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In the figure, if \(\dfrac{AE}{EB}=\dfrac{AF}{FC} \), then we can conclude that :
If then using converse of basic proportionality theorem EF  BC
In the triangle ABC,DE  BC, then the length of DB is :
(approx)
In \(\triangle \)ABC, if DE  BC, then the value of x is :
In the trapezium ABCD, ABCD, then the value of x is :
In the given figure, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm, then :
EF  QR (by converse of BPT)
In the given figure, if \(\triangle ADE\sim\triangle ABC \), then \(BC \) is equal to:
In the given figure, \(\triangle ACB\sim\triangle APQ \). If \(BC=8\ cm \). \(PQ=4\ cm.\ BA=6.5\ cm \) and \(AP=2.8\ cm \), then the length of \( AQ\) is
If \(\triangle ABC\sim\triangle PQR \) and \(\angle P=50^o,\ \angle B=60^o \), then \( \angle R\) is:
\(\triangle ABC\sim\triangle DEF \) and be perimeters of \(\triangle ABC \) and \(\triangle DEF \) are \(30\ cm \) and \(18\ cm \) respectively. If \(BC=9\ cm\), then \(EF\) is equal to :
*** QuickLaTeX cannot compile formula: \mathrm{\dfrac{Perimeter\ of\ \triangle ABC}{Perimeter\ of\ \triangle DEF}=\dfrac{30}{18}=\dfrac{BC}{DE}\\ \\ \Rightarrow \dfrac{30}{18}=\dfrac{9}{DE}\\ \\ \Rightarrow DE=\dfrac{9\times18}{30}=5.4\ cm *** Error message: File ended while scanning use of \math@egroup. Emergency stop.
\(\triangle ABC\sim\triangle DEF \) such that \(AB=9.1\ cm \) and \(DE=6.5\ cm \). If the perimeter of \(\triangle DEF \) is \(25\ cm \), then perimeter of \(\triangle ABC \) is :
If \(\triangle ABC\sim\triangle EDF \) and \(\triangle ABC \) is not similar to \(\triangle DEF \), then which of the following is not true ?
If
or
or
If in two triangles ABC and PQR. \(\dfrac{AB}{QR}=\dfrac{BC}{PR}=\dfrac{CA}{PQ} \), then
If
then
In the given figure, two line segments, \(AC \) and \(BD \) intersect each other at the point \(P \) such that \(PA=6\ cm,\ PB=3\ cm,\ PC=2.5\ cm,\ PD=5\ cm,\angle APB=50^o\) and \( \angle CDP=30^o\). Then \(\angle PBA \) is equal to
(given)
(VOA)
(SAS)
The areas of two similar triangles are \(169\ cm^2\) and \(121\ cm^2\), if the longest side of the larger triangle is \(26\ cm\), then the longest side of the other triangle is :
In the following trapezium \(ABCD,\ ABCD \) and \(CD=2\ AB \). If area \((\triangle AOB)=84\ cm^2 \), then area \((\triangle COD) \) is:
If \(\triangle ABC\sim\triangle PQR \) area \((\triangle ABC)=80\ cm^2 \) and area \( (\triangle PQR)=245\ cm^2\), then \(\dfrac{AB}{PQ} \) is equal to:
In the similar triangles, \(\triangle ABC \) and \(\triangle DEF,\dfrac{ar(\triangle ABC)}{ar(\triangle DEF)}=\dfrac{3}{4} \). If the median \(AL=6\ cm \), then the median \(DM \) of \( \triangle DEF\) is:
In the figure, if AC = DE, then the value of EB is :
In the quadrilateral \(ABCD \), if \(\angle B=90^o \) and \(\angle ACD=90^o \) then \(AD^2 \) is
If diagonals of a rhombus are 12 cm and 16 cm, then the perimeter of the rhombus is :
Side
Perimeter
In triangle \(ABC \) and \(DEF,\ \angle A\ \ne\ \angle C,\angle B=\angle E,\angle F=\angle C \) and \(AB=EF \). Then , the two triangle are:
In triangle and then two triangle are similar but not congruent.
If \(\triangle ABC\sim\triangle QRP,\dfrac{ar(\triangle ABC)}{ar(\triangle PQR)}=\dfrac{9}{4} \), perimeter \(\triangle ABC=48\ cm,\ AB=18\ cm \) and \(BC=18\ cm \), then \(PQ \) is equal to:
D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD=2 cm, BD=4 cm, BC=9 cm and DEBC. Then, length of DE (in cm) is :
It is given that \(\triangle ABC\sim\triangle PQR \) with \(\dfrac{BC}{QR}=\dfrac{1}{4} \), then \(\dfrac{ar(PRQ)}{ar(BCA)} \) is equal to:
The lengths of the diagonals of a rhombus are 24 cm and 32 cm. The perimeter of the rhombus is :
Side
Perimeter
\(\triangle ABC\sim \triangle PQR,\ M \) is the midpoint of \(BC \) and \(N \) is the midpoint of \(QR \). If the area of \(\triangle ABC =100\ sq.\ cm \), the area of \(\triangle PQR=144\ sq.\ cm \) and \(AM=4\ cm \) then \(PN \) is:
\(\triangle ABC \) is such that \(AB=3\ cm,\ BC=2\ cm \) and \(CA=2.5\ cm \). If \(\triangle DEF\sim\triangle ABC \) and \(EF=4\ cm \), then perimeter of \(\triangle DEF \) is:
Perimeter of
A vertical stick 30 m long casts a shadow 15 m long on the ground. At the same time, a tower casts a shadow 75 m long on the ground. The height of the tower is :
Height
\(\triangle ABC \) and \(\triangle PQR \) are similar triangles such that \(\angle A=32^o \) and \(\angle R=65^o \), then \(\angle B \) is:
In the figure \(\triangle ABC\sim\triangle PQR \), then \(y+z \) is:
In an isosceles \(\triangle ABC \), if \(AC=BC \) and \(AB^2=2AC^2 \), then \( \angle C\) is equal to:
The areas of two similar triangles \(ABC \) and \(PQR \) are \(25\ cm^2\) and \(49\ cm^2\) respectively. If \(QR=9.8\ cm\), then \(BC \) is :
If the ratio of the corresponding sides of two similar triangles is 2 : 3, then the ratio of their corresponding altitude is :
In two similar triangles, if the corresponding sides are in a particular ratio then altitude will also be in the same ratio.
In the figure PQBC and AP : PB = 1 : 2. Find \(\dfrac{ar(\triangle APQ)}{ar(\triangle ABC)} \):