Math, asked by prarthanachinky, 9 months ago

7. If the length of one of the diagonals of a rhombus is 7cm more than that of the other diagonal, and its. area is 15cm2 then find the length of the two diagonals

Answers

Answered by Anonymous
26

Solution :-

Let the other diagonal ( d1 ) be 'x' cm

One of the diagonal ( d2 ) = 7 cm more than the other = (x + 7)

Area of the rhombus = 15 cm²

Also, Area of the diagonal = d1 * d2 / 2 sq.units

⇒ d1 * d2 / 2 = 15

⇒ x(x + 7) / 2 = 15

⇒ x² + 7x = 15 * 2

⇒ x² + 7x = 30

⇒ x² + 7x - 30 = 0

⇒ x² + 10x - 3x - 30 = 0

⇒ x(x + 10) - 3(x + 10) = 0

⇒ (x - 3)(x + 10) = 0

⇒ x - 3 = 0 or x + 10 = 0

⇒ x = 3 or x = - 10

Length of diagonals cannot be negative

⇒ x = 3

Another diagonal ( d1 ) = x = 3 cm

One of the diagonal ( d2 ) = x + 7 = 3 + 7 = 10 cm

Therefore the lengths of the diagonals are 3 cm and 10 cm.

Verification :-

Area of rhombus = d1 * d2 /2 = 3 * 10/2 = 30/2 = 15 cm².

Answered by MrBhukkad
7

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 \bf{Let,} \\  \:  \:  \:  \:  \:  \bf{First \: diagonal( d_{1}) } = d \: cm \\   \:  \:  \:  \:  \: \bf{Then, \: second \: diagonal( d_{2}) = (d + 7) \: cm }  \\  \\  \bf{Given,} \\  \:  \:  \:  \:  \: \bf{Area \: of \: rhombus(A) = 15 \:  {cm}^{2} } \\  \\  \bf{We \: know \: that, } \\   \:  \:  \:  \:  \: \bf{Area \: of \: rhombus(A) \:  = \frac{ d_{1} \times  d_{2}  }{2}   } \\  \bf{or, \:  \frac{ d_{1} \times  d_{2}  }{2}  = A } \\  \bf{or,  \:  \frac{d \times (d + 7)}{2}  = 15} \\  \bf{or, \: d \times (d + 7) = 30} \\  \bf{or, \:  {d}^{2}   + 7d =30 } \\  \bf{or, \:  {d}^{2} + 7d - 30 = 0 } \\   \\ \bf{From \: middle \: term \: splitting \: method,} \\  \bf{or, \:  {d}^{2} + 10d - 3d - 30 = 0 } \\  \bf{or, \: d(d + 10) - 3(d + 10) = 0} \\  \bf{or, \: (d  + 10)(d - 3) = 0} \\  \\  \bf{ \underline{Either} \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \: { \underline{</p><p>Or}}} \\  \bf{d + 10 = 0 \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \: d - 3 = 0} \\  \bf{or, \: d =  - 10 \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  or, \: d  = 3}  \\  \\  \bf{But \: diagonal \: can \: never \: be \: negative.} \\  \bf{So, \:diagonal \: is \: 3 \: cm }  \\  \\  \bf{So, \: first \: diagonal( d_{1} ) = 3 \: cm} \\  \bf{Then, \: second \: diagonal( d_{2}) = (3 + 7 ) = 10 \: cm} \\  \\  \\  \\  \\  \huge{ \underline{ \underline{ \pink{ \mathfrak{Verification}}}}}:-  \\  \bf{Area  \: of \:rhombus \:  =  (\frac{1}{2}  \times  d_{1}  \times  d_{2} )\: {cm}^{2} } \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:\bf{ =  (\frac{1}{2} \times 3 \times 10) } \:  {cm}^{2}  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:\: \bf{ = (3 \times 5) \:  {cm}^{2} } \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \: \bf{ = 15 \:  {cm}^{2} } \\  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \huge{ \red{ \mathcal{ \boxed{Verified}}}}

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