in triangle ABC~ APQ A- P-B and A-Q-C if A(∆ABC) =16A(∆APQ) .find PQ/BC.10th examples for practice
Answers
Answer:
Given
⎩
⎪
⎪
⎨
⎪
⎪
⎧
InaCylinder
Volume=1232cm
2
Height=8cm
To find:-
Curved\:surface\:area{}_{(CSA)}Curvedsurfacearea
(CSA)
Total\:Surface\:area {}_{(TSA)}TotalSurfacearea
(TSA)
Solution:-
Let radius=r
as we know that in a Cylinder
{\boxed{Volume={\pi}r {}^{2}h}}
Volume=πr
2
h
Substitute the values
{:}\longrightarrow:⟶ {\dfrac {22}{7}}×{r}^{2}×8=1232
7
22
×r
2
×8=1232
{:}\longrightarrow:⟶ 176{r}^{2}=1232176r
2
=1232
{:}\longrightarrow:⟶ {r}^{2}={\dfrac {1232}{176}}r
2
=
176
1232
{:}\longrightarrow:⟶ {r }^{2}=49r
2
=49
{:}\longrightarrow:⟶ r={\sqrt{49}}r=
49
{:}\longrightarrow:⟶ r=7cmr=7cm
CSA:-
As we know that
{\boxed{CSA=2{\pi}rh}}
CSA=2πrh
Substitute the values
{:}\longrightarrow:⟶ CSA=2×{\dfrac{22}{7}}×7×8CSA=2×
7
22
×7×8
{:}\longrightarrow:⟶ CSA=44×8CSA=44×8
\therefore∴ {\boxed {CSA=352cm {}^{2}}}
CSA=352cm
2
TSA:-
as we know that
{\boxed{TSA=2 {\pi}r (h+r)}}
TSA=2πr(h+r)
Substitute the values
{:}\longrightarrow:⟶ TSA=2×{\dfrac{22}{7}}×7 (8+7)TSA=2×
7
22
×7(8+7)
{:}\longrightarrow:⟶ TSA=2×22×15TSA=2×22×15
\therefore∴ {\boxed{TSA=660cm {}^{2}}}
TSA=660cm
2