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\(D \) is a point of the side \(BC \) of a triangle ABC such that \(\angle ADC=\angle BAC \). Show that \(CA^2=CB.CD. \)
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In and
(common angle)
(given)
(Corresponding sides are proportional)
\(S \) and \( T\) are points on sides \(PR \) and \(QR \) of \(\triangle PQR \) such that \(\angle P=\angle RTS \). Show that \(\triangle RPQ\sim\triangle RTS \).
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In and
(given)
(given)
(AA similarly)
In the given figure, AB||QR. Find the length of PB.
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(AA Similarly)
In the given figure, if AB||CD, find the value of x.
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In and
(alternate angle)
(alternate angle)
(AA)
In the given figure, \(\angle ABC=90^o \) and \(BD\bot AC \). If \(BD=8\ cm \) and \(AD=4\ cm \), find \(CD \).
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Let
Also
In and
(each
)
(each 90-x)
(AA)
In a right angled triangle with sides a and b and hypotenuse c, the altitude drawn on the hypotenuse is x. Prove that ab=cx.
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In and
(common)
(each 90)
(AA Similarly)
In the given figure, \(\dfrac{AD}{DB}=\dfrac{AE}{EC} \) and \(\angle ADE=\angle ACB \). Prove that \(\triangle ABC \) is an isosceles triangle.
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(converse of BPT)
(corresponding angles) ….(1)
…(2)
From (1) and (2)
(opposite sides of opposite angle are equal)
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that \(\mathrm{\triangle ABE\sim\triangle CFB} \).
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(opposite angles of a parallelogram are equal) …(1)
(alternate angles) …(2)
In and
(From (1))
(From (2))
P and Q are points on sides AB and AC respectively of \(\triangle \)ABC . If AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ.
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(SS)
In a triangle ABC, altitudes AL and BM intersect in O. Prove that \(\dfrac{AO}{BO}=\dfrac{OM}{OL} \).
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In and
(AA)
A wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
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ABCD is a trapezium with AB||DC. If E and F are on non-parallel sides AD and BC respectively such that EF is parallel to AB, then show that \(\dfrac{AE}{ED}=\dfrac{BF}{FC} \).
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and
So
In …(1)
In …(2)
From (1) and (2)
In the given figure, OC.OD = OA.OB. Show that \(\angle \)A = \(\angle \)C and \(\angle \)B = \(\angle \)D.
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…(1)
In and
…(1)
(VOA)
(SAS)
( If two triangles are similar their corresponding angles are equal)
An aeroplane leaves an airport and flies due north at a speed of 500 km per hour. At the same time, another aeroplane leaves the same airport and flies due to west at a speed of 1000 km per hour. How far apart will be the two planes after 2 hours?
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(approx)
In the figure, A, B and C are points on OP,OQ and OR respectively such that AB||PQ and AC||PR. Show that BC||QR.
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(using BPT) ….(1)
….(2) (using BPT)
From (1) and (2)
(converse of BPT)
In the figure, two triangles ABC and DBC are on the same base BC in which \(\angle A=\angle D=90^o \). If CA and BD meet each other at E show that \(\mathrm{AE\times CE=BE\times DE} \).
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In and
(VOA)
(AA)
In the figure, PQR and SQR are two triangles on the same base QR. If PS intersect QR at O, then show that : \(\dfrac{ar\ (PQR)}{ar\ (SQR)}=\dfrac{PO}{SO} \).
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….(1)
In and
(AA)
….(2)
From (1) and (2)
In the figure, \(O \) is a point inside \(\triangle PQR \) such that \(\angle POR=90^o\ OP=6\ cm \) and \(OR=8\ cm \). If \(PR=24\ cm,QR=26\ cm \), prove that \(\triangle QPR \) is a right angled triangle.
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In
(by converse of pythagoras theorem)
Hence is a height angled triangle.
Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.
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(AAA similar condition)
In the figure, DE || AC and DF || AE. Prove that \(\dfrac{EF}{BF}=\dfrac{EC}{BE} \).
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In
….(1)
In
….(2)
From (1) and (2)
(Proved)
In the figure, if PQ || CB and PR || CD, prove that \(\dfrac{AR}{AD}=\dfrac{AQ}{AB} \)
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In
….(1) (by basic proportionality theorem)
In
…(2)
From (1) and (2)
In the figure, PM = 6 cm, MR = 8 cm and QR = 26 cm, find the length of PQ.
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In the figure, \(ABCD \) is a rhombus. Prove that \(4AB^2=AC^2+BD^2 \).
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In
In the figure, D,E,F, are mid-point of sides BC, CA, AB respectively of \(\triangle \)ABC. Find the ratio of area of \(\triangle \)DEF to area of \(\triangle \)ABC.
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We know that the line segment joining the mid points of any two sides of a triangle is half the third side and parallel to it.
In and
….(1)
(SSS)
If one diagonal of a trapezium divides the other diagonal in the ratio 1 : 2, prove that one of the parallel sides is double the other.
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In and
(VOA)
(alt
)
(AA)
In the figure DE || BC. Find x.
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In the figure, \(AB||DE \) and \(BD||EF \). Prove that \(DC^2=CF\times AC \).
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In
(By BPT) …(1)
In
(By BPT) …(2)
From (1) and (2)
If \(D,E \) are points on the sides \(AB \) and \(AC \) of a \(\triangle ABC \) such that \(AD=6\ cm,\ BD=9\ cm,\ AE=8\ cm\ EC=12\ cm \). Prove that \(DE||BC \).
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By converse of BPT
In the figure, DE || OQ and DF || OR. Show that EF || OR.
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In
….(1)
In
….(2)
From (1) and (2)
In the figure \(\angle ACB=90^o \) and \(CD\bot AB \). Prove that \(\dfrac{BC^2}{AC^2}=\dfrac{BD}{AD} \).
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If a perpendicular is drawn from the vertex of the right angle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
..(1)
Also
….(2)
Dividing eqn (2) by (1)
(Proved)
In the figure, \(XN || CA\) and \(XM || BA.\ T\) is a point on \(CB\) produced. Prove that \(TX^2=TB.TC. \)
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In
(By BPT) ….(1)
In
(by BPT) ….(2)
From (1) and (2)
In the figure. \(AB\bot BC,\ DE\ \bot\ AC \) and \(GF\ \bot\ BC \). Prove that \(\triangle ADE\sim\triangle GCF \).
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In and
(AA)
In and
(AA)
So
From the given figure, find \(\angle \)MLN.
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(SSS)
In a quadrilateral ABCD, \(\angle B=90^o \). If \(AD^2=AB^2+BC^2+CD^2 \), prove that \( \angle ACD=90^o\).
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(given)
But (by pythagoras theorem)
By converse of pythagoras theorem
\(ABC \) is an isosceles triangle with \(AC=BC\). If \(AB^2=2AC^2 \), prove that \(ABC \) is a right angled triangle.
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So is the hypotenuse
Hence is a right angled triangle
D,E and F are respectively the mid-point of sides AB, BC and CA of \(\triangle \)ABC . Determine the ratio of the areas of \(\triangle \)DEF and \(\triangle \)ABC .
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(by mid point theorem)
and
So is a parallelogram
Similarly is a parallelogram
In the given figure, \(PQ=24\ cm,\ QR=26\ cm,\ \angle PAR=90^o,\ PA=6\ cm \) and \(AR=8\ cm \). Find \(\angle QPR \).
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In
Hence by converse of pythagoras theorem
A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point?
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AC is 17m from his starting point