Sol : Option B Explanation: Since Z is the centroid, NZ = 2YZ ⇒ NZ = 8 and YZ = 4. So, area ∆NMX = 1/2 * 8 * 8 = 32. Since the median divides the triangle into two equal areas, area ∆LMN = 64 cm2
3. The sum of all the three sides of an equilateral triangle is 72 &redic;3 cm. Find the height of this triangle.
A. 124 cm
B. 48 cm
C. 36 cm
D. 144 cm
E. 224 cm
Sol : Option C Explanation: Sum of three sides of the triangle is 72&redic;3.
Each side of the equilateral triangle is therefore72&redic;3 / 3 = 24&redic;3
The altitude of this equilateral triangle will be &redic;3 / 2 * 24&redic;3 = 36 cm
4. In triangle PQR, S and T are midpoints of sides PR and PQ respectively. The medians QS and RT intersect at U. Then area PQU: area PQR is
A. 1 : 3
B. 1 : 4.
C. 1 : 5
D. 2 : 5
E. 3 : 4
Sol : Option A Explanation: The intersection of the 3 medians trisects each other.
So the height will be 1 / 3rd the height of the triangle.
The base of the two triangles is the same, so the ratio of the areas depends on the ratio of the heights = 1: 3
5. If there is an equilateral triangle having area A and there is a square having an area B and they have the same perimeter, then
A. X>Y
B. X<Y
C. X = Y
D. X ≥ Y
E. X ≤ Y
Sol : Option B Explanation: √3 a2/4 = X
b2 = Y
Also 3a = 4b
a = 4b/3
√3 /4 (4b2/3)2 = X
So X = 0.77 b2.
Hence X is less than Y.
Q.6. The sum of three sides of an isosceles triangle equals 14m; the ratio of lateral side to the base is 5 is to 4. Find the area of the triangle in square meters.
A. 4&redic;21
B. 3&redic;3
C. 4&redic;3
D. 5&redic;3
E. 2&redic;21
Sol : Option E
Perimeter = 14 m.
Lateral side: Base = 5: 4.
∴ Sides of the triangle are 5 m, 5 m and 4 m.
∴ Area of the triangle thus formed is &redic;7*2*2*3 = 2&redic;21 sq m.
Q.7. A flag pole, which is 36.2 ft high, casts a shadow of 5 ft at 11 a.m. The shadow cast by a tower at the same time is 35 ft long. The height of the tower is
A. 153.4 ft
B. 203.4 ft
C. 200 ft
D. 186.3 ft
E. 253.4 ft
Sol : Option E
Height of the tower will be
35/5 * 36.2 = 253.4ft
(Height of an object is proportional to its shadow)
Q8. PQR is a triangle. If S is a point in the plane of the triangle such that the perpendicular distance from S to the three sides of the triangle are all equal, then there exist(s)
A. Just one such point S
B. Three such points S
C. Two such points S
D. Four such points S
E. None of the above
Sol : Option A
The point that is referred to in this question is the centre of the in-circle.
Since triangles only have 1 in-circle, there can exist only one such point.
Q9. If the area of triangle ABC is 1/4 (a2+b2) where a and b are the lengths of two sides, find the angles of the triangle.
A. 30°, 60°, 90°
B. 60°, 60°, 60°
C. 90° , 45°, 45°
D. 45° , 60°, 45°
E. None of these
Sol : Option C
Here the best way to go is to look at options,
In option B, we are looking at equilateral triangles, in that case area is &redic;3/4 a2 – substituting a = b, we don't get the answer.
Now, moving on to option C
We take a = b, and area = 1/2 * ab = 1/2 * a2. Substituting a = b in the 1/4 * (a2 + b2), we find that it matches.
So, option C is the right choice.
Q10. The length of the three sides of a triangle is in the ratio of 5:12:13. The difference between largest side of this triangle and the smallest side of this triangle is 1.6 centimeters. Find the area of the triangle?
A. 2.4 cm2
B. 30 cm2
C. 1.2 cm2
D. 1.8 cm2
E. 1.5 cm2
Sol : Option C
As lengths of the sides of a triangle are proportional to the numbers 5, 12 and 13,
∴ This triangle is a right angled triangle.
Difference between largest side and smallest side = 1.6cm.
Also if the diff. is 13 – 5 = 8, then smallest side =5.
∴ If difference is 1.6 then smallest side = 5/8 × 1.6 = 1cm
Second will be 12/5 × 1 = 2.4 cm
∴ Area of the triangle = 1/2 * 1 * 2.4 = 1.2cm2
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