Geometry Test

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Geometry Test - 1

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Question 1
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A vertical stick 20 m long casts a shadow 10 m long on the ground. At the same time, a tower casts a shadow 50 m long on the ground. Find the height of the tower.

A
80 m

B
100 m

C
25 m

D
200 m
Solution
When the length of stick = 20 m, then length of shadow = 10 m i.e. in this case length =2 * shadow
With the same angle of inclination of the sun, the length of the tower that casts a shadow of 50 m is: 2 * 50m = 100m
⇒Height of tower = 100 m
Question 2
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The area of similar triangles, ABC and DEF are 144cm² and 81 cm² respectively. If the longest side of the larger △ABC be 36 cm, then the longest side of the smaller △DEF is:

A
27 cm

B
26 cm

C
29 cm

D
30 cm

Solution
Question 3
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Find the value of x in the figure, if it is given that AC and BD are diameters of the circle.

A
60 °

B
30 °

C
15 °

D
45 °
Solution
The triangle BOC is an isosceles triangle with sides OB and OC both being equal as they are the radii of the circle. Hence, the angle OBC = angle OCB = 30 °.
Hence, the third angle of the triangle BOC i.e. Angle BOC would be equal to 120 °.
⇒BOC = AOD = 120 °
Also, in the isosceles triangle DOA:
Angle ODA = Angle DAO = x = 30 °
Question 4
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Find the value of x in the given figure.

A
16 cm

B
7 cm

C
12 cm

D
9 cm

Solution

By the rule of tangents, we get:
⇒12² = (x + 7)x
⇒144 = x² + 7x
⇒ x² + 7x –144 = 0
⇒ x² +16x –9x –144 = 0
⇒ x(x + 16) –9(x + 16) = 0
⇒ x = 9 or –16
–16 can ’t be the length, hence this value is discarded. Thus,x = 9
Question 5
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Find the value of x in the given figure.

A
16 cm

B
12 cm

C
13 cm

D
15 cm

Solution

By the rule of chords, cutting externally, we get:
(9 + 6) * 6 = (5 + x) * 5
90 = 25 + 5x
5x = 65
x = 13 cm
Question 6
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AB is the diameter of the circle and ∠PAB=40 ∘
what is the value of ∠PCA?

A
50 ∘

B
55 °

C
70 °

D
45 °
Solution
In △PAB
⇒  ∠PAB=40o      [ Given ]
⇒  ∠BPA=90o    [ angle inscribed in a semi-circle ]
⇒  ∠PAB+∠PBA+∠BPA=180o
∴  40o+∠PBA+90o=180o
∴  ∠PBA=180o −130o
∴  ∠PBA=50o
⇒  ∠PBA=∠PCA=50o    [ angles inscribed in a same arc PA ]
∴  ∠PCA=50o
Question 7
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In the figure, AB is parallel to CD and RD || SL || TM || AN, and BR : RS : ST : TA = 3 : 5 : 2 : 7. If it is known that CN = 1.333 BR. Find the ratio of BF : FG : GH : HI : IC

A
3 : 5 : 2 : 7 : 4

B
3 : 7 : 2 : 5 : 4

C
4 : 7 : 2 : 5 : 3

D
4 : 5 : 2 : 7 : 3
Solution
Since the lines, AB and CDare parallel to each other, and the lines RD and AN are parallel , it means that the triangles RBF and NCI are similar to each other. Since the ratio ofCN : BR = 1.333 , if we take BR as 3, we will get CN as 4.
This means that the ratio of BF : CI would also be 3 : 4 .
Also, the ratio of BR : RS : ST : TA = BF : FG : GH : HI = 3 : 5 : 2 : 7 (given).
Hence, the correct answer is3 : 5 : 2 : 7 : 4
Question 8
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In the following figure, it is given that O is the centre of the circle and ㄥAOC = 140 °. Find ㄥABC.

A
120 °

B
110 °

C
115 °

D
130 °

Solution

∠AOC of minor sector = 140 °
∠AOC of major sector=360 °- 140 °= 220 °
Theorem: The angle subtended at the centre is twice the angle formed at the circumference of the circle.
Hence,

∴The measure of ∠x = 110 °
Question 9
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In the figure below, PQ = QS, QR = RS and angle SRQ = 100 °. How many degrees is angle QPS?

A
30 °

B
15 °

C
40 °

D
20 °

Solution

In ΔQRS, QR = RS
⇒ ㄥRQS = ㄥRSQ (because angles opposite to equal sides are equal).
Thus:
ㄥRQS + ㄥRSQ = 180 °- 100 °= 80 °
ㄥRQS = ㄥRSQ = 40 °
ㄥPQS = 180 °–40 °= 140 ° (sum of angles on a line = 180 °)
Then again, ㄥQPS = ㄥQSP (since angles opposite to equal sides are equal)
ㄥQPS + ㄥQSP = 180 °–140 °= 40 °
ㄥQPS = ㄥQSP = 20 °
Question 10
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In the given figure, AD is the bisector of ∠BAC, AB = 6 cm, AC = 5 cm and BD = 3 cm. Find DC.

A
11.3 cm

B
4 cm

C
3.5 cm

D
2.5 cm

Solution
Question 11
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In the given figure, straight line AB and CD intersect at O. IF ∠δ=3 ∠v, then ∠v = ?

A
40 ⁰

B
50 ⁰

C
55 ⁰

D
45 ⁰

Solution

COD is a straight line
∴∠δ+ ∠v =180 ⁰⇒3v +v =180 ⇒4v = 180 ⇒v =45 ⁰.
Question 12
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In a triangle ABC, the incentre is at 0. If ㄥBOC = 100 °, find ㄥBAC.

A
10 °

B
20 °

C
30 °

D
40 °

Solution

In ∠BOC
⇒x + y = 80 °
⇒2x + 2y. = 160 °
Also, 2x + 2y + 2z = 180 °
⇒160 °+ 2z = 180 °
⇒ ∠BAC = 2z = 20 °
Question 13
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A rectangular enclosure 40 m x 36 m has a horse tethered to a corner with a rope of 14 m in length. What is the ratio of the respective areas it can graze if it is outside the enclosure and if it is inside the enclosure?

A
2 : 1

B
2 : 3

C
1 : 4

D
3 : 1
Solution
Imagine a circle on the corner of the rectangle .
3 quarters of the circle lie outside the rectangle and 1 quarter lies inside.
Hence, Required ratio = 3 : 1
Question 14
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Read the passage below and solve the questions based on it.
The area of a square is equal to the area of a rectangle. Moreover, the perimeter of the square is also equal to the perimeter of the rectangle.
Q. The length of the rectangle is equal to the:

A
Breadth of the rectangle

B
Side of the square

C
Cannot be determined

D
Both A and B

Solution

This is possible only when both the length and breadth of the rectangle areequal to the side of the square.
Question 15
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A quadrilateral is inscribed in a circle. If an angle is inscribed in each of the four segments outside the quadrilateral, then what is the sum of these four angles?

A
270 °

B
360 °

C
540 °

D
720 °

Solution