Well, basically, I got this question wrong on a homework and I don't understand?
There are two equal sides of 10cm and the base is 6cm, I did 6x10x1/2 and it was wrong.
Apparently I need to use pythagoras but I don't get where. Could someone help please?
There are two equal sides of 10 and the base is 6. If you split it you can make 2 right angled triangles.m
There are two equal sides of 10cm and the base is 6cm, I did 6x10x1/2 and it was wrong.
Apparently I need to use pythagoras but I don't get where. Could someone help please?
There are two equal sides of 10 and the base is 6. If you split it you can make 2 right angled triangles.m
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The area A of an isoceles triangle is:
A = (1/2)*( b the base length) *(h perpendicular height)
That is what you tried to use but we don't know the perpendicular height. The mistake you made was to use the length of one of the sides.
To find the height we use pythagorean theorem.
The perpendicular height cuts an isoceles traingle in to two equal right angled traingles. Hre, the hypotenuse of each = 10 cm. The shorter (base side) side = 6/2 =3cm. Then:
10^2 = 3^ + h^2
=> h = sqrt(100- 9)
=> h = sqrt(91) ~ 9.539 cm. Hence the area of the isosceles triangle is:
A = (1/2)*(6)*sqrt(91)28.618...
=> A = 28.7 cm^2 (3 sig. fig)
A = (1/2)*( b the base length) *(h perpendicular height)
That is what you tried to use but we don't know the perpendicular height. The mistake you made was to use the length of one of the sides.
To find the height we use pythagorean theorem.
The perpendicular height cuts an isoceles traingle in to two equal right angled traingles. Hre, the hypotenuse of each = 10 cm. The shorter (base side) side = 6/2 =3cm. Then:
10^2 = 3^ + h^2
=> h = sqrt(100- 9)
=> h = sqrt(91) ~ 9.539 cm. Hence the area of the isosceles triangle is:
A = (1/2)*(6)*sqrt(91)28.618...
=> A = 28.7 cm^2 (3 sig. fig)
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the legs of an isosceles triangle are NOT THE SAME AS THE HEIGHT/ALTITUDE.
to calculate the altitude, you need to use the Pythagorean theorem...
the altitude bisects the base, so...
3^2 + h^2 = 10^2
solve for h:
h = √91 cm
now, calculate the area...
A = (1/2) * 6 * (√91) = 3√91 cm^2
@ß
to calculate the altitude, you need to use the Pythagorean theorem...
the altitude bisects the base, so...
3^2 + h^2 = 10^2
solve for h:
h = √91 cm
now, calculate the area...
A = (1/2) * 6 * (√91) = 3√91 cm^2
@ß
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Height is not 10 cm
Height is \/ (10^2 - 3^2) = \/91
Area = 1/2 X 6 X \/91 = 3 X \/91 = 28.62 cm^2 ANSWER
ANOTHER METHOD
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Use HERON'S FORMULA
AREA = \/ s ( s-a) ( s- b) ( s-c)
where s = semi perimeter
AREA = \/ 13 ( 3) (3) ( 7) = 3\/91 = 28.62 cm^2 ANSWER
Height is \/ (10^2 - 3^2) = \/91
Area = 1/2 X 6 X \/91 = 3 X \/91 = 28.62 cm^2 ANSWER
ANOTHER METHOD
-------------------------------
Use HERON'S FORMULA
AREA = \/ s ( s-a) ( s- b) ( s-c)
where s = semi perimeter
AREA = \/ 13 ( 3) (3) ( 7) = 3\/91 = 28.62 cm^2 ANSWER
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The equal sides measure 10 cm each.
The base measures 6 cm.
The height of the triangle is sqrt (10^2 - 3^2)
The area of the triangle is 1/2 x base x height
= 1/2*6*sqrt(91) = 3sqrt(91) = 3*9.54 = 28.62 cm^2.
The base measures 6 cm.
The height of the triangle is sqrt (10^2 - 3^2)
The area of the triangle is 1/2 x base x height
= 1/2*6*sqrt(91) = 3sqrt(91) = 3*9.54 = 28.62 cm^2.
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Hi
see if this page helps you
http://www.ajdesigner.com/phptriangle/is…
or this one
http://mathworld.wolfram.com/IsoscelesTr…
see if this page helps you
http://www.ajdesigner.com/phptriangle/is…
or this one
http://mathworld.wolfram.com/IsoscelesTr…
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a = 10 ; b = 6
A = b { [ square root of (4a^2 - b^2) ] / 4}
A = 6 [square root of 364 / 4]
A = 28.62cm
A = b { [ square root of (4a^2 - b^2) ] / 4}
A = 6 [square root of 364 / 4]
A = 28.62cm