CBSE Set Qa5 Maths Class X 2000 Sample Test Papers For Class 10th for students online

Q28) Prove that ratio of area of two similar triangles is equal to the ratio of squares of corresponding sides. Using the above do the following :-
In fig ABC and PQR are isosceles in which A = P
If ar(ABC)/ar(PQR) = 9/16, find AD/PS (Marks 6)
 
Ans28)
 
Given ABC ~ DEF
To prove ar (ABC)/ ar (DEF) = AB²/DE² = BC²/EF² = AC²/DF²
Const. AG  |  BC, DH  |  EF
Proof  ar (ABC)/ar (DEF) = (1/2 x BC x AG)/(1/2 x EF x DH)  - (i)   (ar. of = 1/2 x base x altitude)
In AGB and DHE
B = E  ( ABC ~ DEF)
AGB = DHE  ( = 90^o)
= AGB ~ DHE   (A.A.)
= AG/DH = AB/DE  (ii)   (corresponding sides proportional)
Also AB/DE = BC/EF  -(iii)   (ABC ~ DEF)
AG/DH = BC/EF  -(iv)   (from (ii) & (iii))
ar (ABC)/ar (DEF) = BC²/EF²    (from (i) & (iv))
Similarly
ar (ABC)/ar (DEF) = AB²/DE² = AC²/DF²
ar (ABC)/ar (DEF) = AB²/DE² = BC²/EF² = AC²/DF²
(ii)
Now in given fig
A = P (given)
AB/AC = PQ/PR = 1/1  (given)
= AB/PQ = AC/PQ
= ABC ~ PQR  (SAS)
= ar (ABC)/ar (PQR) = AD²/PS²
9/16 = AD²/PS²   (^..^. ar (ABC)/ar (PQR) = 9/16)
= 3/4 = AD/PS

Q29) Prove that sum of either pair of opposite angles of a cyclic quadrilateral is 180^o. Using the above solve the following:

In fig POQ is a diameter and PQRS is a cyclic quadrilateral. If PSR = 150^o, find RPQ.  (Marks 6)
Ans29)  

Given A cyclic quadrilateral ABCD in a circle with centre O
To prove A + C = 180^o
B + D = 180^o
Const. Join OB, OC
Proof. BCD subtends 1 at the centre  (degree measure of an arc is twice the angle subtended
and BAD on the circumference            by it at any point in the alternate segment w.r.t. that arc)
= BAD = 1/21  -(i)
BAD subtends 2 (reflex 1)  at the centre  (degree measure of an arc is twice the angle subtended
and BCD on the circumference            by it at any point in the alternate segment w.r.t. that arc)
= BCD = 1/2(reflex 1)  -(ii)
BAD + BCD = 1/2(1 + reflex 1)   (adding (i) & (ii))
= 1/2 x 360^o   (angles at a point)
= 180^o
= A + C = 180^o
Similarly B + D = 180^o
(ii) In the given figure
PQR = 180^o - 150^o = 30^o (using the above theorem)
PRQ = 90^o   ( in a semi circle)
RPQ = 180^o - (90^o + 30^o) = 60^o  (angle sum property)

Q30) A man on the roof of a home, which is 10 m high, observes the angle of elevation of the top of a building as 42^o and angle of depression of the base of the building as 40^o. Find the height of the building and its distance from the home.  (Marks 6)
Ans30)  

BCE
cot 40^o = CE/BE
= CE = 10 x 1.1918
= 11.918 m
ACE
tan 42^o = AE/CE
= 0.9004 = (h - 10)/11.918
= h - 10 = 10.73
= h = 20.73
.^.. height of the building = 20.73 m
Distance of the building from the home = 11.918 m