Question

the area of a triangle is 48^2. If its height is greater than the base by 4cm . Find the base .
please give the answer faster .I need it very importantly .​

Answer 1

Area of triangle =48*48
Let the base be x cm
Height = x+4 cm
Using formula
Area of triangle =48*48
1/2*b*h=48*48
1/2*x*x+4=48*48
x*x+4=48*48*2
X^2 +4x= 4608
X^2+4x-4608=0
Using MTS
We need two numbers,multiplying those numbers should give 4608 and adding those same numbers should give sum 4
.When we will get those numbers arrange those numbers like greater number should come first +or -smaller number then group those numbers into a pair of 2 and find common between them.The common expression should be same.Then outer numbers should be taken into brackets together and using either and or find the value of x ie base and find height after putting the value of x in x+4.

Answer 2

Answer:

$\huge{\red{\boxed{\green{\sf{Height=12cm}}}}}$

$\huge{\red{\boxed{\green{\sf{Base=8cm}}}}}$

Step-by-step explanation:

$\huge{\red{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$

□Area of triangle=48cm^2

□Let base be x cm

□Height= x+4cm

To find:

○Height=?

○Base=?

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$area \: of \: triangle = \frac{1}{2} \times base \times altitude \\ \to \: 48 {cm}^{2} = \frac{1}{2} \times x \times (x + 4) \\ \to \: 48 \times 2= {x}^{2} + 4x \\ \to \: 96 = {x}^{2} + 4x \\ \to \: {x}^{2} + 4x - 96 = 0 \\ \\ using \: middle \: term \: spliting \: to \: find \: value \: of \: x \\ \to \: {x}^{2} + 4x - 96 = 0 \\ \to \: {x}^{2} + 12x -8x - 96 = 0 \\ \to \: x(x + 12) - 8(x + 12) = 0 \\ \to \: (x - 8)(x + 12) = 0 \\ \\ \to so \: we \: take \: value \: of \: x \: in \: positive \\ \to \: bcz \: area \: cannot \: be \: in \: negative \\ \\ \to \: x - 8 = 0 \\ \to \: x = 8 \\ \\ height = x + 4 = 8 + 4 = 12 \\ base = x = 8 \\ \\ \: \: \: \: \: \: {\red{ \boxed{verification}}} \\ area \: of \: triangle = \frac{1}{2} \times base \times height \\ \to \: \frac{1}{2} \times 8 \times 12 \\ \to4 \times 12 \\ \to \: 48 {cm}^{2} \\ \: \: \: \: \: {\green{ \boxed{verified}}}$

$\huge{\red{\boxed{\green{\sf{Height=12cm}}}}}$

$\huge{\red{\boxed{\green{\sf{Base=8cm}}}}}$

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