Posts

Showing posts with the label Similarity and Pythagoras theorem

In ▢ABCD points P, Q, R and S lies on sides AB, BC, CD and AD respectively such that seg PS is parallel to seg BD parallel to seg QR and seg PQ is parallel to seg SR. Then prove that seg PQ is parallel to seg AC.

Image
Post a Comment
Read more

In △ABC, m∠BAC=90°. Seg DE is perpendicular to side AB, seg DF is perpendicular to AC. Prove that Area of quadrilateral AEDF= √(AE x EB x AFx FC)

Image
Post a Comment
Read more

In △BAC, Angle BAC= 90°, segment AD, seg BE and seg CF are medians. Prove: 2(AD²+BE²+CF²)=3BC²

Image
Post a Comment
Read more

Find the radius of a circle drawn by a compass when angle between two arms of compass is 120° and length of each arm is 24cm.

Image
Post a Comment
Read more

In △PQR, angle PQR= 90°, As shown in figure, seg QS is perpendicular to side PR. Seg QM is angle bisector of angle PQR. Prove that: PM²/MR² = PS/SR.

Image
Post a Comment
Read more

Bisector of angle B and angle C in △ABC meet each other at P. Line AP cuts the side BC at Q. Prove that: AP/PQ=(AB+AC)/BC.

Image
Post a Comment
Read more

Through the midpoint M of the side CD of parallelogram ABCD, the line BM is drawn intersecting AC in L and AD produced in E. Prove that EL=2BL.

Image
Post a Comment
Read more

In Quadrilateral ABCD, M is the midpoint of diagonal AC and N is the midpoint of diagonal BD. Prove that: AB²+BC²+CD²+DA²=AC²+BD²+4MN².

Image
Post a Comment
Read more

△PQR is a right angled at Q such that QR=b and a=A(△PQR). If QN is perpendicular to PR then S.T: QN = 2ab/√(b^4+4a²)

Image
Post a Comment
Read more

In △ABC, angle ABC= 135°. Prove that AC² = AB² + BC² + 4A(△ABC).

Image
Post a Comment
Read more

In the adjoining figure, each of segments PA, QB, RC and SD is perpendicular to line l. If AB=6, BC=9 , CD=12 and PS=36, then determine PQ, QR and RS.

Image
Post a Comment
Read more

In an equilateral triangle ABC, the side BC is trisected at D. Prove that 9AD² = 7AB².

Image
Post a Comment
Read more

In triangle ABC, Angle ACB=90°, seg CD is perpendicular to seg AB, seg DE is perpendicular to seg CD. Show that: CD² x AC = AD x AB x DE.

Image
Post a Comment
Read more

In the adjoining figure, AD is the bisector of the exterior angle A of triangle ABC. Seg AD intersects the side BC produced in D. Prove that: BD/CD=AB/AC.

Image
Post a Comment
Read more

In Quadrilateral ABCD, side BC is parallel to side AD. Side AC and side BD intersect in point Q. If AQ=(1/3)AC then show that, DQ=(1/2)BQ.

Image
Post a Comment
Read more