SAS Similarity
Trending Questions
If two sides of a triangle are proportional to two sides of another triangle then the triangles are similar.
True
False
In the given figure, PSSQ= PTTR and∠PST= ∠PRQ, then ΔPQR is _______triangle.
an Isosceles
a Right angled
a Scalene
an Equilateral
- ΔPSQ∼ΔTQR
- ΔQPS∼ΔTQR
- QSP∼ΔTQR
- ΔPQS∼ΔTQR
In the adjoining figure, two triangles FGH and QPR are:
Similar using AA criteria
Similar using SAS criteria
Not similar
None of these
It is given that ΔABC≅ΔRPQ. It is true to say that BC = QR ? why?
In the figure, PQ || AB, CQ = 6 cm, CB = 8 cm. If CP = 7 cm, then what is the length of AC?
- 8 cm
- 9 cm
- 254 cm
- 283 cm
- ABEB=BCED
- ABEB=ACED
- ΔABC∼ΔEBD
- ΔBAC∼ΔEBD
if ∠ZXY = 25° ?
- 10°
- 25°
- 15°
- 35°
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP. show that:
(ii)ar(RQC)=38ar(ABC)
A(△ABC):A(△BCD)
In the adjoining figure, the lines l, m and n are parallel to each other, and G is mid-point of CD. Calculate:
(i) BG if AD = 6 cm
(ii) CF if GE = 2.3 cm
(iii) AB if BC = 2.4 cm
(iv) ED if FD = 4.4 cm
In an isosceles ΔABC, the base AB has produced both ways in P and Q such that AP×BQ=AC2.
Prove that ΔACP∼ΔBCQ.
(3 Marks)
In the figure, if ∠B=∠Q, then:
△ABC∼△PQR
△ABC∼△RQP
△CBA∼△QPR
△CBA∼△QPR
In the figure below, if the line segment ST is parallel to line segment QR such that PSSQ=PTTR. This provided data is not sufficient to prove that
STQR=PTTR
False
True
In the figure, AC = 10cm, PC = 15cm, PQ = 12cm, find PB.
6cm
7cm
8cm
9cm
Prove that ar(△AOD)=ar(△BOC).
In the following figure, AB = AC = AD, △ABC ≅ △ACD if ___.
BC = CD
BC = AD
BC = AC
CD = AC
In ΔABC, AB=BC=6cm. If a circle is drawn with center at B and radius 2 cm which intersects AB and BC at E and D, then ΔABC∼ΔEBD.
True
False
- True
- False
- SSS
- SAS
- AAA
- None of these
In the figure, PQ || AB, CQ = 6 cm, CB = 8 cm. If CP = 7 cm, then what is the length of AC?
- 8 cm
- 9 cm
- 254 cm
- 283 cm
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:
(v) ar(BFE) = 2ar(FED)
- 1:2
- 3:5
- 3:2
- 3:8
Prove that ar(△AOD)=ar(△BOC).