Question

The altitude of a triangle is 1cm less that the base to which it is drawn. the area of the of the triangle is 21 square cm. find the length of thebase

Answer 1

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GIVEN :-

✥ The altitude of a triangle is 1 cm less that the base to which it is drawn.

✥ Area of the of the triangle is 21 cm².

TO FIND :-

✥ The length of the base.

SOLUTION :-

Let the length of the base be x cm.

Then the length of the altitude drawn to it is (x - 1) cm.

We know,

Area of a triangle = 1/2 × base × height

⇒21 = 1/2 * x * (x - 1)

⇒21 = 1/2 * x(x - 1)

⇒21 * 2 = x² - x

⇒42 = x² - x

⇒x² - x - 42 = 0

⇒x² -7x + 6x - 42 = 0

⇒x(x - 7) + 6(x - 7) = 0

⇒(x - 7)(x + 6) = 0

⇒(x - 7) = 0    or    (x + 6) = 0

⇒x = 7           or     x = -6

✥ Length of something can never negative.

∴ So, the length of the base = x cm = 7 cm.

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✥ The length of the altitude = x - 1 = 6 cm.

Answer 2

Answer :

The length of the base of the triangle is 7cm

Given :

• The altitude of a triangle is 1cm less than the base to which it is drawn.
• The area of the triangle is 21 sq.cm

To Find :

• The length of the base of a triangle.

Formula to be Used :

$\sf Area \: \: of \: \: a \: \: triangle = \dfrac{1}{2} \times base \times altitude$

Solution :

Let us consider the base of the triangle be x and altitude be y

According to first condition :

$\sf \implies y = x - 1 \: \: ..........(1)$

And by the next condition :

$\sf area \: of \: triangle \: = 21 \\ \\ \sf \implies \dfrac{1}{2} \times base \times altitude \\ \\ \sf \implies \sf \dfrac{1}{2} xy = 21 \\ \\ \implies \sf\frac{1}{2} x(x - 1) = 21 \: \: \: \{ from \: \: (1)\} \\ \\ \implies \sf x(x - 1) = 42 \\ \\ \implies \sf {x}^{2} - x = 42 \\ \\ \sf \implies {x}^{2} - x - 42 = 0 \\ \\ \sf \implies {x}^{2} + 6x - 7x - 42 = 0 \\ \\ \sf \implies x(x + 6) - 7(x + 6) = 0 \\ \\ \sf \implies(x + 6)(x - 7) = 0$

Now we have :

$\sf \: x + 6 = 0 \: \: and \: \: x - 7 = 0 \\ \\ \implies \sf x = - 6 \: \: and \implies x = 7$

Since x is the length of the base and length can never be negative so x ≠ -6 .

Therefore, the length of the base is 7cm

again

$\sf \implies y = 7 - 1 \\ \\ \sf \implies y = 6$

The altitude is 6cm