Math, asked by ahujanaina1, 8 days ago

find area of shared portion​

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Answered by viperisbackagain
0

 \huge \red{answer}

 \large \bf \green{ \underline {given :} \: ag = 7cm. \: gb = 6cm. \: bf = 3cm. \: fc = 4cm.ae = 4cm}

 \bf \blue{now \: ab = ag + gb = 7 + 6 = 13cm} \\  \bf \blue{bc = bf + fc = 3 + 4 = 7cm} \\  \bf \blue{ad = bc = 7cm \: and \: ab = dc = 13cm} \\  \bf \blue{ \boxed{aera \: of \: rectangle = (l  \times  b)}} \\  \bf \blue{aera \: of \: rectangle = 13 \times 7 = 91cm {}^{2}  -  -  - (1)} \\  \\  \bf \purple{now \:  \triangle{aeg} \: is \: right \: angle \: triangle } \\  \bf \purple{ \boxed {aera \: of \: right \: angle \: triangle =  \frac{1}{2} \times b \times h }} \\  \bf \purple{aera \: of \: right \: angle \: triangle =  \frac{1}{2} \times 4 \times 7 } \\  \bf \purple{aera \: of \: right \: angle \: triangle =  \frac{1}{2} \times 28  = 14cm {}^{2}  -  - (2)} \\  \\  \bf \pink{also  \:  \: \triangle{gbf = right \: angle}} \\  \bf \pink{ \boxed{aera \: of \: right \: angle \: triangle =  \frac{1}{2}  \times b \times h}} \\  \bf \pink{aera \: of \: right \: angle \: triangle =  \frac{1}{2} \times6 \times 3  } \\  \bf \pink{aera \: of \: right  \: angle \: triangle = \frac{1}{2} \times 12   = 6cm {}^{2}  -  - (3)} \\  \\  \bf \color{yellow}{ \boxed{then \: aera \: of \: shaded \: poration = aera \: of \: square - aera \: of \: both \: triangles}} \\  \bf \color{yellow}{aera \: of \: shaded \: region = 91 - (14 + 6)} \\  \bf \color{yellow}{aera \: of \: shaded \: region = 91 - 18} \\  \large \bf \color{yellow}{ \underline{aera \: of \: shaded \: region = 73cm {}^{2} }}

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