Answer:
![V=400\ cm^(3)](https://img.qammunity.org/2020/formulas/mathematics/college/o9uionyxmpsf11fy7715y8om4elk6ih20k.png)
Explanation:
we know that
The volume of a pyramid is equal to
![V=(1)/(3)BH](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1npkck5tezb4g42bc6m0koc5jua6d5elt6.png)
where
B is the area of the square base
H is the height of the pyramid
Find the area of the square base B
![B=b^(2)](https://img.qammunity.org/2020/formulas/mathematics/college/667534ej1e02hlaw03w4wz9bve2szpb3ql.png)
![b=10\ cm](https://img.qammunity.org/2020/formulas/mathematics/middle-school/lvril1kapxwbouxfynd7ruv0en96qk15el.png)
![B=10^(2)=100\ cm^(2)](https://img.qammunity.org/2020/formulas/mathematics/college/n2tvcv26ulfaocen1o9ohwy9xuh9uehecp.png)
Find the height H of the pyramid
Applying the Pythagoras Theorem
![l^(2)=H^(2)+(b/2)^(2)](https://img.qammunity.org/2020/formulas/mathematics/college/lb6eg99ifr12oonxuwu88jjrlz6gs2awo9.png)
we have
![b=10\ cm](https://img.qammunity.org/2020/formulas/mathematics/middle-school/lvril1kapxwbouxfynd7ruv0en96qk15el.png)
![l=13\ cm](https://img.qammunity.org/2020/formulas/mathematics/college/e438vfez62w9jjof3txmt0yu6k17poz8tu.png)
substitute
![13^(2)=H^(2)+(10/2)^(2)](https://img.qammunity.org/2020/formulas/mathematics/college/61bih8al683wm5cw35ijyhfbo99ik2ykhk.png)
![169=H^(2)+25](https://img.qammunity.org/2020/formulas/mathematics/college/1mtly3qjmdw77s3qhe0was0p2gw8jbev98.png)
![H^(2)=169-25](https://img.qammunity.org/2020/formulas/mathematics/college/b3pe9id7d5956t7o25xaxwcxr7pte202ta.png)
![H^(2)=144](https://img.qammunity.org/2020/formulas/mathematics/college/uoyw4ecth7pn9mdzb7lkx6fmsha7eb5ql5.png)
![H=12\ cm](https://img.qammunity.org/2020/formulas/mathematics/college/uy2zjb0yan9wa4a2t16cljkbye5l2nw4ni.png)
Find the volume
![V=(1)/(3)(100)(12)](https://img.qammunity.org/2020/formulas/mathematics/college/6t6z72r6yhstjjw2yjh72yltywazm4ovxn.png)
![V=400\ cm^(3)](https://img.qammunity.org/2020/formulas/mathematics/college/o9uionyxmpsf11fy7715y8om4elk6ih20k.png)