Olympiad Test: Elementary Mensuration-I -2 - Class 7 MCQ

From II, base : height = 5 : 12
Let base = 5x and height = 12x
Then, hypotenuse

From I, perimeter of the triangle = 30 cm.
∴ 5x + 12x + 13x = 30 ⇒ x = 1
So, base = 5x = 5 cm, height = 12x = 12 cm.
∴ Area = (1/2 × 5 × 12) = 30 cm²
Thus, I and II together give the answer.
Clearly III is redundant, since the breadth of the rectangle is not given.
∴ Correct answer is (a).

I. 2(l + b) = 110 l + b = 55
II. l = (b + 5) l – b = 5
III. l/b  = 6/5 5l – 6b = 0
These are three equations in l and b. We may solve them pair-wise.
∴  Any two of the three will give the answer.
∴ Correct answer is (b).

From I and II, we can find the length and breadth of the rectangle and therefore the area can be obtained.
So, III is redundant.
Also, from II and III, we can find the length and breadth and therefore the area can be obtained.
So, I is redundant.
∴ Correct answer is “II and either I or III”.

From II, let l = 4x, b = 6x; and h = 5x.
Then, area of the hall = (24x2) m2.
From I. Area of the hall = 24 m2.
From II and I, we get 24x2 = 24 x = 1.
∴ l = 4 m, b = 6 and h = 5 m.
Thus, the area of two adjacent walls
= [(l × h) + (b × h)] m² can be found out and so the cost of painting two adjacent walls may be found out.
Thus, III is redundant.
∴ Correct answer is (c).

By the formula:
Required area = π (14/2)² = 22/7 × 7²
= 154 cm²

Area of quadrilateral  any diagonal × (sum of perpendiculars drawn on diagonal from two vertices)

Circumference of the circle = 2πr

∴ Side of the inscribed square

% increase in its area
= 2 × 5 + 52/100 
= 10 + 0.25 =10.25%

Required percentage increase
= 2 × 2 + 2²/100 
= 4 + 0.04 
= 4.04%

Diameter is rarely used as the measuring side of a circle. Thus % increase in circumference =12%