Mathematics Academy

Showing posts with label Similar Triangles. Show all posts

Saturday, October 5, 2019

Pythagoras Theorem-Proof I

पाइथागोरस प्रमेय

किसी समकोण त्रिभुज में कर्ण की लंबाई का वर्ग अन्य दो भुआओं के वर्ग के योग के बराबर होता है। यदि त्रिभुज ABC में कोण B समकोण हो तो AB² + BC² = AC² । इस प्रमेय को सिद्ध करने के लिए हम समरूप त्रिभुज की अवधारणाओं की मदद ले सकते हैं।

त्रिभुज ABC में बिन्दु B से यदि AC पर लंब BD डाला जाए तो

अवलोकन 1 : △ABC ~ △ADB , अत: AB/AC = AD/AB, AB²=AC.AD ... (1)

अवलोकन 2 : △ABC ~ △BDC , अत: AC/BC = BC/DC, BC²=AC.DC … (2)

समीकरण (1) और (2) से AB² + BC² = AC.AD + AC.DC

AB² + BC² = AC.(AD+DC) = AC.AC = AC²

नीचे दिए एपलेट में बिन्दुओं A , B , C की स्थिति को माउस की मदद से बदला जा सकता है और AB , BC तथा AC के अलग-अलग मानों के लिए पाइथागोरस प्रमेय की जाँच की जा सकती है।

Pythagoras Theorem

In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs (other two sides). IF in △ ABC, angle B is right angle then AB² + BC² = AC². For proving this theorem, we will use the concept of similar triangles.

In △ ABC, if we draw a perpendicular BD from point B to side AC then,

Observation 1: △ ABC ~ △ ADB, so AB/AC = AD/AB, AB²=AC.AD ….. (1)

Observation 2: △ ABC ~ △ BDC, so AC/BC = BC/DC, BC²=AC.DC …...(2)

From equation (1) and (2), AB² + BC² = AC.AD + AC.DC

AB² + BC² = AC.(AD+DC) = AC.AC = AC²

In the applet shown below, points A, B, C can be moved with the help of a mouse to see the verification of the Pythagoras Theorem for different values of AB, BC, and AC.

Saturday, August 18, 2012

Practice Questions - Basic Proportionality Theorem

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.

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.” This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Monday, July 30, 2012

Similar Triangles and Medians

If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median of another triangle, then the two triangles are similar.

Two triangles ABC and DEF , in which AP and DM are the medians , such that AB/DE = AC/DF = AP/DM , then ΔABC ∼ ΔDEF