CBSE Set Qa3 Maths Class X 1999 Sample Test Papers For Class 10th for students online

Q14) Determine the length of an altitude of an equilateral triangle of side 2a cm.  (Marks 2)
Ans14) 

In ADB & ADC ADB = ADC (each = 90o) AB = AC (each = 2a) AD = AD (common) = ADB ~ ADC (R.H.S.) = BD = DC (CPCT) = BD = DC = a (... BC = 2a)

In ADB, D = 90^o
= AB² = AD² + BD² (Pythagoras Theorem)
= (2a)² = AD² + a²
= 3a² = AD²
= 3 a = AD
.^.. AD = 3 a units 

Q15) In the figure, a circle touches all the four sides of a quadrilateral ABCD where sides AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.  (Marks 2)

Ans15) Let the circle touches the side AB, BC, CD and DA at P, Q, R, S respectively.
AP = AS     (Length os tangent
BP = BQ     segment drawn from
DR = DS     an external point to a
CR = CQ     circle are equal)
Adding them
(AP + BP) + (CR + RD) = (BQ + QC) + (DS + SA)
= AB + CD = BC + DA
= 6 + 4 = 7 + AD
= 10 - 7 = AD
= 3 = AD
.^.. AD = 3 cm

Q16) Solve graphically:
2x - y = 1, x + 2y = 8. Also, find the co-ordinates of the points where the line meet the axis of y.  (Marks 4)
Ans16)  

Solution (2, 3)
Line 2x - y = 1 meets y - axis at (0, -1)
Line x + 2y = 8 meets y - axis at (0, 4)

Q17) Simplify :- (2x² + x - 3)/(x - 1)² (2x² + 5x - 3)/(x² - 1)  (Marks 4)
Ans17) (2x² + x - 1)/(x - 1)² (2x² + 5x - 3)/(x² - 1)  
= (2x² + x - 1)/(x - 1)² x (x² - 1)/(2x² + 5x - 3)
=  (2x + 3)(x - 1) / (x - 1)(x - 1) x (x + 1)(x - 1) / (2x + 3)(x + 1)
= 1/1
= 1 

Q18) Cyclic factorization. Deleted from the syllabus.  (Marks 4)

Q19) The sum of radius of base and the height of a solid cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 cm², find the volume of the cylinder ( = 22/7).  (Marks 4)
Ans19) Let the radius be x
height be h
.^.. r + h = 37 (given)
= h = 37 - r
Total surface area of the cylinder = 1628
= 2r(r + h) = 1628
= 2 x 22/7 x r (r + 37 - r) = 1628
= r = 1628 x 7 / 2 x 22 x 37
= r = 7 cm
.^.. h = 37 - r
= 37 - 7
= 30 cm
Volume of the cylinder = r²l
= 22/7 x 7 x 7 x 30
= 4620 cm³

Q20) A plane left 30 minutes late than the scheduled time and in order to reach the destination 1500 km away in time, it has to increase the speed by 250 km/hr from the usual speed. Find the usual time.  (Marks 4)
Ans20) Let the usual speed of plane = x km/hr
The increased speed of the plane = (x + 250) km/hr
Distance travelled = 1500 km
According to the question
1500/x - 1500/(x + 250) = 1/2
= (1500x + 37500 - 1500x)/(x² + 250x) = 1/2
= x² + 250x = 750000
= x² + 250x - 750000 = 0
= (x + 1000)(x - 750) = 0
= x = -1000 or x = 750
Speed cannot be -ve
.^.. usual speed of the plane is 750 km/hr

Q21) The arc of a circle subtending a right angle at any point of the circle in its alternate segment is a semicircle - prove.  (Marks 4)
Ans21)  

Given. Circle with centre O
AB is any acr. C is any point in the alternate segment with respect to that arc such that ACB = 90^o
To find. AB is a semi-circular arc.
Const. Join OA and OB
Proof. arc (AB) = 2 ACB  (Degree measure of an arc of a circle is twice the angle subtended by it at any point in the alternate segment with respect to that arc)
= arc (AB) = 2 x 90^o  (.^.. ACB = 90^o)
= 180^o
= AB is a semi-circular arc   ( Definition of semi-circular arc)

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