how in the crap do i find h
#Isoceles traingle someone helpppppppppppppppp
stable nacelle
ocean meteor
Oh, it's a right triangle! You neglected to say that.
Well, suppose an isosceles right triangle has lateral sides a, hypotenuse b and height h onto the hypotenuse.
Then, remembering the two formulas for the area of the triangle, we get:
S = (1/2)a^2 = (1/2)bh
So:
bh = a^2
h = a^2/b
But by thd Pythagorean theorem we have b = √(a^2 + a^2) = a√(2). So:
h = a^2/(a√(2)) = a/√(2)
stable nacelle
as a year 8, i can say i understood absolutely nothing from that
im sorry man im too dum for this
vapid hare
as a year 8, i can say i understood absolutely nothing from that
Do you agree with this much?
west remnant
1- first find b using Pythagoras: sqrt(7^2+7^2) = sqrt(98)
2-find area of triangle : (7 * 7)/2= 24.5 (since the angle between 7 cm sides is 90)
3-since H is also 90, you can find area of triangle by (h*b)/2 too.
4- solve the equation : (sqrt(98) * h)/2=24.5
5- h equals (7√2)/2
sick hedge
Since b/2 and h are equal, and angle BH is 90deg,
$sqrt(2h^2) = 7.
2h^2 = 49.
h^2 = 24.5.
h = sqrt(24.5) = ±(7 sqrt(2))/2$.
Because the height cannot be negative, the answer is $(7 sqrt(2))/2$
brave crescentBOT
JakeTheNoob
vivid marlin
how in the crap do i find h
Another solution. Using Phytagorean theorem, it's known that $b = 7\sqrt{2}$ cm.
Since $h$ is perpendicular to $b$ and it's an isosceles triangle; therefore, it applies that $h = \frac{b}{2}$. So, $h = \frac{7}{2} \sqrt{2}$