PE 1 - Triangles | Geometry - Triangles
Two poles, of height 4 m and 6 m, are 15 m apart. The height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is:
- (a)
2.4 m
- (b)
2.0 m
- (c)
10.0 m
- (d)
6.0 m
- (e)
None of these
Answer: Option A
Explanation :
∆BCD ~ ∆BFE (A-A Rule)
∴ BF/BC = h/6 …(1)
Also, ∆CBA ~ ∆CFE (A-A Rule)
∴ CF/CB = h/4 …(2)
Adding (1) and (2), we get
BF/BC + CF/CB = h/6 + h/4
⇒ (BF + FC)/BC = h/6 + h/4
Now, BF + FC = BC
∴ h/6 + h/4 = BC/BC = 1
⇒ h = 12/5 = 2.4 meters.
Hence, option (a).
Workspace:
PQRS is a trapezium in which the length of the parallel sides PQ and RS is in the ratio 3 : 4. If the diagonals intersect at O, then find the ratio of area of ΔPOQ to ΔROS.
- (a)
16 : 9
- (b)
9 : 16
- (c)
3 : 4
- (d)
None of these
Answer: Option B
Explanation :
In ∆POQ and ∆ROS
∠OPQ = ∠ORS (Corresponding angles)
∠OQP = ∠OSR (Corresponding angles)
∴ ∆POQ ~ ∆ROS
⇒
(Since triangles are similar.)
Hence, option (b).
Workspace:
Consider the triangle PQR as shown in the following figure where QR = 8 cm, QS = 6 cm, SR = 4 cm and ∠SPR = ∠SRQ. What is the ratio of the perimeter of ΔPSR to that of the ΔQSR?
- (a)
7/9
- (b)
8/9
- (c)
6/9
- (d)
5/9
Answer: Option A
Explanation :
We can redraw ∆QSR in the following manner.
In ∆QPR and ∆QRS
∠Q is common
∠QPR = ∠QRS
∴ ∆QPR and ∆QRS (A-A Rule)
⇒ PQ/RQ = QR/QS = PR/RS
⇒ PQ/8 = 8/6 = PR/4
⇒ PQ = 32/3 and PR = 16/3
∴ PS = 32/3 – 6 = 14/3
Perimeter of ∆PSR = 14/3 + 4 + 16/3 = 14
Perimeter of ∆QSR = 6 + 4 + 8 = 18
∴ P(∆PSR)/P(∆QSR) = 14/18 = 7/9
Hence, option (a).
Workspace:
In the given figures, angles B, C and E are all right angles. AB and BC are sides with integral length. If the length of side AC is 17 cm then find the possible value of the ratio of areas of triangle ABC and triangle BCD.
- (a)
225 : 64
- (b)
64 : 225
- (c)
Both (a) and (b)
- (d)
Cannot be determined
Answer: Option D
Explanation :
In ΔABC
if ∠A = θ then ∠ACB = 90 – θ
In ΔBEC, ∠EBC = θ (∵ ∠BEC = 90 and ∠ECB = 90 - θ)
In ΔBCD, ∠D = 90 – θ (∵ ∠DBC = θ)
In ΔABC and ΔBCD
∠CAB = ∠DBC = θ, and
∠ABC = ∠BCD = 90°
∴ ΔABC ~ ΔBCD (A-A Rule)
If AC = 17 and AB & BC are integers then their possible values are 15 & 8 or 8 & 15 respectively.
Case I: AB = 15, BC = 8
Case II: AB = 8, BC = 15
Hence, option (d).
Workspace:
ABC is a right angled triangle with AB = 21 units and BC = 28 units. If PQRB is a square, find its area?
- (a)
81
- (b)
256
- (c)
144
- (d)
169
Answer: Option C
Explanation :
In ∆ABC and ∆QRC
∠B = ∠R = 90°
∠C is common
∴ ∆ABC ~ ∆QRC (A-A Rule)
⇒
⇒
⇒
Let RC = 4x and QR = 3x = BR
∴ BR + RC = 7x = 28
⇒ x = 4
∴ Area of PQRB = 12 × 12 = 144.
Hence, option (c).
Workspace:
ABC is a right angled triangle with AB = 15 unit and BC = 20 unit. If AM = 4 unit and CO = 6 unit, find the length of LN?
- (a)
12.5
- (b)
17.8
- (c)
21.6
- (d)
19.4
Answer: Option B
Explanation :
∆ABC ~ ∆ALM (A-A Rule)
⇒
⇒
⇒ AL = 2.4
Similarly, ∆ABC ~ ∆OCN (A-A Rule)
⇒
⇒
⇒ AL = 4.8
∴ LN = AC – AL – CN = 25 – 2.4 – 4.8 = 17.8.
Hence, option (b).
Workspace:
ABCD is a square of side 10 units. P, Q, R and S are midpoints of sides AD, DC, CB and BA respectively. Find the length of JK?
- (a)
5
- (b)
2√5
- (c)
√10
- (d)
None of these
Answer: Option B
Explanation :
In ASCQ,
AS = QC and also AS || QC
⇒ ASCQ is a parallelogram.
Since the figure is symmetric we get
AJ = DM = CL = BK, and
PJ = QM = RL = SK, and
JM = ML = LK = KJ
Now, in ∆DCL,
DQ = QC and QM || CL
⇒ DM = ML (By Basic Proportionality Theorem)
and also, MQ = ½ LC
Let LC = 2x ⇒ QM = x
∴ DR = DM + ML + LR = 2x + 2x + x = 5x
In right triangle DCR,
DR2 = DC2 + CR2
⇒ (5x)2 = 102 + 52
⇒ 25x2 = 125
⇒ x2 = 5
⇒ x = √5
∴ JK = 2x = 2√5
Hence, option (b).
Workspace:
If the length of the sides of a triangle are in the ratio 5 : 12 : 13, find the ratio of the length of the altitudes of the respective sides.
- (a)
5 : 12 : 13
- (b)
13 : 12 : 5
- (c)
156 : 65 : 60
- (d)
Cannot be determined
Answer: Option C
Explanation :
Let the three sides are 5a, 12a and 13a.
Area of triangle = × side1 × height1 = × side2 × height2 = × side3 × height3
Hence, 5a × altitude1 = 12a × altitude2 = 13a × altitude3.
altitude1 : altitude2 : altitude3 = = 156 ∶ 65 ∶ 60
Hence, option (c).
Workspace:
How many differently shaped triangles exist in which no two sides are of the same length, each side is of integral unit length and the perimeter of the triangle is less than 14 units?
- (a)
3
- (b)
4
- (c)
5
- (d)
6
- (e)
None of these
Answer: Option C
Explanation :
In a triangle we know that sum of any two sides is always greater than the third side.
Also, no two sides can be of same length. (Given in the question)
Case 1: Perimeter = 13
(6, 5, 2) or (6, 4, 3)
Case 2: Perimeter = 12
(5, 4, 3)
Case 3: Perimeter = 11
(5, 4, 2)
Case 4: Perimeter = 10
Not possible
Case 5: Perimeter = 9
(4, 3, 2)
Perimeter 8 or less will not be possible with given conditions.
∴ Possible sides of different triangles can be as follows: (4, 3, 2), (5, 4, 2), (5, 4, 3), (6, 4, 3) and (6, 5, 2)
Hence, option (c).
Workspace:
The area of a triangle with side length a, b and c (a ≥ b ≥ c) is 1 unit. Then b cannot be less than
- (a)
2√2
- (b)
2
- (c)
√3
- (d)
√2
Answer: Option D
Explanation :
Let the angle between sides c and b be α.
So, area of triangle is 1/2 × c × b × sinα = 1
Here, b = 2/(c × sinα)
To find minimum value of b, c and sinα must be maximum.
Sinα can be maximum 1 and c can maximum be equal to b.
∴ b2 = 2 and b = √2 is the minimum value of b.
Hence, option (d).
Workspace:
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