Question

The area of
a triangle is 48 cm² . If its
height is
greater then the base by 4 cm find
the base?​

Answer 1

Answer:

Base is 8

Step-by-step explanation:

1/2*b*(b+4)=48

b*(b+4)=96

b² +4b=96

From here you can solve quadratic or put value, so put a value which can take nearer to 96 so I took 8 and it will satisfy the equation.

Answer 2

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• Area of the triangle = 48 cm

• It’s height is grater than the base

by 4.

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• Base of the triangle

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

Let the base be x cm.

Given that,

• Its height is greater than its base by 4.

Then, the height of the triangle will be

= (x +4) cm

$\underline{\:\textsf{ As we know that :}}$

$\tt{\small{\boxed{\bold{\bold{\green{\sf{Area\:of\:Triangle=\dfrac{1}{2}\times{b}\times{h}}}}}}}}$

$\underline{\:\textsf{ Now, put the given values in the formula :}}$

$\longrightarrow \tt{48=\dfrac{1}{2}\times{x}\times{(x+4)}}$

$\longrightarrow \tt{48=\dfrac{1}{2}\times{{x}^{2}}+4x}$

$\longrightarrow \tt{48\times{2}={x}^{2}+4x}$

$\longrightarrow \tt{96={x}^{2}+4x}$

$\longrightarrow \tt{{x}^{2}+4x-96}$

(By splitting Middle Term)

$\longrightarrow \tt{{x}^{2}+4x-96=0}$

$\longrightarrow \tt{{x}^{2}+12x-8x-96=0}$

$\longrightarrow \tt{x(x+12)-8(x+12)=0}$

$\longrightarrow \tt{(x-8)(x+12)=0}$

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\bf{ \longrightarrow x = 8}$

Or,

$\bf{ \longrightarrow x = -12}$

Here, remember that base can't be negative. Therefore, base of the triangle will be 8 cm.

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