. DE is the hypotenuse of ∆CDE, so DE=√(CD²+CE²)=√(64+225)=17. DE is the height of ∆ABD, so the area of ∆ABD=½(AB)(DE)=½×40×17=340. Areas of ∆s BCD and ACD are the same so their combined area is 8×25=200. The area of ∆ABC=½×40×15=300.
The total surface area of the pyramid is 340+200+300=840 square units.
(2) ∆s ACD and ABD are two right triangles. We can find the squares of the lengths of their hypotenuses:
If DE is the perpendicular from D on to BC, then we have two right triangles, BDE and CDE. DE²=BD²-BE²=CD²-CE². But CE=BC-BE=14-BE.
425-481=28BE-196, BE=(425+196-481)/28=5 and CE=9.
Therefore, DE²=425-25=481-81=400, DE=20, which is the height of ∆BCD.
If AF is the perpendicular from A on to BC, we have two right triangles, ABF and ACF. AF²=AB²-BF²=AC²-CF², and BF+CF=BC=14.
169-225+196=28BF, BF=5 (so E and F are the same point), CF=CE=9.
AF²=169-25=225-81=144, AF=12, height of ABC with base BC=14.
Area of ∆ABC=½×14×12=84.
Area of ∆ABD=½×13×16=104.
Area of ∆ACD=½×15×16=120.
Area of ∆BCD=½×14×20=140.
S1=84+104+120+140=448 square units.