SSC Papers for Practice Geometry Maharashtra Board
Q.1)
Solve any Five (5)
1) ∆ABC ~ ∆PQR. If AB : PQ = 4:5. Find BC:QR
2) ∆PQR ∠P= ∠R = 45o. ∠Q = 90o
PQ = 7. Find PR
3) In ABC PQ ||
BC. If AP:PB = 3:4 find AQ:QC
4) Find slope of
line having inclination of 45o.
5) Draw a
segment BC of length 5.5 cm and bisect it.
6) Find slope of
point (4,1) and (3,7)
Q.2)
Solve any Four (8)
1) In the figure ray YM is the bisector of ∠XYZ, where XY = YZ. Find relation between XM and
MZ..
2) Check whether
9,40,41 are sides of right angled triangle.
3) In ∆PQR ∠P = 30o, ∠Q = 60o, ∠R = 90o
and PQ = 10. Then find PR and QR
4) Check whether following points are collinear P(-7,8), Q (-5,2) and R (3,6)
5) Draw a circle
of radius 3.4cm. Take a point K on the circle and draw tangent using centre.
6) Find equation
of line passing through P(0,6) and m = 6
7
Q.3.
Solve any Three (9)
1) Adjacent sides of a
parallelogram are 11cm and 17cm. if the length of one of its diagonal is 26cm.
find length of the other diagonal.
2) Prove Basic
Proportionality Theorem.
3) Construct the circumcircle of ∆SIM in which SI = 6.5 cm., ∠I = 1250,
IM = 4.4 cm.
4) Draw a circle of radius 5.5 cm.
Take ‘any point K on it. Draw a tangent to the circle at K without using centre
of the circle.
5) Find k if
(-3,11)(6,2) and (k,4) are collinear
Q.4.
Solve any Two (8)
1) Find equation of line passing through (-3,-5)
and parallel to x -2y = 7
2) Draw a tangent to the circle from the point L with radius 2.8 cm.
Point ‘L’ is at a distance 5 cm from the centre ‘M’.
3) In ABC DE||BC. If 2A(∆ADE) = A(□DBCE) find AB:AD and show that BC = √3 × DE
Q.5.
Solve any Two (10)
1) The bisector of interior LA of
ΔABC meets BC in D. the bisector of
exterior LA meets BC produced in E.
prove that BD = CD
BE CE
2) From the information given in the
figure. Show that PM = PN = √3a, where QR =a.
3) ∆AMT ~ ∆AHE where MA = 6.3cm, AT = 4.9cm, LMAT = 120o and MA = 7
HA 5