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 DE2 = 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.
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
“ 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.”
∠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)2
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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