Question

The base edge of the regular triangular pyramid is b=10 cm and altitude of the base hb ≈ 8.66 cm. The slant height of the pyramid is k=8 cm. Find:
Lateral area and Surface area of the pyramid

Answer 1

Lets LA lateral area and SA surface area, b base and l slant height, h the height:


LA= 3(1/2 b l)= 3/2 (10)(8)= 120 cm²


SA= 1/2 bh+3/2 bl


SA=1/2 (8.66)(10)+120cm²= 163.3 cm²

Answer 2

Answer:

Step-by-step explanation:

It is given that the base edge of the regular triangular pyramid is b=10 cm and altitude of the base h =8.66 cm. The slant height of the pyramid is k=8 cm.

Now, the lateral surface area of the pyramid is given as:

$LSA={\frac{3}{2}}(b)(k)$

Substituting the given values, we have

$LSA=\frac{3}{2}(10)(8)$

$LSA=120cm^2$

Thus, the Lateral surface area of the pyramid is $120cm^2$.

Now, the surface area is given as:

$SA=\frac{1}{2}bh+LSA$

$SA=\frac{1}{2}bh+120$

$SA=\frac{1}{2}(10)(8.66)+120$

$SA=43.3+120$

$SA=163.3cm^2$

Thus, the surface area of the pyramid will be $163.3cm^2$.