# Help me out these easy geometry problems.?

• Jan 14th 2013, 07:39 PM
kohila
Help me out these easy geometry problems.?
I want the explanation too, cause i don't understand the question properly and then put it clearly, cause i am just 8th class student. best of answers will be given 5 *'s, plus a thumbs up.

1.In ▲ PQR, PD ┴ QR and PO is the bisector of ∠ QPR. If ∠ PQR = 65° and ∠ PQR = 23.5°,then
∠ DPO in degrees=
2.In a parallelogram PQRS,the bisectors of ∠ P and ∠ Q meet on RS.If the perimeter PQRS is 13.5 cm, then the measure of QR =
3.A circle is passing through three vertices of a rhombus of side 8cm and its centre is the forth vertex of the rhombus. Find the length of the longest diagonal of the rhombus.
4.In a rhombus ABCD, the diagonals intersect each other at O.If ∠ A = 60° and OA = 2 cm, then the side of the rhombus is =
5.In a rhombus PQRS,the diagonals intersect at O.Given that, ∠ P = 120° and OP = 3 cm. What is the side of the rhombus is =
6.P is an interior point of quadrilateral ABCD and B = 3.5 cm,BC = 4 cm,CD = 4.8 cm and AD = 3.7 cm.Then the possible value of ( AP+BP+CP+DP) =
7. MN and PS are 2 equal chords of a circle drawn on the either side of centre O of the circle. Both the chords are produced to meet at point A.If the radius of the circle is 10 cm, MN = 12 cm and OA= 17 cm, then NA =
8. In ▲ ABC , if ∠ A < ∠ B < 45°, then ▲ ABC is a/an .................................... triangle.
9. If G is the centroid of ▲ ABC, then the area of ▲ BGC is ............................... thimes the area of the quadrilateral ABCG.

If you can pr
ovide a diagram also, it is good.... but if you can't it's not even a silly problem.. Please answer this question.I am trying for many Days ( nearly about 4 days.) Help me out.. ♣♣(Bow)
• Jan 14th 2013, 08:49 PM
Soroban
Re: Help me out these easy geometry problems.?
Hello, kohila!

Here is some help . . .

Quote:

3. A circle is passing through three vertices of a rhombus of side 8cm
and its centre is the forth vertex of the rhombus.
Find the length of the longest diagonal of the rhombus.

We have rhombus $\displaystyle ABCO$.
The center of the circle is $\displaystyle O.$
Since $\displaystyle OA = OB = OC$, the rhombus
. . is made of two equilateral triangles.
Hence: $\displaystyle \angle AOC = 120^o.$
Code:

                  B                   *           8  *  :  *  8             *    :    *         *        :        *     A *  *  *  *  *  *  *  *  * C         *        :        *             *    :    *           8  *  :  *  8                   *                   O
Law of Cosines: .$\displaystyle AC^2 \:=\:8^2 + 8^2 - 2(8)(8)\cos120^o$

. . . . . . . . . . . . $\displaystyle AC^2\:=\: 64 + 64 - 128(\text{-}\tfrac{1}{2}) \:=\:192$

Therefore: .$\displaystyle AC \:=\:\sqrt{192} \:=\:8\sqrt{3}$

Quote:

4. In a rhombus ABCD, the diagonals intersect each other at O.
If ∠ A = 60° and OA = 2 cm, find the length of a side of the rhombus.

Once again, the rhombus is comprised of two equilateral triangles.
Code:

                  B                   *               *  :  *             *    :    *         *    2  :        *     A *  -  -  -  +  -  -  -  * C         *        :O      *             *    :    *           8  *  :  *  8                   *                   D
The altitude of the equilateral triangle is 2.
Code:

            *           *|*           * | * x         *  |2 *         *  |  *       *    |    *       * * * * * * *               x/2
Pythagorus: .$\displaystyle x^2 \;=\;\left(\tfrac{x}{2}\right)^2 + 2^2 \quad\Rightarrow\quad x^2 \:=\:\tfrac{x^2}{4} + 4$

. . . . . . . . . $\displaystyle \tfrac{3}{4}x^2 \:=\:4 \quad\Rightarrow\quad x^2 \:=\:\tfrac{16}{3}$

. . . . . . . . . . $\displaystyle x \:=\:\sqrt{\frac{16}{3}} \:=\:\frac{4}{\sqrt{3}} \:=\:\frac{4\sqrt{3}}{3}$
• Jan 16th 2013, 04:19 AM
kohila
Re: Help me out these easy geometry problems.?
thank you.. you can answer all but y u delayed.