## Friday, August 31, 2012

## Wednesday, August 29, 2012

## Tuesday, August 28, 2012

## Sunday, August 26, 2012

### Construction of Circle

**To construct a circle with a given radius and touching the two given intersecting lines**

- Draw two lines AB and AC at some angle say α.
- Draw bisector of ∠BAC.
- Construct a line parallel to line AB at a distance R (radius of the circle to be constructed(http://mathematicsbhilai.blogspot.in/2012/08/construction-of-parallel-line.html). Let this line intersect the angle bisector at point O.
- Draw a circle with center O and Radius R. This is the required circle.

## Saturday, August 25, 2012

### Construction of Parallel Line

Given line AB and point P, not on AB. Construct a line through P, parallel to AB.

- Draw a line through P, so that it intersects line AB at point D.
- Open the compass to a comfortable distance, set the compass at D, and make an arc so that it intersects DP at E and line AB at F.
- Using the same compass opening, make an identical arc with the compass set at P, so that the arc intersects DP at G.
- Set the compass at E, and make an arc so that it intersects line AB at point F.
- Using the same compass opening, make an identical arc with the compass set at G so that it intersects the arc you made in step 3 at point H.
- Draw a line through P and H , line PH is parallel to AB.

## Thursday, August 23, 2012

## Wednesday, August 22, 2012

## Monday, August 20, 2012

### Appolonius Theorem

In a triangle , the sum of the square of two sides of a triangle is equal to twice the sum of the square of the median which bisects the third side and the square of half the third side. In triangle ABC , if AD is a median then AB

^{2}+ AC^{2}= 2(AD^{2}+DC^{2}) or 2(AD^{2}+BD^{2})## Saturday, August 18, 2012

## Thursday, August 16, 2012

## Wednesday, August 15, 2012

## Monday, August 13, 2012

## Sunday, August 12, 2012

## Saturday, August 11, 2012

## Wednesday, August 8, 2012

## Tuesday, August 7, 2012

## Sunday, August 5, 2012

### Practice Questions - Similar Triangles

1. In the following figure , DEFG is a square and ∠BAC = 90° .Prove that DE

2. In the following figure , D divides AB such that AD : DB = 3 :2. E is a point on BC such that DE || AC. Find the ratio of the areas of a) ΔABC and ΔBDE b) Trapezium ACED and ΔBED

3. In the following figure , DE || BC and AD : DB = 5 : 4 , find area(ΔDEF)/area(ΔCFB)

4. There is a stair case as shown in the following figure. Measurements of steps are marked in the figure. Find the straight line distance between A and B.

5. A right triangle has hypotenuse of length p cm and one side of length q cm. If p-q = 1, express the length of third side of the right triangle in terms of p.

6. By using the Pythagoras Theorem , calculate ar(ΔPQR) from the following figure.

7. Equilateral triangles are drawn on the sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.

8. If two triangles are equiangular , prove that the ratio of the corresponding sides is same as the ratio of the corresponding medians.

9. If two triangles are equiangular , prove that the ratio of the corresponding sides is same as the ratio of the corresponding angle bisector segments.

10. If two triangles are equiangular , prove that the ratio of the corresponding sides is same as the ratio of the corresponding altitudes.

^{2}= BD x EC.2. In the following figure , D divides AB such that AD : DB = 3 :2. E is a point on BC such that DE || AC. Find the ratio of the areas of a) ΔABC and ΔBDE b) Trapezium ACED and ΔBED

3. In the following figure , DE || BC and AD : DB = 5 : 4 , find area(ΔDEF)/area(ΔCFB)

4. There is a stair case as shown in the following figure. Measurements of steps are marked in the figure. Find the straight line distance between A and B.

5. A right triangle has hypotenuse of length p cm and one side of length q cm. If p-q = 1, express the length of third side of the right triangle in terms of p.

6. By using the Pythagoras Theorem , calculate ar(ΔPQR) from the following figure.

7. Equilateral triangles are drawn on the sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.

8. If two triangles are equiangular , prove that the ratio of the corresponding sides is same as the ratio of the corresponding medians.

9. If two triangles are equiangular , prove that the ratio of the corresponding sides is same as the ratio of the corresponding angle bisector segments.

10. If two triangles are equiangular , prove that the ratio of the corresponding sides is same as the ratio of the corresponding altitudes.

## Friday, August 3, 2012

### Right Similar Triangles

In the following applet , let triangle PQR be a right triangle, right angled at Q. Let QS be the perpendicular to the hypotenuse PR.
From ΔPSQ and ΔPQR, we have ∠P = ∠P,

∠PSQ=∠PQR(both 90°) ,

so ΔPSQ ∼ ΔPQR ( By AA Criteria)

Similarly , ΔQSR ∼ ΔPQR.

So , ΔPSQ ∼ ΔQSR , thus we can say that

∠PSQ=∠PQR(both 90°) ,

so ΔPSQ ∼ ΔPQR ( By AA Criteria)

Similarly , ΔQSR ∼ ΔPQR.

So , ΔPSQ ∼ ΔQSR , thus we can say that

*“ If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.”*## Wednesday, August 1, 2012

### Similar Triangles , Area and Perimeter

If two triangles ABC and PQR are similar (ΔABC ∼ ΔPQR) , then their corresponding sides are proportional and corresponding angles are equal i.e. PQ/AB = QR/BC = RP/CA.

The ratio of areas of similar triangles is the square of the ratio of their sides i.e. Area PQR / Area ABC = (PQ/AB)

The ratio of perimeters of similar triangles is the ratio of their sides i.e.

Perimeter PQR / Perimeter ABC = PQ/AB

The ratio of areas of similar triangles is the square of the ratio of their sides i.e. Area PQR / Area ABC = (PQ/AB)

^{2}The ratio of perimeters of similar triangles is the ratio of their sides i.e.

Perimeter PQR / Perimeter ABC = PQ/AB

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