Question

ABC is a right-angled triangle with right angle at B. If the semi-circle on AB with AB as diameter encloses an area of 81 sq.cm and semi-circle on BC with BC as diameter encloses an area of 36 sq.cm then the area of the semi-circle on AC with AC as diameter will be

Answer 1

$\huge\mathcal\red{SolutiOn:}$

$\sf{Given-}$

• ABC is a right-angled triangle with right angle at B.
• If the semi-circle on AB with AB as diameter encloses an area of 81 sq.cm.
• Semi-circle on BC with BC as diameter encloses an area of 36 sq.cm then the area of the semi-circle on AC with AC.

$\sf\red{We~have~to~find~the~diameter~and~semicircle~on~AC:}$

$\sf{Required~area=}\:\frac{1}{2}\:\pi \sf{r^2=\frac{1}{2}}\pi × (\frac{AC}{2})^2\\~~~~~~~~~~~~~=\frac{\pi}{8}\:\:AC^2\\~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{\pi}{8}\:(AB^2+BC^2)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{2}\:\pi\:(\frac{AB}{2})^2+\frac{1}{2}\:\pi(\frac{BC}{2})^2\\~~~~~~~~~~~~~~~~~~=\mathtt{81+36}\\~~~~~~~~~~~~~~~~~~=\mathtt{117cm^2}$

$\sf{You~can ~see~ the~ first~part} \:\frac{1}{2}\pi(\frac{AB}{2})^2\:\sf{this~is~the~area~}\\\sf{and~AB~as~the~diameter}\\\bold{Similarly,}$

$\sf{The~second~part}~\frac{1}{2}\pi(\frac{BC}{2})^2\sf{~is~the~area~of~the~semi-circle~}\\\sf{and,~BC~as~the~diameter}$

$\sf\purple{∴~the~area~of~the~semi-circle,}$

$\sf\purple{~on~AC~with~AC~as ~diameter~will~be~177cm^2.}$

_______________________________________

$\huge\underline\mathfrak\pink{Brainliest~plzz!}$

Answer 2

Given:

A right-angled triangle ABC.

Area of semi-circle AB = 81 sq.cm

Area of semi-circle BC = 36 sq.cm

To find:

Area of semi-circle AC.

Solution:

ABC is a right triangle with right angle at B. Hence, by Pythagoras theorem,

$AC^{2} =AB^{2}+BC^{2}...(1)$

Let semicircle AB be $S_{1}$, semicircle BC be $S_{2}$ and semicircle AC be $S_{3}$.

Let the diameters of $S_{1},S_{2},S_{3}$ be $d_{1},d_{2},d_{3}$ respectively.

Area of a semi-circle is given by:

$A=\frac{1}{2}\pi r^{2}$

where $A$ is the area and $r$ is the radius of the semi-circle.

Let $A_{1},A_{2},A_{3}$ be the areas of semicircles $S_{1},S_{2},S_{3}$ respectively.

Here, areas of semicircles $S_{1}$ and $S_{2}$ are given as $81sq.cm$ and $36 sq.cm$.

$A_{1}=\frac{1}{2}\pi r_{1}^{2}$

where $r_{1}$ is the radius of semicircle AB and is given by $r_{1}=\frac{d_{1}}{2}$ and $d_{1}=AB$

$81=\frac{1}{2}\pi (\frac{d_{1}}{2} )^{2}$

$81=\frac{1}{2}\pi \frac{d_{1}^{2}}{4}$

$81=\frac{\pi }{8} AB^{2}$

$AB^{2}=81*\frac{8}{\pi }...(2)$

$A_{2}=\frac{1}{2} \pi r_{2}^{2}$

where $r_{2}$ is the radius of semicircle BC and is given by $r_{2}=\frac{d_{2}}{2}$ and $d_{2}=BC$

$36=\frac{1}{2}\pi (\frac{d_{2}}{2} )^{2}$

$36=\frac{1}{2}\pi \frac{d_{2}^{2}}{4}$

$36=\frac{\pi }{8} BC^{2}$

$BC^{2}=36*\frac{8}{\pi }...(3)$

Here, area of semicircle AC is unknown. Its diameter is AC, i.e., $d_{3}=AC$ and

$AC=\sqrt{AB^{2}+BC^{2}}$

$A_{3}=\frac{1}{2}\pi r_{3}^{2}$

$A_{3}=\frac{1}{2}\pi ( \frac{d_{3}}{2} )^{2}$

$A_{3}=\frac{1}{2}\pi \frac{d_{3}^{2}}{4}$

$A_{3}=\frac{\pi }{8} (AC)^{2}$

From equation (1),

$A_{3}=\frac{\pi }{8}(AB^{2}+BC^{2})$

From equations (2) and (3),

$A_{3}=\frac{\pi }{8}(81*\frac{8}{\pi } +36*\frac{8}{\pi } )$

Taking $\frac{8}{\pi }$ common from the brackets,

$A_{3}=\frac{\pi }{8}*\frac{8}{\pi } (81+36)$

$A_{3}=81+36=117$

∴ Area of semicircle on AC is $117sq.cm$ with AC as diameter.

Area of semicircle on AC is $117sq.cm$ with AC as diameter.