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marshall27 [118]

2 years ago

11

He height of a triangle is 9 m less than is base the area is 56 m^2 what is the length of the base?

1 answer:

zloy xaker [14]2 years ago

4 0

Formula for area of triangle:
A= 1/2 bh
Where b is the base and h is the height of the triangle.
Here, h= b-9
A=56 m^2
Put values in formula.
56= 1/2 (b) (b-9)
56=b(b-9)/2
Multiply both sides by 2.
112=b^2-9b
or
b^2-9b-112=0
Factorise the equation.
b^2-16b+7b-112=0
b(b-16)+7(b-16)=0
(b+7)(b-16)=0
Using zero factor theorem.
b+7=0 or b=-7
and
b-16=0 or b=16
Ignore the value b=-7, because length cannot be negative.
So length of base=16 m

Answer: Base=16m

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2 years ago

Se tiene un rectángulo cuya base mide el doble que su altura y su área es 12

irina [24]

Answer:

Part 1) The perimeter is $P=6\sqrt{6}\ cm$

Part 2) The diagonal is $d=\sqrt{30}\ cm$

Step-by-step explanation:

<u><em>The question in English is</em></u>

You have a rectangle whose base is twice the height and its area is 12

square centimeters. Calculate the perimeter of the rectangle and its diagonal

step 1

Find the dimensions of rectangle

we know that

The area of rectangle is equal to

$A=bh$

$A=12\ cm^2$

so

$bh=12$ ----> equation A

The base is twice the height

so

$b=2h$ ----> equation B

substitute equation B in equation A

$(2h)h=12\\2h^2=12\\h^2=6\\h=\sqrt{6}\ cm$

Find the value of b

$b=2\sqrt{6}\ cm$

step 2

Find the perimeter of rectangle

The perimeter is given by

$P=2(b+h)$

substitute

$P=2(2\sqrt{6}+\sqrt{6})\\P=6\sqrt{6}\ cm$

step 3

Find the diagonal of rectangle

Applying the Pythagorean Theorem

$d^2=b^2+h^2$

substitute

$d^2=(2\sqrt{6})^2+(\sqrt{6})^2$

$d^2=30$

$d=\sqrt{30}\ cm$

3 0

2 years ago