Geometry and Mensuration: Level 2 Test 2

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Geometry and Mensuration: Level 2 Test 2

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Question 1

If the lengths of the sides of a triangle are in the ratio 4: 5: 6 and the in radius of the triangle is 3 cm, then the altitude of the triangle corresponding to the largest side as base is:

A	7.5 cm
B	6 cm
C	10 cm
D	8 cm

Question 1 Explanation:

Let the side of the triangle be 4x, 5x and 6x respectively. Semi-perimeter (s) =7.5x
The area = 3 x 7.5 sq. cm.
Thus, ½ X6x = 3 x 7.5
=> x= 7.5 cm

Question 2

ABC is an isosceles right angled triangle with ∠B= 90^o. On the sides AC and AB, two equilateral triangles ACD and ABE have been constructed. The ratio of areas of ABE and ACD is

A	1: 3
B	2: 3
C	1: 2
D	1: √2

Question 2 Explanation:

The ratio of the hypotenuse AC to AB = 1:√2
Since both the triangles are equilateral their area are in the ratio
= 1²:(√2)² =1:2

Question 3

Length of the floor of a rectangular auditorium is 6 metre more than the radius of a circle with a circumference of 572 m. The perimeter of the floor of the rectangular auditorium is 356 m. What will be cost of flooring the auditorium (only the floor of the auditorium), if the cost of flooring is Rs.121m²?

A	Rs.87, 954
B	Rs.91, 236
C	Rs.94, 284
D	Rs.75, 490

Question 3 Explanation:

The radius = 572/2Π = 91 m
The length = 91+6 = 97 cm
The breadth = 356/2 – 91 = 81 m
Since the cost of flooring = Rs. 12 / m²the total cost = Rs.97x81x12 = 94,284

Question 4

O and C are respectively the orthocenter and circumcentre of an acute-angled triangle PQR. The points P and O are joined and produced to meet the side QR at S. If PQS= 60^o and QCR= 130^o, then RPS=

A	30^o
B	35^o
C	100^o
D	60^o

Question 4 Explanation:

∠PQS= 60^o
∠QCR = 130 , QC= CR .
Thus ∠CQR= ∠QRC = (180 -130)/2 =2
PQC = 60 -25 = 35 , QPC = 35
Thus,
∠CPR+∠PRC + 60+35+25 = 180
Thus ∠RPS= 30 +∠CPS
Now, ∠QPS =90-60 = 30.
Thus ∠CPS = 35-30 = 5.
∠RPS = 35.
Correct option is (b)

Question 5

ABC is a triangle, The bisectors of the internal angle B and external angle C intersect at D. If BDC= 50^o, then ∠A is