# Needed some hints to complete my solutions on geometry

• Sep 13th 2005, 05:05 AM
totalnewbie
Needed some hints to complete my solutions on geometry
Drawing 1:
parallelogram
d1=40
d2=74
Find the area of the parallelogram of the drawing 1.
I got a= 51 and b=sqrt(937)
I am not able to find the area because I don't know the angles.

Drawing 2:
d1=14
d2=48
Find the height (h) of the rhombus
• Sep 13th 2005, 06:10 AM
hemza
Hi,

Ok, your drawing is not easy to work with since d1 < d2 which is not as it is in your drawings

In both figures, I will name the parrallelogram ABCD (clockwise, A at the bottom left) with AC the diagonal D1 and BD the diagonal D2. Fianlly, E is the intersection of d1 and d2.

You know that d1 and d2 intersect at their middle. So m(AE)=m(EC) and m(BE)=m(ED).

Let's beggin with the second parallelogram. We got m(AB)=m(BC)=m(CD)=m(DA). So in particular, the triangle ABD is isosceles with m(AB)=m(DA). And since AE cuts the side BD at its middle so it is also perpendicular to BD (because ABD is isosceles). So, the triangle ABE is right-angled triangle. We have m(AE)=7 and m(BE)=24 so m(AB)=sqrt(49+24^2) = 25. Let's find the measure of the angle BAE so sin(BAE)=24/25 so m(BAE)=73.74 degrees. Since AE bisects the angle BAD so m(BAE)=m(EAD)=73.74 degrees.

Ok, I'll name the foot of your height F (So the height is BF). So the angle BAF measures 180-2*73.74 = 32.52 degrees. Now, we are almost done. The triangle BAF is right-angled so sin(BAF) = m(BF)/25 so sin(32.52)=m(BF)/25 so m(BF)=13.44. Done !

__________________________________________________ ____________

For the first one, since you have a and b, it is easy. You don't need the angles. Heron formula for the area of a triangle is sufficient. Use it 1 time : to find the area of the triangle ABD. So ABD has sides 74, sqrt(937) and 51 so the p=semiperimeter=(74+sqrt(937)+51)/2=77.805
Area(ABD)=sqrt(p*(p-74)*(p-sqrt(937))*(p-51))=612. And since the triangle BDC has the same length of sides so its area is also 612 and so the area of the parallelogram is Area(BDC)+Area(ABD)=2*612=1224.

Done !

If you don't understand how to find a and b just reply and ask for that.
• Sep 13th 2005, 08:40 AM
totalnewbie

Isosceles trapezium 1:
the longer base is 44.
both sides which are aslant (i don't know what is the name of it in english) are 17.
the diagonal is 39.
what is the area of the trapezium ?
I have been trying to solve it. The diagonal is something for but I can't think out how to use this advantage.

Isosceles trapezium 2:
bases are 10 and 26.
diagonals are perpendicular (across) with both sides which are aslant (i don't know what is the name of it in english)
what is the area of the trapezium ?
• Sep 13th 2005, 09:40 AM
hemza
This is not as easy as the ones before (it will take you some patience) :

1) I will name my trapezium ABCD (clockwise) again with A at the left bottom and the larger base at the bottom (AD). I draw the diagonal AC.

I take the triangle ACD (I have all lengths of the sides) and find its area by Heron formula : p=1/2 (39+44+17)=50 so Area(ACD)=...=330.
Here, it is magnificiant because we can use the usual formula for the Area of a triangle. A(ACD) = b*h/2 where we can take the base being the side AD so out height CH (H is the foot of the height) will be also the height of the trapezium. Cool !

so A(ACD)=330=b*h/2. We know b=44 so h=15. So m(CH)=15 and the triangle CHD is right-angled so by Pytagorus we can find m(HD)=sqrt(CD^2-CH^2) = 8 (beautiful numbers in this kind of calculations with sqrt usually are a sign that you are on the right track, I say usually).

So, since the trapezium is isosceles, we have a symetry and then m(BC)=m(AD)-2*m(HD)=44-2*8=28